Phy5646: Difference between revisions

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'''Abelian Group'''
'''Abelian Group'''


If all the elements commute with each other, then we shall call the group is "Abelian", otherwise it's non-abelian.
If all the elements commute with each other, then the group is "Abelian."


In a group <math>G</math>,if <math>AB=BA \forall A,B\in G</math>
In a group <math>G</math>,if <math>AB=BA \forall A,B\in G</math>
Line 3,859: Line 3,859:


''Example:''The addition of real numbers yields an abelian group. However, the group of square matrices with dimension N under multiplication is not abelian.
''Example:''The addition of real numbers yields an abelian group. However, the group of square matrices with dimension N under multiplication is not abelian.
'''Non-Abelian Group'''
If all the elements of the group do not commute with each other, then the group is "Non-Abelian."
''Example:'' Matrices, in general do not commute, and can form a non-abelian group. For example, <math>SU\left(2\right)</math>: the group of <math>2\times2</math> matrices with unit determinant (special unitary group)


'''Continuously Connected'''
'''Continuously Connected'''

Revision as of 20:48, 23 April 2010

Welcome to the Quantum Mechanics B PHY5646 Spring 2009

Schrodinger equation. The most fundamental equation of quantum mechanics which describes the rule according to which a state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi\rangle} evolves in time.

This is the second semester of a two-semester graduate level sequence, the first being PHY5645 Quantum A. Its goal is to explain the concepts and mathematical methods of Quantum Mechanics, and to prepare a student to solve quantum mechanics problems arising in different physical applications. The emphasis of the courses is equally on conceptual grasp of the subject as well as on problem solving. This sequence of courses builds the foundation for more advanced courses and graduate research in experimental or theoretical physics.

The key component of the course is the collaborative student contribution to the course Wiki-textbook. Each team of students (see Phy5646 wiki-groups) is responsible for BOTH writing the assigned chapter AND editing chapters of others.

This course's website can be found here.

Team assignments: Spring 2010 student teams

Outline of the course:


Stationary state perturbation theory in Quantum Mechanics

Very often, quantum mechanical problems cannot be solved exactly. An approximate technique can be very useful since it gives us quantitative insight into a larger class of problems which do not admit exact solutions. One technique is the WKB approximation, which holds in the asymptotic limit Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar\rightarrow 0 } .

Perturbation theory is another very useful technique, which is also approximate, and attempts to find corrections to exact solutions in powers of the terms in the Hamiltonian which render the problem insolvable. The basic idea of perturbation theory deals with the notion of continuity such that you must be able to write the given Hamiltonian in a way that involves the solvable part of the Hamiltonian with very small additional terms that represent the insolvable parts. In the case of non-degenerate perturbation theory the following assumption must hold: both the energy and the wavefunctions of the insolvable Hamiltonian have analytic expansion in powers of the real parameter Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda\!} -- to insure no jump discontinuities -- equals zero where the perturbing term is taken to be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda\mathcal{H}'} . The quantity Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda\!} , which is taken to be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0 < \lambda < 1 \!} , has no physical significance, and is merely used as a way to keep track of order.

The Hamiltonian is taken to have the following structure:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}=\mathcal{H}_0+\lambda\mathcal{H}'}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}_0} is exactly solvable and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}'} makes it insolvable by analytical methods. Therefore the eigenvalue problem becomes:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathcal{H}_0+\lambda\mathcal{H}')|\psi> = E_n|\psi> }

At the end of the calculation we set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda=1\!} .

It is important to note that perturbation theory tends to yield fairly accurate energies, but usually yields very poor wavefunctions.

Rayleigh-Schrödinger Perturbation Theory

We begin with an unperturbed problem, whose solution is known exactly. That is, for the unperturbed Hamiltonian, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}_0} , we have eigenstates, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n\rangle } , and eigenenergies, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_n \!} , that are known solutions to the Schrodinger equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}_0 |n\rangle = \epsilon_n |n\rangle \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad (1.1.1) }


To find the solution to the perturbed hamiltonian, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , we first consider an auxiliary problem, parameterized by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal \lambda} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H} = \mathcal{H}_0 + \lambda \mathcal{H}^' \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad (1.1.2) }

The only reason for doing this is that we can now, via the parameter Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} , expand the solution in powers of the component of the hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}'} , which is presumed to be relatively small in comparison with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}_0} . In nature we do not know a priori that this will work, and choosing the correct perturbation for a particular problem will likely require some insight into the problem or by a numerical solution.

We attempt to find eigenstates Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N(\lambda)\rangle} and eigenvalues Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n\!} of the Hermitian operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H}} , and assume that they can be expanded in a power series of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal\lambda} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n(\lambda) = E_n^{(0)} + \lambda E_n^{(1)} + ... + \lambda^j E_n^{(j)} + ... }


Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N(\lambda)\rangle = |\Psi_n^{(0)}\rangle + \lambda|\Psi_n^{(1)}\rangle + \lambda^2 |\Psi_n^{(2)}\rangle + ... \lambda^j |\Psi_n^{(j)}\rangle + ... \qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\; (1.1.3)}

Where the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi_n^{(j)}\rangle} signify the j-th order correction to the unperturbed eigenstate Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n\rangle} , upon perturbation. Then we must have,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H} |N(\lambda)\rangle = E(\lambda) |N(\lambda)\rangle \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\; (1.1.4)}

Which upon expansion, becomes: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\mathcal{H}_0 + \lambda \mathcal{H}')\left(\sum_{j=0}^{\infty}\lambda^j |\Psi_n^{(j)}\rangle \right) = \left(\sum_{l=0}^{\infty} \lambda^l E_n^{(l)}\right)\left(\sum_{j=0}^{\infty}\lambda^j |\Psi_n^{(j)}\rangle \right) \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; (1.1.5)}

In order for this method to be useful, the perturbed energies must vary continuously with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} . Knowing this we can see several things about our as yet undetermined perturbed energies and eigenstates. For one, as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda \rightarrow 0, |N(\lambda)\rangle \rightarrow |\Psi_n^{(0)}\rangle = |n\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n(\lambda) \rightarrow E_n^{(0)} = \epsilon_n} for some unperturbed state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n\rangle} .

For convenience, assume that the unperturbed states are already normalized: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n | n \rangle = 1} , and choose normalization such that the exact states satisfy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n|N(\lambda)\rangle=1} . Then in general Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N\rangle} will not be normalized, and we must normalize it after we have found the states (see Phy5646#Renormalization).

Thus, we have:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n|N(\lambda)\rangle= 1 = \langle n |\Psi_n^{(0)}\rangle + \lambda \langle n |\Psi_n^{(1)}\rangle + \lambda^2 \langle n |\Psi_n^{(2)}\rangle + ... \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\;(1.1.6)}

Coefficients of the powers of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda\!} must match, so,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n | \Psi_n^{(i)} \rangle = 0, i = 1, 2, 3, ... \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;(1.1.7)}


Which shows that, if we start with the unperturbed state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n\rangle } , upon perturbation, then we add to this initial state a set of perturbation states, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\Psi_n^{(0)}\rangle, |\Psi_n^{(1)}\rangle, ... } which are all orthogonal to the original state -- so the unperturbed states become mixed together.


We equate coefficients in the above expanded form of the perturbed Hamiltonian (eq. #1.1.5), we are provided with the corrected eigenvalues for whichever order of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} we want. The first few are as follows,

0th Order Energy Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda^0 \rightarrow E_n^{(0)} = \epsilon_n \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (1.1.8)}
which we already had before.

1st Order Energy Corrections Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda^1 \rightarrow \mathcal{H}_0 |\Psi_n^{(1)}\rangle + \mathcal{H}' |\Psi_n^{(0)}\rangle = E_n^{(1)} |\Psi_n^{(0)}\rangle + E_n^{(0)} |\Psi_n^{(1)}\rangle } , taking the scalar product of this result of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n\rangle} , and using our previous results, we get: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n^{(1)} = \langle n|\mathcal{H}'|n\rangle \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;(1.1.9)}

2nd Order Energy Corrections

Taking the terms in eq #1.1.5 that are second order in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda\!} and operating on them with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n\rangle} provides us with up to the second order:

One interesting thing to note about this is that is positive definite. Therefore, since , the second order energy correction will always lower the ground state energy.

kth order Energy Corrections In general,

This result provides us with a recursive relation for the eigenenergies of the perturbed state, so that we have access to the eigenenergies for an state of arbitrary order in .

What about the eigenstates? Express the perturbed states in terms of the unperturbed states:

Go back to equation #1.1.5 and take the scalar product from the left with and then, compare orders of to find:

1st order Eigenkets

The first order contribution is then the sum of this equation over all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m\!} , and adding the zeroth order we get the eigenstates of the perturbed hamiltonian to the 1st order in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda\!} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N\rangle = |n\rangle + \lambda\sum_{k \not= n} |m\rangle \frac{\langle m |V| n\rangle}{\epsilon_n - \epsilon_m} + ...\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;(1.1.11)}

Going beyond order one in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda} gets increasingly messy, but can be done by the same procedure.

Renormalization

Earlier we assumed that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n|N(\lambda)\rangle=1} , which means that our Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N(\lambda)\rangle} states are not normalized themselves. To reconcile this we introduce the normalized perturbed eigenstates, denoted Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bar{N}} . These will then be related to the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N(\lambda)} :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\bar{N}\rangle = \frac{|N\rangle}{\sqrt{\langle N|N\rangle}} =z^{1/2}|N\rangle \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(1.1.12)}

Thus Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z} gives us a measure of how close the perturbed state is to the original state.

To second order in

Renorm.jpg

Where we use a taylor expansion to arrive at the final result (noting that ).

Then, interestingly, we can show that is related to the energies by employing equation 1.1.10:

Where the derivative is taken with respect to , while holding constant. Using the Brillouin-Wigner perturbation theory (see next section) it can supposedly be shown that this relation holds exactly, without approximation.

Problem examples of non-degenerate perturbation theory :

-Problem 1: demonstrating how linear algebra can be used to solve for the exact eigenstates, exact eigenevalues, first and second order corrections to the eigenvalues, and first order corrections to the eigenstates of a given Hamiltonian

-Problem 2

-Problem 3

-Problem 4

Brillouin-Wigner Perturbation Theory

Brillouin-Wigner perturbation theory is an alternative perturbation method based on treating the right hand side of

as a known quantity. This method is not strictly an expansion in lambda.

Using a basic formula derived from the Schrodinger equation, you can find an approximation for any power of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda \!} required using an iterative process. This theory is less widely used as compared to the RS theory. At first order the two theories are equivalent. However,the BW theory extends more easily to higher order and avoid the need for separate treatment of non degenerate and degenerate levels. In addition, if we have a good approximation for the value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n \!} , the BW series should converge more rapidly than the RS series.

Starting with the Schrodinger equation:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} ({\mathcal H}_o+\lambda {\mathcal H}')|N\rangle &= E_n|N\rangle \\ \lambda {\mathcal H}'|N\rangle &= (E_n-{\mathcal H}_o)|N\rangle \\ \langle n|(\lambda {\mathcal H}'|N\rangle) &= \langle n|(E_n-{\mathcal H}_o)|N\rangle \\ \lambda \langle n|{\mathcal H}'|N\rangle &= (E_n-\epsilon_n)\langle n|N\rangle \\ \end{align} }

If we choose to normalize Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n|N \rangle = 1 } , then so far we have: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (E_n-\epsilon_n) = \lambda\langle n|{\mathcal H}'|N\rangle \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\; (1.2.1)}

which is still an exact expression (no approximation have been made yet). The wavefunction we are interested in, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N\rangle } can be rewritten as a summation of the eigenstates of the (unperturbed, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\mathcal H}_o } ) Hamiltonian: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} |N\rangle &= \sum_m|m\rangle\langle m|N\rangle\\ &= |n\rangle\langle n|N\rangle + \sum_{m\neq n}|m\rangle\langle m|N\rangle\\ &= |n\rangle + \sum_{m\neq n}|m\rangle\frac{\lambda\langle m|{\mathcal H}'|N\rangle}{(E_n-\epsilon_m)}\\ \end{align} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\; (1.2.2)}

The last step has been obtained by using eq #1.2.1. So now we have a recursive relationship for both Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n \!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N\rangle }

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n = \epsilon_n+\lambda\langle n|{\mathcal H}'|N\rangle } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N\rangle } can be written recursively to any order of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda \!} desired

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N\rangle = |n\rangle+\lambda \sum_{m\neq n}|m\rangle\frac{\lambda\langle m|{\mathcal H}'|N\rangle}{(E_n-\epsilon_m)} } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n } can be written recursively to any order of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda \!} desired

For example, the expression for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N\rangle } to a third order in Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda \!} would be:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} |N\rangle &= |n\rangle + \lambda\sum_{m\neq n}|m\rangle\frac{\langle m|{\mathcal H}'}{(E_n-\epsilon_m)}\left(|n\rangle + \lambda\sum_{j\neq n}|j\rangle\frac{\langle j|{\mathcal H}'}{(E_n-\epsilon_j)}\left(|n\rangle + \lambda\sum_{k\neq n}|k\rangle\frac{\langle k|{\mathcal H}'|n\rangle}{(E_n-\epsilon_k)}\right)\right)\\ &= |n\rangle + \lambda\sum_{m\neq n}|m\rangle\frac{\langle m|{\mathcal H}'|n\rangle}{(E_n-\epsilon_m)} + \lambda^2\sum_{m,j\neq n}|m\rangle\frac{\langle m|{\mathcal H}'|j\rangle\langle j|{\mathcal H}'|n\rangle}{(E_n-\epsilon_m)(E_n-\epsilon_j)} + \lambda^3\sum_{m,j,k\neq n}|m\rangle\frac{\langle m|{\mathcal H}'|j\rangle\langle j|{\mathcal H}'|k\rangle\langle k|{\mathcal H}'|n\rangle}{(E_n-\epsilon_m)(E_n-\epsilon_j)(E_n-\epsilon_k)}\\ \end{align} } ,

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{j} |j \rangle \langle j |} is unity.

Note that we have chosen Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n|N \rangle = 1} , i.e. the correction is perpendicular to the unperturbed state. That is why at this point Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |N \rangle} is not normalized. The normalized wave function can be written as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\bar{N} (\lambda) \rangle = \frac{|N(\lambda) \rangle}{\sqrt{\langle N (\lambda) | N (\lambda) }} \equiv \sqrt{Z (\lambda)} | N(\lambda) \rangle }

Interestingly, the normalization constant Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Z} turns out be exactly equal to the derivative of the exact energy with respect to the unperturbed energy, ie

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{\partial E_{n}(\lambda)}{\partial \epsilon_{n}} = Z}

The calculation for the normalization constant can be found through this link.

We can expand #1.2.1 with the help of #1.2.2, this gives:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n = \epsilon_n + \lambda\langle n|H'|n\rangle + \lambda^2\sum ' \frac{|\langle m|H'|n\rangle|^2}{E_n - \epsilon_m} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\; (1.2.3)} .

Notice that if we replaced Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n\!} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon_n\!} we would recover the Raleigh-Schrodinger perturbation theory. By itself #1.2.2 provides a transcendental equation of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n\!} , since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n\!} appears in the denominator of the right hand side. If we have some idea of the value of a particular Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n\!} , then we could use this as a numerical method to iteratively get better and better values for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n\!} .

Degenerate Perturbation Theory

Degenerate perturbation theory is an extension of standard perturbation theory which allows us to handle systems where one or more states of the system have non-distinct energies. Normal perturbation theory fails in these cases because the denominators of the expressions for the first-order corrected wave function and for the second-order corrected energy become zero. If more than one eigenstate for the Hamiltonian Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\mathcal H}_o } has the same energy value, the problem is said to be degenerate. If we try to get a solution using perturbation theory, we fail, since Rayleigh-Schroedinger PT includes terms like Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac {1}{\mathcal(\epsilon_n-\epsilon_m)} \!} .

Instead of trying to use these degenerate eigenstates with perturbation theory, we start with the non-degenerate linear combinations of the original eigenstates so that regular perturbation theory may be applied. In other words, the first, and only, extra step of degenerate perturbation theory is to find linear combinations by diagonalizing the perturbation within the set of degenerate states and then proceeding as usual in non-degenerate perturbation.

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{|n_a\rangle,|n_b\rangle,|n_c\rangle,\dots\} } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \longrightarrow } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{|n_{\alpha}\rangle,|n_{\beta}\rangle,|n_{\gamma}\rangle,\dots\} } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n_{\alpha}\rangle = \sum_iC_{\alpha,i}|n_i\rangle } etc

The general procedure for doing this type of problem is to create the matrix with elements Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle n_a|{\mathcal H}'|n_b\rangle } formed from the degenerate eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\mathcal H}_o } . This matrix can then be diagonalized, and the eigenstates of this matrix are the correct linear combinations to be used in non-degenerate perturbation theory. In other words, we choose to manipulate the expression for the Hamiltonian so that goes to zero for all cases . One can then apply the standard equation for the first-order energy correction, noting that the change in energy will apply to the energy states described by the new basis set. (In general, the new basis will consist of some linear superposition of the existing state vectors of the original system.)

One of the well-known examples of an application of degenerate perturbation theory is the Stark Effect. If we consider a Hydrogen atom with in the presence of an external electric field . The Hamiltonian for this system is . The eigenstates of the system are . The matrix of the degenerate eigenstates and the perturbation is:


To briefly summarize how most of the terms in this matrix work out to be zero (the full arguments as to how most of these terms are zero is worked out in G Baym's "Lectures on Quantum Mechanics" in the section on Degenerate Perturbation Theory) first note that the hydrogen atom is degenerate under parity, and as a result, all the elements on the diagonal become zero. The other elements vanish because of angular momentum. Matrix elements of between states with different eigenvalues of vanish, since commutes with . For example,

which means that

The correct linear combination of the degenerate eigenstates ends up being

Because of the perturbation due to the electric field, the and states will be unaffected. However, the energy of the and states will have a shift due to the electric field.

Example: 1D harmonic oscillator

Consider 1D harmonic oscillator perturbed by a constant force.

The energy up to second order is given by

Let's see the matrix elements

We see that:

  • The first order term in eq. #1.3.1 is:
  • The Second order term is:


Finally the energy is given by

This results is exactly the same as when we solve the problem without perturbation theory.

Time dependent perturbation theory in Quantum Mechanics

Formalism

Previously, we learned time independent perturbation theory in which a little change in the Hamiltonian generates a correction in the form of a series expansion for the energy and wave functions. The problem for a time independent can be solved by finding a solution to the equation . And then changes in time can be modeled by constructing the states where . In principle this describes any closed system, and there would never be a reason for time-dependent problems if it were practical to consider all systems as closed. However, there are many examples in nature of systems that are easier described as not being closed. For example, while the stationary approach can be used to describe the interaction of electromagnetic field with atoms (i.e. photon with Hydrogen atom), it is more practical to describe it as an open system with an explicitly time dependent term (due to EM radiation). Therefore we explore Time Dependent Perturbation Theory.


One of the main tasks of this theory is the calculation of transition probabilities from one state to another state that occurs under the influence of a time dependent potential. Generally, the transition of a system from one state to another state only makes sense if the potential acts on within a finite time period from to . Except for this time period, the total energy is a constant of motion which can be measured. We start with the Time Dependent Schrodinger Equation,

where a is the Bohr radius. Then, to answer any questions about the behavior of the system at a later time we must find its state , assuming that the perturbation acts after time , we get

The problem therefore consists of finding the solution with boundary condition for . However, such a problem is usually impossible to solve completely in closed form.
Therefore, we limit ourselves to the problems in which is small. In that case we can treat as a perturbation and seek it's effect on the wavefunction in powers of .

Since is small, the time dependence of the solution will largely come from . So we use

,

which we substitute into the Schrodinger Equation to get

.

In this equation and the operator are in the interaction representation. Now, we integrate equation #(2.1.4) to get

or

Equation #(2.1.5) can be iterated by inserting this equation itself as the integrand in the r.h.s. We can then write equation #(2.1.5) as

which can be written compactly as

This is the general solution. is called the time ordering operator, which ensures that the series is expanded in the correct order. For now, we consider only the correction to the first order in .

First Order Transitions

If we limit ourselves to the first order, we use

We want to see the system undergoes a transition to another state, say . So we project the wave function to . From now on, let
for brevity. In other words, what is the probability of a state making a transition into a state at a given time ?

Projecting into state and letting if , we get

Expression #(2.1.9) is the probability amplitude of transition. Therefore, we square the final expression to get the probability of having the system in state at time .
Squaring, we get

For example, let us consider a potential which is turned on sharply at time , but independent of thereafter. Furthermore, we let for convenience. Therefore :

The plot of the probability vs. is given in the following plot:

Amplitude.JPG

,where . So we conclude that as the time grows, the probability has a very narrow peak and approximate energy conservation is required for a transition with appreciable probability. However, this "uncertainty relation" is not the same as the fundamental uncertainty relation because while and are both observable variables, time in non-relativistic quantum mechanics is just a parameter, not an observable.

Now, we imagine shining a light of a certain frequency on a hydrogen atom. We probably ended up getting the atom at a certain bound state. However, it might be ionized as well. The problem with ionization is the fact that the final state is a continuum, so we cannot just simply pick up a state to end with i.e. a plane wave with a specific .

Furthermore, if the wave function is normalized, the plane waves states will contain a factor of which goes to zero if is very large. But, we know that ionization exists, so there must be something missing. Instead of measuring the probability to a transition to a pointlike wavenumber, , we want to measure the amplitude of transition to a group of states around a particular , i.e., we want to measure the transition amplitude from to .

Let's suppose that the state is one of the continuum state, then what we could ask is the probability that the system makes transition to a small group of states about , not to a specific value of . For example, for a free particle, what we can find is the transition probability from initial state to a small group of states, viz. , or in other words, the transition probability to an element of phase space

The next step is a mathematical trick. We use

Applying this to our result from above, we see that as ,

If this result used in the equation #2.1.11, it gives

or as a rate of transition,  :

which is called The Fermi Golden Rule. Using this formula, we should keep in mind to sum over the entire continuum of final states.

To make things clear, let's try to calculate the transition probability for a system from a state to a final state due to a potential .

What we want is the rate of transition, or actually scattering in this case, into a small solid angle . So we must sum over the momentum states in this solid angle:

The sum over states for continuum can be calculated using an integral,

Therefore,

The flux of particles per incident particle of momentum in a volume is , so

, in the Born Approximation.

This result makes sense since our potential does not depend on time, so what happened here is that we sent a particle with wave vector through a potential and later detect a particle coming out from that potential with wave vector . So, it is a scattering problem solved using a different method.

This is another simple example of transition probability calculation in time dependent perturbation theory with different potential.

Here another example example


An example of the of the first excited state of the hydrogen atom.

Harmonic Perturbation Theory

Harmonic perturbation is one of the main interests in perturbation theory. We know that in experiment, we usually perturb the system using a certain signal to extract information about it, for example the difference between the energy levels. We could send a photon with a certain frequency to a Hydrogen atom to excite the electron and let it decay to observe the difference between two energy levels by measuring the frequency of the photon emitted from it. The photon acts as an electromagnetic signal, and it is harmonic (if we consider it as an electromagnetic wave).

In general, we write down the harmonic perturbation as


where specifies the rate at which the perturbation is turned on. Since we assume the perturbation is turned on very slowly is a very small positive number which at the end of the calculation, when , is set to be zero.

We start from . Since there's no perturbation at that time. We want to find the probability that there will be a transition from the initial state to some other state, . The transition amplitude is,

To the first order of V we write

Now we calculate the probability as usual:

Where all oscillatory terms have been averaged to zero. The transition rate is given by :

Now, if the response is immediate or if the potential is turned on suddenly, we take . Therefore:


Which is the Fermi Golden Rule. This result shows that there will be a non-zero transition probability for cases where - Roughly speaking, there will be significant transitions only when is a "resonant frequency" for a particular transition. The Fermi Golden Rule also shows that it doesn't matter how the potential is turned on -- fast or slow -- the transition rate is not really affected.

Second Order Transitions

Sometimes the first order matrix element is identically zero (parity, Wigner Eckart, etc.) but other matrix elements are nonzero—and the transition can be accomplished by an indirect route.

where is the probability amplitude for the second-order process,

Taking the gradually switched-on harmonic perturbation , and the initial time , as above

The integrals are straightforward, and yield

Exactly as in the section above on the first-order Golden Rule, we can find the transition rate:


The in the denominator goes to on replacing the frequencies with energies E, both in the denominator and the delta function, remember that


This is a transition in which the system gains energy from the beam, in other words two photons are absorbed, the first taking the system to the intermediate energy , which is short-lived and therefore not well defined in energy—there is no energy conservation requirement into this state, only between initial and final states.


Of course, if an atom in an arbitrary state is exposed to monochromatic light, other second order processes in which two photons are emitted, or one is absorbed and one emitted (in either order) are also possible.

Example of Two Level System : Ammonia Maser

Ammonia.JPG

This is a very complicated quantum system and there is no way to solve it in a formal form, however we can take some assumptions to solve the problem. In this model, we assume that the Nitrogen atom, being heavier than Hydrogen, is motionless. The Hydrogen atoms form a rigid equilateral triangle whose axis is always passes through the Nitrogen Atom.

Since there are two significant and different states (the position of the Hydrogen triangle), we write down the wave function as a superposition of both states. Of course it is a function of time.

Then we operate on this state the time dependent Schrodinger equation to find the eigenvalues:

In the presence of electric field, the additional energy enters only on the diagonal part of the Hamiltonian matrix

Typically, which gives the frequency of the movement of the Hydrogen triangle and the wavelength (microwave region).

Solving for the Schrodinger equation we have above, we find the energy of the two states

The followings are the graphs of the eigenenergy as a function of the applied electric field

Crossing.JPG

Because of these two different states, ammonia molecule is separable in the electric field. This can be used to select molecule with certain value of energy.

Ammonia maser.JPG

It should be clear that if , our eigenstates are with energies

Let , then we find

Now, let

Also, we define the electric field as a function of time so the above expression can be written as

Now, we observe that as the first term in the right hand side of the first equation will oscillate very rapidly compared to the second term of the same equation. The average of this rapid oscillating term will be zero. So we don't need to consider these oscillating terms in the next calculation. Therefore we get a result

At resonance, , these equations are simplified to

We can then differentiate the first equation with respect to time and substitute the second equation into it to get

With solution (also for with substitution)

Let's assume that at time , the molecule is in state (experimentally, we can prepare the molecule to be in this state) so that and . This assumption gives

Therefore the each probability that the molecule remains in the state and is :

Note that the probability depends on time. The molecules enter in upper energy state. If the length of the cavity is chosen appropriately, the molecules will come out surely in lower energy state . If that is the case, the molecules lost some energy and, in reverse, the cavity gains the same amount of energy. The cavity is therefore excited and then produces stimulated emission. That is the mechanism of a MASER which stands for Microwave Amplification by Stimulated Emission of Radiation.

Interaction of radiation and matter

The conventional treatment of quantum mechanics uses time-independent wavefunctions with the Schrödinger equation to determine the energy levels (eigenvalues) of a system. To understand the interaction of radiation (electromagnetic radiation) and matter, we need to consider the time-dependent Schrödinger equation.

Quantization of electromagnetic radiation

Classical view

Let's use transverse gauge (sometimes called Coulomb gauge) which give us:

In this gauge the electromagnetic fields are given by:

The energy in this radiation is

The rate and direction of energy transfer are given by Poynting vector

The radiation generated by classical current is

Where is the d'Alembert operator. Solutions in the region where are given by

where and , as we are considering EM waves in vacuum. The and are the two general polarization vectors, perpendicular to . Note that, in general,

Here the plane waves are normalized with respect to some volume . This is just for convenience and the physics won't change. Note that , as the polarization vectors are unit vectors. Notice that in this writing is a real vector.

Let's compute . For this

Taking the average, the oscillating terms will disappear. Then we have

It is well known that for plane waves , where is the direction of . This clearly shows that . However, let's see this explicitly:

Each component is given by

Then

Again, taking the average the oscillating terms vanish. Then we have

Finally the energy of this radiation is given by

So far, we have treated the potential as a combination of two waves with the same frequency. Now let's extend the discussion to any form of . To do this, we can sum over all values of and :

To calculate the energy with useing the fact that any exponential time-dependent term is on average zero. Therefore, in the previous sum all cross terms with different vanishes. Then, it is clear that

Then, the energy is given by

Let's define the following quantities:

Notice that

Adding

Then the energy (in this case the Hamiltonian) can be written as

This has the same form as the familiar Hamiltonian for a harmonic oscillator.

Note that,

The makeshift variables, and are canonically conjugate.

We see that the classical radiation field behaves as a collection of harmonic oscillators, indexed by and , whose frequencies depends on .

From classical mechanics to quatum mechanics for radiation

As usual we proceed to do the canonical quantization:



Where last are quantum operators. The Hamiltonian can be written as



The classical potential can be written as



Notice that the quantum operator is time dependent. Therefore we can identify it as the field operator in interaction representation. (That's the reason to label it with int). Let's find the Schrodinger representation of the field operator:



COMMENTS

  • The meaning of is as following: The classical electromagnetic field is quantized. This quantum field exist even if there is not any source. This means that the vacuum is a physical object who can interact with matter. In classical mechanics this doesn't occur because, fields are created by sources.
  • Due to this, the vacuum has to be treated as a quantum dynamical object. Therefore we can define to this object a quantum state.
  • The perturbation of this quantum field is called photon (it is called the quanta of the electromagnetic field).


ANALYSIS OF THE VACUUM AT GROUND STATE

Let's call the ground state of the vacuum. The following can be stated:

  • The energy of the ground state is infinite. To see this notice that for ground state we have
  • The state represent an exited state of the vacuum with energy . This means that the extra energy is carried by a single photon. Therefore represent the creation operator of one single photon with energy . In the same reasoning, represent the annihilation operator of one single photon.
  • Consider the following normalized state of the vacuum: . At the first glance we may think that creates a single photon with energy . However this interpretation is forbidden in our model. Instead, this operator will create two photons each of the carryng the energy .

    Proof

    Suppose that creates a single photon with energy . We can find an operator who can create a photon with the same energy . This means that

    Let's see if this works. Using commutation relationship we have

    Replace the highlighted part by



    Since , the initial assumption is wrong, namely:

    This means that cannot create a single photon with energy . Instead it will create two photons each of them with energy


ALGEBRA OF VACUUM STATES

A general vacuum state can be written as

where is the number of photons in the state which exist in the vacuum. Using our knowledge of harmonic oscillator we conclude that this state can be written as

Also it is clear that

Matter + Radiation

Hamiltonian of Single Particle in Presence of Radiation (Gauge Invariance)

The Hamiltonian of a single charged particle in presence of E&M potentials is given by

where the vector potential in the first term and the scalar potential in the second term is external E-M interaction and the third term is related to other potentials.

The time dependent Schrödinger equation is

Since a gauge transformation,

left invariant the E&M fields, we expect that which is an observable is also gauge independent. Since is independent of the phase choice, we can relate this phase with the E&M gauge transformation. In other words, the phase transformation with E&M transformation must leave Schrödinger equation invariant. This phase transformation is given by:

Let's see this in detail. We want to see if:

Let's put the transformations:

Replacing

Finally let's write the Hamiltonian in the following way

Where is the Hamiltonian without external fields (say hydrogen atom) and is the interaction part with the radiation.

Example: electron on helium surface

Hamiltonian of Multiple Particles in Presence of Radiation

If we have a system of particles we have the following Hamiltonian

(Where and are the charge and the mass of the i-th particle respectively and and are its coordinate and momentum operators.)

Let's assume all particles having same mass and same charge. Then we have

Using delta function operator we can write

Then


COMMENTS

  • can be interpreted as the particle density operator.
  • is called paramagnetic current. It is just a piece of the total current . Explicitly we have

Light Absorption and Induced Emmission

Generally for atomic fields . Using the transverse gauge we can approximate the interaction Hamiltonian as

Let's write using the Fourier expansion as described above:

Where


Let's use golden rule to calculate transition rates for this time-dependent interaction. The evolution of the state in first approximation is

where is the initial state and is the usual slow "switch" factor. The transition amplitude to a state is

and are eigenstates of . Then we have

The transition probability is given by

Where all oscillatory terms have been averaged to zero. Taking a time derivative we obtain the transition rate

The above equation says that the transition rate between two states is composed by two possibilities: absorption or induced emission . Let's analyze the matrix elements between states.

Absorption

Let's suppose that initial and final states are:

Where are the initial and final states of (say hydrogen atom) with energies and are the initial and final states of (the vacuum).

The matrix element of isgiven by:

The last shows how in the absorption process, the system absorbs a single photon from the radiation. Namely the final state is given by:

Finally we can write the transition rate absorption as following


Induced Emission

Let's suppose that initial and final states are:

Where are the initial and final states of (say hydrogen atom) with energies and are the initial and final states of (the vacuum).

The matrix element of isgiven by:

The last shows how in the emmision process, the system release a single photon from the radiation. Namely the final state is given by:

Finally we can write the transition rate absorption as following


Important Phenomena: Spontaneous Emission

Let's suppose that initial is a single Hydrogen atom in the 2P state in the vacuum (and nothing else!!!). The state can be written as

According to induced emission, there could be a process in which the final state is:

Where a single photon has been emitted without any external perturbation. This is emission process is called Spontaneous emission. For an experimental observation of a Lamb-like shift in a solid state setup see here.

Einstein's Model of Absorption and Induced Emmision

Let's use Statistical Mechanics to study a cavity with radiation. For this we need to use the Plank distribution:

This is just the occupation number of the state . Let's suppose the following situation:

  • Our cavity is made up with atoms with two quantum levels with energies and such that .
  • The walls are emitting and absorbing radiation (Thermal Radiation) such that system is at equilibrium. Since there is just two levels, the photons emitted by atoms must have energy equal to .

The Boltzmann distribution tells us that the probabilities to find atoms at energies and are respectively

Let's call the number of photons at equilibrium. At equilibrium we have

It is natural to express the absorption and emission rate as:

Where B is some constant. Since we have

This means that eventually all photons will be absorbed and then . This of course is not a physical situation. Einstein realized that there is another kind of process of emission that balances the rates in such way that . This emission is precisely the spontaneous emission and can be written as

Then we have

And solving for we have

As conclusion we obtain for the emission rate the follwing:

Notice that the factor matches with our previous result.

Example

Details of Spontaneous Emission

Power of the emitted light

Using our previous result for , we can calculate the power of the light with polarization per unit of solid angle that the spontaneus emission produce:

Then


Conservation of Momentum

Consider a matter in the eigenstate of the momentum . Suppose that it make a transition to eigenstate with momentum via spontaneus emission. The momentum must conserve. Therefore we have a process where:

Initial Momenta


Final Momenta


Let's calculate the matrix element for two cases.


Case 1: Single free charged particle

This result is very interesting!!!. It says that the emitted light must be

However this is impossible from the point of view of conservation of energy:

This means that a single charged particle can not make transitions. So a single charged particle doesn't see the vacuum fluctuations.


Case 2: General Case (System of particles)


We can use the total momentum of the system as generator of translations for . So that we can write


Then

The last shows that

Electric Dipole Transitions

Let's consider a nucleus (say hydrogen atom) well localized in space. Typically the wave length of the emitted light is much bigger than electron's orbit around nucleus (say Bohr radius ). For example the wavelength of blue light is on the order of 100 nm or 1000 Angstrom, while the wavelength of the electron orbiting the nucleus in the Hydrogen atom is of the order of 1 Angstrom. This means that:

The matrix element is then

Notice that is the off diagonal elements of the dipole moment operator. The power per unit of solid angle for a given polarization is given by

Selection Rules

Let's assume that initial and final states are eigenstates of and . Using commutation relationships we can obtain the following selection rules the vector :

1. Selection Rules for

1.1. From this we have

This means that if .

1.2

  • . From this we have
  • . From this we have

Combining

From here we see that

This means that if

2. Selection Rule for

Consider the following commutator proposed by Dirac

After some algebra we can see that

Since is non negative . There are two possibilities:

  • if . However for , which corresponds to . This possibility is trivial and it doesn't say anything new.
  • if


Summary

If the initial and final states are eigenstates for and then the possible transitions that can occur in the dipole approximation are


Example: Transitions Among Levels n=1,2,3 of Hydrogen Atom

Let's consider the levels n=1,2,3 of Hydrogen Atom. The possible transitions to the state 1S according to the sharp selection rules are the following


1s.jpg


The possibles transitions to the state are the following


2p0.jpg


Power & Polarization of Emitted Light

Case : In this case the selection rules tell us that:

Then we can say

  • The light is always plane polarized in the plane defined by .

Planepolarization.png

  • The power is given by

Case : In this case the selection rules tell us that:

From the previous result we have

Then

Then we can say

  • rest at the XY plane. The polarization of the emitted light is circular.
  • Lets put a detector to see the light coming toward positive Z axis. Since right circular polarized light has angular momentum while negative circular polarized light has angular momentum we can state the following:
    • If we see a circular polarized light then by conservation of angular momentum we know that
      the transition was
    • If we see a negative circular polarized light then by conservation of angular momentum we know that
      the transition was

    Scattering of Light

    ( Notes and LaTex code, courtesy of Dr. Oskar Vafek)



    We can analyze how a charged system interact with photons and scatter it. The problem of light scattering can be considered as a transition from initial state, to a final state . For this transition we can calculate the transition amplitude. Let us deal with some basics first. First of all we can write the Schrodinger equation for an electron in a potential interacting with quantized EM radiation as:

    where

    We are considering the transverse gauge, in which the vector potential operator can be defined as:

    where

    Let us define,

    where


    and


    We can use the Dirac picture to represent the wavefunction as:

    Therefore,


    More precisely,

    where the vector potential operator which is now time dependent can be defined as,

    Using second order time dependent perturbation theory up to to order, we can write the wavefunction is Dirac picture as,

    where the perturbation is slowly switched on at .

    As mentioned before,we need to calculate the transition probability from

    to the final state

    Therefore we need to calculate that following transition probability,

    Using second order time dependent perturbation theory the probability for such a transition is

    The required transition can be made by term proportional to )(the diamagnetic term)in first order, while term proportional to (paramagnetic term) gives non-zero overlap in second order perturbation theory. Therefore we have:


    We can ignore the dependence in gauge field by using dipole approximation, that is we can say

    Let's define

    where

    The second order term is


    Therefore,

    The time dependent probability is


    and the transition rate is


    We observe that,


    Taking (as ) we get,

    where in the second line we have used the energy conserving function, giving . Using the above commutation relation again we finally find

    Therefore


    To get the total transition rate we need to sum over all wavevectors in a solid angle .

    where . Finally the differential cross-section is found by dividing by the photon flux to yield


    Therefore the scattering cross-section is inversely proportional to the fourth power of wavelength ( for elastic scattering). This explains why sky is blue since blue light having lower wavelength, gets scattered more.

    Non-perturbative methods

    Apart from the conventional perturbative methods, there also exist non-perturbative methods to approximately determine the lowest energy eigenstate or ground state, and some excited states, of a given system. Superconductivity and the fractional quantum Hall effect are examples of problems that were solved using non-perturbative methods. One of the important methods in the approximate determination of the wave function and eigenvalues of a system is the Variational Method, which is based on the variational principle. The variational method is a very general one that can be used whenever the equations can be put into the variational form. The variational method is now a pringboard to many numerical computations.

    Principle of the Variational Method

    Consider a completely arbitrary system with time independent Hamiltonian and we assume that it's entire spectrum is discrete and non-degenerate.

    = ;

    Let's apply the variational principle to find the ground state of the system.Let be an arbitrary ket of the system. We can define the expectation value of the Hamiltonian as

    Of course, if the wavefunction is normalized so that , then the expectation value of the hamiltonian is just:

    The variational principle states that,

    is only true if the wave functions used in the expectation value are the exact wave functions of the true ground state for the Hamiltonian; they can not be unperturbed or approximate wave functions.

    Because the expectation value of the Hamiltionian is always greater than or equal to the ground state energy, this gives an upper bound for the ground state energy when using unperturbed wavefunctions to calculate the expectation value.

    If you are making a guess at the wavefunction, but do not know it explicitly, you can write it up to a parameter and then minimize the expectation value of the hamiltonian with respect to that parameter. For example, we can write the ground state wavefunction of the hydrogen atom, up to a parameter as:

    You would then minimize the expectation value of with respect to , lowering your upper bound as far as possible so that you have a better idea of the true value of the energy.

    In some cases a lower bound can also be found by a similar method. In the case that , is said to be a positive operator because . Therefore, is a lower bound for the energy.


    Since the exact eigenfunctions of form a complete set, we can express our arbitrary ket as a linear combination of the exact wavefunction.Therefore,we have

    Multiplying both sides by we get


    However, . So, we can write the above equation as

    Or

    with , thus proving eq. #4.1.2.


    Thus eq. #4.1.2 gives an upper bound to the exact ground state energy. For the equality to be applicable in the all coefficients except should be zero and then will be the eigenvector of the Hamiltonian and the ground state eigenvalue.

    Generalization of Variational Principle: The Ritz Theorem.

    We claim that the expectation value of the Hamiltonian is stationary in the neighborhood of its discrete eigenvalues. Let us again consider the expectation value of the Hamiltonian eq.#4.1.1.

    Here is considered as a functional of . Let us define the variation of such that goes to where is considered to be infinetly small. Let us rewrite eq.#4.1.1 as

    .


    Differentiating the above relation,


    However, is just a c-number, so we can rewrite eq #4.2.2 as

    .


    If , then the mean value of the Hamiltonian is stationary.

    Therefore,

    .


    Define, .

    Hence,eq. #4.2.3 becomes .


    We can define the variation of as

    ,


    with being a small (real) number. Therefore eq #4.2.5 can be written as

    Since the norm is zero, the wave function itself should be zero. Keeping this in mind, if we analyze eq #4.2.4, it's clear that we can rewrite it as an eigenvalue problem.

    .


    Finally, we can say that expectation value of the Hamiltonian is stationary if the arbitrary wavefunction is actually the eigenvector of the Hamiltonian with the stationary values of the expectation values of the Hamiltonian, being precisely the eigen values of the Hamiltonian.

    The general method is to find an approximate trial wavefunction that contain one or more parameters . If the expectation value can be differentiated with respect to these parameters, the extrema of can be found using the following equation.

    The absolute minimum of the expectation value of the Hamiltonian obtained by this method correspond to the upper bound on the ground state energy. The other relative, extrema corresponds to excited states. There are many virtues of using the Variational method. Even a poor approximation to the actual wave function can yield an excellent approximation to the actual energy.

    Upper Bound on First Excited State

    We claim that if , then where is the energy of the first excited state and is the exact ground state of the Hamiltonian.

    From it is clear that if the above condition is satisfied then, . Therefore,we can write the expectation value of the hamiltonian as

    Thus if we can find a suitable trial wavefunction that is orthogonal to the exact ground state wavefunction, then by calculating the expectation value of the Hamiltonian, we get an upperbound on the first excited state. The trouble is that we might not know the exact ground state( which is one reason why we implement the variational principle). However if we have a Hamiltonian which is an even function, then the exact ground state will be an even function and hence any odd trial function will be a right candidate as the first excited state wavefunction.


    Trial wavefunction with linear parameter and the Hylleraas- Undheim Theory

    Let us consider a set of orthonormal wavefunctions . So any arbitrary trial wavefunction may be constituted by the linear combination of . That is

    where are linear parameters. The energy expectation value is now defined as,

    ...............................(A)



    Since we have not told about the symmetry or orthonormality of , lets us consider,

    and


    Now if are orthonormal then,

    Therefore equation (A) stands as,

    = 0


    Now minimizing the above equation with respect to , we have


    = 0

    omiting the subscript we can write

    = 0

    For = 1,2,3,...... we have

    = 0

    = 0 . .

    .

    .

    = 0


    Now in Matrix form,

    For the non-trivial values of we have


    The solution of this NxN determinant will give N roots say . From the values the first one will give the upper bound to the ground state energy.

    Now substituting in 1st equation of the set of linear equations and then solving them we get the values of and then the wave function . Now if we add one more wave function in then the secular equation will give (N+1) values E, say

    The previous energy level will now spirit to

    This theorem is known as the Hylleraas- Undheim Theorem.

    A Special Case where The Trial Functions form a Subspace

    Assume that we choose for the trial kets the set of kets belonging to a vector subspace of . In this case, the variational method reduces to the resolution of the eigenvalue equation of the Hamiltonian inside , and no longer in all of .

    To see this, we simply apply the argument of Sec. , limiting it to the kets of the subspace . The maxima and minima of , characterized by , are obtained when is an eigen vector of in . The corresponding eigenvalues constitute the variational method approximation for the true eigenvalues of in .

    We stress the fact that the restriction of the eigenvalue equation of to a subspace of the state space can considerably simplify its solution. However, if is badly chosen, it can also yield results which are rather far from true eigenvalues and eigenvectors of in . The subspace must therefore be chosen so as to simplify the problem enough to make it soluble, without too greatly altering the physical reality. In certain cases, it is possible to reduce the study of a complex system to that of a two-level system, or at least, to that of a system of a limited number of levels. Another important example of this procedure is the method of the linear combination of atomic orbitals, widely used in molecular physics. This method essentially consists of the determination of the wave functions of electrons in a molecule in the form of linear combination of the eigenfunctions associated with the various atoms which constitute the molecule, treated as if they were isolated. It, therefore, limits the search for the molecular states to a subspace chosen using physical criteria. Similarly, in complement, we shall choose as a trial wave function for an electron in a solid a linear combination of atomic orbitals relative to the various ions which constitute this solid.

    Applications of Variational Method

    One Dimensional Infinite Well Potential

    The One-D Infinite potential well is defined by:

    V = \left\{\begin{matrix}0, for \left | x \right |<a 

    & \\\infty, for \left | x \right |>a 
&

    \end{matrix}\right.

    The exact solutions are: , and . Now let us suppose that we do not know the exact solution, but try to guess some tiral wavefunction and use variational method to arrive at the approximate answer. Now the wavefunction must vanish at . the ground state wavefunction cannot have any nodes or wiggles.The simplest trial solution satisfying these requirements is a parabola passing through . So our trial solution becomes: \phi(x) = ,upto a normalization factor.

    For this trial function,

    or = = .

    Even with such a simple trial function, with no variational parameter, we come within

    Harmonic Potential

    Armed with the variational method let us apply it first to a simple Hamiltonian. Consider the following Hamiltonian with harmonic potential whose eigenvalues and eigenfunctions are known exactly. We will determine how close we can get with a suitable trial function.



    The above hamiltonian is even therefore, to find the ground state upper bound we need to use an even trial function. Let us consider the following state vector with one parameter



    where is the normalization constant.

    Let us normalize the trial wavefunction to be unity


    While,

    Minimizing the expectation value with respect to the parameter we get,


    Putting this value back in the expectation value, we get

    Due to our judicious selection of trial wavefunction, we were able to find the exact ground state energy. If we want to find the first excited state, a suitable candidate for trial wavefunction would be an odd function.


    Rational wave functions

    The calculations of the previous sections enabled us to familiarize ourselves with the variational method, but they do not really allow us to judge its effectiveness as a method of approximation, since the families chosen always included the exact wave function. Therefore, we shall now choose trial functions of a totally different type, for example

    A simple calculation then yields:

    and finally:

    The minimum value of this function is obtained for:

    and is equal to:

    The minimum value is therefore equal to times the exact ground state energy . To measure the error committed, we can calculate the ratio of to the energy quantum :


    Discussions

    The example of the previous section shows that it is easy to obtain the ground state energy of a system, without significant error, starting with arbitrary chosen trial kets. This is one of the principal advantages of the variational method. Since the exact eigenvalue is a minimum of the mean value , it is not surprising that does not vary much near this minimum.

    On the other hand, as the same reasoning shows, the "approximate" state can be rather different from the true eigenstate. Thus, in the example of the previous section, the wave function decreases too rapidly for small values of and much too slowly when becomes large. The table below gives quantitative support for this qualitative assertion. It gives, for various values of , the values of the exact normalized eigenfunction:

    and of the approximate normalized eigenfunction of the wave function  :

    ,

    where .

    0 0.893 1.034
    1/2 0.696 0.606
    1 0.329 0.270
    3/2 0.094 0.141
    2 0.016 0.084
    5/2 0.002 0.055
    3 0.0001 0.039

    Therefore, it is necessary to be very careful when physical properties other than the energy of the system are calculated using the approximate state obtained from the variational method. The validity of the result obtained varies enormously depending on the physical quantity under consideration. In the particular problem which we are studying here, we find, for example, that the approximate mean value of the operator is not very different from the exact value:

    which is to be compared with . On the other hand, the mean value of is infinite for the approximate normalized eigenfunction, while it is, of course, finite for the real wave function. More generally, the table shows that the approximation will be very poor for all properties which depend strongly on the behavior of the wave function for .

    The drawback we have just mentioned is all the more serious as it is very difficult, if not impossible, to evaluate the error in a variational calculation if we do not know the exact solution of the problem (and, of course, if we use the variational method, it is because we do not know this exact solution).

    The variational method is therefore a very flexible approximation method, which can be adapted to very diverse situations and which gives great scope to physical intuition in the choice of trial kets. It gives good values for the energy rather easily, but the approximate state vectors may present certain completely unpredictable erroneous features, and we can not check these errors. This method is particularly valuable when physical arguments give us an idea of the qualitative or semi-qualitative form of the solutions.


    Here is another problem related to the energy of the ground state and first excited state of a harmonic potential. -problem1

    Delta Function Potential

    As another example lets consider the delta function potential.

    Suppose . Use as a trial wave function, a gaussian function.

    First normalizing this:

    First calculate <T> then <V>.

    Evidently

    Minimizing with respect to the parameter b:

    So, plugging b back into the expression for the expectation value, we get

    Ground State of Helium atom

    Let us use variational principle to determine the ground state energy of a Helium atom with a stationary nucleus. Helium atom has two electrons and two protons. For simplicity, we ignore the presence of neutrons. We also assume that the atom is non relativistic and ignore spin.

    The Hamiltonian can be written as

    where are the coordinates of the two electrons.

    If we ignore the mutual interaction term, then the wavefunction will be the product of the two individual electron wavefunction which in this case is that of a hydrogen like atom. Therefore, the ground state wavefunction can be written as

    where we ignored spin and

    where

    Therefore we can write


    We can write the lowest unperturbed energy for this situation with as

    The first order correction to the energy is

    Therefore, the ground state energy in first approximation is

    However, the ground state energy has been experimentally determined accurately to . Therefore, our model is not a good one. Now, let us apply variational method using a trial wavefunction. The trial wavefucntion we are going to use is #4.5.3.2 itself but we will allow to remain a free parameter. This argument is perfectly valid since each electron screens the nuclear charges seen by the other electron and hence the effective atomic number is less than .

    We can manipulate the Hamiltonian with going to and rewriting the potential term as . So the Hamiltonian becomes

    Now we can use the variational principle. The expectation value of the Hamiltonian is


    The first two terms give

    The fourth and fifth term will give the same. The third term and sixth term will give

    The seventh term will give an expectation value of

    Adding all this we get,


    where


    Excercise 18.22 of E. Merzbacher's Quantum Mechanics (3rd Ed.)

    Since in #4.5.3.7 is the variational parameter, we can minimize the energy, with respect to . That is,

    This will give us

    Therefore, putting this value in #4.5.3.7 , then we have

    where

    Putting , we get Substituting this, in #4.5.3.5 instead of , we get

    which is very close to the experimental value. Thus, using variational principle we were able to calculate the ground state energy of the helium atom very close to the experimental value.

    A sample problem related Rayleigh_Ritz variational principle: Exercise 18.23 of Quantum Mechanics, 3rd Ed., which is written by Eugen Merzbacher: [1]

    A problem related to the variational principle and non-degenerate perturbation theory: problem

    Spin

    General Theory of Angular Momentum

    Up to now we have been working with the rotational degree of freedom using the orbital angular momentum. Namely we use the operator (the generator of rotations in ) to construct wave functions which carry the rotational information of the system.

    To clarify, all that we have done consists of quantizing everything that we know from classical mechanics. Specifically:

    • Invariance of time translation Conservation of the Hamiltonian
    • Invariance of Spatial translation Conservation of the Momentum
    • Invariance of Spacial Rotations Conservation of Orbital Angular Momentum

    However nature shows that there are other kinds of degrees of freedom that don't have classical analog. The first one was observed for the first time in 1922 by Stern and Gerlach (see Cohen-Tannoudji Chap 4). They saw that electrons have one additional degree of freedom of the angular momentum. This degree of freedom was called "Spin 1/2" since they exhibit just two single results: Up and Down. It is interesting to note that from the algebra of angular momenta, it is necessary that spins take on either half-integer or integer values; there is no continuous range of possible spins. (For example, one will never find a spin 2/3 particle.)

    Spin 1/2 is the first truly revolutionary discovery of quantum mechanics. The properties of this physical quantity in itself, the importance of its existence, and the universality of its physical effects were totally unexpected. The physical phenomenon is the following. In order to describe completely the physics of an electron, one cannot use only its degrees of freedom corre- sponding to translations in space. One must take into account the existence of an internal degree of freedom that corresponds to an intrinsic angular mo- mentum. In other words, the electron, which is a pointlike particle, “spins” on itself. We use quotation marks for the word “spins”. One must be cautious with words, because this intrinsic angular momentum is purely a quantum phenomenon. It has no classical analogue, except that it is an angular mo- mentum. One can use analogies and imagine that the electron is a sort of quantum top. But we must keep in mind the word “quantum”. The electron is a pointlike object down to distances of m.One must admit that a pointlike object can possess an intrinsic angular momentum.

    The goal of this section is to extend the notion of orbital angular momentum to a general case. For this we use the letter for this abstract angular momentum. As we will see, orbital angular momentum is just a simple case of the general angular momentum.

    Experimental results

    Experimentally, this intrinsic angular momentum, called spin, has the follow- ing manifestations (we do not enter in any technical detail):

    1. If we measure the projection of the spin along any axis, whatever the state of the electron, we find either of two possibilities:

    or

    There are two and only two possible results for this measurement.

    2.Consequently, if one measures the square of any component of the spin, the result is with probability one.

    3. Therefore, a measurement of the square of the spin gives the result

    4. A system that has a classical analogue, such as a rotating molecule, can rotate more or less rapidly on itself. Its intrinsic angular momentum can take various values. However, for the electron, as well as for many other particles, it is an amazing fact that the square of its spin S2 is always the same. It is fixed: all electrons in the universe have the same values of the square of their spins . The electron “spins” on itself, but it is not possible to make it spin faster.

    One can imagine that people did not come to that conclusion immediately. The discovery of the spin 1/2 of the electron is perhaps the most breathtaking story of quantum mechanics.

    The elaboration of the concept of spin was certainly the most difficult step of all quantum theory during the first quarter of the 20th century. It is a real suspense that could be called the various appearances of the number 2 in physics. There are many numbers in physics; it is difficult to find a simpler one than that.

    And that number 2 appeared in a variety of phenomena and enigmas that seemed to have nothing to do a priori with one another, or to have a common explanation. The explanation was simple, but it was revolutionary. For the first time people were facing a purely quantum effect, with no classical analogue. Nearly all the physical world depends on this quantity, the spin 1/2.

    The challenge existed for a quarter of a century (since 1897). Perhaps, there was never such a long collective effort to understand a physical structure. It is almost impossible to say who discovered spin 1/2, even though one personality dominates, Pauli, who put all his energy into finding the solution.

    We show that in order to manipulate spin 1/2, and understand technical- ities we essentially know everything already. We have done it more or less on two-state systems.


    Note on Hund's Rules

    Hund's Rules describe coupling approximation as long as there is convergence for an atom with a given configuration. Below list the steps.

    1. Choose a maximum value of (total spin) consistent with Pauli exclusion principle

    2. Choose a maximum value of (angular momentum)

    3. If the shell is less than half full, choose

    4. If the shell is more than half full, choose

    For example, we use Silicon: then for 2 spin particles. Angular momentum is and since it is less than half full, .


    Since the equations has the same form (same commutation relationships) as in the case of orbital angular momentum, we can easily extend everything:

    One important feature is that the allowed values for are integers or half-integers (See Shankar). Therefore the possible values for are

    We can construct the following table:


    Each of these columns represent subspaces of the basis that diagonalize and . For orbital angular momentum the allowed values for are integers. This is due to the periodicity of the azimuthal angle.

    For electrons, they have an additional degree of freedom which takes values "up" or "down". Physically this phenomena appears when the electron is exposed to magnetic fields. Since the coupling with the magnetic field is via magnetic moment, it is natural to consider this degree of freedom as internal angular momentum. Since there are just 2 states, therefore, the angular momentum is represented by the subspace .

    It is important to see explicitly the representation of this group. Namely we want see the matrix elements of the operators , , and . The procedure is as follow:

    • and are diagonal since the basis are their eigenvectors.
    • To find and , we use the fact that

    And the matrix elements of are given by

    Let's find the representations for the subspaces , and


    Subspace : (matrix 1x1)


    Subspace : (matrix 2x2)

    • For and are given by


    • The matrices for and are given by
    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} J_x & = \frac{1}{2}[J_+ + J_- ] =\frac{1}{2}\left [ \hbar\begin{pmatrix} 0 & 1 \\ 0 & 0\\ \end{pmatrix} + \hbar\begin{pmatrix} 0 & 0 \\ 1 & 0\\ \end{pmatrix} \right ]\\ &=\frac{\hbar}{2}\begin{pmatrix} 0 & 1 \\ 1 & 0\\ \end{pmatrix}\\ J_y & = \frac{1}{2i}[J_+ - J_- ] =\frac{1}{2i}\left [ \hbar\begin{pmatrix} 0 & 1 \\ 0 & 0\\ \end{pmatrix} - \hbar\begin{pmatrix} 0 & 0 \\ 1 & 0\\ \end{pmatrix} \right ]\\ &=\frac{\hbar}{2i}\begin{pmatrix} 0 & 1 \\ -1 & 0\\ \end{pmatrix} =\frac{\hbar}{2}\begin{pmatrix} 0 & -i \\ i & 0\\ \end{pmatrix} \end{align}}


    Subspace Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle j=1} : (matrix 3x3)

    • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf{J}^{2}= \begin{array}{r|c|c|c} & |1,1\rangle & |1,0\rangle & |1,-1\rangle \\ \hline \langle 1,1| & 2\hbar^{2} & 0 &0 \\ \hline \langle 1,0| & 0 & 2\hbar^{2} &0 \\ \hline \langle 1,-1| & 0 & 0 &2\hbar^{2} \\ \hline \end{array} =2\hbar^{2} \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix} }
    • Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_z = \begin{array}{r|c|c|c} & |1,1\rangle & |1,0\rangle & |1,-1\rangle \\ \hline \langle 1,1| & \hbar & 0 &0 \\ \hline \langle 1,0| & 0 & 0 &0 \\ \hline \langle 1,-1| & 0 & 0 &-\hbar \\ \hline \end{array} =\hbar \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1\\ \end{pmatrix} }
    • For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_+ \!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_- \!} are given by
    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} J_+ & = \hbar \sqrt{1(1+1)-m(m+1)}\;\;\;\delta_{1,1} \delta_{m',m+1}\\ &=\begin{array}{r|c|c|c} & |1,1\rangle & |1,0\rangle & |1,-1\rangle \\ \hline \langle 1,1| & 0 & \hbar \sqrt{1(1+1)-0(0+1)} &0 \\ \hline \langle 1,0| & 0 & 0 &\hbar \sqrt{1(1+1)-(-1)((-1)+1)} \\ \hline \langle 1,-1| & 0 & 0 &0 \\ \hline \end{array} =\hbar \begin{pmatrix} 0 & \sqrt{2} & 0\\ 0 & 0 & \sqrt{2}\\ 0 & 0 & 0\\ \end{pmatrix} \end{align} }
    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} J_- & = \hbar \sqrt{1(1+1)-m(m-1)}\;\;\;\delta_{1,1} \delta_{m',m-1}\\ &=\begin{array}{r|c|c|c} & |1,1\rangle & |1,0\rangle & |1,-1\rangle \\ \hline \langle 1,1| & 0 & 0 &0 \\ \hline \langle 1,0| &\hbar \sqrt{1(1+1)-1(1-1)} & 0 &0 \\ \hline \langle 1,-1| & 0 &\hbar \sqrt{1(1+1)-0(0-1)} &0 \\ \hline \end{array} =\hbar \begin{pmatrix} 0 & 0 & 0\\ \sqrt{2} & 0 & 0\\ 0 & \sqrt{2} &0 \\ \end{pmatrix} \end{align}}


    • The matrices for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_x \!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_y \!} are given by
    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} J_x & = \frac{1}{2}[J_+ + J_- ] =\frac{1}{2}\left [ \hbar \begin{pmatrix} 0 & \sqrt{2} & 0\\ 0 & 0 & \sqrt{2}\\ 0 & 0 & 0\\ \end{pmatrix} + \hbar \begin{pmatrix} 0 & 0 & 0\\ \sqrt{2} & 0 & 0\\ 0 & \sqrt{2} &0 \\ \end{pmatrix} \right ]\\ &=\frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \\ \end{pmatrix}\\ J_y & = \frac{1}{2i}[J_+ - J_- ] =\frac{1}{2i}\left [ \hbar \begin{pmatrix} 0 & \sqrt{2} & 0\\ 0 & 0 & \sqrt{2}\\ 0 & 0 & 0\\ \end{pmatrix} - \hbar \begin{pmatrix} 0 & 0 & 0\\ \sqrt{2} & 0 & 0\\ 0 & \sqrt{2} &0 \\ \end{pmatrix} \right ]\\ &=\frac{\hbar}{\sqrt{2}i} \begin{pmatrix} 0 & 1 & 0\\ -1 & 0 & 1\\ 0 & -1 & 0 \\ \end{pmatrix} =\frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \\ \end{pmatrix} \end{align}}

    Summary

    The following table is the summary of above calculations.

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{r|c|c|c|c|c|c|c|c} & j=0 & j=1/2 &j=1 \\ \hline \mathbf{J}^{2} &0 &\frac{3}{4}\hbar^{2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} &2\hbar^{2} \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ \end{pmatrix}\\ \hline J_z &0 &\frac{\hbar}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} &\hbar \begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1\\ \end{pmatrix}\\ \hline J_x &0 &\frac{\hbar}{2} \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} &\frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0\\ \end{pmatrix}\\ \hline J_y &0 &\frac{\hbar}{2} \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} &\frac{\hbar}{\sqrt{2}} \begin{pmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0\\ \end{pmatrix}\\ \hline \end{array} }

    Spin 1/2 Angular Momentum

    Many particles, such as the electron, proton and neutron, exhibit an intrinsic angular momentum, which unlike orbital angular momentum, has no relation with the spatial degrees of freedom. These are called spin 1/2 particles. An important concept about spin is that it is a purely quantum mechanical construct, with no classical analog, and it cannot be described by a differential operator. The angular momentum of a stationary spin 1/2 particle is found to be quantized to the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pm\frac{\hbar}{2}} regardless of the direction of the axis chosen to measure the angular momentum. This means that there is a vector operator Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{S}=(S_x, S_y, S_z)} and when it projected along an arbitrary axis satisfies the following equations:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{S}\cdot\hat{m}|\hat{m}\uparrow\rangle = \frac{\hbar}{2}|\hat{m}\uparrow\rangle}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{S}\cdot\hat{m}|\hat{m}\downarrow\rangle = -\frac{\hbar}{2}|\hat{m}\downarrow\rangle}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\hat{m}\uparrow\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\hat{m}\downarrow\rangle} form a complete basis, which means that any state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\hat{n}\uparrow\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\hat{n}\downarrow\rangle} with different quantization axis can be expanded as a linear combination of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\hat{m}\uparrow\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\hat{m}\downarrow\rangle} .

    The spin operator obeys the standard angular momentum commutation relations

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [S_{\mu}, S_{\nu}]=i\hbar\epsilon_{\mu\nu\lambda}S_{\lambda}\Rightarrow [S_{x}, S_{z}]=-i\hbar S_{y}}

    The most commonly used basis is the one which diagonalizes Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{S}\cdot \hat{z} = S_{z}} .

    By acting on the states Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\hat{z}\uparrow\rangle} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\hat{z}\downarrow\rangle \!} with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_z \!} , we find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{z}|\hat{z}\uparrow\rangle = \frac{\hbar}{2}|\hat{z}\uparrow\rangle} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{z}|\hat{z}\downarrow\rangle = -\frac{\hbar}{2}|\hat{z}\downarrow\rangle}

    Now by acting to the left with another state, we can form a 2x2 matrix.

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} S_{z} & =\left( \begin{array}{ll} \langle\hat{z}\uparrow|S_{z}|\hat{z}\uparrow\rangle & \langle\hat{z}\uparrow|S_{z}|\hat{z}\downarrow\rangle \\ \langle\hat{z}\downarrow|S_{z}|\hat{z}\uparrow\rangle & \langle\hat{z}\downarrow|S_{z}|\hat{z}\downarrow\rangle \end{array} \right)\\ & =\left(\begin{array}{ll} \hbar/2 & 0 \\ 0 & -\hbar/2 \end{array}\right)\\ & =\dfrac{\hbar}{2}\left( \begin{array}{ll} 1 & 0 \\ 0 & -1 \end{array}\right)\\ &=\dfrac{\hbar}{2}\sigma_{z} \end{align}}

    where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal\sigma_{z}} is the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle z \!} component of Pauli spin matrix. Repeating the steps (or applying the commutation relations), we can solve for the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x \! } and components.



    In this basis, . It should be noted that a spin lying along an axis may be rotated to any other axis using the proper rotation operator.


    Properties of the Pauli Spin Matrices

    Each Pauli matrix squared produces the unity matrix

    The commutation relation is as follows

    and the anticommutator relation

    For example,

    The above equation is true for spins only!!

    In general,

    Finally, any matrix can be written in the form

    For infinitesimal

    Spin.JPG

    Note that using the previous developed formulas, we find that

    To this order in , this equation is equivalent to

    .

    This equation is exact for any not just infinitesimal just as in hte case of orbital angular momentum.

    Consider acts on which is an eigenstate of ,


    Another way of expressing the rotation of the spin basis by an angle about some axis (and the one derived in class) is the following.

    Consider an operator from the previous equation. This can also be written as

    .

    Consider, .

    The non-squared terms vanish because of the anti-commutation property of the Pauli matrices.

    Therefore, ( is an integer), thus the above equation can be split:

    This form may be more convenient when performing rotations. A solved problem for spins

    A Solved Problem on General Spin Vectors.

    Addition of angular momenta

    Formalism

    In order to consider the addition of angular momentum, consider two angular momenta, and which belong to two different subspaces. has a Hilbert space of states, and has a Hilbert space of states. The total angular momentum is then given by:

    where and are the identity matrices of 's and 's Hilbert spaces, and where the dimension Hilbert space is .

    The components of and obey the commutation relation:

    And since and belong to different Hilbert spaces:

    Given the simultaneous eigenkets of and denoted by , and of and denoted by we have the following relations:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_1^2|j_1 m_1\rangle = j_1(j_1+1)\hbar|j_1 m_1\rangle }

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{1z}|j_1 m_1\rangle = m_1\hbar|j_1 m_1\rangle }

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_2^2|j_2 m_2\rangle = j_2(j_2+1)\hbar|j_2 m_2\rangle }

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{2z}|j_2 m_2\rangle = m_2\hbar|j_2 m_2\rangle }

    Now looking at the two subspaces together, the operators Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_1^2} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!J_{1z}} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_2^2} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!J_{2z}} can be simultaneously diagonalized by their join eigenstates. These eigenstates can be formed by the direct products of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_1 m_1\rangle } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_2 m_2\rangle } :

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_1 m_1\rangle \otimes |j_2 m_2\rangle = |j_1,j_2; m_1,m_2\rangle }

    This basis for the total system diagonalizes Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_1^2 \!} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!J_{1z}} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_2^2 \! } , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!J_{2z}} , but these four operators DO NOT define the total angular momentum of the system. Therefore it is useful to relate these direct product eigenstates to the total angular momentum Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{J} = \vec{J}_{1} + \vec{J}_{2}} .

    Recall that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{z} = J_{1z} + J_{2z} \!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[J_{\mu}, J_{\nu}\right] = i\hbar\epsilon_{\mu\nu\lambda} J_{\lambda}} .

    We also know the relations: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[J_{1,2}^2, J^2\right]=0 } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[J_{1,2}^2, J_{z}\right] = 0 } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left[J^{2}, J_{z}\right] = 0 }

    This tells us that we have a set of four operators that commute with each other. From this we can specify Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{1}^2 , J_{2}^2 , J^2} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{z}\!} simultaneously. The joint eigenstates of these four operators denoted by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j m j_1 j_2\rangle} . These four operators operate on the base kets according to:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J^2|j m j_1 j_2\rangle = \hbar^2 j(j + 1)|j m j_1 j_2\rangle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\;\;\;\; (6.1.7)}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{z}|j m j_{1} j_{2}\rangle=\hbar m |j m j_{1} j_{2}\rangle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad\;\;\;\;\;\ (6.1.8)}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{1,2}^2|j m j_{1} j_{2}\rangle=\hbar^2 j_{1,2}(j_{1,2}+1) |j m j_{1} j_{2}\rangle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad\;\;\ (6.1.9)}

    The choice of basis is now dictated by the specific problem being solved because we can find the relationship between the direct product basis and total-Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J \!} basis.

    For example, consider two spin 1/2 particles with basis Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\uparrow\uparrow\rangle, |\uparrow\downarrow\rangle, |\downarrow\uparrow\rangle ,} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\downarrow\downarrow\rangle } . These states are eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{1}^2, J_{2}^2, J_{1z},\!} andFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{2z}\!} , but are they eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!J^2} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_z^2} ?
    Let us see what happens with the state Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\uparrow\downarrow\rangle} :

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J^2 |\uparrow\downarrow\rangle = \left(J_{x}^2+J_{y}^2+J_{z}^2\right)|\uparrow\downarrow\rangle = \left((J_{x}+iJ_{y})(J_{x}-iJ_{y})+i[J_{x}, J_{y}] + J_{z}^2 \right)|\uparrow\downarrow\rangle} .

    Let's define , then

    Now

    Also,

    Which means that is not an eigenstate of . Similarly, it can be shown that the other three states are also not eigenstates of .


    To find a relationship between the direct product basis and the total- basis, begin by finding the maximum total state:

    This must be true because is the only state with . Now we can lower this state using to yield:

    And then lower this state to yield:

    All we are missing now is the antisymmetric combination of and :

    We now have a relationship between the two bases. Also, we can write where and represent triplet state and single state respectively.

    Problem: CG coefficients[[2]]

    Another problem: CG coefficients[[3]]

    Clebsch-Gordan Coefficients

    Now that we have constructed two different bases of eigenkets, it is imperative to devise a way such that eigenkets of one basis may be written as linear combinations of the eigenkets of the other basis. To achieve this, we write:

    In above, we have used the completeness of the basis , given by:

    The coefficients are called Clebsch-Gordan coefficients (for an extensive list of these coefficients, see here), which have the following properties, giving rise to two "selection rules":

    1. If , then the coefficients vanish.

    Proof: , we get
    . Q.E.D.

    2. The coefficients vanish, unless

    This follows from a simple counting argument. Let us assume, without any loss of generality, that . The dimensions of the two bases should be the same. If we count the dimensions using the states, we observe that for any value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \! j } , the values of run from to . Therefore, for and , the number of eigenkets is . Now, counting the dimensions using the eigenkets, we observe that, again, runs from to . Therefore, the number of dimensions is . It is easy to see that for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \! a = j_1 - j_2 } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \! b = j_1 + j_2.} Therefore, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N = (2j_1 + 1)(2j_2 +1)\!} .

    Further, it turns out that, for fixed Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!j_1} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!j_2} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!j} , coefficients with different values for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!m_1} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!m_2} are related to each other through recursion relations. To derive these relations, we first note that: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{\pm}|j m j_1 j_2\rangle = \sqrt{(j \mp m)(j \pm m + 1)}\hbar |j m \pm 1 j_1 j_2\rangle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\;\ (6.2.3) }

    Now we write, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{\pm}|j m j_1 j_2 \rangle = (J_{1 \pm} + J_{2 \pm}) \sum_{m_1,m_2}|j_1 j_2 m_1 m_2 \rangle \langle j_1 j_2 m_1 m_2|j m j_1 j_2 \rangle \qquad \qquad \qquad \qquad (6.2.4)}

    Using equation #(6.2.4), we get (with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_1 \to m'_1 } , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_2 \to m'_2 } ):

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} & \sqrt{(j \mp m)(j \pm m + 1)}|j m \pm 1 j_1 j_2 \rangle \\ & = \sum_{m'_1,m'_2} \left( \sqrt{(j_1 \mp m'_1)(j_1 \pm m'_1 + 1)}|j_1 j_2 m'_1 \pm 1 m'_2 \rangle + \sqrt{(j_2 \mp m'_2)(j_2 \pm m'_2 + 1)}|j_1 j_2 m'_1 m'_2 \pm 1 \rangle \right)\langle j_1 j_2 m'_1 m'_2|j m j_1 j_2 \rangle \end{align}}


    The Clebsch-Gordan coefficients form a unitary matrix, and by convention, they are all taken real. Any real unitary matrix is orthogonal, as we study below.


    Additionally, another way to calculate the Clebsch-Gordan coefficients would be to go to www.volya.net, which is Dr. Alex Volya's website. Simply click the link on the side under the menu "Science Tools" for "Vector Coupling" and follow the directions for entering the angular momenta. The website will then calculate the Clebsch-Gordan coefficients for you.

    Example

    As an example lets calculate some Clebsch-Gordan coefficients through applications of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{\plusmn}=S_x+iS_y} on the states Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |Sm_s> } .


    Let Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S=S_1+S_2} be the total angular momentum of two spin 1/2 particles Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (S_1=S_2=1/2) } . Calculate the Clebsch-Gordan coefficients

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <m_1m_2|Sm_s> } by successive applications of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{\plusmn}=S_x+iS_y } on the states Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |Sm_s> } . Work separately in the two subspaces S=1 and S=0.

    In order to find the coefficients for the addition of spin 1/2, we shall use the following relations:

    I Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{\plusmn}|Sm_s> = \hbar\sqrt{S(S+1)-m_s(m_s\plusmn 1)}|Sm_s\plusmn 1>}

    II Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{1\plusmn}|m_1m_2> = \hbar\sqrt{S_1(S_1+1)-m_1(m_1\plusmn 1)}|m_1\plusmn 1 m_2>}

    IIIFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{2\plusmn}|m_1m_2> = \hbar\sqrt{S_2(S_2+1)-m_2(m_2\plusmn 1)}|m_1 m_2\plusmn 1>}

    We shall also use the phase condition

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |S=S_1+S_2,m_s=\plusmn(S_1+S_2)> = |m_1=\plusmn S_1,m_2=\plusmn S_2> }

    Note: The states Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |S=S_1=S_2,m_s=\plusmn (S_1+S_2)> } are eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S^2 } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_z } , with nondegenerate eigenvalues Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lambda_{\plusmn} = \plusmn \hbar(S_1+S_2) } , repsectively. Therefore,

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |S=S_1+S_2,m_s=\plusmn(S_1+S_2)> = e^{i \phi}|m_1=\plusmn S_1,m_2=\plusmn S_2> }

    and the phase may be chosen to be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi =0} .


    i Subspace S=1: From the phase condition we immediately have Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,1>=|1/2,1/2>=|++> }

    Then, operating with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_-=S_{1-}+S_{2-} } on both sides of this, and using II and III, we obtain

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{-}|1,1>=\hbar\sqrt{1(1+1)-1(1-1)}|1,0>=\hbar\sqrt{2}|1,0>} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{-}|1,1>=(S_{1-}+S_{2-})|1/2,1/2>=\hbar\sqrt{1}|-1/2,1/2>+\hbar|1/2,-1/2>}

    Thus, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,0>=\frac{1}{\sqrt{2}}(|1/2,-1/2>+|-1/2,1/2>) = \frac{1}{\sqrt{2}}(|+->+|-+>) }

    Similarly, operating with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{-} } once again on the state |1,0>, we find

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{-}|1,0> = \hbar\sqrt{1(1+1)-0(0-1)}|1,-1>=\hbar\sqrt{2}|1,-1>} Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle S_{-}|1,0> = \frac{1}{\sqrt{2}}S_{1-}(|1/2,-1/2>+|-1/2,1/2>)+\frac{1}{\sqrt{2}}S_{2-}(|1/2,-1/2>+|-1/2,1/2>) } Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle = \frac{\hbar}{\sqrt{2}}(|-1/2,-1/2>+|-1/2,-1/2>) = \frac{2}{\sqrt{2}}|-1/2,-1/2>}

    Therefore, in accordance with the phase condition, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1,-1> = |-1/2,-1/2> = |--> }


    ii Subspace S=0: Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_s = m_1 + m_2} (in this case Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m_s=0} ), we have

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |0,0> = \alpha|-1/2,1/2>+ \beta |-1/2,1/2> }

    Next, due to orthonormality of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |Sm_s> } basis we get

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <1,0|0,0> = 0 \rightarrow \frac{1}{\sqrt{2}}(\alpha+\beta)=0 \rightarrow \beta=-\alpha }

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle <0,0|0,0>=1 \rightarrow |\alpha|^2+|\beta|^2=1 \rightarrow 2|\alpha|^2=1 \rightarrow \alpha = \frac{1}{\sqrt{2}} }

    Therefore, we find Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |0,0> = \frac{1}{\sqrt{2}}(|1/2,-1/2>-|-1/2,1/2>) }

    Orthogonality of Clebsch-Gordan Coefficients

    We have the following symmetry:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle j_1 j_2 m_1 m_2 | j_1 j_2 j m \rangle = (-1)^{j-j_1-j_2} \langle j_2 j_1 m_2 m_1 | j_2 j_1 j m \rangle = \langle j_2 j_1, -m_2, -m_1 |j_2 j_1 j, -m \rangle}

    If we put the coefficients into a matrix, it is real and unitary, meaning Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \langle j m j_1 j_2 |j_1 j_2 m_1 m_2 \rangle = \langle j_1 j_2 m_1 m_2 |j m j_1 j_2 \rangle ^*}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle | j_1 j_2 m_1 m_2 \rangle = \sum_{j,m} |j m j_1 j_2 \rangle \langle j m j_1 j_2 |j_1 j_2 m_1 m_2 \rangle } .

    For example,

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\uparrow_1 \downarrow_2 \rangle = \dfrac{1}{\sqrt{2}}(|10 \rangle + |00 \rangle )}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |\downarrow_1 \uparrow_2 \rangle = \dfrac{1}{\sqrt{2}}(|10 \rangle - |00 \rangle )}

    We have the following orthogonality relations:

    Hydrogen atom with spin orbit coupling given by the following hamiltonian

    Recall, the atomic spectrum for bound states

    where

    The ground state, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1s\rangle} , is doubly degenerate: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \dfrac{\uparrow\downarrow}{1s}}

    First excited state is 8-fold degenerate: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \dfrac{\uparrow\downarrow}{1s}\dfrac{\uparrow\downarrow}{}\dfrac{\uparrow\downarrow}{2p}\dfrac{\uparrow\downarrow}{}}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n \!} -th state is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2n^2\!} fold degenerate.

    We can break apart the angular momentum and spin into its Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x, y, z \!} -components

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{L}\cdot\vec{S} = L_x S_x + L_y S_y + L_z S_z }

    Define lowering and raising operators

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow L_\pm = L_x \pm iL_y}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow S_\pm = S_x \pm iS_y}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{L}\cdot\vec{S} = L_z S_z + \dfrac{1}{2} L_{+} S_{-} + \dfrac{1}{2} L_{-} S_{+} }

    For the ground state, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (|1s, \uparrow\rangle, |1s, \downarrow\rangle )} , nothing happens. Kramer's theorem protects the double degeneracy.

    For the first excited state, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (|2s, \uparrow\rangle, |2s, \downarrow\rangle )} , once again nothing happens.

    For Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (|2p, \uparrow\rangle, |2p, \downarrow\rangle )} , there is a four fold degeneracy.

    We can express the solutions in matrix form

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( \begin{array}{llllll} \dfrac{\hbar^2}{2} & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & \dfrac{\hbar^2}{2} \end{array} \right)}

    But there is a better and more exact solution, which we can solve for by adding the momenta first.

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{L}\cdot\vec{S} = \frac{1}{2} \left(\vec{L} + \vec{S}\right)^2 -\frac{1}{2}\vec{L}^2 -\frac{1}{2}\vec{S}^2 = \frac{1}{2}\left(J^2 -L^2 - S^2\right)}

    add the angular momenta:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |1s\rangle : l=0, s=\dfrac{1}{2}: 0\otimes \dfrac{1}{2}= \dfrac{1}{2}}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |2s\rangle : l=0, s=\dfrac{1}{2}: 0\otimes \dfrac{1}{2}= \dfrac{1}{2}}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |2p_m, 0 \rangle : l=1, s=\dfrac{1}{2}: 1\otimes \dfrac{1}{2}= \dfrac{3}{2} \oplus \dfrac{1}{2}}

    So that

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{L}\cdot\vec{S} \left|j=\dfrac{3}{2}, m, l=1, s=\dfrac{1}{2} \right\rangle =\dfrac{1}{2} \left(\hbar^2\dfrac{3}{2}\dfrac{5}{2}-2 \hbar^2 - \dfrac{3}{4} \hbar^2\right) \left|j=\dfrac{3}{2}, m, l=1, s=\dfrac{1}{2} \right\rangle = \dfrac{\hbar^2}{2} \left| j=\dfrac{3}{2}, m, l=1, s=\dfrac{1}{2} \right\rangle }

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{L}\cdot\vec{S} \left|j=\dfrac{1}{2}, m, l=1, s=\dfrac{1}{2} \right\rangle =\dfrac{-\hbar^2}{2} \left| j=\dfrac{1}{2}, m, l=1, s=\dfrac{1}{2} \right\rangle }

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|j=\dfrac{3}{2}, m= \dfrac{3}{2}, l=1, s=\dfrac{1}{2} \right\rangle = \left|l=1, m_l =1 \right\rangle \left|s=\dfrac{1}{2}, m_s = \dfrac{1}{2} \right\rangle }

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|j= \dfrac{3}{2}, m= \dfrac{3}{2} \right\rangle = \left|m_l =1 \right\rangle \left|m_s = \dfrac{1}{2} \right\rangle }

    Define Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{-} = L_{-} + S_{-} \!}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle J_{-} \left|\dfrac{3}{2}, \dfrac{1}{2} \right\rangle = \left(L_{-} + S_{-} \right)\left|l=1, m=1 \right\rangle \left| S=\frac{1}{2}, m_s=\frac{1}{2} \right\rangle }


    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \hbar \sqrt{\dfrac{3}{2} \dfrac{5}{2}- \dfrac{3}{2}\dfrac{1}{2}} \left|\dfrac{3}{2}, \dfrac{1}{2} \right\rangle = \hbar \sqrt{2} \left|l=1, m=0 \right\rangle \left|s=\frac{1}{2}, m_s=\frac{1}{2} \right\rangle + \hbar \sqrt{\dfrac{1}{2} \dfrac{3}{2} - \frac{1}{2}\left(\frac{1}{2}-1\right)}|l=1,m=1\rangle \left| s=\frac{1}{2}, m_s=-\frac{1}{2} \right\rangle}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \sqrt{3} \left|\dfrac{3}{2}, \dfrac{1}{2} \right\rangle = \sqrt{2}\left|1,0 \right\rangle \left|\frac{1}{2}, \frac{1}{2}\right\rangle + \left|1,1 \right\rangle \left| \frac{1}{2}, -\frac{1}{2} \right\rangle}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \left|\frac{3}{2}, \frac{1}{2} \right\rangle = \sqrt{\frac{2}{3}}\left|1,0 \right\rangle \left|\dfrac{1}{2}, \frac{1}{2} \right\rangle + \sqrt{\dfrac{1}{3}}\left|1,1 \right\rangle \left| \frac{1}{2},- \dfrac{1}{2} \right \rangle}

    As the same,

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|\dfrac{3}{2}, -\dfrac{1}{2} \right\rangle = \sqrt{\dfrac{2}{3}} \left|1,0 \right\rangle \left| \frac{1}{2},-\dfrac{1}{2} \right\rangle + \sqrt{\dfrac{1}{3}}\left|1,-1 \right\rangle \left| \frac{1}{2}, \dfrac{1}{2} \right\rangle} ,

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|\dfrac{3}{2}, \pm \dfrac{3}{2} \right\rangle = \left|1, \pm 1 \right \rangle \left| \frac{1}{2}, \pm \dfrac{1}{2} \right\rangle}

    We can express as follows:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|j=\dfrac{1}{2}, m =\dfrac{1}{2} \right\rangle = \alpha \left|1,0 \right\rangle \left| \frac{1}{2}, \dfrac{1}{2} \right\rangle + \beta \left|1,1 \right\rangle \left|\frac{1}{2}, -\dfrac{1}{2} \right\rangle } ,

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left|j=\dfrac{1}{2}, m = -\dfrac{1}{2} \right \rangle = \alpha ' \left|1,0 \right\rangle \left|\frac{1}{2},-\dfrac{1}{2} \right\rangle + \beta ' \left|1,-1 \right\rangle \left| \frac{1}{2}, \dfrac{1}{2} \right\rangle } .

    When we project these states on the previously found states, we find that

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha = \dfrac{1}{\sqrt{3}}} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta = - \sqrt{\dfrac{2}{3}}} ,

    and

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \alpha' = - \dfrac{1}{\sqrt{3}}} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \beta' = \sqrt{\dfrac{2}{3}}} .

    For a more detailed account of these and other related results, see here.

    Addition of three angular momenta

    To add three angular momenta Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold J_1, \bold J_2, \bold J_3} , first we add Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold J_{12}=\bold J_1 + \bold J_2} , and construct the simultaneous eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold J_{1} ^{2}, \bold J_2 ^{2}, \bold J_{12} ^{2}, \bold J_{12z}, \bold J_3 ^{2}, \bold J_{3z}} . We write such states as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_1 j_2 j_{12} m_{12} j_3 m_3 \rangle} . Such states can be given in terms of Clebsch-Gordan coefficients and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_1 j_2 j_3 m_1 m_2 m_3 \rangle} (eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold J_1^{2}, \bold J_2^{2}, \bold J_3^{2}, \bold J_{1z}, \bold J_{2z}, \bold J_{3z}} ):

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_1 j_2 j_{12} m_{12} j_3 m_3 \rangle = \sum_{m_1,m_2} |j_1 j_2 j_3 m_1 m_2 m_3 \rangle \langle j_1 j_2 j_3 m_1 m_2 m_3|j_1 j_2 j_{12} m_{12} j_3 m_3\rangle}

    Next we add Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold J_{12}} to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold J_3} , forming simultaneous eigenstates Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_1 j_2 j_{12} j_3 j m \rangle} of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold J_1^{2}, \bold J_2^{2}, \bold J_{12}^{2}, \bold J_3^{2}, \bold J^{2}, \bold J_{z}} . These are given in terms of the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle | j_1 j_2 j_{12} m_{12} j_3 m_3 \rangle} by

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_1 j_2 j_{12} j_3 j m \rangle = \sum_{m_{12},m_{3}} | j_1 j_2 j_{12} m_{12} j_3 m_3 \rangle \langle j_{12} m_{12} j_3 m_3 | j_{12} j_3 j m \rangle }

    Therefore, we can construct eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold J_1^{2}, \bold J_2^{2}, \bold J_{12}^{2}, \bold J_3^{2}, \bold J^{2}, \bold J_{z}} in terms of eigenstates of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \bold J_1^{2}, \bold J_2^{2}, \bold J_3^{2}, \bold J_{1z}, \bold J_{2z}, \bold J_{3z}} as follows:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |j_1 j_2 j_{12} j_3 j m \rangle = \sum_{m_{1},m_{2},m_{3}} |j_1 j_2 j_3 m_1 m_2 m_3 \rangle \sum_{m_{12}} \langle j_1 j_2 m_1 m_2|j_1 j_2 j_{12} m_{12}\rangle \langle j_{12} m_{12} j_3 m_3 | j_{12} j_3 j m \rangle }

    Thus the analogous addition coefficients for three angular momenta are products of Clebsch-Gordan coefficients.

    Note that for addition of two angular momenta, the dimension of Hilbert space is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!(2J_1 + 1)(2J_2 + 1)} . For three angular momenta, it is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!(2J_1 + 1)(2J_2 + 1)(2J_3 + 1)} .

    Elementary applications of group theory in Quantum Mechanics

    Symmetry

    Mathematically, A Group is consisted as a set of elements in which a operation is defined to combine any two of the elements to form a third element. Also, the group should satisfy some axioms such as closure, associativity, identity and invertibility.

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G=\lbrace A,B,C \dots\rbrace}

    G is a group under operation AB if

    . ,forFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \forall A,B\in G} (closure)

    . Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left(AB\right)C=A\left(BC\right)} (associativity)

    . Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \exists \mathbf I} , such that , forFailed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \forall A} (identity)

    . Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \forall A, \exists A^{-1}} , Such that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle AA^{-1}=A^{-1}A=\mathbf I } (invertibility)


    Broad characteristics of various groups

    Discrete and Continuous Groups

    G can be discrete(isolated elements) or it can be continuous(rotation)

    Examples(discrete group):

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf I_2=\lbrace\mathbf I,A\rbrace} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A^{2}=\mathbf I}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathbf I_n=\lbrace n \rbrace} , The operation Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle “\circ”} is defined as Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n\circ m=n+m}

    And continuous group:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle U\left(1\right)=\lbrace e^{i\theta}; \theta\in\lbrack 0,2\pi\rbrack\rbrace} , in which Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle e^{i\theta}e^{i\phi}=e^{i\left(\theta+\phi\right)}}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle SU\left(2\right)} : group of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\times2} matrices with unit determinant (special unitary group)

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle O\left(3\right)} : group of all rotations about the origin in 3D. i.e. set of all orthogonal transformation in a 3D vector space, or a group of all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3\times3} orthogonal matrices.

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle SU\left(3\right)} : a group of all unitary matrices with determinant Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle +1}

    Abelian Group

    If all the elements commute with each other, then the group is "Abelian."

    In a group Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle G} ,if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle AB=BA \forall A,B\in G}

    Then the group is said to be abelian.

    Example:The addition of real numbers yields an abelian group. However, the group of square matrices with dimension N under multiplication is not abelian.

    Non-Abelian Group

    If all the elements of the group do not commute with each other, then the group is "Non-Abelian."

    Example: Matrices, in general do not commute, and can form a non-abelian group. For example, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle SU\left(2\right)} : the group of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2\times2} matrices with unit determinant (special unitary group)

    Continuously Connected

    A groups is called continuously connected if a continuous variation of the group parameters leads from any arbitrary element of the group to any other.

    Example: The translation group of the elements Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {a = a_{x}\varepsilon_{1}+a_{y}\varepsilon_{2}+a_{z}\varepsilon_{3}}} possesses three continuous parameters Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (a_{x},a_{y},a_{z})} .We can generate each displacement vector in space by continuous variation of these parameters. However, rotations combined with reflection in space [called Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle O[3]} ], form a continuous,not connected group. similarly, the Lorentz transformation group in relativity for a not connected group.


    Definition of conjugate elements: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B} is conjugate to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} if Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B=XAX^{-1}} for some Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle X\in G} . This property is reciprocal since that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=X^{-1}BX}

    Collecting all conjugate elements gives a conjugacy class. We can divide G into conjugate classes.

    Compact Group In each sequence within a compact group there exists an infinite partial sequence Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a_{n}} of group elements, which converges to an element of the group, i.e Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\to\infty}a_{n}=a,a\in G}

    Examples, (a) The group of the translation vectors on a lattice Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T = {a_{n} = n_{1}\varepsilon_{1}+n_{2}\varepsilon_{2}+n_{3}\varepsilon_{3}}} , with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n_{1},n_{2},n_{3} \in N}

    Theory of group representations

    Group representations are the description on abstract objects using linear transformations of vector spaces. Usually, the elements in the group are represent by matrices so that the group operation can be represents by matrix multiplication.

    Associate a matrix Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma\left(A\right)} with each Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A\in G}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Gamma\left(A\right)\Gamma\left(B\right)=\Gamma\left(AB\right)}


    Group theory and Quantum Mechanics

    Irreducible tensor representations and Wigner-Eckart theorem

    Representation of rotations

    If is the total angular momentum of a system (), the operator acting to the right on a state of the system, rotates it in a positive sense about the axis by an angle . This is similar to how rotates in the plane, and rotates spin states. Suppose that we act with on an eigenstate of and . This generates a superposition of states. Under the rotation, the state is generally no longer an eigenstate of . However, the rotated state remains an eigenstate of , so the value of remains the same while the value of will change. This is because commutes with every component of , (,,), and therefore commutes with . Indeed

    Therefore, when we act with the rotation operator on a state, we are only mixing the multiplet. For example, acting on a 3d state with the rotation operator will result in a mixture of only the five 3d states. There will be no mixing of the 3p or 3s states.

    Considering , we know will not commute with because does not commute with either or :

    Therefore the rotated state can be expressed as a linear combination of as follows:

    Multiplying this equation on the left by a state , and using the orthonormality of the angular momentum eigenstates we find

    Thus we can associate with each rotation a by dimension matrix whose matrix elements are . These matrix elements don't depend on dynamics of the system; they are determined entirely by the properties of the angular momentum.

    The matrices have a very important property. Two rotations performed in a row, say followed by , are equivalent to a single rotation, . Thus

    Taking matrix elements of both sides and putting in a complete set of states between and we find

    ,

    or

    or equivalently

    A set of matrices associated with rotations having this property is call a representation of the rotation group.

    The rotation operators act on the set of states for fixed j, in an irreducible fashion. To see what this means, let's consider the effect of rotations on the set of eight states for and . Under any rotation a state becomes a linear combination of states, with no components, conversely a state becomes a linear combination of states with no components. Thus this set of eight states which transform each among themselves under rotation with no mixing. One says that the rotations act on these eight states reducibly. On the other hand, for a set of states all with the same j, there is no smaller subset of states that transforms privately among itself under all rotations; the rotations are said to act irreducibly. Put another way, if we start with any state, , then we can rotate it linearly independent states, and therefore there can't be any subspace of j states that transforms among itself under rotations. One can prove this in detail starting from the fact that one can generate all the states starting form by applying enough times.

    Tensor operators

    The types of operators having simple transform properties under rotations are known as tensor operators. By an irreducible tensor operator of order k we shall mean a set of 2k+1 operators that transform among themselves under rotation according to the transformation law:

    If we consider and infinitesimal rotation ,then

    to the first order in

    Comparing coefficients of we see that tensor operators must obey the commutation relation with the angular momentum:

    The z component of this relation is

    while

    Tensor operators have many simple properties. For example, acts on state of a system ( refers to other quantum number), creating a state whose z component of angular momentum is . To prove this, let us consider the transformation properties of the state under rotation about the z axis by

    but

    so that

    This is exactly the transformation law for an eigenstate of with eigenvalue . Thus is an operator that increases the eigenvalue of by q.

    Wigner-Eckart theorem

    The Wigner-Eckart theorem postulates that in a total angular momentum basis, the matrix element of a tensor operator can be expressed as the product of a factor that is independent of and a Clebsch-Gordan coefficient. To see how this is derived, we can start with the matrix element , where the represents all the properties of the state not related to angular momentum:

    Using the orthogonality of rotation matrices, this reduces to

    Finally, using the fact that and the orthogonality of the Clebsch-Gordan coefficients, we obtain

    Historically, this is written as

    where is referred to as the reduced matrix element.

    In summary, the Wigner-Eckart theorem states that the matrix elements of spherical tensor operators with respect to the total- eigenstates can be written in terms of the Clebsch-Gordan coefficients, , and the reduced matrix elements of , which do not depend on the orientation of the system is space, i.e., no dependence on , , and :

    As an example of how this theory can be useful, consider the example of the matrix element with hydrogen atom states . Because of the Clebsch-Gordan coefficients, the matrix element is automatically zero unless and or . Also, because z is odd under parity, we can also eliminate the transition.

    Also, for , the Wigner-Eckart Theorem reads

    The result vanishes since the CG coefficients on the right hand side are zero.

    Problem [4]

    EXAMPLE PROBLEM [5]

    Application [6]

    Elements of relativistic quantum mechanics

    The description of phenomena at high energies requires the investigation of the relativistic wave equations, the equations which are invariant under Lorentz transformations. The translation from a nonrelativistic to a relativistic description, implies that several concepts of the nonrelativistic theory have to be reinvestigated, in particular:

    (1) Spatial and temporal coordinates have to be treated equally within the theory.

    (2) Since, from the Uncertainty principle, we know

    ,

    a relativistic particle can not be localized more accurately than ; otherwise pair creation occurs for . Thus, the idea of a free particle only makes sense if the particle is not confined by external constraints to a volume which is smaller than approximately than the Compton wavelength . Otherwise, the particle automatically has companions due to particle-antiparticle creation.

    (3) If the position of the particle is uncertain, i.e. if

    ,

    then the time is also uncertain, because

    .

    In a nonrelativistic theory, can be arbitrary small, because . Thereby, we recognize the necessity to reconsider the concept of probability density, which describes the probability of finding a particle at a definite place at a fixed time .

    (4) At high energies i.e. in the relativistic regime, pair creation and annihilation processes occur, ususlly in the form of creating particle-antiparticle pairs. Thus, at relativistic energies, particle conservation is no longer a valid assumption. A relativistic theory must be able to describe the phenomena like pair creation, vacuum polarization, particle conservation etc.

    In nonrelativistic quantum mechanics, states of particles are described by Schrodinger equation of states:

    Schrodinger equation is a first order differential equation in time. However, it is second order in space and therefore, it is not invariant under the Lorentz transformations. As mentioned above, in relativistic quantum mechanics, the equation describing the states must be invariant under the Lorentz transformations. In order to satisfy this condition, equation of state must contain the derivatives with respect to time and space of the same order. Equations of states in relativistic quantum mechanics are Klein-Gordon equation (for spinless particles) and Dirac equation (for spin particles). The former contains second ordered derivatives while the latter contains first ordered derivatives with respect to both time and space. The way to derive these equations is similar to that of Schrodinger equation: making use of the correspondence principle, starting from the equation connecting the energy and momentum with the substitution by and by .

    Follow this link to learn about Klein-Gordon equation.

    Follow this link to learn about Dirac equation.

    Here is a worked problem for a free relativistic particle.

    Here is a worked problem to review the use of relativistic 4-vectors: relativistic 4-vectors

    The Adiabatic Approximation and Berry Phase

    The adiabatic approximation can be applied to systems in which the Hamiltonian evolves slowly with time. The Hamiltonian of an adiabatic system contains several degrees of freedom. The basic idea behind the adiabatic approximation is to solve the Schrodinger equation for the "fast" degree of freedom and only then allow the "slow" degree of freedom to evolve slowly. For example, imagine a molecule with a heavy nucleus and an electron. In this system there is a "slow" degree of freedom (the nucleus) and a "fast" degree of freedom (the electrons). Imagine that the nucleus is stationary, and the electrons align themselves. Now that the electrons have aligned themselves, allow the nucleus to move very slowly - which will cause the electrons to realign. This is the adiabatic approximation.

    Adiabatic Process

    The gradual change in the external conditions characterizes an adiabatic process. In another word, let be the internal characteristic frequency, be the external characteristic frequency. The adiabatic process is one for which .


    The Adiabatic Theorem

    The adiabatic theorem states that if a system is initially in the nth state and if its Hamiltonian evolves slowly with time, it will be found at a later time in the nth state of the new Hamiltonian. (Proof: Messiah Q.M. (wiley NY 1962) Vol II ch. XVII)

    Application(Born-Oppenheimer Approximation)[7]

    Geometric Phase (Berry Phase)

    The phase of a wave function is often considered arbitrary, and it is canceled out for most physics quantities, such as . For that reason, the time-dependent phase factor on the wave function of a particle going from the nth eigenstate of to the nth eigenstate of was ignored. However, Berry showed that if the Hamiltonian is evolved over a closed loop the relative phase is not arbitrary, and cannot be gauged away. For more information on this discovery, see this paper. This is called the Berry Phase.

    If ,

    , where , is called dynamic phase, and is geometric phase.

    Do you want to know what geometric phase looks like? If so, let' s begin our work :

    Substitute into Schrodinger equation,

    So,

    Since


    This is the expression of geometric phase.

    If it's a 1 D problem : Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma _n(t)=0} , there is no geometric phase change.

    If more than 1 D: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma _n(t)=\int _{R_i}^{R_f}\langle n\left(t'\right)|\nabla _R\left|n\left(t'\right)\rangle dR.\right.}

    The larger number of dimensions allows for the geometric phase change.


    Berry' s phase : If the Hamiltonian returns to its oringinal form after a time T, the net geometric phase change is :

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma _n(t)=\oint \langle n\left(t'\right)|\nabla _R\left|n\left(t'\right)\rangle dR.\right.}


    Why geometric phase is special? Because it does have physical meaning. We can observe it from interference experiment.

    Berry Potentials

    It is possible to construct potentials that give rise to this phase, by carefully considering a general Hamiltonian of two interacting particles, where one is much larger (and hence slower) than the other. (This can also be done for more particles, but the construction in very similar.)

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H} = \frac{P^2}{2m_n} + \frac{p^2}{2m_e} + V(\vec{R},\vec{r}) } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{R}} refers to the coordinate of the larger particle, and not the center of mass.

    After some work, it can be shown that terms similar to both a vector and scalar potential can be found that explicitly create the Berry Phase.

    The final result is:

    for the Vector Potential,

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A^{(n)} = i\hbar \langle n(R)|\vec{\nabla_R}|n(R)\rangle }

    and for the Scalar Potential,

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Phi^{(n)} = \frac{\hbar^2}{2m_n}\left(\langle\vec{\nabla_R} n(R)|\vec{\nabla_R} n(R) \rangle - \langle \vec{\nabla_R} n(R)|n(R)\rangle \langle n(R)|\vec{\nabla_R}n(R)\rangle \right) }

    where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |n(R)\rangle } is the wavefunction of the smaller particle, depending on the position of the larger. More generally, it would be the wavefunction of the object with the 'fast' degree of freedom, depended on the state of the slower degree of freedom.

    Once these are found, an effective Hamiltonian may be constructed:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \mathcal{H} = \frac{1}{2m_n}\left(\vec{P}-\vec{A^{(n)}}\right)^2 + \Phi^{(n)} }

    Time Reversal Symmetry (& Kramer's Degeneracy)

    The Schrodinger equation is: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\partial }{\partial (t)}\psi(r,t) =H\psi(r,t)}

    Taking Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle t \rightarrow -t} yeilds: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -i\hbar\frac{\partial }{\partial (t)}\psi(r,-t) =H\psi(r,-t)}

    This is obviously not a symmetric transformation. In addition, take the complex conjugate: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle i\hbar\frac{\partial }{\partial (-t)}\psi ^*=H^*\psi ^*}

    Whether or not this equation is symmetric depends on the form of H we are working with.

    To find an expression for the time reversal operator, we consider the specific Hamiltonian for an electron:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H = \frac{p^2}{2m} + V(r) + (\frac{1}{2m^2c^2}\frac{1}{r}\frac{dV}{dr})\mathbf{L}\cdot \mathbf{S}}

    The time reversal operator for a one electron system is:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{K} = i \sigma _{y} C } where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle C\!} indicates to take the complex conjugate

    Prove:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K\psi=i\sigma_y\psi^*\!}

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle KHK^{\dagger}=(i\sigma_yC) H (-i\sigma _yC^*)=H^*}

    To get degeneracy from the time reversal. (Kramers Degeneracy)

    For n-electrons:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat{K}=i^n\sigma _{y_1}\sigma _{y_2}...\sigma _{y_n}C}

    For time reversal invariant H:

    So, and have same energy.

    Assume they are linear dependent:

    It requires .

    However,

    And thus we have arrived at Kramer's Degeneracy Theoreom: For an odd number of electrons, the energy levels of the system are at least doubly degenerate, as long as is time reversal invariant.

    Many Particle Systems and Identical Particles

    At this point it is second nature to write down the hamiltonian for a system if the potential and kinetic energy of the particle is known. The hamiltonian is then denoted by:

    The next natural step is to investigate what the hamiltonian would look like and the resulting wavefunctions and energies for systems with more than one particle. The easiest place to start is with two identical particles.

    Two Identical Particles

    It is straight forward to generalize a Hamiltonian for two identical particles:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat H = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + V(\vec r_1) + V(\vec r_2) + (u(\vec r_1, \vec r_2) + u(\vec r_2, \vec r_1)), }

    where the potential and kinetic energy are written down for each particle individually and there is an additional term which represents the interaction between the two particles. For simplicity we will treat the interaction potential as a central force and from this point on write it as .

    The above Hamiltonian for the two identical particles is invariant under exchange symmetry (or the permutation of particle labels) and as such it is either even or odd under permutation. Likewise the eigenfunctions can be chosen to be even or odd under the exchange of particle labels. Ignoring spin orbit coupling the general solution will therefore be of the form:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (\eta_1, \eta_2) = \phi (\vec r_1, \vec r_2) \chi (\sigma_1, \sigma_2), }

    where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_1 \!} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_2 \!} are just spin labels not Pauli spin matrices. If Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (\eta_1, \eta_2) \! } is a solution, then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (\eta_2, \eta_1) \! } is also a solution and as such there are two possible states, one is symmetric and the other is anti-symmetric solution:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \frac{1}{\sqrt{2}} \left(\psi (\eta_1, \eta_2) + \psi (\eta_2, \eta_1)\right) }

    and

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \Rightarrow \frac{1}{\sqrt{2}} \left(\psi (\eta_1, \eta_2) - \psi (\eta_2, \eta_1)\right) }

    Although mathematically this formula will result in symmetric and anti-symmetric solutions, in nature that is not the case, and the solution must be chosen to be one or the other. If the system deals with two fermions, which have half-integer spin, then only the anti-symmetric solution appears in nature. Likewise if the system deals with two bosons, which have integer spin, then only the symmetric solution appears in nature.

    N Particles

    If the Hamiltonian were for a three particle system, it would be:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat H = \frac{p_1^2}{2m} + \frac{p_2^2}{2m} + \frac{p_3^2}{2m} + V(\vec r_1) + V(\vec r_2) + V(\vec r_3) +(u(|\vec r_1 - \vec r_2 |) + u(| \vec r_1 - \vec r_3 |) + u(| \vec r_2 - \vec r_3 |).}

    In general, the Hamiltonian for a system with N particles can be written as:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hat H = \sum _{j=1} ^{N} \left( \frac{p^2_j}{2m} + V(\vec r_j)\right) + \frac{1}{2} \sum_{j,k}^N u(r_j, r_k). }

    In general, it is difficult to solve this problem with interaction terms but assuming you could do so. The only physically admissible states are either symmetric or antisymmetric under exchange of any two particle labels as before, therefore, the wavefunction is given by: Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi (\eta_1, \eta_2, \eta_3, ... , \eta_N) = \phi (\vec r_1, \vec r_2, \vec r_3, ... , \vec r_N) \chi (\sigma_1, \sigma_2, \sigma_3, ..., \sigma_N) } ,

    and it follows the same rules as before for bosons and fermions.

    It is important to note that if a solution doesn't satisfy the proper symmetry, then a linear combination of all permutations will result in a properly symmetrized solution that will be an eigenstate.

    Constructing Admissible Eigenstates

    As stated above, if a solution does not satisfy the necessary symmetry properties, then a linear combination of the different permutations of product states (that are completely symmetric for bosons and anti-symmetric for fermions) must be made.

    For spin-less bosons the normalized wavefunction is:

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_{\mbox{bosons}} (1,2,....N) = \sqrt{\frac{N_a! N_b! .... N_n!}{N!}} \sum_P P \varphi_a (1) \varphi_b(2) ....... \varphi_n (N)}

    where the sum is over all Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle N!\!} permutations of indices 1 through .

    For spin-less fermions the normalized wavefunction is:

    where is if a permutation can be decomposed into an even number of two particle exchanges and for odd.

    Another way of writing the sum to form an anti-symmetric matrix is through the use of the Slater determinant.

    Second Quantization

    Consider now a wavefunction pertaining to a many-particle system, , which is considered to be a field variable. For the many-particle system, this field variable must also quantized by a process known as second quantization.

    In order to perform this quantization of the field variable, we must construct special raising and lowering operators, associated with the individual energy levels of the system, and , which add and subtract particles from the energy level, respectively. In the presence of spin, an additional subscript is added to separate the creation and annihilation operators for each case of spin, so that each operator only acts on particles with the same spin attributed to said operator. In the simple, although rather non-physical, case of spinless particles, this extra factor can be ignored for simplicity in examining how the operators work on the quantized field:

    For the case of fermions, an additional constraint on the operators is placed due to the exclusion principle:

    Given the two classes of particles, fermions and bosons, two sets of relations result to relate the creation and annihilation operators.

    For the case of bosons, the operators obey a commutator relationship of the form:

    The state of the system, , where refers to the number of particles in the state, is therefore of the form:

    Fermions, however, obey anti-commutator relationships, of the following form:

    For this type of system, the state can be written as:

    Furthermore, for both classes of particles, we can create an operator that, upon acting on the total state of the system, returns the number of particles in a given state (for fermions this will obviously be 0 or 1). This operator is of the form . Therefore, . From this it is easy to obtain an operator, that returns the total number of particles in the system:

    From this operator, the average number of particles, and therefore the average flux number, can be calculated by the following:

    To continue the analysis of these second quantization operators, lets consider the projection of on the state , that is to say, define a function such that

    Rearranging this yields an expression for the position state of a particular particle, weighted over the different energy levels by the function

    This new field operator acting on the ground state is called the field creation operator:

    Therefore, the position state of a system containing particles can be expressed as the following:

    Here it is important to note the permutation relationships differ for fermions and bosons, that is to say the following:

    For bosons:

    Whereas for fermions:

    Manipulation of the field creation and annihilation operators yield results congruent with those associated with the particle creation and annihilation operators discussed above, recognizing that is a delta function. Therefore, we obtain the following expressions:

    For fermions,

    And for bosons,

    From these relations, we show that adding a particle at position is expressed by the automatically correctly symmetrized expression:

    The annihilation operator also results in a correctly symmetrized expression, however, it is only valid if . The result is a correctly symmetrized combination of particle states:

    Finally, we will consider certain preexisting operators, expressed in the new second quantization format. Consider first the density operator which was previously in the first quantization. Recalling our earlier commutator relations, this operator is refined as:

    In order to obtain the total kinetic energy and interaction potential operators, one must first express the annihilation and creation operators in momentum space to simplify the mathematics. Therefore, define . Therefore, rearranging this yields:

    Therefore, for , we can rewrite this as the following:

    The interaction potential, taking care to include the term for symmetry, is thereby expressed as:

    Note that this entire analysis has been done for the spinless particle case, however the addition of spin only requires that we sum over the possible spins of the system, namely:

    Thus finally the total Hamiltonian for the system can be written as:


    relativistic Quantum mechanics and Dirac Equation

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