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<math>J_{1z,2z}|j_1 j_2 m_1 m_2\rangle = \hbar m_{1,2}|j_1 j_2 m_1 m_2\rangle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad (6.1.6)</math>
<math>J_{1z,2z}|j_1 j_2 m_1 m_2\rangle = \hbar m_{1,2}|j_1 j_2 m_1 m_2\rangle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad (6.1.6)</math>


<b>2</b>. The simultaneous eigenkets of <math>\!J^2</math>, <math>J_1^2</math>, <math>J_2^2</math> and <math>J_z</math>, denoted by <math>|j m j_1 j_2\rangle</math>. These four operators operate on the base kets according to:
<b>2</b>. The simultaneous eigenkets of <math>\!J^2</math>, <math>J_1^2</math>, <math>J_2^2</math> and <math>\!J_z</math>, denoted by <math>|j m j_1 j_2\rangle</math>. These four operators operate on the base kets according to:


<math>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)</math>
<math>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)</math>

Revision as of 13:31, 1 May 2009

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.

Outline of the course:


Stationary state perturbation theory in Quantum Mechanics

Very often, quantum mechanical problems cannot be solved exactly. We have seen last semester that 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. The technique we used last semester was WKB, 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 insoluble.All perturbative methods depends on few simple assumptions.The first of these that we have a mathematical expression for a physical quantity for which we are unable to obtain a exact solution.The next assumption is that this physical quantity may be broken down into a part which can be solved exactly and the troublesome part which has no analytic solution.

Typically, the (Hamiltonian) problem has 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+\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 soluble 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 insoluble.

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 \mathcal \epsilon_n } , that are known solutions to the Schrodinger eq:

Failed to parse (SVG (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.

If 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\;\;\;\; (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^{(0)}\rangle} signify the nth 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_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 |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^{(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 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 | N_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 out 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, the 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.


If 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 } , which we already had from before (1.1.8)

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 Failed to parse (SVG (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} up to the second order:

Failed to parse (SVG (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 m|\mathcal{H}'|n\rangle + \lambda^2 \sum_{m \not= n} \frac{|\langle n|\mathcal{H}'|m\rangle|^2}{\epsilon_n - \epsilon_m}+ O(\lambda^3)}

One interesting thing to not about this is 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 |V_{mn} = \langle m |V| n\rangle|^2} is positive definite. Therefore, 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 \epsilon_0 - \epsilon_m < 0} , the second order energy correction will lower the ground state energy.

kth order Energy Corrections 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 E_n^{(k)} = \langle n | \mathcal{H}' | N_n^{(k - 1)} \rangle \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;(1.1.10)}

This result provides us with a recursive relationship for the Eigenenergies of the perturbed state, so that we have access to the eigenenergies for an state of arbitrary 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} .

What about the eigenstates? Express the perturbed states in terms of the unperturbed 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^{(k)}\rangle = \sum_{m \not= n}|m\rangle\langle m|\Psi^{(k)}\rangle}

Go back to equation 1.1.5 and taking the scalar product from the left with and taking orders of we find:

1st order Eigenkets

The first order contribution is then the sum of this equation over all , and adding the zeroth order we get the eigenstates of the perturbed hamiltonian to the 1st order in :

Going beyond order one in gets increasingly messy, but can be done by the same procedure.

Renormalization Earlier we assumed that , which means that our states are not normalized themselves. To reconsile this we introduce the normalized perturbed eigenstates, denoted . These will then be related to the :

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 2nd 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 \frac{1}{z (\lambda )} = \langle N(\lambda )|N(\lambda )\rangle = ( \langle n| + \lambda \langle\Psi_n^{(1)}| + ...)(|n\rangle + |\Psi_n^{(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 z(\lambda ) = \frac{1}{1 + \lambda^2\sum_{ n \not= m}\frac{|\langle m|V|n\rangle|^2}{(\epsilon_n - \epsilon_m)^2} + ...} = 1 - \lambda^2\sum_{ n \not= m}\frac{|\langle m|V|n\rangle|^2}{(\epsilon_n - \epsilon_m)^2} + ... \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;\;\;\;\;(1.1.14)}

Renorm.jpg

Where we use a taylor expansion to arrive at the final result (noting 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 \frac{|\langle m|V|n\rangle|^2}{(\epsilon_n - \epsilon_m)^2} < 1} ).

Then, interestingly, we can show 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 |n\rangle} is related to the energies by employing equation 1.1.10:

Failed to parse (SVG (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(\lambda ) = \frac{\partial E_n}{\partial \epsilon_n}\Big|_{\epsilon_m}{\langle m|V|n\rangle} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;\;\;(1.1.15)}

Where the derivative is taken with respect 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 \epsilon_n} , while holding Failed to parse (SVG (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 m|V|n\rangle} constant. Using the Brillouin-Wigner perturbation theory (see next section) it can supposedly be shown that this relation holds exactly, without approximation.

Brillouin-Wigner Perturbation Theory

This is another type of perturbation theory. 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.

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_m) = \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 exact state, therefore, is 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 Z|N \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. 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 go to 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 1/\mathcal(\epsilon_n-\epsilon_m) } .

Instead of trying to use these (degenerate) eigenstates with perturbation theory, if we start with the correct linear combinations of eigenstates, regular perturbation theory will no longer fail! So the issue now is how to find these linear combinations.

Failed to parse (SVG (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 thatFailed to parse (SVG (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|H'|n_b\rangle } goes to zero for all cases r ≠ n. 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 Failed to parse (SVG (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=2 } 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:

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. 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 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 the time independent perturbation theory which can be applied on various systems in which a little change in the Hamiltonian appears as a correction in the form of a series for the energy and wave functions. However, this stationary approach cannot be used to describe the interaction of electromagnetic field with atoms (i.e. photon with Hydrogen atom), since such systems would be inherently time dependent. Therefore we must 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 only within a finite time period from Failed to parse (SVG (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 = 0}
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 \!t = T} . 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,

Failed to parse (SVG (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_t^0 \rangle = H_0 |\psi_t^0\rangle, \qquad t<t_0. \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (2.1)}

Then, assuming that the perturbation acts after time Failed to parse (SVG (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_0} , 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 i\hbar\frac{\partial}{\partial t}|\psi_t \rangle = (H_0 + V_t)|\psi_t\rangle, \qquad t>t_0 \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad\;\;\; (2.2)}

The problem therefore consists of finding the 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 |\psi(t)\rangle} with boundary 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 |\psi(t)\rangle = |\psi_t^0\rangle} 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 t \leq t_0} . However, such a problem is not generally soluble.
Therefore, we limit ourselves to the problems 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 \!V_t} is small. In that case we can treat Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \!V_t} as a perturbation and seek it's effect on the wavefucntion in 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 \!V_t} .

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 \!V_t} is small, the time dependence of the solution will largely come from Failed to parse (SVG (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_0} . So we use

Failed to parse (SVG (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_t\rangle = e^{-i H_0 t/\hbar}|\psi(t)\rangle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\; (2.3)} ,

which we substitute into the Schrodinger Equation to 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 i\hbar\frac{\partial}{\partial t}|\psi(t)\rangle=V(t)|\psi(t)\rangle \quad \text{where}\quad V(t) = e^{i H_0 t/\hbar}V_te^{-i H_0 t/\hbar}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad (2.4)} .

In this 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 \psi(t)} and the 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 V(t)} are in the interaction representation. Now, we integrate equation #(2.4) to 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 \int_{t_o}^{t}dt \frac{\partial}{\partial t}|\psi(t)\rangle = \psi(t) - \psi(t_0) = \frac{1}{i\hbar}\int_{t_0}^{t}dt' V(t')|\psi(t')\rangle}

or

Failed to parse (SVG (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(t)\rangle = |\psi(t_0)\rangle + \frac{1}{i\hbar}\int_{t_0}^{t}dt' V(t')|\psi(t')\rangle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \;\;\ (2.5)}

Equation #(2.5) can be iterated by inserting this equation itself as the integrand in the r.h.s. We can then write equation #(2.5) 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 |\psi(t)\rangle = |\psi(t_0)\rangle + \frac{1}{i\hbar}\int_{t_0}^{t}dt' V(t')\left(|\psi(t_0)\rangle + \frac{1}{i\hbar}\int_{t_0}^{t}dt'' V(t'')|\psi(t'')\rangle\right), \qquad t''<t'\qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad\;\ (2.6)}

which can be written compactly 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 |\psi(t)\rangle = T e^{\frac{i}{t}\int_{t_0}^{t}V(t')dt'}\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad\;\;\ (2.7)}

This is the general solution. T is called the time ordering operator, which ensures it the series is expanded in the correct order. For now, we consider only the correction to the first 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 \!V(t)} .

First Order Transitions

If we limit ourselves to the first order we use,

Failed to parse (SVG (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(t)\rangle = |\psi(t_0)\rangle + \frac{1}{i\hbar}\int_{t_0}^{t}dt'V(t')|\psi(t_0)\rangle \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad\;\;\ (2.8)}

We want to see the system undergoes a transition to another state, say Failed to parse (SVG (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} . So we project the wave function Failed to parse (SVG (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(t)\rangle} 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 |n\rangle} . From now on, 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 |\psi(t_0)\rangle = |0\rangle}
for brevity. Projecting into state and assuming we get,

Expression #(2.9) is the probability amplitude of transition. Therefore, we square the final expression to get the probability of having the system in state at time t.
Squaring, we get

For example, let us consider a potential which is turned on sharply at time , but independent of t 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 is the largest for the transition to conserve the energy to within an amount given in that relation.

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

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,

Which, if used in the equation #(2.11) 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 an using 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.

Harmonic Perturbation Theory

Harmonic perturbation is one of the main interest 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 specify the rate at which the perturbation is turned on, is a very small positive number which at the end of the calculation 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 has been averaged to zero. Transition rate is given by :

now, if the response is immediate, or 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.

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

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 different state (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:

Which 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 we have above, we find the energy of the two states

Which graphically, we can draw the graphics of Electric field vs Energy

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

So, following above equation, we find

Now, let

And we define the electric field as a function

so the above expression can be written as

Now, we observe that as the 1st term in the r.h.s of the 1st equation will oscillate very rapidly compared to the 2nd term of the same equation. This rapidly oscillating term will average to zero. So we don't include those oscillating term in the next calculation. Therefore we get a result

At the resonance, so these equations are simplified to

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

With solution (also for b with substitution)

Let's assume that at time , the molecule is in state 1 (experimentally, we prepare the molecule to be in this state) so that and this assumption gives

The probability amplitude that the molecule remains in the state is

and the probability is just the square of this probability amplitude :

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):

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 in order to satisfy the transversality. Here the plane waves are normalized with respect to some volume . This is just for convenience and the physics won't change. We can choose . 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 use the fact that any exponential time-dependent term is in 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 adn , 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

The Schrödinger equation is then

Since a gauge transformation

left invariant the E&M fields, we expect that which is an observable it 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.

Hamiltonian of Multiple Particles in Presence of Radiation

If we have a system of N particles we have the following hamiltonian

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

Using delta function operator we can write

Then


COMMENTS

  • can be interpreted as density of particles 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 proximate 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 a experimental observation of a Lamb-like shift in a solid state setup see Science 322, 1357 (2008). DOI: 10.1126/science.1164482.

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 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 <N> 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.

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 ). 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 tells as 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 tells as that:

From the previews 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 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 exists non-perturbative methods to approximately determine the lowest energy eigenstate or ground state, and some excited states of a given system. One of the important method in the approximate determination of the wave function and eigenvalues of a system is the Variational Method, which is based on the variational principle. Variational method is a very general one that it can be used whenever the equations can be put into the variational form. Variational method is now the springboard to many numerical computation.

    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.

    = ; n = 0,1,2

    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

    The variational principle states that,


    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


    Thus 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

    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 as

    .


    Differentiating the above relation,


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


    .


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

    Therefore,

    .


    Define,

    .

    Hence, become

    .


    We can define the variation of as

    ,


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

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

    .


    Finally we can say that expectation value of the Hamiltonian is stationary iff 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 a 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 ground state exact 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.

    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 consists essentially of the determination of the wave functions of electrons in a molecule in the form of linear combinations 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

    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.

    Ground State of Helium

    Let us use variational principle to determine the ground state energy of Helium. Helium has two electrons and two protons. For simplicity, we ignore the presence of neutrons.

    We can write 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,


    where we ignored spin and


    Therefore we can write



    We can write the ground state energy for this situation with as


    However, the ground state energy has been experimentally determined accurately to -78.6 eV. 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 is itself but with as variable. This argument is perfectly valid since each electron screens the nuclear charge seen by the other electron and hence the atomic number is less than 2.

    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 nwill give

    The seventh term will give an expectation value of



    Adding all this we get,


    Since is the variational parameter, we can minimize the Energy with respect to . This will give us

    Putting we get which is an improvemnet.


    Substituting this in 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.

    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 very 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:

    0 0.893 1.034
    1/2 0.696 0.605
    1 0.329 0.270
    3/2 0.094 0.140
    2 0.016 0.083
    5/2 0.002 0.055
    3 0.0001 0.039

    It is therefore 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.

    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 what we have done consist in quantise everything that we know from classical mechanics. Specifically this is:

    • 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 1992 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.

    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.

    Note on Hund's Rules

    Hund's Rules describe L-S 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 S(total spin) consistent with Pauli exclusion principle

    2. Choose a maximum value of L(angular momentum)

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

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

    For example is we use Silicon: then for 2 spin particles. Angular momentum is L=1 and since it is less than half full, J=0.


    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 this 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 it 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

    Subspace : (matrix 1x1)


    Subspace : (matrix 2x2)

    • For and are given by


    • The matrices for and are given by


    Subspace : (matrix 3x3)

    • For and are given by


    • The matrices for and are given by

    Summary

    The following table is the summary of above calculations.

    Spin 1/2 Angular Momentum

    The angular momentum of a stationary spin 1/2 particle is found to be quantized to the regardless of the direction of the axis chosen to measure the angular momentum. There is a vector operator when projected along an arbitrary axis satisfies the following equations:

    and form a complete basis, which means that any state and can be expanded as a linear combination of and .

    The spin operator obeys the standard angular momentum commutation relations

    The most commonly used basis is the one which diagonalizes .

    By acting on the states and with , we find

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

    where is the z Pauli spin matrix. Repeating the steps (or applying the commutation relations), we can solve for the x and y components.



    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, if

    Then,

    The above equation is true for 1/2 spins only!!

    In general,

    Finally, any 2x2 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 :

    for finite (correct for all orders)


    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 the operator from the previous equation. This can also be written,

    Because of the commutation relations of , it can easily be shown that (n is an integer), thus the above equation can be split:

    This form may be more convenient when performing rotations.

    Addition of angular momenta

    Formalism

    Classically, the sum of two angular momentum is the sum of the C-number vectors. Quantum mechanically we have to be careful about the eigenstates.

    For two angular momenta, and , such that, for their respective eigenstates, and

    and similarly for .

    We define their total angular momentum as,
    where and are the identity matrices of 's 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 \overrightarrow{J_{2}}} 's Hilbert spaces, and where Hilbert space size is .

    We have the following commutation relations:

    And consequently,

    However, . Therefore, to construct a basis, one can not take a direct product between the set of eigenkets of and those of . For example,
    assume two spin 1/2 particles with basis . These states are eigenstates of , but are they eigenstates of and ?
    Let us see what happens with the state :

    define

    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 . As a result, there are two
    choices for sets of base kets which can be used:

    1. The simultaneous eigenkets of , , , , denoted by . These four operators commute with each other, and they operate on the base kets
    according to:

    2. The simultaneous eigenkets of , , and , denoted by . These four operators operate on the base kets according to:

    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 , 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

    and .

    Further, it turns out that, for fixed , and , coefficients with different values for and are related to each other through recursion

    relations. To derive these relations, we first note that:

    Now we write,

    Using equation #(6.2.4), we get (with , ):


    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.

    Orthogonality of Clebsch-Gordan Coefficients

    We have the following symmetry:

    If we put the coefficients into a matrix, it is real and unitary, meaning

    . E.g.,

    We have the following orthogonality relations:

    Hydrogen atom with spin orbit coupling given by the following hamiltonian

    Recall, the atomic spectrum for bound states

    The ground state, , is doubly degenerate:

    First excited state is 8-fold degenerate:

    nth state is fold degenerate

    We can break apart the angular momentum and spin into its x, y, z-components

    Define lowering and raising operators

    For the ground state, , nothing happens. Kramer's theorem protects the double degeneracy.

    For the first excited state, , once again nothing happens.

    For , there is a four fold degeneracy.

    We can express the solutions in matrix form

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

    add the angular momenta:

    So that

    Define J_



    Can express as

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

    and

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

    Addition of three angular momenta

    To add three angular momenta first we add , construct the simultaneous eigenstates of . We write such states as . Such states can be given in terms of Clebsch-Gordan coefficients and (eigenstates of ):

    Next we add to , forming simultaneous eigenstates of . These are given in terms of the by

    Therefore, we can construct eigenstates of in terms of eigenstates of as follows:

    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 . For three angular momenta, it is .


    Elementary applications of group theory in Quantum Mechanics

    Group theory is related with symmetry G={A,B,C....} G is a group under operation AB if . (closure)Bold text . (AB)C=A(BC)associated lawBold text .Bold text . Such that Bold text

    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 . Suppose that we act with on an eigenstate of and . Under the rotation, the state is generally no longer an eigenstate of with eigenvalue m. However rotated state remains an eigenstate of with the same eigenvalue because and commute. Indeed

    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

    Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle R_{\alpha }=e^{-\frac{i}{\hbar}\overrightarrow{s}.\overrightarrow{\alpha }}}


    eigenstates of

    for

    only mixing same j states,mix m, j not change


    j=1/2 d will be a 2*2 matrices will be (2j+1)*(2j+1)matrices

    ==

    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.

    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.

    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 Failed to parse (SVG (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}{2}} 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 Failed to parse (SVG (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} 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 i\hbar \frac {\partial}{\partial t}} 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 \bold p} 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 -i\hbar \nabla} .

    Follow this link to learn about Klein-Gordon equation.

    Follow this link to learn about Dirac equation.

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