Phy5646

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.

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.

Raleigh-Shrödinger Peturbation Theory

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

To find the sulution 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 \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}^'}

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 \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 + ...}

then they must obey the equation,

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

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)}

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. 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 + ... }

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, ... }

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 original state is added to 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.

If we equate coefficients in the above expanded form of the perturbed Hamiltonian, 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} that we want. The first few are as follows,

1st 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 \lambda = 0 \rightarrow E_n^{(0)} = \epsilon_n } , which we already had from before,

2nd 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 \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\langle} , 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 }

kth order 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 }

I have skipped a few steps since they are covered in Baym anyways. 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?

Brillouin-Wigner Peturbation Theory

This is another type of perturbation theory. Using a basic formula derived from the Schroedinger 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. Starting with the Schroedinger 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 } , 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} }

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} }

Degenerate Perturbation Theory

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/(\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.

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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \vec{\mathcal E}={\mathcal E}\hat{z} } . The Hamiltonian for this system is Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle {\mathcal H}={\mathcal H}_o-e{\mathcal E}z } . The eigenstates of the system are Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{|2S\rangle,|2P_{-1}\rangle,|2P_0\rangle,|2P_{+1}\rangle\} } . The matrix of the degenerate eigenstates and the perturbation is:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} \langle n_i|{\mathcal H}'|n_j\rangle &\longrightarrow \left(\begin{array}{cccc}\langle2S|-e{\mathcal E}z|2S\rangle&\langle2S|-e{\mathcal E}z|2P_{-1}\rangle&\langle2S|-e{\mathcal E}z|2P_0\rangle&\langle2S|-e{\mathcal E}z|2P_{+1}\rangle\\\langle2P_{-1}|-e{\mathcal E}z|2S\rangle&\langle2P_{-1}|-e{\mathcal E}z|2P_{-1}\rangle&\langle2P_{-1}|-e{\mathcal E}z|2P_0\rangle&\langle2P_{-1}|-e{\mathcal E}z|2P_{+1}\rangle\\\langle2P_0|-e{\mathcal E}z|2S\rangle&\langle2P_0|-e{\mathcal E}z|2P_{-1}\rangle&\langle2P_0|-e{\mathcal E}z|2P_0\rangle&\langle2P_0|-e{\mathcal E}z|2P_{+1}\rangle\\\langle2P_{+1}|-e{\mathcal E}z|2S\rangle&\langle2P_{+1}|-e{\mathcal E}z|2P_{-1}\rangle&\langle2P_{+1}|-e{\mathcal E}z|2P_0\rangle&\langle2P_{+1}|-e{\mathcal E}z|2P_{+1}\rangle\\\end{array}\right)\\ &\longrightarrow \left(\begin{array}{cccc}0&0&\langle2S|-e{\mathcal E}z|2P_0\rangle&0\\0&0&0&0\\\langle2P_0|-e{\mathcal E}z|2S\rangle&0&0&0\\0&0&0&0\\\end{array}\right)\\ &\longrightarrow \left(\begin{array}{cccc}0&0&-3e{\mathcal E}a_B&0\\0&0&0&0\\-3e{\mathcal E}a_B&0&0&0\\0&0&0&0\\\end{array}\right)\\ \end{align} }

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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{|2P_{-1}\rangle,|2P_{+1}\rangle,\frac{1}{\sqrt{2}}\left(|2S\rangle+|2P_0\rangle\right),\frac{1}{\sqrt{2}}\left(|2S\rangle-|2P_0\rangle\right)\} }

Because of the perturbation due to the electric field, 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 |2P_{-1}\rangle } and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |2P_{+1}\rangle } states will be unaffected. However, the energy of the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |2S\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 |2P_0\rangle } states will have a shift due to the electric field.

Time dependent perturbation theory in Quantum Mechanics

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. This leads us to the Time Dependent Perturbation Theory.

One of the main tasks of this theory is the calculation of transition probabilities from one 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_n \rangle} to another 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_m \rangle} that occurs under the influence of time dependent potential. Generally, 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(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(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.

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(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(4)}

In this equation we work using interaction representation. Now, we integrate equation (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(5)}

Equation (5) can be iterated by inserting this equation itself as the integrand in the r.h.s. We can then write equation (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(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 (7)}

With T as the time ordering operator to ensure it can be expanded in series 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)} . If we limit ourselves to the first order we use

$|\psi (t)\rangle =|\psi (t_{0})\rangle +{\frac {1}{i\hbar }}\int _{t_{0}}^{t}dt'V(t')|\psi (t_{0})\rangle \qquad (8)$

We want to see the system undergoes a transition to another state, say $|n\rangle$. So we project the wave function $|\psi (t)\rangle$ to $|n\rangle$. From now on, let $|\psi (t_{0})\rangle =|0\rangle$ for brevity. Projecting into state $|n\rangle$, we get

${\begin{aligned}\langle n|\psi (t)\rangle &=\langle n|0\rangle +{\frac {1}{i\hbar }}\int _{t_{0}}^{t}dt'\langle n|V(t')|0\rangle \\&={\frac {1}{i\hbar }}\int _{t_{0}}^{t}dt'\langle n|e^{{\frac {1}{\hbar }}H_{0}t}V_{t'}e^{-{\frac {1}{\hbar }}H_{0}t}|0\rangle \\&={\frac {1}{i\hbar }}\int _{t_{0}}^{t}dt'e^{{\frac {i}{\hbar }}(\epsilon _{n}-\epsilon _{0})t'}\langle n|V_{t'}|0\rangle \end{aligned}}\qquad (9)$

Expression (9) is the probability amplitude of transition. Therefore, we square the final expression to get the probability of having the system in state $|n\rangle$ at time t. Squaring, we get

${\underset {0\rightarrow n}{P_{(t)}}}=|\langle n|\psi (t)\rangle |^{2}=\left|{\frac {1}{i\hbar }}\int _{t_{0}}^{t}dt'e^{{\frac {i}{\hbar }}(\epsilon _{n}-\epsilon _{0})t'}\langle n|V_{t'}|0\rangle \right|^{2}\qquad (10)$

For example, let us consider a potential $V_{t}$ which is turned on sharply at time $t_{0}$, but independent of t thereafter. Furthermore, we let $t_{0}=0$

Interaction of radiation and matter

Quantization of electromagnetic radiation

Additional Reading

• Experimental observation of a Lamb-like shift in a solid state setup Science 322, 1357 (2008).

Classical view

---

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

$\varphi (\mathbf {r} ,t)=0$

$\nabla \cdot \mathbf {A} =0$

In this gauge the electromagnetic fields are given by:

$\mathbf {E} (\mathbf {r} ,t)=-{\frac {1}{c}}{\frac {\partial \mathbf {A} }{\partial t}}$

$\mathbf {B} (\mathbf {r} ,t)=\nabla \times \mathbf {A}$

The energy in this radiation is

$\varepsilon ={\frac {1}{8\pi }}\int d^{3}\mathbf {r} (\mathbf {E} ^{2}+\mathbf {B} ^{2})$

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

$\mathbf {P} ={\frac {c}{4\pi }}\mathbf {E} \times \mathbf {B}$

The radiation generated by classical current is

$\Box \mathbf {A} =-{\frac {4\pi }{c}}\mathbf {j}$

Where $\Box$ is the d'Alembert operator. Solutions in the region where $\mathbf {j} =0$ are given by

$\mathbf {A} (\mathbf {r} ,t)=\alpha {\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+\alpha ^{*}{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}$

where $\omega =c|\mathbf {k} |$ and ${\boldsymbol {\lambda }}\cdot \mathbf {k} =0$ in order to satisfy the transversality. Here the plane waves are normalized respect to a volume $V$. This is just for convenience and the physics wont change. We can choose ${\boldsymbol {\lambda }}\cdot {\boldsymbol {\lambda }}^{*}=1$. Notice that in this writing $\mathbf {A}$ is a real vector.

Let's compute $\varepsilon$. For this

${\begin{aligned}\mathbf {E} (\mathbf {r} ,t)&=-{\frac {1}{c}}{\frac {\partial \mathbf {A} }{\partial t}}\\&=-{\frac {1}{c{\sqrt {V}}}}{\frac {\partial }{\partial t}}\left[\alpha {\boldsymbol {\lambda }}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}+\alpha ^{*}{\boldsymbol {\lambda }}^{*}e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\right]\\&=-{\frac {i\omega }{c{\sqrt {V}}}}\left[-\alpha {\boldsymbol {\lambda }}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}+\alpha ^{*}{\boldsymbol {\lambda }}^{*}e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\right]\\\mathbf {E} ^{2}(\mathbf {r} ,t)&=-{\frac {\omega ^{2}}{c^{2}V}}\left[\alpha ^{2}{\boldsymbol {\lambda }}^{2}e^{2i(\mathbf {k} \cdot \mathbf {r} -\omega t)}-\alpha \alpha ^{*}{\boldsymbol {\lambda }}\cdot {\boldsymbol {\lambda }}^{*}-\alpha ^{*}\alpha {\boldsymbol {\lambda }}^{*}\cdot {\boldsymbol {\lambda }}+\alpha ^{*2}{\boldsymbol {\lambda }}^{*2}e^{-2i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\right]\\\end{aligned}}$

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

${\begin{aligned}\mathbf {E} ^{2}(\mathbf {r} )&={\frac {\omega ^{2}}{c^{2}V}}\left[\alpha \alpha ^{*}+\alpha ^{*}\alpha \right]\\&=2{\frac {\omega ^{2}}{c^{2}V}}|\alpha |^{2}\\\end{aligned}}$

It is well known that for plane waves $\mathbf {B} =\mathbf {n} \times \mathbf {E}$, where $\mathbf {n}$ is the direction of $\mathbf {k}$. This clearly shows that $\mathbf {B} ^{2}=\mathbf {E} ^{2}$. However let's see this explicitly:

${\begin{aligned}\mathbf {B} (\mathbf {r} ,t)&=\nabla \times \mathbf {A} \\&=\nabla \left[\alpha {\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+\alpha ^{*}{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]\\\end{aligned}}$

Each component is given by

${\begin{aligned}\mathbf {B} _{i}(\mathbf {r} ,t)&={\frac {1}{\sqrt {V}}}\left[\alpha \varepsilon _{ijk}\partial _{j}\left({\boldsymbol {\lambda }}_{k}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\right)+\alpha ^{*}\varepsilon _{ijk}\partial _{j}\left({\boldsymbol {\lambda }}_{k}^{*}e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\right)\right]\\&={\frac {i}{\sqrt {V}}}\left[\alpha \varepsilon _{ijk}\mathbf {k} _{j}{\boldsymbol {\lambda }}_{k}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}-\alpha ^{*}\varepsilon _{ijk}\mathbf {k} _{j}{\boldsymbol {\lambda }}_{k}^{*}e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\right]\\\end{aligned}}$

Then

${\begin{aligned}\mathbf {B} (\mathbf {r} ,t)&={\frac {i}{\sqrt {V}}}\left[\alpha \mathbf {k} \times {\boldsymbol {\lambda }}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}-\alpha ^{*}\mathbf {k} \times {\boldsymbol {\lambda }}^{*}e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\right]\\\mathbf {B} ^{2}(\mathbf {r} ,t)&={\frac {-1}{V}}\left[\alpha \mathbf {k} \times {\boldsymbol {\lambda }}e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}-\alpha ^{*}\mathbf {k} \times {\boldsymbol {\lambda }}^{*}e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\right]\\&={\frac {-1}{V}}\left[\alpha ^{2}(\mathbf {k} \times {\boldsymbol {\lambda }})^{2}e^{2i(\mathbf {k} \cdot \mathbf {r} -\omega t)}-\alpha \alpha ^{*}(\mathbf {k} \times {\boldsymbol {\lambda }})\cdot (\mathbf {k} \times {\boldsymbol {\lambda }}^{*})-\alpha ^{*}\alpha (\mathbf {k} \times {\boldsymbol {\lambda }}^{*})\cdot (\mathbf {k} \times {\boldsymbol {\lambda }})+\alpha ^{*2}(\mathbf {k} \times {\boldsymbol {\lambda }}^{*})^{2}e^{-2i(\mathbf {k} \cdot \mathbf {r} -\omega t)}\right]\\\end{aligned}}$

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

${\begin{aligned}\mathbf {B} ^{2}(\mathbf {r} )&={\frac {1}{V}}\left[\alpha \alpha ^{*}+\alpha ^{*}\alpha \right](\mathbf {k} \times {\boldsymbol {\lambda }})\cdot (\mathbf {k} \times {\boldsymbol {\lambda }}^{*})\\&={\frac {1}{V}}\left[\alpha \alpha ^{*}+\alpha ^{*}\alpha \right][\mathbf {k} ^{2}({\boldsymbol {\lambda }}\cdot {\boldsymbol {\lambda ^{*}}})-(\mathbf {k} \cdot {\boldsymbol {\lambda ^{*}}})\cdot (\mathbf {k} \cdot {\boldsymbol {\lambda ^{*}}})]\\&={\frac {2}{V}}|\alpha |^{2}\mathbf {k} ^{2}\\&=2{\frac {\omega ^{2}}{c^{2}V}}|\alpha |^{2}\\&=\mathbf {E} ^{2}(\mathbf {r} ,t)\\\end{aligned}}$

Finally the energy of this radiation is given by

${\begin{aligned}\varepsilon &={\frac {1}{8\pi }}\int d^{3}\mathbf {r} (\mathbf {E} ^{2}+\mathbf {B} ^{2})\\&={\frac {1}{4\pi }}\int d^{3}\mathbf {r} \;\mathbf {E} ^{2}\\&={\frac {1}{4\pi }}\int d^{3}\mathbf {r} \left(2{\frac {\omega ^{2}}{c^{2}V}}|\alpha |^{2}\right)\\&={\frac {\omega ^{2}}{2\pi c^{2}}}|\alpha |^{2}\\\end{aligned}}$

So far we have treated the potential $A(\mathbf {r} ,t)$ as a combination of two waves with the same frequency. Now let's extend the discussion to any form of $A(\mathbf {r} ,t)$. To do this we can expand $A(\mathbf {r} ,t)$ in a infinite series of Fourier:

${\begin{aligned}A(\mathbf {r} ,t)=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]\\\end{aligned}}$

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 $\mathbf {k}$ vanishes. Then it is clear that

${\begin{aligned}\mathbf {E} ^{2}(\mathbf {r} )&=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}{\frac {\omega ^{2}}{c^{2}V}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}A_{\mathbf {k} {\boldsymbol {\lambda }}}\right]\\\mathbf {B} ^{2}(\mathbf {r} )&=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}{\frac {\mathbf {k} ^{2}}{V}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}A_{\mathbf {k} {\boldsymbol {\lambda }}}\right]\\\end{aligned}}$

Then the energy is given by

${\begin{aligned}\varepsilon &={\frac {1}{8\pi }}\int d^{3}\mathbf {r} (\mathbf {E} ^{2}+\mathbf {B} ^{2})\\&={\frac {1}{4\pi }}\int d^{3}\mathbf {r} \;\mathbf {E} ^{2}\\&={\frac {1}{4\pi }}\int d^{3}\mathbf {r} \sum _{\mathbf {k} {\boldsymbol {\lambda }}}{\frac {\omega ^{2}}{c^{2}V}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}A_{\mathbf {k} {\boldsymbol {\lambda }}}\right]\\&={\frac {1}{4\pi }}\sum _{\mathbf {k} {\boldsymbol {\lambda }}}{\frac {\omega ^{2}}{c^{2}}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}A_{\mathbf {k} {\boldsymbol {\lambda }}}\right]\\\end{aligned}}$

Let's define the following quantities:

${\begin{aligned}Q_{\mathbf {k} {\boldsymbol {\lambda }}}&={\frac {1}{{\sqrt {4\pi }}c}}(A_{\mathbf {k} {\boldsymbol {\lambda }}}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*})\\P_{\mathbf {k} {\boldsymbol {\lambda }}}&={\frac {-i\omega }{{\sqrt {4\pi }}c}}(A_{\mathbf {k} {\boldsymbol {\lambda }}}-A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*})\\\end{aligned}}$

Notice that

${\begin{aligned}\omega ^{2}Q_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}&={\frac {\omega ^{2}}{4\pi c^{2}}}(A_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}+A_{\mathbf {k} {\boldsymbol {\lambda }}}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*2})\\P_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}&={\frac {-\omega ^{2}}{4\pi c^{2}}}(A_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}-A_{\mathbf {k} {\boldsymbol {\lambda }}}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}-A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*2})\\\end{aligned}}$

Adding

${\begin{aligned}P_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}+\omega ^{2}Q_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}&={\frac {\omega ^{2}}{2\pi c^{2}}}(A_{\mathbf {k} {\boldsymbol {\lambda }}}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}\cdot A_{\mathbf {k} {\boldsymbol {\lambda }}})\\\end{aligned}}$

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

${\begin{aligned}H={\frac {1}{2}}\sum _{\mathbf {k} {\boldsymbol {\lambda }}}P_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}+\omega ^{2}Q_{\mathbf {k} {\boldsymbol {\lambda }}}^{2}\end{aligned}}$

We end up with a collection of harmonic oscillators, each labeled by $\mathbf {k}$ adn ${\boldsymbol {\lambda }}$, whose frequencies depends on $|\mathbf {k} |$.

From classical mechanics to quatum mechanics for radiation

---

As usual we proceed to do the canonical quantization:

${\begin{aligned}P_{\mathbf {k} {\boldsymbol {\lambda }}}&\to \mathbf {P} _{\mathbf {k} {\boldsymbol {\lambda }}}\\Q_{\mathbf {k} {\boldsymbol {\lambda }}}&\to \mathbf {Q} _{\mathbf {k} {\boldsymbol {\lambda }}}\\\end{aligned}}$

${\begin{aligned}A_{\mathbf {k} {\boldsymbol {\lambda }}}&\to {\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\;\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}\;,\;[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}},\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }]=\delta _{\mathbf {kk'} }\delta _{\boldsymbol {\lambda \lambda '}}\\\end{aligned}}$

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

${\begin{aligned}\mathbf {H} _{radiation}&=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\hbar \omega _{\mathbf {k} }(\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}+{\frac {1}{2}})&={\frac {1}{2}}\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\hbar \omega _{\mathbf {k} }(\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}+\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger })\\\end{aligned}}$

The classical potential can be written as

$\underbrace {A(\mathbf {r} ,t)=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[A_{\mathbf {k} {\boldsymbol {\lambda }}}{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+A_{\mathbf {k} {\boldsymbol {\lambda }}}^{*}{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]} _{\textrm {ClassicalVectorpotential}}\;\;\;\longrightarrow \;\;\;\underbrace {\mathbf {A} _{int}(\mathbf {r} ,t)={\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]} _{\textrm {QuantumOperator}}$

Notice that the quantum operator is time dependent. Therefor 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:

${\begin{aligned}\mathbf {A} (\mathbf {r} )&=e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\mathbf {A} _{int}(\mathbf {r} ,t)e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\\&=e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\left[{\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]\right]e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\\&={\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[\left[e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\right]{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+\left[e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\right]{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]\\&={\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}e^{i\omega t}\right]{\boldsymbol {\lambda }}{\frac {e^{i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}+\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }e^{-i\omega t}\right]{\boldsymbol {\lambda }}^{*}{\frac {e^{-i(\mathbf {k} \cdot \mathbf {r} -\omega t)}}{\sqrt {V}}}\right]\\&={\sqrt {\frac {2\pi \hbar }{\omega _{\mathbf {k} }}}}c\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\left[\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}{\boldsymbol {\lambda }}{\frac {e^{i\mathbf {k} \cdot \mathbf {r} }}{\sqrt {V}}}+\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }{\boldsymbol {\lambda }}^{*}{\frac {e^{-i\mathbf {k} \cdot \mathbf {r} }}{\sqrt {V}}}\right]\\\end{aligned}}$

COMMENTS

• The meaning of $\mathbf {H} _{radiation}$ 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 $|0\rangle$ 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 ${\begin{aligned}\mathbf {H} _{radiation}&=\sum _{\mathbf {k} {\boldsymbol {\lambda }}}{\frac {1}{2}}\hbar \omega _{\mathbf {k} }=\infty \end{aligned}}$
• The state $\;\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle$ represent an exited state of the vacuum with energy $\hbar \omega _{\mathbf {k} }(1+1/2)$. This means that the extra energy $\hbar \omega _{\mathbf {k} }$is carried by a single photon. Therefore $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }$ represent the creation operator of one single photon with energy $\hbar \omega _{\mathbf {k} }$. In the same reasoning, $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}$ represent the annihilation operator of one single photon.
• Consider the following normalized state of the vacuum: ${\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle$. At the first glance we may think that $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }$ creates a single photon with energy $2\hbar \omega _{\mathbf {k} }$. However this interpretation is forbidden in our model. Instead, this operator will create two photons each of the carryng the energy $\hbar \omega _{\mathbf {k} }$.

  Proof

  Suppose that $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }$ creates a single photon with energy $2\hbar \omega _{\mathbf {k} }$. We can find an operator $\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }$ who can create a photon with the same energy $2\hbar \omega _{\mathbf {k} }$. This means that

  ${\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle {\overset {\underset {\mathrm {?} }{}}{=}}\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle \;\;\;\longrightarrow \;\;\;{\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }{\overset {\underset {\mathrm {?} }{}}{=}}\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }\;\;\;\longrightarrow \;\;\;{\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}{\overset {\underset {\mathrm {?} }{}}{=}}\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}$

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

  $\left[\underbrace {\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}} ,\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }\right]=0$

  Replace the highlighted part by $\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}$

  
  $\left[\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}},\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }\right]=0$

  
  

  Since $\left[\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}},\mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }\right]=1$, the initial assumption is wrong, namely:

  ${\frac {1}{\sqrt {2}}}\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle \neq \mathbf {a} _{\mathbf {k'} {\boldsymbol {\lambda }}}^{\dagger }|0\rangle$

  This means that $\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }$ cannot create a single photon with energy $2\hbar \omega _{\mathbf {k} }$. Instead it will create two photons each of them with energy $\hbar \omega _{\mathbf {k} \blacksquare }$

ALGEBRA OF VACUUM STATES

A general vacuum state can be written as

$|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}};...\rangle$

where $n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}$ is the number of photons in the state $\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}$ which exist in the vacuum. Using our knowledge of harmonic oscillator we conclude that this state can be written as

$|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}};...\rangle =\prod _{\mathbf {k_{j}} {\boldsymbol {\lambda _{j}}}}{\frac {(\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger })^{n_{\mathbf {k_{j}} {\boldsymbol {\lambda _{j}}}}}}{\sqrt {n_{\mathbf {k_{j}} {\boldsymbol {\lambda _{j}}}}!}}}|0\rangle$

Also it is clear that

$\mathbf {a} _{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}^{\dagger }|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}};...\rangle ={\sqrt {n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}+1}}|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}+1;...\rangle$

$\mathbf {a} _{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}};...\rangle ={\sqrt {n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}}}|n_{\mathbf {k_{1}} {\boldsymbol {\lambda _{1}}}};n_{\mathbf {k_{2}} {\boldsymbol {\lambda _{2}}}};...;n_{\mathbf {k_{i}} {\boldsymbol {\lambda _{i}}}}-1;...\rangle$

Matter + Radiation: Electron bounded to a nucleus with transverse radiation field

---

The Hamiltonian of the system can be written as:

${\begin{aligned}\mathbf {H} &=\mathbf {H} _{particle+radiation}+\mathbf {H} _{radiation}\\&={\frac {(\mathbf {P} -{\frac {e}{c}}\mathbf {A} (\mathbf {r} ))^{2}}{2m}}+\mathbf {V(\mathbf {r} )} +\sum _{\mathbf {k} {\boldsymbol {\lambda }}}\hbar \omega _{\mathbf {k} }(\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}^{\dagger }\mathbf {a} _{\mathbf {k} {\boldsymbol {\lambda }}}+{\frac {1}{2}})\end{aligned}}$

Notice that in this writing the Hamiltonian is time independent. The field operator $\mathbf {A} (\mathbf {r} )$ must be time independent. The state of whole system can be written as

$|\psi \rangle =|m;\{n_{\mathbf {k} {\boldsymbol {\lambda }}}\}\rangle =|m\rangle \otimes |\{n_{\mathbf {k} {\boldsymbol {\lambda }}}\}\rangle$

The Schrodinger equation can be transformed to interaction representation (all operators with be time dependent). This transformation is give by:

${\begin{aligned}|\psi \rangle &=e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle \\|\chi \rangle &=e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle \;\;\;\longleftarrow \;\;\;\mathbf {H} _{par}={\frac {\mathbf {P} ^{2}}{2m}}+\mathbf {V(\mathbf {r} )} \\\end{aligned}}$

Then

${\begin{aligned}i\hbar {\frac {\partial }{\partial t}}|\psi \rangle &=[\mathbf {H} _{par+rad}+\mathbf {H} _{rad}]|\psi \rangle \\i\hbar {\frac {\partial }{\partial t}}e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle &=[\mathbf {H} _{par+rad}+\mathbf {H} _{rad}]e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle \\\mathbf {H} _{rad}e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle +i\hbar e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}{\frac {\partial }{\partial t}}|\chi \rangle &=[\mathbf {H} _{par+rad}+\mathbf {H} _{rad}]e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\chi \rangle &=e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\mathbf {H} _{par+rad}e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\chi \rangle &=e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\left[{\frac {(\mathbf {P} -{\frac {e}{c}}\mathbf {A} (\mathbf {r} ))^{2}}{2m}}+\mathbf {V(\mathbf {r} )} \right]e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\chi \rangle &=\left[{\frac {e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}(\mathbf {P} -{\frac {e}{c}}\mathbf {A} (\mathbf {r} ))^{2}e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}}{2m}}+\mathbf {V} _{int}(\mathbf {r} )\right]|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\chi \rangle &=\left[{\frac {(\mathbf {P} -{\frac {e}{c}}\mathbf {A} _{int}(\mathbf {r} ,t))^{2}}{2m}}+\mathbf {V} (\mathbf {r} )\right]|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\chi \rangle &=\left[\left({\frac {\mathbf {P} ^{2}}{2m}}+\mathbf {V} (\mathbf {r} )\right)-2{\frac {e}{2mc}}\mathbf {A} _{int}(\mathbf {r} ,t)\mathbf {P} +{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} ,t)\right]|\chi \rangle \\i\hbar {\frac {\partial }{\partial t}}e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle &=\left[\left({\frac {\mathbf {P} ^{2}}{2m}}+\mathbf {V} (\mathbf {r} )\right)-2{\frac {e}{2mc}}\mathbf {A} _{int}(\mathbf {r} ,t)\mathbf {P} +{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} ,t)\right]e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle \\\mathbf {H} _{par}e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}|\varphi \rangle +i\hbar e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}{\frac {\partial }{\partial t}}|\varphi \rangle &=\left[\mathbf {H} _{par}-{\frac {e}{mc}}\mathbf {A} _{int}(\mathbf {r} ,t)\mathbf {P} +{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} ,t)\right]e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle \\i\hbar e^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}{\frac {\partial }{\partial t}}|\varphi \rangle &=\left[-{\frac {e}{mc}}\mathbf {A} _{int}(\mathbf {r} ,t)\mathbf {P} +{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} ,t)\right]e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\varphi \rangle &=e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}\left[-{\frac {e}{mc}}\mathbf {A} _{int}(\mathbf {r} ,t)\mathbf {P} +{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} ,t)\right]e^{-{\frac {i}{\hbar }}\mathbf {H} _{par}t}|\varphi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\varphi \rangle &=\left[-{\frac {e}{mc}}\mathbf {A} _{int}(\mathbf {e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}re^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}} ,t)\mathbf {P} _{int}(t)+{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {e^{{\frac {i}{\hbar }}\mathbf {H} _{rad}t}re^{-{\frac {i}{\hbar }}\mathbf {H} _{rad}t}} ,t)\right]|\varphi \rangle \\i\hbar {\frac {\partial }{\partial t}}|\varphi \rangle &=\left[-{\frac {e}{mc}}\mathbf {A} _{int}(\mathbf {r} _{int}(t),t)\mathbf {P} _{int}(t)+{\frac {e^{2}}{2mc^{2}}}\mathbf {A} _{int}^{2}(\mathbf {r} _{int}(t),t)\right]|\varphi \rangle \\\end{aligned}}$

Non-perturbative methods

One important method in approximate determination of wave function and eigenvalue is Variational Principle. Variation method is very general one that it can be used whenever the equations can be put into variational form.

Spin

Spin 1/2 Angular Momentum

The angular momentum of a stationary spin 1/2 particle is found to be quantized to the $\pm {\frac {\hbar }{2}}$ regardless of the direction of the axis chosen to measure the angular momentum. There is a vector operator ${\textbf {\overrightarrow {S}}}=(S_{x},S_{y},S_{z})$ when projected along an arbitrary axis satisfies the following equations:

${\overrightarrow {S}}\cdot {\hat {m}}|{\hat {m}}\uparrow \rangle ={\dfrac {\hbar }{2}}|{\hat {m}}\uparrow \rangle$

${\overrightarrow {S}}\cdot {\hat {m}}|{\hat {m}}\downarrow \rangle ={\dfrac {-\hbar }{2}}|{\hat {m}}\downarrow \rangle$

$|{\hat {m}}\uparrow \rangle$ and $|{\hat {m}}\downarrow \rangle$ form a complete basis, which means that any state $|{\hat {n}}\uparrow \rangle$ and $|{\hat {n}}\downarrow \rangle$ can be expanded as a linear combination of $|{\hat {m}}\uparrow \rangle$ and $|{\hat {m}}\downarrow \rangle$.

The spin angular momentum operator obeys the standard commutation relations

$[S_{\mu },S_{\nu }]=i\hbar \epsilon _{\mu \nu \lambda }S_{\lambda }\Rightarrow [S_{x},S_{z}]=-i\hbar S_{y}$

The most commonly used basis is the one which diagonalizes ${\overrightarrow {S}}\cdot {\hat {z}}=S_{z}$.

By acting on the states $|{\hat {z}}\uparrow \rangle$ and $|{\hat {z}}\downarrow \rangle$ with $S_{z}$, we find $S_{z}|{\hat {z}}\uparrow \rangle ={\dfrac {\hbar }{2}}|{\hat {z}}\uparrow \rangle$ $S_{z}|{\hat {z}}\downarrow \rangle ={\dfrac {-\hbar }{2}}|{\hat {z}}\downarrow \rangle$

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

$S_{z}=\left({\begin{array}{ll}\langle {\hat {z}}\uparrow |S_{z}|{\hat {z}}\uparrow \rangle &\langle {\hat {z}}\uparrow |S_{z}|{\hat {z}}\downarrow \rangle \\\langle {\hat {z}}\downarrow |S_{z}|{\hat {z}}\uparrow \rangle &\langle {\hat {z}}\downarrow |S_{z}|{\hat {z}}\downarrow \rangle \end{array}}\right)=\left({\begin{array}{ll}\hbar /2&0\\0&-\hbar /2\end{array}}\right)={\dfrac {\hbar }{2}}\left({\begin{array}{ll}1&0\\0&-1\end{array}}\right)={\dfrac {\hbar }{2}}\sigma _{z}$

where $\sigma _{z}$ is the z Pauli spin matrix. Repeating the steps (or commutator relations), we can solve for the x and y directions.

$S_{x}={\dfrac {\hbar }{2}}\left({\begin{array}{ll}0&1\\1&0\end{array}}\right)={\dfrac {\hbar }{2}}\sigma _{x}$

$S_{y}={\dfrac {\hbar }{2}}\left({\begin{array}{ll}0&-i\\i&0\end{array}}\right)={\dfrac {\hbar }{2}}\sigma _{y}$

Properties of the Pauli Spin Matrices

Each Pauli matrix squared produces the unity matrix

$\sigma _{x}^{2}=\sigma _{y}^{2}=\sigma _{z}^{2}=\left({\begin{array}{ll}1&0\\0&1\end{array}}\right)$

The commutation relation is as follows

$[\sigma _{\mu },\sigma _{\nu }]=2i\epsilon _{\mu \nu \lambda }\sigma _{\lambda }$

and the anticommutator relation

${\sigma _{\mu },\sigma _{\nu }}=\sigma _{\mu }\sigma _{\nu }+\sigma _{\nu }\sigma _{\mu }=2\delta _{\mu \nu }\left({\begin{array}{ll}0&1\\1&0\end{array}}\right)$

For example, if $\sigma _{\mu }\sigma _{\nu }=-\sigma _{\nu }\sigma _{\mu }$

Then

$\sigma _{\mu }\sigma _{\nu }={\dfrac {1}{2}}[\sigma _{\mu },\sigma _{\nu }]+{\dfrac {1}{2}}{\sigma _{\mu },\sigma _{\nu }}=i\epsilon _{\mu \nu \lambda }\sigma _{\lambda }+\delta _{\mu \nu }$

$S_{\mu }S_{\nu }={\dfrac {\hbar ^{2}}{4}}\delta _{\mu \nu }+{\dfrac {i\hbar }{2}}\epsilon _{\mu \nu \lambda }S_{\lambda }$

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

In general

$(a\cdot \sigma )(b\cdot \sigma )=(a_{x}\sigma _{x}+a_{y}\sigma _{y}+a_{z}\sigma _{z})(b_{x}\sigma _{x}+b_{y}\sigma _{y}+b_{z}\sigma _{z})=a_{\mu }\sigma _{mu}b_{\nu }\sigma _{nu}=a_{\mu }b_{\nu }\sigma _{mu}\sigma _{nu}=a_{\mu }b_{\nu }\left(\left({\begin{array}{ll}1&0\\0&1\end{array}}\right)\delta _{\mu \nu }+i\epsilon _{\mu \nu \lambda }\sigma _{\lambda }\right)=\left({\begin{array}{ll}1&0\\0&1\end{array}}\right)a\cdot b+i(a\times b)\cdot \sigma$

Finally, any 2x2 matrix can be written in the form

$M=\alpha \left({\begin{array}{ll}1&0\\0&1\end{array}}\right)+\beta \cdot \sigma =\left({\begin{array}{ll}M_{11}&M_{12}\\M_{21}&M_{22}\end{array}}\right)$

$\Rightarrow \alpha ={\frac {1}{2}}(M_{11}+M_{22})$

$\Rightarrow \beta _{x}={\frac {1}{2}}(M_{12}+M_{21})$

$\Rightarrow \beta _{y}={\frac {i}{2}}(M_{12}-M_{21})$

$\Rightarrow \beta _{z}={\frac {1}{2}}(M_{11}-M_{22})$

for infinitesimal $\alpha$

${\hat {n}}={\hat {m}}+{\overrightarrow {\alpha }}x{\hat {m}}+O(\alpha ^{2})$

${\overrightarrow {S}}\cdot {\hat {n}}={\overrightarrow {S}}\cdot {\hat {m}}+{\overrightarrow {S}}\cdot ({\overrightarrow {\alpha }}\times {\overrightarrow {m}})$

$S_{\mu }{\hat {n_{\mu }}}=S_{\mu }{\hat {m_{\mu }}}+S_{\mu }\epsilon _{\mu \nu \lambda }\alpha _{\nu }{\hat {m_{\lambda }}}$

Note that using the previous developed formulas, we find that

$S_{\mu }\epsilon _{\mu \nu \lambda }={\dfrac {1}{i\hbar }}[S_{\nu },S_{\lambda }]\Rightarrow {\overrightarrow {S}}\cdot {\hat {n}}={\overrightarrow {S}}\cdot {\hat {m}}+{\dfrac {1}{i\hbar }}[{\overrightarrow {\alpha }}\cdot {\overrightarrow {S}},{\hat {m}}\cdot {\overrightarrow {S}}]={\overrightarrow {S}}\cdot {\hat {m}}+{\dfrac {1}{i\hbar }}[{\overrightarrow {S}}\cdot {\hat {m}},{\overrightarrow {S}}\cdot {\overrightarrow {\alpha }}]$

To this order in ${\overrightarrow {\alpha }}$:

${\overrightarrow {S}}\cdot {\hat {n}}=e^{-{\frac {i}{\hbar }}{\overrightarrow {S}}\cdot {\overrightarrow {\alpha }}}({\overrightarrow {S}}\cdot {\hat {m}})e^{{\frac {i}{\hbar }}{\overrightarrow {S}}\cdot {\overrightarrow {\alpha }}}$

for finite $\alpha$ (correct for all orders)

${\overrightarrow {S}}\cdot {\hat {n}}\left(e^{-{\frac {i}{\hbar }}{\overrightarrow {S}}\cdot {\overrightarrow {\alpha }}}|{\hat {m}}\uparrow \rangle \right)={\dfrac {\hbar }{2}}\left(e^{-{\frac {i}{\hbar }}{\overrightarrow {S}}\cdot {\overrightarrow {\alpha }}}|{\hat {m}}\uparrow \rangle \right)$

Addition of angular momenta

Total angular momentum is defined as

${\overrightarrow {J}}={\overrightarrow {J_{1}}}+{\overrightarrow {J_{2}}}={\overrightarrow {J_{1}}}\otimes \left({\begin{array}{ll}1&0\\0&1\end{array}}\right)+\left({\begin{array}{ll}1&0\\0&1\end{array}}\right)\otimes {\overrightarrow {J_{2}}}\Rightarrow |j_{1}m_{1}\rangle \otimes |j_{2}m_{2}\rangle =|j_{1}m_{1}j_{2}m_{2}\rangle$

where Hilbert space size is $\!(2j_{1}+1)(2j_{2}+1)$.

We have the following commutation relations:

$[J_{\mu },J_{\nu }]=i\hbar \epsilon _{\mu \nu \lambda }J_{\lambda }$ which can be used to find the eigenstate of $J^{2}$ and $J_{z}$.

$[J_{1,2}^{2},J^{2}]=[J_{1,2}^{2},J_{z}]=0$.

$[J_{2},J_{z}]=0$

There are two sets of base kets which can be used:

1. The simultaneous eigenkets of $J_{1}^{2}$, $J_{2}^{2}$, $\!J_{1z}$, $J_{2z}$, denoted by $|j_{1}j_{2}m_{1}m_{2}\rangle$. These four operators commute with each other, and they operate on the base kets according to:

$J_{1,2}^{2}|j_{1}j_{2}m_{1}m_{2}\rangle =\hbar _{2}j_{1,2}(j_{1,2}+1)|j_{1}j_{2}m_{1}m_{2}\rangle$

$J_{1z,2z}|j_{1}j_{2}m_{1}m_{2}\rangle =\hbar m_{1,2}|j_{1}j_{2}m_{1}m_{2}\rangle$

2. The simultaneous eigenkets of $J_{2}$, $J_{1}^{2}$, $J_{2}^{2}$ and $J_{z}$, denoted by $|jmj_{1}j_{2}\rangle$. These four operators operate on the base kets according to:

$J^{2}|jmj_{1}j_{2}\rangle =\hbar ^{2}j(j+1)|jmj_{1}j_{2}\rangle$

$J_{z}|jmj_{1}j_{2}\rangle =\hbar m|jmj_{1}j_{2}\rangle$

$J_{1,2}^{2}|jmj_{1}j_{2}\rangle =\hbar ^{2}j_{1,2}(j_{1,2}+1)|jmj_{1}j_{2}\rangle$

For example, assume two spin 1/2 particles with basis ${|\uparrow \uparrow \rangle ,|\uparrow \downarrow \rangle ,|\downarrow \uparrow \rangle ,|\downarrow \downarrow \rangle }$. These states are eigenstates of $J_{1}^{1},J_{2}^{2},J_{1z},J_{2z}$, but are they eigenstates of $J^{2}$ and $J_{z}^{2}$?

$J^{2}|\uparrow \downarrow \rangle =(J_{x}^{2}+J_{y}^{2}+J_{z}^{2})|\uparrow \downarrow \rangle =((J_{x}+iJ_{y})(J_{x}-iJ_{y})+i[J_{x},J_{y}]+J_{z}^{2})|\uparrow \downarrow \rangle$

define $J_{\pm }=(J_{x}\pm iJ_{y})$

$J^{2}|\uparrow \downarrow \rangle =(J_{+}J_{-}-\hbar J_{z}+J_{z}^{2})|\uparrow \downarrow \rangle =((J_{1+}+J_{2+})(J_{1-}+J_{2-})+(J_{1z}+J_{2z})^{2}-\hbar (J_{1z}+J_{2z}))|\uparrow \downarrow \rangle$

$\Rightarrow (J_{1z}+J_{2z})|\uparrow \downarrow \rangle =({\dfrac {\hbar }{2}}-{\dfrac {\hbar }{2}})|\uparrow \downarrow \rangle =0$

$\Rightarrow J_{1+}|\uparrow \downarrow \rangle =J_{2+}|\downarrow \uparrow \rangle =J_{2-}|\uparrow \downarrow \rangle =0$

$J^{2}|\uparrow \downarrow \rangle =(J_{1+}+J_{2+})(J_{1-}+J_{2-})|\uparrow \downarrow \rangle =(J_{1+}+J_{2+})|\downarrow \downarrow \rangle =(|\uparrow \downarrow \rangle +|\downarrow \uparrow \rangle )$

Which means that $|\uparrow \downarrow \rangle$ is not an eigenstate of $J^{2}$.

Orthogonality (with Clebsch Gordon Coefficients)

$\sum _{jm}\langle j_{1}m'_{1}j_{2}m'_{2}|jmj_{1}j_{2}\rangle \langle jmj_{1}j_{2}|j_{1}m_{1}j_{2}m_{2}\rangle =\delta _{m_{1}m'_{1}}\delta _{m_{2}m'_{2}}$

$\sum _{m_{1}m_{2}}\langle jmj_{1}j_{2}|j_{1}m_{1}j_{2}m_{2}\rangle \langle j_{1}m_{1}j_{2}m_{2}|j'm'j_{1}j_{2}\rangle =\delta _{jj'}\delta _{mm'}$

Hydrogen atom with spin orbit coupling given by the following hamiltonian

$H'={\dfrac {c^{2}}{2m^{2}c^{2}r^{3}}}{\overrightarrow {L}}\cdot {\overrightarrow {S}}$

Recall, the atomic spectrum for bound states

$E_{n}={\dfrac {-e^{2}}{2a_{o}}}{\dfrac {1}{n^{2}}}$ $;n=1,2,3,...$

The ground state, $|1s\rangle$, is doubly degenerate: ${\dfrac {\uparrow \downarrow }{1s}}$

First excited state is 8-fold degenerate: ${\dfrac {\uparrow \downarrow }{1s}}{\dfrac {\uparrow \downarrow }{}}{\dfrac {\uparrow \downarrow }{2p}}{\dfrac {\uparrow \downarrow }{}}$

nth state is $2n^{2}$ fold degenerate

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

${\overrightarrow {L}}\cdot {\overrightarrow {S}}=L_{x}S_{x}+L_{y}S_{y}+L_{z}S_{z}$

Define lowering and raising operators

$\Rightarrow L_{\pm }=L_{x}\pm iL_{y}$

$\Rightarrow S_{\pm }=S_{x}\pm iS_{y}$

${\overrightarrow {L}}\cdot {\overrightarrow {S}}=L_{z}S_{z}+{\dfrac {1}{2}}L_{+}S_{-}+{\dfrac {1}{2}}L_{-}S_{+}$

For the ground state, $(|1s,\uparrow \rangle ,|1s,\downarrow \rangle )$, nothing happens. Kramer's theorem protects the double degeneracy.

For the first excited state, $(|2s,\uparrow \rangle ,|2s,\downarrow \rangle )$, once again nothing happens.

For $(|2p,\uparrow \rangle ,|2p,\downarrow \rangle )$, there is a four fold degeneracy.

For the exact solution,

${\overrightarrow {L}}\cdot {\overrightarrow {S}}={\dfrac {1}{2}}({\overrightarrow {L}}+{\overrightarrow {S}})^{2}-{\dfrac {1}{2}}{\overrightarrow {L^{2}}}-{\dfrac {1}{2}}{\overrightarrow {S^{2}}}={\dfrac {1}{2}}(J^{2}-L^{2}-S^{2})$

add the angular momenta:

$|1s\rangle :l=0,s={\dfrac {1}{2}}:0\otimes {\dfrac {1}{2}}={\dfrac {1}{2}}$

$|2s\rangle :l=0,s={\dfrac {1}{2}}:0\otimes {\dfrac {1}{2}}={\dfrac {1}{2}}$

$|2p_{m},0\rangle :l=1,s={\dfrac {1}{2}}:1\otimes {\dfrac {1}{2}}={\dfrac {3}{2}}\oplus {\dfrac {1}{2}}$

So that

${\overrightarrow {L}}\cdot {\overrightarrow {S}}|j={\dfrac {3}{2}},m,l=1,s={\dfrac {1}{2}}\rangle ={\dfrac {1}{2}}(\hbar ^{2}{\dfrac {3}{2}}{\dfrac {5}{2}}-2\hbar ^{2}-{\dfrac {3}{4}}\hbar ^{2})|j={\dfrac {3}{2}},m,l=1,s={\dfrac {1}{2}}\rangle ={\dfrac {\hbar ^{2}}{2}}|j={\dfrac {3}{2}},m,l=1,s={\dfrac {1}{2}}\rangle$

${\overrightarrow {L}}\cdot {\overrightarrow {S}}|j={\dfrac {1}{2}},m,l=1,s={\dfrac {1}{2}}\rangle ={\dfrac {-\hbar ^{2}}{2}}|j={\dfrac {1}{2}},m,l=1,s={\dfrac {1}{2}}\rangle$

$|j={\dfrac {3}{2}},m={\dfrac {3}{2}},l=1,s={\dfrac {1}{2}}\rangle =|l=1,m_{l}=1\rangle |s={\dfrac {1}{2}},m_{s}={\dfrac {1}{2}}\rangle$

$|j={\dfrac {3}{2}},m={\dfrac {3}{2}}\rangle =|m_{l}=1\rangle |m_{s}={\dfrac {1}{2}}\rangle$

Define J_

$J_{|}{\dfrac {3}{2}},{\dfrac {1}{2}}\rangle =(L_{+}S_{)}|1\rangle |{\dfrac {1}{2}}\rangle$

$\hbar {\sqrt {{\dfrac {3}{2}}{\dfrac {5}{2}}-{\dfrac {3}{2}}{\dfrac {1}{2}}}}|{\dfrac {3}{2}},{\dfrac {1}{2}}\rangle =\hbar {\sqrt {2}}|0\rangle |{\dfrac {1}{2}}\rangle +\hbar {\sqrt {{\dfrac {1}{2}}{\dfrac {3}{2}}+{\dfrac {1}{4}}}}|1,-{\dfrac {1}{2}}\rangle$

${\sqrt {3}}|{\dfrac {3}{2}},{\dfrac {1}{2}}\rangle ={\sqrt {2}}|0\rangle |{\dfrac {1}{2}}\rangle +|1,-{\dfrac {1}{2}}\rangle$

$|{\dfrac {3}{2}},{\dfrac {1}{2}}\rangle ={\sqrt {\dfrac {2}{3}}}|0\rangle |{\dfrac {1}{2}}\rangle +{\sqrt {\dfrac {1}{3}}}|1\rangle |-{\dfrac {1}{2}}\rangle$

$|{\dfrac {3}{2}},{\dfrac {-1}{2}}\rangle ={\sqrt {\dfrac {2}{3}}}|0\rangle |{\dfrac {-1}{2}}\rangle +{\sqrt {\dfrac {1}{3}}}|-1\rangle |{\dfrac {1}{2}}\rangle$

$|{\dfrac {3}{2}},{\dfrac {-3}{2}}\rangle =|-1\rangle |-{\dfrac {1}{2}}\rangle$

Can express as

$|j={\dfrac {1}{2}},m={\dfrac {1}{2}}\rangle =\alpha |0\rangle |{\dfrac {1}{2}}\rangle +\beta |1\rangle |-{\dfrac {1}{2}}\rangle$

$|j={\dfrac {1}{2}},m=-{\dfrac {1}{2}}\rangle =\alpha '|0\rangle |-{\dfrac {1}{2}}\rangle +\beta '|-1\rangle |{\dfrac {1}{2}}\rangle$

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

$\alpha =-{\sqrt {\dfrac {1}{3}}}$ and $\beta ={\sqrt {\dfrac {2}{3}}}$