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## PHY456H1F: Quantum Mechanics II. Lecture 5 (Taught by Prof J.E. Sipe). Pertubation theory and degeneracy. Review of dynamics

Posted by peeterjoot on September 26, 2011

[Click here for a PDF of this post with nicer formatting (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Issues concerning degeneracy.

## When the perturbed state is non-degenerate.

Suppose the state of interest is non-degenerate but others are

FIXME: diagram. states designated by dashes labeled $n1$, $n2$, $n3$ degeneracy $\alpha = 3$ for energy $E_n^{(0)}$.

This is no problem except for notation, and if the analysis is repeated we find

\begin{aligned}E_s &= E_s^{(0)} + \lambda {H_{ss}}' + \lambda^2 \sum_{m \ne s, \alpha} \frac{{\left\lvert{{H_{m \alpha ; s}}'}\right\rvert}^2 }{ E_s^{(0)} - E_m^{(0)} } + \cdots\\ {\lvert {\bar{\psi}_s} \rangle} &= {\lvert {{\psi_s}^{(0)}} \rangle} + \lambda\sum_{m \ne s, \alpha} \frac{{H_{m \alpha ; s}}'}{ E_s^{(0)} - E_m^{(0)} } {\lvert {{\psi_{m \alpha}}^{(0)}} \rangle}+ \cdots,\end{aligned} \hspace{\stretch{1}}(2.1)

where

\begin{aligned}{H_{m \alpha ; s}}' ={\langle {{\psi_{m \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(2.3)

## When the perturbed state is also degenerate.

FIXME: diagram. states designated by dashes labeled $n1$, $n2$, $n3$ degeneracy $\alpha = 3$ for energy $E_n^{(0)}$, and states designated by dashes labeled $s1$, $s2$, $s3$ degeneracy $\alpha = 3$ for energy $E_s^{(0)}$.

If we just blindly repeat the derivation for the non-degenerate case we would obtain

\begin{aligned}E_s &= E_s^{(0)} + \lambda {H_{s1 ; s1}}' + \lambda^2 \sum_{m \ne s, \alpha} \frac{{\left\lvert{{H_{m \alpha ; s1}}'}\right\rvert}^2 }{ E_s^{(0)} - E_m^{(0)} } + \lambda^2 \sum_{\alpha \ne 1} \frac{{\left\lvert{{H_{s \alpha ; s1}}'}\right\rvert}^2 }{ E_s^{(0)} - {red} } + \cdots\\ {\lvert {\bar{\psi}_s} \rangle} &= {\lvert {{\psi_s}^{(0)}} \rangle} + \lambda\sum_{m \ne s, \alpha} \frac{{H_{m \alpha ; s}}'}{ E_s^{(0)} - E_m^{(0)} } {\lvert {{\psi_{m \alpha}}^{(0)}} \rangle}+ \lambda\sum_{\alpha \ne s1} \frac{{H_{s \alpha ; s1}}'}{ E_s^{(0)} - {red} } {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle}+ \cdots,\end{aligned} \hspace{\stretch{1}}(2.4)

where

\begin{aligned}{H_{m \alpha ; s1}}' ={\langle {{\psi_{m \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s1}}^{(0)}} \rangle}\end{aligned} \hspace{\stretch{1}}(2.6)

Note that the $E_s^{(0)} -{red}$ is NOT a typo, and why we run into trouble. There is one case where a perturbation approach is still possible. That case is if we happen to have

\begin{aligned}{\langle {{\psi_{m \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s1}}^{(0)}} \rangle} = 0.\end{aligned} \hspace{\stretch{1}}(2.7)

That may not be obvious, but if one returns to the original derivation, the right terms cancel so that one will not end up with the $0/0$ problem.

FIXME: do this derivation.

## Diagonalizing the perturbation Hamiltonian.

Suppose that we do not have this special zero condition that allows the perturbation treatment to remain valid. What can we do. It turns out that we can make use of the fact that the perturbation Hamiltonian is Hermitian, and diagonalize the matrix

\begin{aligned}{\langle {{\psi_{s \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s \beta}}^{(0)}} \rangle} \end{aligned} \hspace{\stretch{1}}(2.8)

In the example of a two fold degeneracy, this amounts to us choosing not to work with the states

\begin{aligned}{\lvert {\psi_{s1}^{(0)}} \rangle}, {\lvert {\psi_{s2}^{(0)}} \rangle},\end{aligned} \hspace{\stretch{1}}(2.9)

both some linear combinations of the two

\begin{aligned}{\lvert {\psi_{sI}^{(0)}} \rangle} &= a_1 {\lvert {\psi_{s1}^{(0)}} \rangle} + b_1 {\lvert {\psi_{s2}^{(0)}} \rangle} \\ {\lvert {\psi_{sII}^{(0)}} \rangle} &= a_2 {\lvert {\psi_{s1}^{(0)}} \rangle} + b_2 {\lvert {\psi_{s2}^{(0)}} \rangle} \end{aligned} \hspace{\stretch{1}}(2.10)

In this new basis, once found, we have

\begin{aligned}{\langle {{\psi_{s \alpha}}^{(0)}} \rvert} H' {\lvert {{\psi_{s \beta}}^{(0)}} \rangle} = \mathcal{H}_\alpha \delta_{\alpha \beta}\end{aligned} \hspace{\stretch{1}}(2.12)

Utilizing this to fix the previous, one would get if the analysis was repeated correctly

\begin{aligned}E_{s\alpha} &= E_s^{(0)} + \lambda {H_{s\alpha ; s\alpha}}' + \lambda^2 \sum_{m \ne s, \alpha} \frac{{\left\lvert{{H_{m \beta ; s \alpha}}'}\right\rvert}^2 }{ E_s^{(0)} - E_m^{(0)} } + \cdots\\ {\lvert {\bar{\psi}_{s \alpha}} \rangle} &= {\lvert {{\psi_{s \alpha}}^{(0)}} \rangle} + \lambda\sum_{m \ne s, \beta} \frac{{H_{m \beta ; s \alpha}}'}{ E_s^{(0)} - E_m^{(0)} } {\lvert {{\psi_{m \beta}}^{(0)}} \rangle}+ \cdots.\end{aligned} \hspace{\stretch{1}}(2.13)

We see that a degenerate state can be split by applying perturbation.

FIXME: do this derivation.

FIXME: diagram. $E_s^{(0)}$ as one energy level without perturbation, and as two distinct levels with perturbation.

guess I’ll bet that this is the origin of the spectral line splitting, especially given that an atom like hydrogen has degenerate states.

# Review of dynamics.

We want to move on to time dependent problems. In general for a time dependent problem, the answer follows provided one has solved for all the perturbed energy eigenvalues. This can be laborious (or not feasible due to infinite sums).

Before doing this, let’s review our dynamics as covered in section 3 of the text [1].

## Schr\”{o}dinger and Heisenberg pictures

Our operator equation in the Schr\”{o}dinger picture is the familiar

\begin{aligned}i \hbar \frac{d}{dt} {\lvert {\psi_s(t)} \rangle} = H {\lvert {\psi_s(t)} \rangle}\end{aligned} \hspace{\stretch{1}}(3.15)

and most of our operators $X, P, \cdots$ are time independent.

\begin{aligned}\left\langle{{O}}\right\rangle(t) = {\langle {\psi_s(t)} \rvert} O_s{\lvert {\psi_s(t)} \rangle}\end{aligned} \hspace{\stretch{1}}(3.16)

where $O_s$ is the operator in the Schr\”{o}dinger picture, and is non time dependent.

Formally, the time evolution of any state is given by

\begin{aligned}{\lvert {\psi_s(t)} \rangle}e^{-i H t/\hbar}{\lvert {\psi_s(0)} \rangle} = U(t, 0) {\lvert {\psi_s(0)} \rangle} \end{aligned} \hspace{\stretch{1}}(3.17)

so the expectation of an operator can be written

\begin{aligned}\left\langle{{O}}\right\rangle(t) = {\langle {\psi_s(0)} \rvert} e^{i H t/\hbar}O_se^{-i H t/\hbar}{\lvert {\psi_s(0)} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.18)

With the introduction of the Heisenberg ket

\begin{aligned}{\lvert {\psi_H} \rangle} = {\lvert {\psi_s(0)} \rangle},\end{aligned} \hspace{\stretch{1}}(3.19)

and Heisenberg operators

\begin{aligned}O_H = e^{i H t/\hbar} O_s e^{-i H t/\hbar},\end{aligned} \hspace{\stretch{1}}(3.20)

the expectation evolution takes the form

\begin{aligned}\left\langle{{O}}\right\rangle(t) = {\langle {\psi_H} \rvert} O_H{\lvert {\psi_H} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.21)

Note that because the Hamiltonian commutes with it’s exponential (it commutes with itself and any power series of itself), the Hamiltonian in the Heisenberg picture is the same as in the Schr\”{o}dinger picture

\begin{aligned}H_H = e^{i H t/\hbar} H e^{-i H t/\hbar} = H.\end{aligned} \hspace{\stretch{1}}(3.22)

### Time evolution and the Commutator

Taking the derivative of 3.20 provides us with the time evolution of any operator in the Heisenberg picture

\begin{aligned}i \hbar \frac{d}{dt} O_H(t) &=i \hbar \frac{d}{dt} \left( e^{i H t/\hbar} O_s e^{-i H t/\hbar}\right) \\ &=i \hbar \left( \frac{i H}{\hbar} e^{i H t/\hbar} O_s e^{-i H t/\hbar}+e^{i H t/\hbar} O_s e^{-i H t/\hbar} \frac{-i H}{\hbar} \right) \\ &=\left( -H O_H+O_H H\right).\end{aligned}

We can write this as a commutator

\begin{aligned}i \hbar \frac{d}{dt} O_H(t) = \left[{O_H},{H}\right].\end{aligned} \hspace{\stretch{1}}(3.23)

### Summarizing the two pictures.

\begin{aligned}\text{Schr\"{o}dinger picture} &\qquad \text{Heisenberg picture} \\ i \hbar \frac{d}{dt} {\lvert {\psi_s(t)} \rangle} = H {\lvert {\psi_s(t)} \rangle} &\qquad i \hbar \frac{d}{dt} O_H(t) = \left[{O_H},{H}\right] \\ {\langle {\psi_s(t)} \rvert} O_S {\lvert {\psi_s(t)} \rangle} &= {\langle {\psi_H} \rvert} O_H {\lvert {\psi_H} \rangle} \\ {\lvert {\psi_s(0)} \rangle} &= {\lvert {\psi_H} \rangle} \\ O_S &= O_H(0)\end{aligned}

# References

[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.