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## A different derivation of the adiabatic perturbation coefficient equation

Posted by peeterjoot on October 27, 2011

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

# Motivation.

Professor Sipe’s adiabatic perturbation and that of the text [1] in section 17.5.1 and section 17.5.2 use different notation for $\gamma_m$ and take a slightly different approach. We can find Prof Sipe’s final result with a bit less work, if a hybrid of the two methods is used.

# Guts

Our starting point is the same, we have a time dependent slowly varying Hamiltonian

\begin{aligned}H = H(t),\end{aligned} \hspace{\stretch{1}}(2.1)

where our perturbation starts at some specific time from a given initial state

\begin{aligned}H(t) = H_0, \qquad t \le 0.\end{aligned} \hspace{\stretch{1}}(2.2)

We assume that instantaneous eigenkets can be found, satisfying

\begin{aligned}H(t) {\left\lvert {n(t)} \right\rangle} = E_n(t) {\left\lvert {n(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.3)

Here I’ll use ${\left\lvert {n} \right\rangle} \equiv {\left\lvert {n(t)} \right\rangle}$ instead of the ${\left\lvert {\hat{\psi}_n(t)} \right\rangle}$ that we used in class because its easier to write.

Now suppose that we have some arbitrary state, expressed in terms of the instantaneous basis kets ${\left\lvert {n} \right\rangle}$

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_n \bar{b}_n(t) e^{-i\alpha_n + i \gamma_n} {\left\lvert {n} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.4)

where

\begin{aligned}\alpha_n(t) = \frac{1}{{\hbar}} \int_0^t dt' E_n(t'),\end{aligned} \hspace{\stretch{1}}(2.5)

and $\gamma_n$ (using the notation in the text, not in class) is to be determined.

For this state, we have at the time just before the perturbation

\begin{aligned}{\left\lvert {\psi(0)} \right\rangle} = \sum_n \bar{b}_n(0) e^{-i\alpha_n(0) + i \gamma_n(0)} {\left\lvert {n(0)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.6)

The question to answer is: How does this particular state evolve?

Another question, for those that don’t like sneaky bastard derivations, is where did that magic factor of $e^{-i\alpha_n}$ come from in our superposition state? We will see after we start taking derivatives that this is what we need to cancel the $H(t){\left\lvert {n} \right\rangle}$ in Schr\”{o}dinger’s equation.

Proceeding to plug into the evolution identity we have

\begin{aligned}0 &={\left\langle {m} \right\rvert} \left( i \hbar \frac{d{{}}}{dt} - H(t) \right) {\left\lvert {\psi} \right\rangle} \\ &={\left\langle {m} \right\rvert} \left(\sum_n e^{-i \alpha_n + i \gamma_n}(i \hbar) \left(\frac{d{{\bar{b}_n}}}{dt}+ \bar{b}_n \left(-i \not{{\frac{E_n}{\hbar}}} + i \dot{\gamma}_m \right)\right) {\left\lvert {n} \right\rangle}+ i \hbar \bar{b}_n \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}- \not{{E_n \bar{b}_n {\left\lvert {n} \right\rangle}}} \right)\\ &=e^{-i \alpha_m + i \gamma_m}(i \hbar) \frac{d{{\bar{b}_m}}}{dt}+e^{-i \alpha_m + i \gamma_m}(i \hbar) i \dot{\gamma}_m \bar{b}_m+ i \hbar \sum_n \bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}e^{-i \alpha_n + i \gamma_n} \\ &\sim\frac{d{{\bar{b}_m}}}{dt}+i \dot{\gamma}_m \bar{b}_m+ \sum_n e^{-i \alpha_n + i \gamma_n}e^{i \alpha_m - i \gamma_m}\bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle} \\ &=\frac{d{{\bar{b}_m}}}{dt}+i \dot{\gamma}_m \bar{b}_m+ \bar{b}_m {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle}+\sum_{n \ne m} e^{-i \alpha_n + i \gamma_n}e^{i \alpha_m - i \gamma_m}\bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}\end{aligned}

We are free to pick $\gamma_m$ to kill the second and third terms

\begin{aligned}0 =i \dot{\gamma}_m \bar{b}_m+ \bar{b}_m {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.7)

or

\begin{aligned}\dot{\gamma}_m = i {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.8)

which after integration is

\begin{aligned}\gamma_m(t)= i \int_0^t dt' {\left\langle {m(t')} \right\rvert} \frac{d}{dt'} {\left\lvert {m(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.9)

as in class we can observe that this is a purely real function. We are left with

\begin{aligned}\frac{d{{\bar{b}_m}}}{dt}=-\sum_{n \ne m} \bar{b}_n e^{-i \alpha_{nm} + i \gamma_{nm}}{\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle} ,\end{aligned} \hspace{\stretch{1}}(2.10)

where

\begin{aligned}\alpha_{nm} &= \alpha_{n} -\alpha_m \\ \gamma_{nm} &= \gamma_{n} -\gamma_m \end{aligned} \hspace{\stretch{1}}(2.11)

The task is now to find solutions for these $\bar{b}_m$ coefficients, and we can refer to the class notes for that without change.

# References

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