Peeter Joot's Blog.

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Second form of adiabatic approximation.

Posted by peeterjoot on December 11, 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.

In class we were shown an adiabatic approximation where we started with (or worked our way towards) a representation of the form

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k c_k(t) e^{-i \int_0^t (\omega_k(t') - \Gamma_k(t')) dt' } {\left\lvert {\psi_k(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.1)

where {\left\lvert {\psi_k(t)} \right\rangle} were normalized energy eigenkets for the (slowly) evolving Hamiltonian

\begin{aligned}H(t) {\left\lvert {\psi_k(t)} \right\rangle} = E_k(t) {\left\lvert {\psi_k(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(1.2)

In the problem sets we were shown a different adiabatic approximation, where are starting point is

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} = \sum_k c_k(t) {\left\lvert {\psi_k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(1.3)

For completeness, here’s a walk through of the general amplitude derivation that’s been used.

Guts

We operate with our energy identity once again

\begin{aligned}0 &=\left(H - i \hbar \frac{d{{}}}{dt} \right) \sum_k c_k {\left\lvert {k} \right\rangle} \\ &=\sum_k c_k E_k {\left\lvert {k} \right\rangle} - i \hbar c_k' {\left\lvert {k} \right\rangle} - i \hbar c_k {\left\lvert {k'} \right\rangle} ,\end{aligned}

where

\begin{aligned}{\left\lvert {k'} \right\rangle} = \frac{d{{}}}{dt} {\left\lvert {k} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.4)

Bra’ing with {\left\langle {m} \right\rvert}, and split the sum into k = m and k \ne m parts

\begin{aligned}0 =c_m E_m - i \hbar c_m' - i \hbar c_m \left\langle{{m}} \vert {{m'}}\right\rangle - i \hbar \sum_{k \ne m} c_k \left\langle{{m}} \vert {{k'}}\right\rangle \end{aligned} \hspace{\stretch{1}}(2.5)

Again writing

\begin{aligned}\Gamma_m = i \left\langle{{m}} \vert {{m'}}\right\rangle \end{aligned} \hspace{\stretch{1}}(2.6)

We have

\begin{aligned}c_m' = \frac{1}{{i \hbar}} c_m (E_m - \hbar \Gamma_m) - \sum_{k \ne m} c_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(2.7)

In this form we can make an “Adiabatic” approximation, dropping the k \ne m terms, and integrate

\begin{aligned}\int \frac{d c_m'}{c_m} = \frac{1}{{i \hbar}} \int_0^t (E_m(t') - \hbar \Gamma_m(t')) dt' \end{aligned} \hspace{\stretch{1}}(2.8)

or

\begin{aligned}c_m(t) = A \exp\left(\frac{1}{{i \hbar}} \int_0^t (E_m(t') - \hbar \Gamma_m(t')) dt' \right).\end{aligned} \hspace{\stretch{1}}(2.9)

Evaluating at t = 0, fixes the integration constant for

\begin{aligned}c_m(t) = c_m(0) \exp\left(\frac{1}{{i \hbar}} \int_0^t (E_m(t') - \hbar \Gamma_m(t')) dt' \right).\end{aligned} \hspace{\stretch{1}}(2.10)

Observe that this is very close to the starting point of the adiabatic approximation we performed in class since we end up with

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k c_k(0) e^{-i \int_0^t (\omega_k(t') - \Gamma_k(t')) dt' } {\left\lvert {k(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.11)

So, to perform the more detailed approximation, that started with 1.1, where we ended up with all the cross terms that had both \omega_k and Berry phase \Gamma_k dependence, we have only to generalize by replacing c_k(0) with c_k(t).

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This entry was posted on December 11, 2011 at 10:06 pm and is filed under Math and Physics Learning.. Tagged: , , . You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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