Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

One more adiabatic pertubation derivation.

Posted by peeterjoot on December 8, 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.)]


I liked one of the adiabatic pertubation derivations that I did to review the material, and am recording it for reference.

Build up.

In time dependent pertubation we started after noting that our ket in the interaction picture, for a Hamiltonian H = H_0 + H'(t), took the form

\begin{aligned}{\left\lvert {\alpha_S(t)} \right\rangle} = e^{-i H_0 t/\hbar} {\left\lvert {\alpha_I(t)} \right\rangle} = e^{-i H_0 t/\hbar} U_I(t) {\left\lvert {\alpha_I(0)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.1)

Here we have basically assumed that the time evolution can be factored into a portion dependent on only the static portion of the Hamiltonian, with some other operator U_I(t), providing the remainder of the time evolution. From 2.1 that operator U_I(t) is found to behave according to

\begin{aligned}i \hbar \frac{d{{U_I}}}{dt} = e^{i H_0 t/\hbar} H'(t) e^{-i H_0 t/\hbar} U_I,\end{aligned} \hspace{\stretch{1}}(2.2)

but for our purposes we just assumed it existed, and used this for motivation. With the assumption that the interaction picture kets can be written in terms of the basis kets for the system at t=0 we write our Schr\”{o}dinger ket as

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{-i H_0 t/\hbar} a_k(t) {\left\lvert {k} \right\rangle}= \sum_k e^{-i \omega_k t/\hbar} a_k(t) {\left\lvert {k} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.3)

where {\left\lvert {k} \right\rangle} are the energy eigenkets for the initial time equation problem

\begin{aligned}H_0 {\left\lvert {k} \right\rangle} = E_k^0 {\left\lvert {k} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.4)

Adiabatic case.

For the adiabatic problem, we assume the system is changing very slowly, as described by the instantanious energy eigenkets

\begin{aligned}H(t) {\left\lvert {k(t)} \right\rangle} = E_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.5)

Can we assume a similar representation to 2.3 above, but allow {\left\lvert {k} \right\rangle} to vary in time? This doesn’t quite work since {\left\lvert {k(t)} \right\rangle} are no longer eigenkets of H_0

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{-i H_0 t/\hbar} a_k(t) {\left\lvert {k(t)} \right\rangle}\ne \sum_k e^{-i \omega_k t} a_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.6)

Operating with e^{i H_0 t/\hbar} does not give the proper time evolution of {\left\lvert {k(t)} \right\rangle}, and we will in general have a more complex functional dependence in our evolution operator for each {\left\lvert {k(t)} \right\rangle}. Instead of an \omega_k t dependence in this time evolution operator let’s assume we have some function \alpha_k(t) to be determined, and can write our ket as

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{-i \alpha_k(t)} a_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.7)

Operating on this with our energy operator equation we have

\begin{aligned}0 &=\left(H - i \hbar \frac{d}{dt} \right) {\left\lvert {\psi} \right\rangle} \\ &=\left(H - i \hbar \frac{d}{dt} \right) \sum_k e^{-i \alpha_k} a_k {\left\lvert {k} \right\rangle} \\ &=\sum_k e^{-i \alpha_k(t)} \left( \left( E_k a_k-i \hbar (-i \alpha_k' a_k + a_k')\right) {\left\lvert {k} \right\rangle}-i \hbar a_k {\left\lvert {k'} \right\rangle}\right) \\ \end{aligned}

Here I’ve written {\left\lvert {k'} \right\rangle} = d{\left\lvert {k} \right\rangle}/dt. In our original time dependent pertubaton the -i \alpha_k' term was -i \omega_k, so this killed off the E_k. If we assume this still kills off the E_k, we must have

\begin{aligned}\alpha_k = \frac{1}{{\hbar}} \int_0^t E_k(t') dt',\end{aligned} \hspace{\stretch{1}}(3.8)

and are left with

\begin{aligned}0=\sum_k e^{-i \alpha_k(t)} \left( a_k' {\left\lvert {k} \right\rangle}+a_k {\left\lvert {k'} \right\rangle}\right).\end{aligned} \hspace{\stretch{1}}(3.9)

Bra’ing with {\left\langle {m} \right\rvert} we have

\begin{aligned}0=e^{-i \alpha_m(t)} a_m' +e^{-i \alpha_m(t)} a_m \left\langle{{m}} \vert {{m'}}\right\rangle+\sum_{k \ne m} e^{-i \alpha_k(t)} a_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.10)


\begin{aligned}a_m' +a_m \left\langle{{m}} \vert {{m'}}\right\rangle=-\sum_{k \ne m} e^{-i \alpha_k(t)} e^{i \alpha_m(t)} a_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.11)

The LHS is a perfect differential if we introduce an integration factor e^{\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle}, so we can write

\begin{aligned}e^{-\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle} ( a_m e^{\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle } )'=-\sum_{k \ne m} e^{-i \alpha_k(t)} e^{i \alpha_m(t)} a_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.12)

This suggests that we want to form a new function

\begin{aligned}b_m = a_m e^{\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle } \end{aligned} \hspace{\stretch{1}}(3.13)


\begin{aligned}a_m = b_m e^{-\int_0^t \left\langle{{m}} \vert {{m'}}\right\rangle } \end{aligned} \hspace{\stretch{1}}(3.14)

Plugging this into our assumed representation we have a more concrete form

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- \int_0^t dt' ( i \omega_k + \left\langle{{k}} \vert {{k'}}\right\rangle ) } b_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.15)


\begin{aligned}\Gamma_k = i \left\langle{{k}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.16)

this becomes

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.17)

A final pass.

Now that we have what appears to be a good representation for any given state if we wish to examine the time evolution, let’s start over, reapplying our instantaneous energy operator equality

\begin{aligned}0 &=\left(H - i \hbar \frac{d}{dt} \right){\left\lvert {\psi} \right\rangle}  \\ &=\left(H - i \hbar \frac{d}{dt} \right)\sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k {\left\lvert {k} \right\rangle} \\ &=- i \hbar \sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } \left(i \Gamma_kb_k {\left\lvert {k} \right\rangle} +b_k' {\left\lvert {k} \right\rangle} +b_k {\left\lvert {k'} \right\rangle} \right).\end{aligned}

Bra’ing with {\left\langle {m} \right\rvert} we find

\begin{aligned}0&=e^{- i\int_0^t dt' ( \omega_m - \Gamma_m ) } i \Gamma_mb_m +e^{- i\int_0^t dt' ( \omega_m - \Gamma_m ) } b_m' \\ &+e^{- i\int_0^t dt' ( \omega_m - \Gamma_m ) } b_m \left\langle{{m}} \vert {{m'}}\right\rangle +\sum_{k \ne m}e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k \left\langle{{m}} \vert {{k'}}\right\rangle \end{aligned}

Since i \Gamma_m = \left\langle{{m}} \vert {{m'}}\right\rangle the first and third terms cancel leaving us just

\begin{aligned}b_m'=-\sum_{k \ne m}e^{- i\int_0^t dt' ( \omega_{km} - \Gamma_{km} ) } b_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(3.18)

where \omega_{km} = \omega_k - \omega_m and \Gamma_{km} = \Gamma_k - \Gamma_m.


We assumed that a ket for the system has a representation in the form

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- i \alpha_k(t) } a_k(t) {\left\lvert {k(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(4.20)

where a_k(t) and \alpha_k(t) are given or to be determined. Application of our energy operator identity provides us with an alternate representation that simplifes the results

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_k e^{- i\int_0^t dt' ( \omega_k - \Gamma_k ) } b_k(t) {\left\lvert {k(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(4.20)


\begin{aligned}{\left\lvert {m'} \right\rangle} &= \frac{d}{dt} {\left\lvert {m} \right\rangle} \\ \Gamma_k &= i \left\langle{{m}} \vert {{m'}}\right\rangle \\ \omega_{km} &= \omega_k - \omega_m \\ \Gamma_{km} &= \Gamma_k - \Gamma_m\end{aligned} \hspace{\stretch{1}}(4.21)

we find that our dynamics of the coefficients are related by

\begin{aligned}b_m'=-\sum_{k \ne m}e^{- i\int_0^t dt' ( \omega_{km} - \Gamma_{km} ) } b_k \left\langle{{m}} \vert {{k'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(4.25)

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

%d bloggers like this: