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## Second order time evolution for the coefficients of an initially pure ket with an adiabatically changing Hamiltonian.

Posted by peeterjoot on November 6, 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 lecture 9, Prof Sipe developed the equations governing the evolution of the coefficients of a given state for an adiabatically changing Hamiltonian. He also indicated that we could do an approximation, finding the evolution of an initially pure state in powers of $\lambda$ (like we did for the solutions of a non-time dependent perturbed Hamiltonian $H = H_0 + \lambda H'$). I tried doing that a couple of times and always ended up going in circles. I’ll show that here and also develop an expansion in time up to second order as an alternative, which appears to work out nicely.

# Review.

We assumed that an adiabatically changing Hamiltonian was known with instantaneous eigenkets governed by

\begin{aligned}H(t) {\left\lvert {\hat{\psi}_n(t)} \right\rangle} = \hbar \omega_n {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \end{aligned} \hspace{\stretch{1}}(2.1)

The problem was to determine the time evolutions of the coefficients $\bar{b}_n(t)$ of some state ${\left\lvert {\psi(t)} \right\rangle}$, and this was found to be

\begin{aligned}{\left\lvert {\psi(t)} \right\rangle} &= \sum_n \bar{b}_n(t) e^{-i \gamma_n(t)} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ \gamma_s(t) &= \int_0^t dt' (\omega_s(t') - \Gamma_s(t')) \\ \Gamma_s(t) &= i {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_s(t)} \right\rangle} \end{aligned} \hspace{\stretch{1}}(2.2)

where the $\bar{b}_s(t)$ coefficient must satisfy the set of LDEs

\begin{aligned}\frac{d{{\bar{b}_s(t)}}}{dt} = - \sum_{n \ne s} \bar{b}_n(t) e^{i \gamma_{sn}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.5)

where

\begin{aligned}\gamma_{sn}(t) = \gamma_{s}(t) - \gamma_{n}(t).\end{aligned} \hspace{\stretch{1}}(2.6)

Solving these in general doesn’t look terribly fun, but perhaps we can find an explicit solution for all the $\bar{b}_s$‘s, if we simplify the problem somewhat. Suppose that our initial state is found to be in the $m$th energy level at the time before we start switching on the changing Hamiltonian.

\begin{aligned}{\left\lvert {\psi(0)} \right\rangle} = \bar{b}_m(0) {\left\lvert {\hat{\psi}_m(0)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.7)

We therefore require (up to a phase factor)

\begin{aligned}\begin{array}{l l}\bar{b}_m(0) = 1 & \\ \bar{b}_s(0) = 0 & \quad \mbox{iflatex s \ne m}.\end{array}\end{aligned} \hspace{\stretch{1}}(2.8)

Equivalently we can write

\begin{aligned}\bar{b}_s(0) = \delta_{ms}\end{aligned} \hspace{\stretch{1}}(2.9)

# Going in circles with a $\lambda$ expansion.

In class it was hinted that we could try a $\lambda$ expansion of the following form to determine a solution for the $\bar{b}_s$ coefficients at later times

\begin{aligned}\bar{b}_s(t) = \delta_{ms} + \lambda \bar{b}^{(1)}_s(t) + \cdots\end{aligned} \hspace{\stretch{1}}(3.10)

I wasn’t able to figure out how to make that work. Trying this first to first order, and plugging in, we find

\begin{aligned}\lambda \frac{d{{}}}{dt} \bar{b}^{(1)}_s(t) = - \sum_{n \ne s} ( \delta_{mn} + \lambda \bar{b}^{(1)}_n(t) ) e^{i \gamma_{sn}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(3.11)

equating powers of $\lambda$ yields two equations

\begin{aligned}\frac{d{{}}}{dt} \bar{b}_s^{(1)}(t) &= - \sum_{n \ne s} \bar{b}^{(1)}_n(t) e^{i \gamma_{sn}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \\ 0 &= - \sum_{n \ne s} \delta_{mn} e^{i \gamma_{sn}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.12)

Observe that the first identity is exactly what we started with in 2.5, but has just replaced the $\bar{b}_n$‘s with $\bar{b}^{(1)}_n$‘s. Worse is that the second equation is only satisfied for $s = m$, and for $s \ne m$ we have

\begin{aligned}0 = - e^{i \gamma_{sm}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_m(t)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(3.14)

So this $\lambda$ power series only appears to work if we somehow had ${\left\lvert {\hat{\psi}_s(t)} \right\rangle}$ always orthonormal to the derivative of ${\left\lvert {\hat{\psi}_m(t)} \right\rangle}$. Perhaps this could be done if the Hamiltonian was also expanded in powers of $\lambda$, but such a beastie seems foreign to the problem. Note that we don’t even have any explicit dependence on the Hamiltonian in the final $\bar{b}_n$ differential equations, as we’d probably need for such an expansion to work out.

# A Taylor series expansion in time.

What we can do is to expand the $\bar{b}_n$‘s in a power series parametrized by time. That is, again, assuming we started with energy equal to $\hbar \omega_m$, form

\begin{aligned}\bar{b}_s(t) = \delta_{sm} + \frac{t}{1!} \left( {\left.{{ \frac{d{{}}}{dt}\bar{b}_s(t) }}\right\vert}_{{t=0}} \right)+ \frac{t^2}{2!} \left( {\left.{{ \frac{d^2}{dt^2} \bar{b}_s(t) }}\right\vert}_{{t=0}} \right)+ \cdots\end{aligned} \hspace{\stretch{1}}(4.15)

The first order term we can grab right from 2.5 and find

\begin{aligned}{\left.{{\frac{d{{\bar{b}_s(t)}}}{dt}}}\right\vert}_{{t=0}} &= - \sum_{n \ne s} \bar{b}_n(0) {\left.{{{\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle}}}\right\vert}_{{t=0}} \\ &= - \sum_{n \ne s} \delta_{nm}{\left.{{{\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle}}}\right\vert}_{{t=0}} \\ &=\left\{\begin{array}{l l}0 & \quad \mbox{latex s = m} \\ – {\left.{{{\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_m(t)} \right\rangle}}}\right\vert}_{{t=0}} \\ & \quad \mbox{$s \ne m$} \\ \end{array}\right.\end{aligned}

Let’s write

\begin{aligned}{\left\lvert {n} \right\rangle} &= {\left\lvert {\hat{\psi}_n(0)} \right\rangle} \\ {\left\lvert {n'} \right\rangle} &= {\left.{{ \frac{d{{}}}{dt}{\left\lvert {\hat{\psi}_n(t)} \right\rangle} }}\right\vert}_{{t=0}}\end{aligned} \hspace{\stretch{1}}(4.16)

So we can write

\begin{aligned}{\left.{{\frac{d{{\bar{b}_s(t)}}}{dt}}}\right\vert}_{{t=0}} =- (1 - \delta_{sm}) \left\langle{{s}} \vert {{m'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(4.18)

and form, to first order in time our approximation for the coefficient is

\begin{aligned}\bar{b}_s(t) =\delta_{sm} - t (1 - \delta_{sm}) \left\langle{{s}} \vert {{m'}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(4.19)

Let’s do the second order term too. For that we have

\begin{aligned}{\left.{{\frac{d^2}{dt^2} \bar{b}_s(t)}}\right\vert}_{{t=0}} &= - \sum_{n \ne s} {\left.{{\left(\left(\frac{d{{}}}{dt} \bar{b}_n(t) +\delta_{nm} i \frac{d{{\gamma_{sn}(t)}}}{dt}\right)\left\langle{{s}} \vert {{n'}}\right\rangle+\delta_{nm} \frac{d{{}}}{dt} \left( {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \right) \right)}}\right\vert}_{{t=0}}\end{aligned}

For the $\gamma_{sn}$ derivative we note that

\begin{aligned}{\left.{{\frac{d{{}}}{dt} \gamma_s(t)}}\right\vert}_{{t=0}} = \omega_s(0) - i\left\langle{{s}} \vert {{s'}}\right\rangle,\end{aligned} \hspace{\stretch{1}}(4.20)

So we have

\begin{aligned}{\left.{{\frac{d^2}{dt^2} \bar{b}_s(t)}}\right\vert}_{{t=0}} &= - \sum_{n \ne s} \Bigl(- (1 - \delta_{nm}) \left\langle{{n}} \vert {{m'}}\right\rangle+\delta_{nm} i (\omega_{sn}(0) - i\left\langle{{s}} \vert {{s'}}\right\rangle + i\left\langle{{n}} \vert {{n'}}\right\rangle)\Bigr)\left\langle{{s}} \vert {{n'}}\right\rangle+\delta_{nm} \Bigl( \left\langle{{s'}} \vert {{n'}}\right\rangle+\left\langle{{s}} \vert {{n''}}\right\rangle\Bigr)\end{aligned}

Again for $s = m$, all terms are killed. That’s somewhat surprising, but suggests that we will need to normalize the coefficients after the perturbation calculation, since we have unity for one of them.

For $s \ne m$ we have

\begin{aligned}{\left.{{\frac{d^2}{dt^2} \bar{b}_s(t)}}\right\vert}_{{t=0}} &= \sum_{n \ne s} \Bigl(\left\langle{{n}} \vert {{m'}}\right\rangle-\delta_{nm} i (\omega_{sn}(0) - i\left\langle{{s}} \vert {{s'}}\right\rangle + i\left\langle{{n}} \vert {{n'}}\right\rangle)\Bigr)\left\langle{{s}} \vert {{n'}}\right\rangle-\delta_{nm} \Bigl( \left\langle{{s'}} \vert {{n'}}\right\rangle+\left\langle{{s}} \vert {{n''}}\right\rangle\Bigr) \\ &= -i (\omega_{sm}(0) - i\left\langle{{s}} \vert {{s'}}\right\rangle + i\left\langle{{m}} \vert {{m'}}\right\rangle)\Bigr)\left\langle{{s}} \vert {{m'}}\right\rangle-\Bigl( \left\langle{{s'}} \vert {{m'}}\right\rangle+\left\langle{{s}} \vert {{m''}}\right\rangle\Bigr) +\sum_{n \ne s} \left\langle{{n}} \vert {{m'}}\right\rangle \left\langle{{s}} \vert {{n'}}\right\rangle.\end{aligned}

So we have, for $s \ne m$

\begin{aligned}{\left.{{\frac{d^2}{dt^2} \bar{b}_s(t)}}\right\vert}_{{t=0}} = (\left\langle{{m}} \vert {{m'}}\right\rangle - \left\langle{{s}} \vert {{s'}}\right\rangle ) \left\langle{{s}} \vert {{m'}}\right\rangle-i \omega_{sm}(0) \left\langle{{s}} \vert {{m'}}\right\rangle-\left\langle{{s'}} \vert {{m'}}\right\rangle-\left\langle{{s}} \vert {{m''}}\right\rangle+\sum_{n \ne s} \left\langle{{n}} \vert {{m'}}\right\rangle \left\langle{{s}} \vert {{n'}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(4.21)

It’s not particularly illuminating looking, but possible to compute, and we can use it to form a second order approximate solution for our perturbed state.

\begin{aligned}\begin{aligned}\bar{b}_s(t) &=\delta_{sm} - t (1 - \delta_{sm}) \left\langle{{s}} \vert {{m'}}\right\rangle \\ &+(1 - \delta_{sm})\left((\left\langle{{m}} \vert {{m'}}\right\rangle - \left\langle{{s}} \vert {{s'}}\right\rangle ) \left\langle{{s}} \vert {{m'}}\right\rangle-i \omega_{sm}(0) \left\langle{{s}} \vert {{m'}}\right\rangle-\left\langle{{s'}} \vert {{m'}}\right\rangle-\left\langle{{s}} \vert {{m''}}\right\rangle+\sum_{n \ne s} \left\langle{{n}} \vert {{m'}}\right\rangle \left\langle{{s}} \vert {{n'}}\right\rangle\right) \frac{t^2}{2}\end{aligned}\end{aligned} \hspace{\stretch{1}}(4.22)