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## Revisiting adiabatic approximation for expansion around an initial pure state.

Posted by peeterjoot on November 10, 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.)]

# New info. How to do the $\lambda$ expansion.

Asking about this, Federico nicely explained. “The reason why you are going in circles when trying the lambda expansion is because you are not assuming the term ${\left\langle {\psi(t)} \right\rvert} (d/dt) {\left\lvert {\psi(t)} \right\rangle}$ to be of order lambda. This has to be assumed, otherwise it doesn’t make sense at all trying a perturbative approach. This assumption means that the coupling between the level $s$ and the other levels is assumed to be small because the time dependent part of the Hamiltonian is small or changes slowly with time. Making a Taylor expansion in time would be sensible only if you are interested in a short interval of time. The lambda-expansion approach would work for any time as long as the time dependent piece of the Hamiltonian doesn’t change wildly or is too big.”

In the tutorial he outlined another way to justify this. We’ve written so far

\begin{aligned}H = \left\{\begin{array}{l l}H(t) & \quad \mbox{latex t > 0} \\ H_0 & \quad \mbox{$t < 0$} \end{array}\right.\end{aligned} \hspace{\stretch{1}}(5.23)

where $H(0) = H_0$. We can make this explicit, and introduce a $\lambda$ factor into the picture if we write

\begin{aligned}H(t) = H_0 + \lambda H'(t),\end{aligned} \hspace{\stretch{1}}(5.24)

where $H_0$ has no time dependence, so that our Hamiltonian is then just the “steady-state” system for $\lambda = 0$.

Now recall the method from [1] that we can use to relate our bra-derivative-ket to the Hamiltonian. Taking derivatives of the energy identity, braketed between two independent kets ($m \ne n$) we have

\begin{aligned}0 &= {\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{}}}{dt} \left(H(t) {\left\lvert {\hat{\psi}_n(t)} \right\rangle} - \hbar \omega_n {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \right) \\ &= {\left\langle {\hat{\psi}_m(t)} \right\rvert}\left(\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} +H(t) \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} -\hbar \frac{d{{\omega_n}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} -\hbar \omega_n \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \right) \\ &= \hbar (\omega_m - \omega_n) {\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} -\not{{\hbar \frac{d{{\omega_n}}}{dt} \delta_{mn}}}+{\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} \end{aligned}

So for $m \ne n$ we find a dependence between the bra-derivative-ket and the time derivative of the Hamiltonian

\begin{aligned}{\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} =\frac{{\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{H(t)}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} }{\hbar (\omega_n - \omega_m) }\end{aligned} \hspace{\stretch{1}}(5.25)

Referring back to 5.24 we see the $\lambda$ dependence in this quantity, coming directly from the $\lambda$ dependence imposed on the time dependent part of the Hamiltonian

\begin{aligned}{\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} =\lambda\frac{{\left\langle {\hat{\psi}_m(t)} \right\rvert}\frac{d{{H'(t)}}}{dt} {\left\lvert {\hat{\psi}_n(t)} \right\rangle} }{\hbar (\omega_n - \omega_m) }\end{aligned} \hspace{\stretch{1}}(5.26)

Given this $\lambda$ dependence, let’s revisit the perturbation attempt of 3.11. Our first order factors of $\lambda$ are now

\begin{aligned}\frac{d{{}}}{dt} \bar{b}_s^{(1)}(t) &= - \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} \\ &= \left\{\begin{array}{l l}0 & \quad \mbox{iflatex m = s} \\ – e^{i \gamma_{sm}(t) } {\left\langle {\hat{\psi}_s(t)} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {\hat{\psi}_m(t)} \right\rangle} &\quad \mbox{if $m \ne s$} \\ \end{array}\right.\end{aligned}

So we find to first order

\begin{aligned}\bar{b}_s(t) =\delta_{ms}(1 + \lambda \text{constant})+ - (1-\delta_{ms}) \lambda\int_0^t dt'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}}(5.27)

A couple observations of this result. One is that the constant factor in the $m = s$ case makes sense. This would likely be a negative contribution since we have to decrease the probability coefficient for finding our wavefunction in the $m = s$ state after perturbation, since we are increasing the probability for finding it elsewhere by changing the Hamiltonian.

Also observe that since $e^{i\gamma_{sm}} \sim 0$ for small $t$ this is consistent with the first order Taylor series expansion where we found our first order contribution was

\begin{aligned}- (1 - \delta_{ms}) 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}}(5.28)

Also note that this $-e^{i \gamma_{sm}(t') } {\left\langle {\hat{\psi}_s(t')} \right\rvert} \frac{d}{dt'} {\left\lvert {\hat{\psi}_m(t')} \right\rangle}$ is exactly the difference from $0$ that was mentioned in class when the trial solution of $\bar{b}_s = \delta_{sm}$ was tested by plugging it into 2.5, so it’s not too surprising that we should have a factor of exactly this form when we refine our approximation.

A question to consider should we wish to refine the $\lambda$ perturbation to higher than first order in $\lambda$: is there any sort of $\lambda$ dependence in the $e^{i \gamma_{sm}}$ coming from the $\Gamma_{sm}$ term in that exponential?

# References

[1] D. Bohm. Quantum Theory. Courier Dover Publications, 1989.