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PHY456H1F: Quantum Mechanics II. Lecture 9 (Taught by Prof J.E. Sipe). Adiabatic perturbation theory (cont.)

Posted by peeterjoot on October 9, 2011

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Peeter’s lecture notes from class. May not be entirely coherent.

Adiabatic perturbation theory (cont.)

We were working through Adiabatic time dependent perturbation (as also covered in section 17.5.2 of the text [1].)

Utilizing an expansion

\begin{aligned}{\lvert {\psi(t)} \rangle} &= \sum_n c_n(t) e^{- i \omega_n^{(0)} t} {\lvert {\psi_n^{(0)} } \rangle} \\ &= \sum_n b_n(t) {\lvert {\hat{\psi}_n(t)} \rangle},\end{aligned} \hspace{\stretch{1}}(2.1)


\begin{aligned}H(t) {\lvert {\hat{\psi}_s(t)} \rangle} = E_s(t) {\lvert {\hat{\psi}_s(t)} \rangle} \end{aligned} \hspace{\stretch{1}}(2.3)

and found

\begin{aligned}\frac{d{{b_s(t)}}}{dt} = -i \left( \omega_s(t) - \Gamma_s(t)\right) b_s(t)- \sum_{n \ne s} b_n(t) {\langle {\hat{\psi}_s(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle}\end{aligned} \hspace{\stretch{1}}(2.4)


\begin{aligned}\Gamma_s(t) =i {\langle {\hat{\psi}_s(t)} \rvert} \frac{d{{}}}{dt} {\lvert {\hat{\psi}_s(t)} \rangle} \end{aligned} \hspace{\stretch{1}}(2.5)

Look for a solution of the form

\begin{aligned}\begin{aligned}b_s(t) &= \bar{b}_s(t) e^{-i \int_0^t dt' (\omega_s(t') - \Gamma_s(t'))} \\ &= \bar{b}_s(t) e^{-i \gamma_s(t)}\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.6)


\begin{aligned}\gamma_s(t) = \int_0^t dt' (\omega_s(t') - \Gamma_s(t')).\end{aligned} \hspace{\stretch{1}}(2.7)

Taking derivatives of \bar{b}_s and after a bit of manipulation we find that things conveniently cancel

\begin{aligned}\frac{d{{\bar{b}_s(t)}}}{dt} &= \frac{d{{}}}{dt} \left( b_s(t) e^{i \gamma_s(t) } \right) \\ &= \frac{d{{b_s(t)}}}{dt} e^{i \gamma_s(t) } +b_s(t) \frac{d{{}}}{dt} e^{i \gamma_s(t) }  \\ &= \frac{d{{b_s(t)}}}{dt} e^{i \gamma_s(t) } +b_s(t) i (\omega_s(t) - \Gamma_s(t)) e^{i \gamma_s(t) }.\end{aligned}

We find

\begin{aligned}\frac{d{{\bar{b}_s(t)}}}{dt} e^{-i \gamma_s(t)} &= \frac{d{{b_s(t)}}}{dt} + i b_s(t) (\omega_s(t) - \Gamma_s(t))  \\ &=\not{{i b_s(t) (\omega_s(t) - \Gamma_s(t)) }}-\not{{i \left( \omega_s(t) - \Gamma_s(t)\right) b_s(t)}}- \sum_{n \ne s} b_n(t) {\langle {\hat{\psi}_s(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle},\end{aligned}


\begin{aligned}\frac{d{{\bar{b}_s(t)}}}{dt} &=- \sum_{n \ne s} b_n(t) e^{i \gamma_s(t)} {\langle {\hat{\psi}_s(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle} \\ &=- \sum_{n \ne s} \bar{b}_n(t) e^{i (\gamma_s(t) - \gamma_n(t))} {\langle {\hat{\psi}_s(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle}.\end{aligned}

With a last bit of notation

\begin{aligned}\gamma_{sn}(t) = \gamma_s(t) - \gamma_n(t)),\end{aligned} \hspace{\stretch{1}}(2.8)

the problem is reduced to one involving only the sums over the n \ne s terms, and where all the dependence on {\langle {\hat{\psi}_s(t)} \rvert} \frac{d{{}}}{dt} {\lvert {\hat{\psi}_s(t)} \rangle} has been nicely isolated in a phase term

\begin{aligned}\frac{d{{\bar{b}_s(t)}}}{dt} =- \sum_{n \ne s} \bar{b}_n(t) e^{i \gamma_{sn}(t) }{\langle {\hat{\psi}_s(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.9)

Looking for an approximate solution.

Try an approximate solution

\begin{aligned}\bar{b}_n(t) = \delta_{nm}\end{aligned} \hspace{\stretch{1}}(2.10)

For s = m this is okay, since we have \frac{d{{\delta_{ns}}}}{dt} = 0 which is consistent with

\begin{aligned}\sum_{n \ne s} \delta_{ns} ( \cdots ) = 0\end{aligned} \hspace{\stretch{1}}(2.11)

However, for s \ne m we get

\begin{aligned}\frac{d{{\bar{b}_s(t)}}}{dt} &=- \sum_{n \ne s} \delta_{nm}e^{i \gamma_{sn}(t) }{\langle {\hat{\psi}_s(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_n(t)} \rangle} \\ &=- e^{i \gamma_{sm}(t) }{\langle {\hat{\psi}_s(t)} \rvert}\frac{d{{}}}{dt} {\lvert {\hat{\psi}_m(t)} \rangle} \\ \end{aligned}


\begin{aligned}\gamma_{sm}(t) = \int_0^t dt' \left( \frac{1}{{\hbar}}( E_s(t') - E_m(t') ) - \Gamma_s(t') + \Gamma_m(t') \right)\end{aligned} \hspace{\stretch{1}}(2.12)

FIXME: I think we argued in class that the \Gamma contributions are negligible. Why was that?

Now, are energy levels will have variation with time, as illustrated in figure (\ref{fig:qmTwoL9fig1})

\caption{Energy level variation with time}

Perhaps unrealistically, suppose that our energy levels have some “typical” energy difference \Delta E, so that

\begin{aligned}\gamma_{sm}(t) \approx \frac{\Delta E}{\hbar} t \equiv \frac{t}{\tau},\end{aligned} \hspace{\stretch{1}}(2.13)


\begin{aligned}\tau = \frac{\hbar}{\Delta E}\end{aligned} \hspace{\stretch{1}}(2.14)

Suppose that \tau is much less than a typical time T over which instantaneous quantities (wavefunctions and brakets) change. After a large time T

\begin{aligned}e^{i \gamma_{sm}(t)} \approx e^{i T/\tau}\end{aligned} \hspace{\stretch{1}}(2.15)

so we have our phase term whipping around really fast, as illustrated in figure (\ref{fig:qmTwoL9fig2}).

\caption{Phase whipping around.}

So, while {\langle {\hat{\psi}_s(t)} \rvert} \frac{d{{}}}{dt} {\lvert {\hat{\psi}_m(t)} \rangle} is moving really slow, but our phase space portion is changing really fast. The key to the approximate solution is factoring out this quickly changing phase term.

Note \Gamma_s(t) is called the “Berry” phase [2], whereas the E_s(t')/\hbar part is called the geometric phase, and can be shown to have a geometric interpretation.

To proceed we can introduce \lambda terms, perhaps

\begin{aligned}\bar{b}_s(t) = \delta_{ms} + \lambda \bar{b}^{(1)}_s(t) + \cdots\end{aligned} \hspace{\stretch{1}}(2.16)


\begin{aligned}- \sum_{n \ne s} e^{i \gamma_{sn}(t)} \lambda (\cdots)\end{aligned} \hspace{\stretch{1}}(2.17)

FIXME: try this, or at least check the text to see if this is done in more detail.


Suppose we have some branching of energy levels that were initially degenerate, as illustrated in figure (\ref{fig:qmTwoL9fig3})

\caption{Degenerate energy level splitting.}

We have a necessity to choose states properly so there is a continuous evolution in the instantaneous eigenvalues as H(t) changes.

Question: A physical example?

FIXME: Prof Sipe to ponder and revisit.

Fermi’s golden rule

See section 17.2 of the text [1].

Fermi originally had two golden rules, but his first one has mostly been forgotten. This refers to his second.

This is really important, and probably the single most important thing to learn in this course. You’ll find this falls out of many complex calculations.

Returning to general time dependent equations with

\begin{aligned}H = H_0 + H'(t)\end{aligned} \hspace{\stretch{1}}(3.18)

\begin{aligned}{\lvert {\psi(t)} \rangle} = \sum_n c_n(t) e^{-i\omega_n t} {\lvert {\psi_n} \rangle}\end{aligned} \hspace{\stretch{1}}(3.19)


\begin{aligned}i \hbar \cdot_n = \sum_n H_{mn}' e^{i \omega_{mn} t} c_n(t)\end{aligned} \hspace{\stretch{1}}(3.20)


\begin{aligned}H_{mn}'(t) &= {\langle {\psi_m} \rvert} H'(t) {\langle {\psi_n} \rvert} \\ \omega_n &= \frac{E_n}{\hbar} \\ \omega_{mn} &= \omega_m - \omega_n\end{aligned} \hspace{\stretch{1}}(3.21)


\begin{aligned}H'(t) = - \boldsymbol{\mu} \cdot \mathbf{E}(t).\end{aligned} \hspace{\stretch{1}}(3.24)

If c_m^{(0)} = \delta_{mi}, then to first order

\begin{aligned}i \hbar \cdot^{(1)}(t) = H_{mi}'(t) e^{i \omega_{mi} t},\end{aligned} \hspace{\stretch{1}}(3.25)


\begin{aligned}c_m^{(1)}(t) = \frac{1}{{i\hbar}} \int_{t_0}^t H_{mi}'(t') e^{i \omega_{mi} t'} dt'.\end{aligned} \hspace{\stretch{1}}(3.26)

Assume the perturbation vanishes before time t_0.

Reminder. Have considered this using 3.26 for a pulse as in figure (\ref{fig:gaussianWavePacket})

\caption{Gaussian wave packet.}

Now we want to consider instead a non-terminating signal, that was zero before some initial time as illustrated in figure (\ref{fig:unitStepSine}), where the separation between two peaks is \Delta t = 2\pi/\omega_0.

\caption{Sine only after an initial time.}

Our matrix element is

\begin{aligned}H_{mi}'(t) = - {\langle {\psi_m} \rvert} \mathbf{u} {\lvert {\psi_i} \rangle} \cdot \mathbf{E}(t) = \left\{\begin{array}{l l}2 A_{mi} \sin(\omega_0 t) & \quad \mbox{if t > 0$} \\ 0 & \quad \mbox{if $t < 0$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(3.27)$

Here the factor of 2 has been included for consistency with the text.

\begin{aligned}H_{mi}'(t) = i A_{mi} \left( e^{-i \omega_0 t}-e^{i \omega_0 t} \right)\end{aligned} \hspace{\stretch{1}}(3.28)

Plug this into the perturbation

\begin{aligned}c_m^{(1)}(t) = \frac{A_{mi}}{\hbar} \int_{t_0}^t dt' \left( e^{i (\omega_{mi} - \omega_0 t) }-e^{i (\omega_{mi} + \omega_0 t) }\right)\end{aligned} \hspace{\stretch{1}}(3.29)

\caption{\omega_{mi} illustrated.}

Suppose that

\begin{aligned}\omega_0 \approx \omega_{mi},\end{aligned} \hspace{\stretch{1}}(3.30)


\begin{aligned}c_m^{(1)}(t) \approx\frac{A_{mi}}{\hbar} \int_{t_0}^t dt' \left( 1-e^{2 i \omega_0 t }\right),\end{aligned} \hspace{\stretch{1}}(3.31)

but the exponential has essentially no contribution

\begin{aligned}{\left\lvert{\int_0^t e^{2 i \omega_0 t'} dt' }\right\rvert} &= {\left\lvert{\frac{e^{2 i \omega_0 t} -1 }{2 i \omega_0}}\right\rvert}  \\ &= \frac{\sin(\omega_0 t)}{\omega_0} \\ &\sim \frac{1}{{\omega_0}}\end{aligned}

so for t \gg \frac{1}{{\omega_0}} and \omega_0 \approx \omega_{mi} we have

\begin{aligned}c_m^{(1)}(t) \approx \frac{A_{mi}}{\hbar} t\end{aligned} \hspace{\stretch{1}}(3.32)

Similarly for \omega_0 \approx \omega_{im} as in figure (\ref{fig:qmTwoL9fig7})



\begin{aligned}c_m^{(1)}(t) \approx\frac{A_{mi}}{\hbar} \int_{t_0}^t dt' \left( e^{-2 i \omega_0 t }-1\right),\end{aligned} \hspace{\stretch{1}}(3.33)

and we have

\begin{aligned}c_m^{(1)}(t) \approx -\frac{A_{mi}}{\hbar} t\end{aligned} \hspace{\stretch{1}}(3.34)


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

[2] Wikipedia. Geometric phase — wikipedia, the free encyclopedia [online]. 2011. [Online; accessed 9-October-2011]. //


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