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Posts Tagged ‘pertubation theory’

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)

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A different derivation of the adiabatic perturbation coefficient equation

Posted by peeterjoot on October 27, 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.)]


Professor Sipe’s adiabatic perturbation and that of the text [1] in section 17.5.1 and section 17.5.2 use different notation for \gamma_m and take a slightly different approach. We can find Prof Sipe’s final result with a bit less work, if a hybrid of the two methods is used.


Our starting point is the same, we have a time dependent slowly varying Hamiltonian

\begin{aligned}H = H(t),\end{aligned} \hspace{\stretch{1}}(2.1)

where our perturbation starts at some specific time from a given initial state

\begin{aligned}H(t) = H_0, \qquad t \le 0.\end{aligned} \hspace{\stretch{1}}(2.2)

We assume that instantaneous eigenkets can be found, satisfying

\begin{aligned}H(t) {\left\lvert {n(t)} \right\rangle} = E_n(t) {\left\lvert {n(t)} \right\rangle}\end{aligned} \hspace{\stretch{1}}(2.3)

Here I’ll use {\left\lvert {n} \right\rangle} \equiv {\left\lvert {n(t)} \right\rangle} instead of the {\left\lvert {\hat{\psi}_n(t)} \right\rangle} that we used in class because its easier to write.

Now suppose that we have some arbitrary state, expressed in terms of the instantaneous basis kets {\left\lvert {n} \right\rangle}

\begin{aligned}{\left\lvert {\psi} \right\rangle} = \sum_n \bar{b}_n(t) e^{-i\alpha_n + i \gamma_n} {\left\lvert {n} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.4)


\begin{aligned}\alpha_n(t) = \frac{1}{{\hbar}} \int_0^t dt' E_n(t'),\end{aligned} \hspace{\stretch{1}}(2.5)

and \gamma_n (using the notation in the text, not in class) is to be determined.

For this state, we have at the time just before the perturbation

\begin{aligned}{\left\lvert {\psi(0)} \right\rangle} = \sum_n \bar{b}_n(0) e^{-i\alpha_n(0) + i \gamma_n(0)} {\left\lvert {n(0)} \right\rangle}.\end{aligned} \hspace{\stretch{1}}(2.6)

The question to answer is: How does this particular state evolve?

Another question, for those that don’t like sneaky bastard derivations, is where did that magic factor of e^{-i\alpha_n} come from in our superposition state? We will see after we start taking derivatives that this is what we need to cancel the H(t){\left\lvert {n} \right\rangle} in Schr\”{o}dinger’s equation.

Proceeding to plug into the evolution identity we have

\begin{aligned}0 &={\left\langle {m} \right\rvert} \left( i \hbar \frac{d{{}}}{dt} - H(t) \right) {\left\lvert {\psi} \right\rangle} \\ &={\left\langle {m} \right\rvert} \left(\sum_n e^{-i \alpha_n + i \gamma_n}(i \hbar) \left(\frac{d{{\bar{b}_n}}}{dt}+ \bar{b}_n \left(-i \not{{\frac{E_n}{\hbar}}} + i \dot{\gamma}_m \right)\right) {\left\lvert {n} \right\rangle}+ i \hbar \bar{b}_n \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}- \not{{E_n \bar{b}_n {\left\lvert {n} \right\rangle}}} \right)\\ &=e^{-i \alpha_m + i \gamma_m}(i \hbar) \frac{d{{\bar{b}_m}}}{dt}+e^{-i \alpha_m + i \gamma_m}(i \hbar) i \dot{\gamma}_m \bar{b}_m+ i \hbar \sum_n \bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}e^{-i \alpha_n + i \gamma_n} \\ &\sim\frac{d{{\bar{b}_m}}}{dt}+i \dot{\gamma}_m \bar{b}_m+ \sum_n e^{-i \alpha_n + i \gamma_n}e^{i \alpha_m - i \gamma_m}\bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle} \\ &=\frac{d{{\bar{b}_m}}}{dt}+i \dot{\gamma}_m \bar{b}_m+ \bar{b}_m {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle}+\sum_{n \ne m} e^{-i \alpha_n + i \gamma_n}e^{i \alpha_m - i \gamma_m}\bar{b}_n {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle}\end{aligned}

We are free to pick \gamma_m to kill the second and third terms

\begin{aligned}0 =i \dot{\gamma}_m \bar{b}_m+ \bar{b}_m {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.7)


\begin{aligned}\dot{\gamma}_m = i {\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {m} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.8)

which after integration is

\begin{aligned}\gamma_m(t)= i \int_0^t dt' {\left\langle {m(t')} \right\rvert} \frac{d}{dt'} {\left\lvert {m(t)} \right\rangle},\end{aligned} \hspace{\stretch{1}}(2.9)

as in class we can observe that this is a purely real function. We are left with

\begin{aligned}\frac{d{{\bar{b}_m}}}{dt}=-\sum_{n \ne m} \bar{b}_n e^{-i \alpha_{nm} + i \gamma_{nm}}{\left\langle {m} \right\rvert} \frac{d{{}}}{dt} {\left\lvert {n} \right\rangle} ,\end{aligned} \hspace{\stretch{1}}(2.10)


\begin{aligned}\alpha_{nm} &= \alpha_{n} -\alpha_m \\ \gamma_{nm} &= \gamma_{n} -\gamma_m \end{aligned} \hspace{\stretch{1}}(2.11)

The task is now to find solutions for these \bar{b}_m coefficients, and we can refer to the class notes for that without change.


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

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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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