D.34 The adiabatic theorem

Consider the Schrö­din­ger equation

{\rm i}\hbar \frac{\partial\Psi}{\partial t} = H \Psi

If the Hamiltonian is independent of time, the solution can be written in terms of the Hamiltonian energy eigenvalues $E_{\vec n}$ and eigenfunctions $\psi_{\vec n}$ as

\Psi = \sum_{\vec n}c_{\vec n}(0) e^{{\rm i}\theta_{\vec n...
... n}
\theta_{\vec n}= - \frac{1}{\hbar} E_{\vec n}t

Here ${\vec n}$ stands for the quantum numbers of the eigenfunctions and the $c_{\vec n}(0)$ are arbitrary constants.

However, the Hamiltonian varies with time for the systems of interest here. Still, at any given time its eigenfunctions form a complete set. So it is still possible to write the wave function as a sum of them, say like

\Psi = \sum_{\vec n}\bar c_{\vec n}e^{{\rm i}\theta_{\vec ...
...\theta_{\vec n}= -\frac{1}{\hbar} \int E_{\vec n}{\,\rm d}t %
\end{displaymath} (D.18)

But the coefficients $\bar{c}_{\vec n}$ can no longer be assumed to be constant like the $c_{\vec n}(0)$. They may be different at different times.

To get an equation for their variation, plug the expression for $\Psi$ in the Schrö­din­ger equation. That gives:

{\rm i}\hbar\sum_{\vec n}\bar c_{\vec n}^{\,\prime}e^{{\rm...
...vec n}\bar c_{\vec n}e^{{\rm i}\theta_{\vec n}} \psi_{\vec n}

where the primes indicate time derivatives. The middle sum in the left hand side and the right hand side cancel against each other since by definition $\psi_{\vec n}$ is an eigenfunction of the Hamiltonian with eigenvalue $E_{\vec n}$. For the remaining two sums, take an inner product with an arbitrary eigenfunction $\langle\psi_{\underline{\vec n}}\vert$:

{\rm i}\hbar \bar c_{\underline{\vec n}}^{\,\prime}e^{{\rm...
...gle\psi_{\underline{\vec n}}\vert \psi_{\vec n}'\rangle
= 0

In the first sum only the term ${\vec n}$ $\vphantom0\raisebox{1.5pt}{$=$}$ $\underline{\vec n}$ survived because of the orthonormality of the eigenfunctions. Divide by ${\rm i}{\hbar}e^{{\rm i}\theta_{\underline{\vec n}}}$ and rearrange to get
\bar c_{\underline{\vec n}}^{\,\prime}
= - \sum_{\vec n}...
...nderline{\vec n}}\vert\psi_{\vec n}'\rangle \bar c_{\vec n} %
\end{displaymath} (D.19)

This is still exact.

However, the purpose of the current derivation is to address the adiabatic approximation. The adiabatic approximation assumes that the entire evolution takes place very slowly over a large time interval $T$. For such an evolution, it helps to consider all quantities to be functions of the scaled time variable $t$$\raisebox{.5pt}{$/$}$$T$. Variables change by a finite amount when $t$ changes by a finite fraction of $T$, so when $t$$\raisebox{.5pt}{$/$}$$T$ changes by a finite amount. This implies that the time derivatives of the slowly varying quantities are normally small, of order 1/$T$.

Consider now first the case that there is no degeneracy, in other words, that there is only one eigenfunction $\psi_{\vec n}$ for each energy $E_{\vec n}$. If the Hamiltonian changes slowly and regularly in time, then so do the energy eigenvalues and eigenfunctions. In particular, the time derivatives of the eigenfunctions in (D.19) are small of order 1/$T$. It then follows from the entire equation that the time derivatives of the coefficients are small of order 1$\raisebox{.5pt}{$/$}$$T$ too.

(Recall that the square magnitudes of the coefficients give the probability for the corresponding energy. So the magnitude of the coefficients is bounded by 1. Also, for simplicity it will be assumed that the number of eigenfunctions in the system is finite. Otherwise the sums over ${\vec n}$ might explode. This book routinely assumes that it is good enough to approximate an infinite system by a large-enough finite one. That makes life a lot easier, not just here but also in other derivations like {D.18}.)

It is convenient to split up the sum in (D.19):

\bar c_{\underline{\vec n}}^{\,\prime}
= - \langle\psi_{...
...nderline{\vec n}}\vert\psi_{\vec n}'\rangle \bar c_{\vec n} %
\end{displaymath} (D.20)

Under the stated conditions, the final sum can be ignored.

However, that is not because it is small due to the time derivative in it, as one reference claims. While the time derivative of $\psi_{\vec n}$ is indeed small of order 1/$T$, it acts over a time that is large of order $T$. The sum can be ignored because of the exponential in it. As the definition of $\theta_{\vec n}$ shows, it varies on the normal time scale, rather than on the long time scale $T$. Therefore it oscillates many times on the long time scale; that causes opposite values of the exponential to largely cancel each other.

To show that more precisely, note that the formal solution of the full equation (D.20) is, [40, 19.2]:

\bar c_{\underline{\vec n}}(t) = e^{{\rm i}\gamma_{\underl...
..._{\underline{\vec n}}\vert\psi_{\underline{\vec n}}'\rangle %
\end{displaymath} (D.21)

To check this solution, you can just plug it in. Note in doing so that the integrands are taken to be functions of $\bar{t}$, not $t$.

All the integrals are negligibly small because of the rapid variation of the first exponential in them. To verify that, rewrite them a bit and then perform an integration by parts:

\int_{\bar t=0}^t - \frac{{\rm i}}{\hbar}(E_{\ve...
..._{\vec n}-E_{\underline{\vec n}})}
{\,\rm d}\bar t

The first term in the right hand side is small of order 1$\raisebox{.5pt}{$/$}$$T$ because the time derivative of the wave function is. The integrand in the second term is small of order 1$\raisebox{.5pt}{$/$}$$T^2$ because of the two time derivatives. So integrated over an order $T$ time range, it is small of order 1$\raisebox{.5pt}{$/$}$$T$ like the first term. It follows that the integrals in (D.21) become zero in the limit $T\to\infty$.

And that means that in the adiabatic approximation

\bar c_{{\vec n}} = c_{{\vec n}}(0) e^{{\rm i}\gamma_{{\ve...
... \langle\psi_{{\vec n}}\vert\psi_{{\vec n}}'\rangle{\,\rm d}t

The underbar used to keep $\underline{\vec n}$ and ${\vec n}$ apart is no longer needed here since only one set of quantum numbers appears. This expression for the coefficients can be plugged in (D.18) to find the wave function $\Psi$. The constants $c_{{\vec n}}(0)$ depend on the initial condition for $\Psi$. (They also depend on the choice of integration constants for $\theta_{{\vec n}}$ and $\gamma_{{\vec n}}$, but normally you take the phases zero at the initial time).

Note that $\gamma_{{\vec n}}$ is real. To verify that, differentiate the normalization requirement to get

\langle\psi_{\vec n}\vert\psi_{\vec n}\rangle = 1
...rangle +
\langle\psi_{\vec n}\vert\psi_{\vec n}'\rangle = 0

So the sum of the inner product plus its complex conjugate are zero. That makes it purely imaginary, so $\gamma_{\vec n}$ is real.

Since both $\gamma_{{\vec n}}$ and $\theta_{{\vec n}}$ are real, it follows that the magnitudes of the coefficients of the eigenfunctions do not change in time. In particular, if the system starts out in a single eigenfunction, then it stays in that eigenfunction.

So far it has been assumed that there is no degeneracy, at least not for the considered state. However it is no problem if at a finite number of times, the energy of the considered state crosses some other energy. For example, consider a three-di­men­sion­al harmonic oscillator with three time varying spring stiffnesses. Whenever any two stiffnesses become equal, there is significant degeneracy. Despite that, the given adiabatic solution still applies. (This does assume that you have chosen the eigenfunctions to change smoothly through degeneracy, as perturbation theory says you can, {D.80}.)

To verify that the solution is indeed still valid, cut out a time interval of size $\delta{T}$ around each crossing time. Here $\delta$ is some number still to be chosen. The parts of the integrals in (D.21) outside of these intervals have magnitudes $\varepsilon(T,\delta)$ that become zero when $T\to\infty$ for the same reasons as before. The parts of the integrals corresponding to the intervals can be estimated as no more than some finite multiple of $\delta$. The reason is that the integrands are of order 1$\raisebox{.5pt}{$/$}$$T$ and they are integrated over ranges of size $\delta{T}$. All together, that is enough to show that the complete integrals are less than say 1%; just take $\delta$ small enough that the intervals contribute no more than 0.5% and then take $T$ large enough that the remaining integration range contributes no more than 0.5% too. Since you can play the same game for 0.1%, 0.01% or any arbitrarily small amount, the conclusion is that for infinite $T$, the contribution of the integrals becomes zero. So in the limit $T\to\infty$, the adiabatic solution applies.

Things change if some energy levels are permanently degenerate. Consider an harmonic oscillator for which at least two spring stiffnesses are permanently equal. In that case, you need to solve for all coefficients at a given energy level $E_{\underline{\vec n}}$ together. To figure out how to do that, you will need to consult a book on mathematics that covers systems of ordinary differential equations. In particular, the coefficient $\bar{c}_{\underline{\vec n}}$ in (D.21) gets replaced by a vector of coefficients with the same energy. The scalar $\gamma_{\underline{\vec n}}$ becomes a matrix with indices ranging over the set of coefficients in the vector. Also, $e^{{\rm i}\gamma_{\underline{\vec n}}}$ gets replaced by a fundamental solution matrix, a matrix consisting of independent solution vectors. And $e^{-{\rm i}\gamma_{\underline{\vec n}}}$ is the inverse matrix. The sum no longer includes any of the coefficients of the considered energy.

More recent derivations allow the spectrum to be continuous, in which case the nonzero energy gaps $E_{\underline{\vec n}}-E_{\vec n}$ can no longer be assumed to be larger than some nonzero amount. And unfortunately, assuming the system to be approximated by a finite one helps only partially here; an accurate approximation will produce very closely spaced energies. Such problems are well outside the scope of this book.