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

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

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

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

$$\bar c_{\underline{\vec n}}^{\,\prime} = - \sum_{\vec n}... ...nderline{\vec n}}\vert\psi_{\vec n}'\rangle \bar c_{\vec n} %$$ (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):

$$\bar c_{\underline{\vec n}}^{\,\prime} = - \langle\psi_{... ...nderline{\vec n}}\vert\psi_{\vec n}'\rangle \bar c_{\vec n} %$$ (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, [39, 19.2]:

$$\bar c_{\underline{\vec n}}(t) = e^{{\rm i}\gamma_{\underl... ...{\underline{\vec n}}\vert\psi_{\underline{\vec n}}'\rangle %$$ (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:

$$\begin{eqnarray*} \lefteqn{ \int_{\bar t=0}^t - \frac{{\rm i}}{\hbar}(E_{\ve... ..._{\vec n}-E_{\underline{\vec n}})} \right)' {\,\rm d}\bar t \end{eqnarray*}$$

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

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

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.