# Adiabatic approximation: higher order corrections

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 10.10.

In deriving the adiabatic theorem, Griffiths (in his section 10.1) shows that the solution to the time-dependent Schrödinger equation can be written as

$\displaystyle \Psi\left(x,t\right)=\sum_{n}c_{n}\left(t\right)\psi_{n}\left(x,t\right)e^{i\theta_{n}\left(t\right)} \ \ \ \ \ (1)$

where the ${\psi_{n}}$ form an orthonormal set of functions that are eigenfunctions of the Hamiltonian at a particular instant of time, and ${\theta_{n}}$ is the dynamic phase. The coefficients ${c_{n}}$ are the usual weighting factors, and they depend only on time.

Later in the derivation, he arrives at a differential equation for the ${c_{m}}$:

$\displaystyle \dot{c}_{m}\left(t\right)=-\sum_{j}c_{j}\left\langle \psi_{m}\left|\dot{\psi}_{j}\right.\right\rangle e^{i\left(\theta_{j}-\theta_{m}\right)} \ \ \ \ \ (2)$

In the adiabatic approximation, this equation has the approximate solution

 $\displaystyle c_{m}\left(t\right)$ $\displaystyle =$ $\displaystyle c_{m}\left(0\right)e^{i\gamma_{m}\left(t\right)}\ \ \ \ \ (3)$ $\displaystyle \gamma_{m}\left(t\right)$ $\displaystyle \equiv$ $\displaystyle i\int_{0}^{t}\left\langle \psi_{m}\left(t'\right)\left|\frac{\partial}{\partial t'}\psi_{m}\left(t'\right)\right.\right\rangle dt' \ \ \ \ \ (4)$

where ${\gamma_{m}}$ is the geometric phase. In particular, if the system starts in a definite eigenstate ${\psi_{n}}$ then ${c_{m}\left(0\right)=\delta_{nm}}$ so

$\displaystyle c_{m}\left(t\right)=\delta_{nm}e^{i\gamma_{n}\left(t\right)} \ \ \ \ \ (5)$

with the result that the overall solution becomes

$\displaystyle \Psi_{n}\left(x,t\right)=\psi_{n}\left(x,t\right)e^{i\theta_{n}\left(t\right)}e^{i\gamma_{n}\left(t\right)} \ \ \ \ \ (6)$

that is, the system stays in the ${n^{th}}$ state over time, although its phase can change.

We can extend the adiabatic approximation recursively by using the first approximation 5 to generate the next approximation. We can do this by inserting 5 into 2 and then solving the resulting differential equation. The sum in 2 is reduced to a single term where ${j=n}$, the eigenstate in which the system starts at ${t=0}$.

 $\displaystyle \dot{c}_{m}\left(t\right)$ $\displaystyle =$ $\displaystyle -e^{i\gamma_{n}\left(t\right)}\left\langle \psi_{m}\left|\dot{\psi}_{n}\right.\right\rangle e^{i\left(\theta_{n}-\theta_{m}\right)}\ \ \ \ \ (7)$ $\displaystyle c_{m}\left(t\right)$ $\displaystyle =$ $\displaystyle c_{m}\left(0\right)-\int_{0}^{t}e^{i\gamma_{n}\left(t\right)}\left\langle \psi_{m}\left|\dot{\psi}_{n}\right.\right\rangle e^{i\left(\theta_{n}-\theta_{m}\right)}dt' \ \ \ \ \ (8)$

This correction to the basic adiabatic approximation now has the ability to predict transitions from the initial state ${\psi_{n}}$ to other states ${\psi_{m}}$ where ${m\ne n}$. We can apply this to the forced oscillator, where we found that in the adiabatic approximation

 $\displaystyle \psi_{n}\left(x,t\right)$ $\displaystyle =$ $\displaystyle \psi_{n}\left(x-f\right)\ \ \ \ \ (9)$ $\displaystyle \theta_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle \frac{m\omega^{2}}{2\hbar}\int_{0}^{t}f^{2}\left(t'\right)dt'-\left(n+\frac{1}{2}\right)\omega t\ \ \ \ \ (10)$ $\displaystyle \gamma_{n}\left(t\right)$ $\displaystyle =$ $\displaystyle \frac{m\dot{f}}{\hbar}\left(x-\frac{f}{2}\right)\approx0 \ \ \ \ \ (11)$

Here, ${m\omega^{2}f\left(t\right)}$ is the forcing term, and the adiabatic approximation is obtained by assuming that ${f}$ changes very slowly, or to be precise:

$\displaystyle \left|\dot{f}\left(t\right)\right|\ll\omega\left|f\left(t\right)\right| \ \ \ \ \ (12)$

To work out the correction, we need to find ${\left\langle \psi_{m}\left|\dot{\psi}_{n}\right.\right\rangle }$ in 8. We can do this using the raising and lowering operators for the harmonic oscillator. In particular, the momentum operator can be written in terms of them as

$\displaystyle p=i\sqrt{\frac{\hbar m\omega}{2}}\left(a_{+}-a_{-}\right) \ \ \ \ \ (13)$

Also, recall that the effects of ${a_{\pm}}$ are

 $\displaystyle a_{+}\psi_{n}$ $\displaystyle =$ $\displaystyle \sqrt{n+1}\psi_{n}\ \ \ \ \ (14)$ $\displaystyle a_{-}\psi_{n}$ $\displaystyle =$ $\displaystyle \sqrt{n}\psi_{n-1} \ \ \ \ \ (15)$

How does this help us? We need to find the derivative ${\partial\psi_{n}\left(x-f\right)/\partial t'}$, so we get, defining ${z\equiv x-f}$:

 $\displaystyle \frac{\partial\psi_{n}\left(x-f\right)}{\partial t'}$ $\displaystyle =$ $\displaystyle \frac{\partial\psi_{n}\left(z\right)}{\partial z}\frac{\partial z}{\partial t'}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial\psi_{n}\left(z\right)}{\partial z}\dot{f}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{\partial\psi_{n}}{\partial x}\dot{f} \ \ \ \ \ (18)$

where the last line follows because ${z=x-f}$ and ${f}$ doesn’t depend on ${x}$.

Now the momentum operator is

$\displaystyle p=\frac{\hbar}{i}\frac{\partial}{\partial x} \ \ \ \ \ (19)$

so our derivative is

 $\displaystyle \frac{\partial\psi_{n}\left(x-f\right)}{\partial t'}$ $\displaystyle =$ $\displaystyle -\frac{i}{\hbar}\dot{f}p\psi_{n}\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \dot{f}\sqrt{\frac{m\omega}{2\hbar}}\left(a_{+}-a_{-}\right)\psi_{n}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \dot{f}\sqrt{\frac{m\omega}{2\hbar}}\left(\sqrt{n+1}\psi_{n+1}-\sqrt{n}\psi_{n-1}\right) \ \ \ \ \ (22)$

Using the orthonormality of the ${\psi_{m}}$ we have

 $\displaystyle \left\langle \psi_{n+1}\left|\dot{\psi}_{n}\right.\right\rangle$ $\displaystyle =$ $\displaystyle \dot{f}\sqrt{\frac{m\omega}{2\hbar}}\sqrt{n+1}\ \ \ \ \ (23)$ $\displaystyle \left\langle \psi_{n-1}\left|\dot{\psi}_{n}\right.\right\rangle$ $\displaystyle =$ $\displaystyle -\dot{f}\sqrt{\frac{m\omega}{2\hbar}}\sqrt{n} \ \ \ \ \ (24)$

with all other matrix elements equal to zero.

Returning to 8 we can work out the phase terms from 10 and 11.

 $\displaystyle \gamma_{n}$ $\displaystyle \approx$ $\displaystyle 0\ \ \ \ \ (25)$ $\displaystyle \theta_{n}-\theta_{n+1}$ $\displaystyle =$ $\displaystyle \omega t\ \ \ \ \ (26)$ $\displaystyle \theta_{n}-\theta_{n-1}$ $\displaystyle =$ $\displaystyle -\omega t \ \ \ \ \ (27)$

Therefore, since ${c_{n+1}\left(0\right)=c_{n-1}\left(0\right)=0}$,

 $\displaystyle c_{n+1}\left(t\right)$ $\displaystyle =$ $\displaystyle -\sqrt{\frac{m\omega}{2\hbar}}\sqrt{n+1}\int_{0}^{t}\dot{f}e^{i\omega t'}dt'\ \ \ \ \ (28)$ $\displaystyle c_{n-1}\left(t\right)$ $\displaystyle =$ $\displaystyle \sqrt{\frac{m\omega}{2\hbar}}\sqrt{n}\int_{0}^{t}\dot{f}e^{-i\omega t'}dt' \ \ \ \ \ (29)$

[These answers aren’t the same as those given in Griffiths’s question (although the square moduli are the same) but I can’t see anything wrong with my derivation. Comments welcome.]

Note that

$\displaystyle \left\langle \psi_{n}\left|\dot{\psi}_{n}\right.\right\rangle =0 \ \ \ \ \ (30)$

so 8 predicts that ${c_{n}\left(t\right)=c_{n}\left(0\right)=1}$, thus the sum of the square moduli of the ${c_{m}}$s is greater than 1. However, these values for the ${c_{m}}$s are correct only to first order in ${\dot{f}}$. To get the second order corrections, we’d need to insert 28 and 29 back into 2 and integrate again to get new values for the ${c_{m}}$s, which would give ${c_{n}\left(t\right)<1}$ for ${t>0}$. The process can be continued as long as we like, giving an adiabatic series.