Daily Archives: Sat, 6 June 2015

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.