# Trying to understand the WKB approximation

1. Jul 1, 2014

### phsopher

I'm trying to understand why the WKB approximation doesn't seem to work in the following case.

Suppose you have a particle of mass $m$ in a potential $V(x)=q m\cos(2mx/\hbar)$, where $q\ll 1$. Consider then the stationary solution with energy $E=m/2$. The Schroedinger equation is then

$$\psi'' + (1-2q\cos 2z)\psi = 0 \qquad (1)$$

where I defined a dimensionless variable $z=mx/\hbar$. With the ansatz $\psi(z) = A(z)e^{iS(z)}$ this reduces to

$$W^2\left[1-\frac{3}{4}\left(\frac{W'}{W^2}\right)^2 + \frac{1}{2}\frac{W''}{W^3}\right] = 1-2q\cos 2z \equiv \omega^2(z)$$

where I have also used the conservation of the Wronskian (current) which gives $(S'/W)'=0$.

Now, if the latter two terms in the square brackets are small then I should be able to solve this iteratively with the leading solution simply being $W=\omega$ so the solution is

$$\psi = \frac{1}{\sqrt{2\omega}}e^{i\int\omega\mathrm dz}$$

This is the gist of the WKB approximation as I understand it. In this case indeed $\omega'/\omega^2, \omega''/\omega^3 \ll 1$ so I don't see why this wouldn't work. However, this produces an oscillating solution with a constant amplitude. In reality, Equation (1) is the Mathieu equation and the amplitude of the oscillations is known to be amplified exponentially (parametric resonance).

I don't quite get why the WKB approximation fails here. Below is a numerical comparison between second order WKB solution and the exact numerical solution of (1).

Last edited: Jul 1, 2014
2. Jul 1, 2014

### Orodruin

Staff Emeritus
What makes you think that it is the WKB approximation that fails rather than your numerical solution?

3. Jul 1, 2014

### phsopher

The Mathieu equation has been studied for over a hundred years. It is known that for certain parameter values there is a parametric resonance where the solutions get exponentially amplified.

4. Jul 1, 2014

5. Jul 2, 2014

### phsopher

That's an interesting paper; however, I don't think it quite answers my question as far as I've been able to tell. What they appear to be considering is a solution which passes through the resonance region, which in the above case would correspond to x-dipendent energy $E(x)$.

What I want to know is why WKB method doesn't produce the solution despite $\omega'/\omega^2, \omega''/\omega^3\ll1$ when we are inside the resonance band.

I also found the following paper:

http://scitation.aip.org/content/aapt/journal/ajp/39/5/10.1119/1.1986212

which sketches out how to draw the instability chart for the Mathieu equation using the WKB approximation but I don't think it answers my question either as far as I'm able to tell.

6. Jul 2, 2014

### Bill_K

Those conditions insure that the WKB approximation will be valid locally, e.g. one cycle, but maybe not globally. The exponential growth in amplitude is a global effect, taking place gradually over many cycles.

Another idea - traditional WKB is the first term in a series. You might try including the next term. ("Higher-order WKB")

7. Jul 2, 2014

### phsopher

If I choose the energy to be $E=m$ (outside the instability band) instead of $E=m/2$ (inside the instability band) then the WKB gives an excellent fit over many cycles:

With respect to the various time scales of the system the two cases are pretty much the same. So my question is how I would know that I can trust WKB in the case $E=m$ but not $E=m/2$.

I did do that by considering iterative solutions, i.e.,

$$\begin{eqnarray} W_0^2 & = & \omega^2 \\ W_1^2 & = & \omega^2 + \frac{3}{4}\left(\frac{\omega'}{\omega}\right)^2 - \frac{1}{2}\frac{\omega''}{\omega} \\ W_2^2 & = & W_1^2 + \frac{3}{4}\left(\frac{W_1'}{W_1}\right)^2 - \frac{1}{2}\frac{W_1''}{W_1} \end{eqnarray}$$

It is this last second order term that is plotted in the Figure in my original post. Is this sort of iterative solution not the way to go? If so, why?

On the other hand if I consider a small perturbation around the "zeroth order" $W=\omega(1+\xi)$ I get a very complicated equation that I don't know how to estimate. Even if linearizing in $\xi$ doesn't help and is doomed to fail anyway because I know from the exact solution that the correction will grow exponentially to much larger values.

So my question is really: how can I know from the equation that I can't trust WKB for $E=m/2$ but can trust it for $E=m$.