Infinite Square Well with an Adiabatic Evolution

[Ateowa]

Homework Statement

I'm trying to find the geometric phase for the adiabatic widening of the infinite square well. Griffiths defines the geometric phase to be:
\gamma=i* \int^{w2}_{w1}<\psi_{n}|\frac{d\psi_{n}}{dR}>dR

Where R is the aspect of the potential that is changing and w1, w2 are the initial and final widths of the well, respectively.

Homework Equations

The wave function of the infinite square well is \frac{d\psi_{n}}{dR}=\sqrt{ \frac{1}{2R^{3}}}Sin[\frac{n \pi x}{R}], sharing the notation above where R is the width of the well.

The Attempt at a Solution

I'm having a bit of trouble with using Latex, so I'll try to describe the process that I'm doing. I think that will suffice for my question, but at the bottom I'll include my LaTex attempt.

When I differentiate the wave function with respect to R (The width of the well), I get a sum of a Sine and Cosine function up to constants. But to find the geometric phase, I integrate over the complex conjugate of the wavefunction times the derivative. This would be fine, but the result of this needs to be purely imaginary, as the integral has a factor of i in front and the phase factor as a whole must be real.

I know I must be missing something. I tried taking the derivative with respect to the wave function multiplied by the dynamic phase in the "complete" wave function, but that doesn't really solve the problem. at best, differentiating with the phase included gives a complex answer-- None of it can be real. Otherwise, the resultant wave function won't be properly normalized and will change magnitude as a result of the adiabatic expansion, which is opposite to the entire rationale behind the method.

\frac{d\psi_n}{dR}=\sqrt{\frac{2}{R}}(\frac{n \pi x}{R})Cos[\frac{n \pi x}{R}]+\sqrt{\frac{1}{2R^{3}}}Sin[\frac{n \pi x}{R}]

But then:

<\psi_n|\frac{d\psi_n}{dR}>=\frac{1}{R}\int\frac{2n\pi}{R^2}Cos[\frac{n \pi x}{R}]Sin[\frac{n \pi x}{R}]-Sin^{2}[\frac{n \pi x}{R}dR=\frac{1}{2R}

Which leads to \gamma=\frac{i}{2}*ln[\frac{w2}{w1}]

This isn't a valid result, as this leads the wave function to be (neglecting the dynamic phase factor for ease of LaTex expression):

\Psi_n(t,x)=e^{\frac{-1}{2}ln[\frac{w2}{w1}]}\psi_n(x)=-\sqrt{\frac{w2}{w1}}\psi_n(x)

 

[Ateowa]

I solved my own problem. For anyone who's interested, I missed a factor of x in my integration, so that the dot product should actually be

<\psi_n|\frac{d\psi_n}{dR}>=\frac{1}{R}\int\frac{2n\pi x}{R^2}Cos[\frac{n \pi x}{R}]Sin[\frac{n \pi x}{R}]-Sin^{2}[\frac{n \pi x}{R}]dR=0

Therefore, gamma is zero and everything is okay.