- #1

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- Homework Statement
- Find the Geometric Phase for an infinite square well with a constant expansion from w1 to w2

- Relevant Equations
- $$\psi = \sqrt{\frac{2}{w(t)}}sin(\frac{n\pi}{w(t)}x)$$

$$\gamma (t) = i\int^{t}_{0}<\psi _n(t')|\frac{\partial}{\partial t'} \psi _n(t')>dt' = i\int^{w_2}_{w_1}<\psi _n | \frac{\partial}{\partial w}\psi _n>dw$$

I took the w derivative of the wave function and got the following. Also w is a function of time, I just didn't notate it for brevity:

$$-\frac{\sqrt{2}n\pi x}{w^{3/2}}cos(\frac{n\pi}{w}x) - \frac{1}{\sqrt{2w^3}}sin^2(\frac{n\pi}{w}x)$$

Then I multiplied the complex conjugate of the wave function to get the first integral and changed the limits from 0 to ##w(t)##:

$$<\psi _n | \frac{\partial}{\partial w}\psi > = -\frac{i}{w^2}\int^{w}_{0}\frac{n\pi}{w}sin(\frac{2n\pi}{2}x) + sin^2(\frac{n\pi}{w}x)dx$$

u-substitution:

##u_1 = \frac{2n\pi x}{w} du_1 = \frac{2n\pi}{w}dx## and ##u_2 = \frac{n\pi x}{w} du_2 = \frac{n\pi}{w}dx##

$$-\frac{1}{4n\pi w}\int^{2n\pi}_0u_1sin(u_1)du_1 - \frac{1}{n\pi w}\int^{n\pi}_0 sin^2(u_2)du_2$$

The result of the integration I got was:

$$\frac{2}{4w} - \frac{1}{4w}$$

Then I plugged it into the gamma integral:

$$\gamma (t) = \frac{i}{4}\int^{w_2}_{w_1}\frac{1}{w}dw$$

I'm not so sure I should be getting a natural log for this result. I mean it will come out of the exponential nicely, but I just have a gut feeling that I did something wrong.

$$-\frac{\sqrt{2}n\pi x}{w^{3/2}}cos(\frac{n\pi}{w}x) - \frac{1}{\sqrt{2w^3}}sin^2(\frac{n\pi}{w}x)$$

Then I multiplied the complex conjugate of the wave function to get the first integral and changed the limits from 0 to ##w(t)##:

$$<\psi _n | \frac{\partial}{\partial w}\psi > = -\frac{i}{w^2}\int^{w}_{0}\frac{n\pi}{w}sin(\frac{2n\pi}{2}x) + sin^2(\frac{n\pi}{w}x)dx$$

u-substitution:

##u_1 = \frac{2n\pi x}{w} du_1 = \frac{2n\pi}{w}dx## and ##u_2 = \frac{n\pi x}{w} du_2 = \frac{n\pi}{w}dx##

$$-\frac{1}{4n\pi w}\int^{2n\pi}_0u_1sin(u_1)du_1 - \frac{1}{n\pi w}\int^{n\pi}_0 sin^2(u_2)du_2$$

The result of the integration I got was:

$$\frac{2}{4w} - \frac{1}{4w}$$

Then I plugged it into the gamma integral:

$$\gamma (t) = \frac{i}{4}\int^{w_2}_{w_1}\frac{1}{w}dw$$

I'm not so sure I should be getting a natural log for this result. I mean it will come out of the exponential nicely, but I just have a gut feeling that I did something wrong.