# Adiabatic Approximation for Infinite Square Well

• rmiller70015
In summary, the final result for the geometric phase is that it is a function of ##u## and ##w## which is equal to ##-\frac{1}{2}\cos u \sin u##.

#### rmiller70015

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.

I noticed the following typographical errors
rmiller70015 said:
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:

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

In the first term the ##w## in the denominator should be raised to the 5/2 power instead of 3/2. There should not be a square on the sine function in the second term

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

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

This looks correct except there should be a factor of ##x## in front of the sine function in the first term of the integrand.

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

The result of the integration I got was:

$$\frac{2}{4w} - \frac{1}{4w}$$
I get ##\frac{2}{4w} - \frac{2}{4w}##

The ##sin^2(kx)## term came from multiplying from the left the complex conjugate, am I wrong there? And I got a ##\frac{2}{4w}## from the first term because I did a trig sub of ##sin(kx)cos(kx) = 2sin(2kx)## that changed my bounds of integration from ##0 \rightarrow w## to ##0 \rightarrow2n\pi## while the second term was changed to ##0 \rightarrow n\pi## because it just had the original ##\frac{n\pi}{w}## inside the sin function.

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rmiller70015 said:
The ##sin^2(kx)## term came from multiplying from the left the complex conjugate, am I wrong there?
Yes, you should get a ##\sin^2(kx)## term after multiplying from the left by the complex conjugate function. But you had the square on the sine function before multiplying. It's not a big deal. I think it's just a typo when you wrote down what you got for the derivative of the wave function with respect to ##w##.

And I got a ##\frac{2}{4w}## from the first term because I did a trig sub of ##sin(kx)cos(kx) = 2sin(2kx)## that changed my bounds of integration from ##0 \rightarrow w## to ##0 \rightarrow2n\pi## while the second term was changed to ##0 \rightarrow n\pi## because it just had the original ##\frac{n\pi}{w}## inside the sin function.
Yes, all of that looks good to me. But when you finally evaluated the integral with respect to ##u_2##, it looks like you got a result of ##\frac {1}{4w}##. I'm getting##\frac {1} {2w}## for this same integral.

Ok, I see the mistake now. It looks like I wrote the solution to the integral as ##\frac{1}{4}(u-cos(kx)sin(kx))## instead of ##\frac{u}{2} - \frac{1}{4}cos(kx)sin(kx)##

Everything else that I have written down is correct those were just typos in the latex. Thank you so much for your help.

rmiller70015 said:
Ok, I see the mistake now. It looks like I wrote the solution to the integral as ##\frac{1}{4}(u-cos(kx)sin(kx))## instead of ##\frac{u}{2} - \frac{1}{4}cos(kx)sin(kx)##

##\int{\sin^2u \, du} = \frac{u}{2} - \frac{1}{4}\sin(2u)= \frac{u}{2} - \frac{1}{2}\cos u \sin u##

So, does your final result for the geometric phase make sense?

I have quite some problems with this argumentation: The hamiltonian is only defined on (a dense subset of ) the square integrable functions over ##0\le x\le L(t)## at each point in time, what means that you can't take the derivative with respect to time as you are comparing functions from different Hilbert spaces. A possible way to proceed would be to consider the infinite well as a limit of a finite well and use a time dependent scaling to make the width of the well independent of t. In a second step, take the limit of the infinite well. It remains to be shown that the Hamiltonian stays self-adjoined in this second step.

I think this would work out, here is a sketch: Scaling can be accomplished with the unitary operator ##U(t)=\exp(-iw(t)\{p,x\}/2)##. Transformation of the hamiltonian, on one hand side, will yield a time dependent mass. On the other hand, from ##-\langle U^+(t)| i\partial_t U(t)\rangle =-w(t)\dot{w}(t) \{p,x\}/2##, this term will appear in the hamiltonian. Now there is no more problem to go to the limit of infinite potential. Completing the square, the new term in the hamiltonian can also be considered to be due to a vector potential ##A(x,t)\sim x## (also, a quadratic term will be introduced in the hamiltonian, which can be neglected in the adiabatic approximation).
The vector potential can be eliminated from the hamiltonian, writing the eigenfunctions as ##\psi_i(x,t)=\exp(i\int^x A(x,t)dx)\psi^0_i(x)##, where ##\psi^0_i## are the usual box eigenfunctions.
Using these functions, the geometric phase can be calculated as ## <i|\int^x (A(x,t_2)-A(x,t_1) dx| i>##.