Variable of integration in geometric phase calculation

[yeshuamo]

Homework Statement

Calculate the geometric phase change when the infinite square well expands adiabatically from width w₁ to w₂.

Homework Equations

Geometric phase:
\gamma_n(t) = i \int_{R_i}^{R_f} \Bigg< \psi_n \Bigg | \frac{\partial \psi_n}{\partial R} \Bigg > dR

Infinite square well wave function:
\psi_n = \sqrt{\frac{2}{w}}sin \Big(\frac{n \pi x}{w}\Big)

The Attempt at a Solution

This is an adiabatic approximation problem, and the variable R(t) here is the width of the well, w.
I took a derivative of the wave function and am integrating a dot product of the wave function with its derivative over w.

\gamma_i (t) = i \int_{w_1}^{w_2} \Big(-\frac{1}{2 w^2}\Big) sin^2 \Big(\frac{n \pi x}{w}\Big) dw - 2 i \int_{w_1}^{w_2} \frac{n \pi x}{w^3} sin\Big(\frac{n \pi x}{w}\Big) cos\Big(\frac{n \pi x}{w}\Big) dw

The first element appears to be unintegrable. I have looked at the solutions to this problem done by other people, and the integration is done over dx instead of dw, which clearly alleviates the integration problem above. Why is it valid to integrate over dx, even though the geometric phase formula above indicates integration over R, i.e. w?

 

[TSny]

\gamma_i (t) = i \int_{w_1}^{w_2} \Big(-\frac{1}{2 w^2}\Big) sin^2 \Big(\frac{n \pi x}{w}\Big) dw - 2 i \int_{w_1}^{w_2} \frac{n \pi x}{w^3} sin\Big(\frac{n \pi x}{w}\Big) cos\Big(\frac{n \pi x}{w}\Big) dw

The first element appears to be unintegrable. I have looked at the solutions to this problem done by other people, and the integration is done over dx instead of dw, which clearly alleviates the integration problem above. Why is it valid to integrate over dx, even though the geometric phase formula above indicates integration over R, i.e. w?

Note that there are two integrations involved in $\gamma_n(t) = i \int_{R_i}^{R_f} \left< \psi_n \Bigg | \frac{\partial \psi_n}{\partial R} \right > dR$. In addition to the integration with respect to R, the bra-ket $\left< \psi_n \Bigg | \frac{\partial \psi_n}{\partial R} \right >$ implies an additional integration (over what variable?).
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There is a nice way to deduce the value of $\int_{R_i}^{R_f} \left< \psi_n \Bigg | \frac{\partial \psi_n}{\partial R} \right> dR$ using just the fact that the wave functions $\psi_n$ are real and normalized. The trick is to relate $\left< \psi_n \Bigg | \frac{\partial \psi_n}{\partial R} \right>$ to $\frac{\partial }{\partial R}\left< \psi_n | \psi_n\right>$ .