Question

(a) Use Equation 10.42 to calculate the geometric phase change when the infinite square well expands adiabatically from width ${\mathit{w}}_{\mathbf{1}}$to width ${\mathit{w}}_{\mathbf{2}}$. Comment on this result.

(b) If the expansion occurs at a constant rate$\left(dw/dt=v\right)$, what is the dynamic phase change for this process?

(c) If the well now contracts back to its original size, what is Berry's phase for the cycle?

Short answer

(a) The geometric phase is${\mathrm{\gamma }}_{\mathrm{n}}\left(\mathrm{t}\right)=0$.

(b) The dynamic phase change for this process islocalid="1656067997518" ${\mathrm{\theta }}_{\mathrm{n}}\left(\mathrm{t}\right)=\frac{{\mathrm{n}}^2{\mathrm{\tau \tau }}^2\mathrm{h}}{2\mathrm{mv}}\left(\frac{1}{{\mathrm{w}}_2}‐\frac{1}{{\mathrm{w}}_1}\right)$.

(c) The Berry’s phase is zero.

Step-by-step solution

Define Geometric phase.

Geometric phase is a phase difference gained during the course of a cycle when a system is subjected to cyclic adiabatic processes, which derives from the geometrical features of the Hamiltonian's parameter space in both classical and quantum mechanics.

Obtain the geometric phase.

(a)

Let an infinite square well whose right wall moves as its position is a function of time be,${\mathrm{\psi }}_{\mathrm{n}}\left(x\right)=\sqrt{\frac{2}{\mathrm{w}}}\sin \left(\frac{\mathrm{n\tau \tau }}{\mathrm{w}}\mathrm{x}\right)$… (1)

The equation 10.42 is given by, … (2)

In this case, the width of the well isrole="math" localid="1656068288003" $\mathrm{R}\left(\mathrm{t}\right)=\mathrm{w}\left(\mathrm{t}\right)$. So,

$\frac{\partial {\mathrm{\psi }}_{\mathrm{n}}}{\partial \mathrm{R}}=\sqrt{2}\left(‐\frac{1}{2}\frac{1}{{\mathrm{w}}^{3/2}}\right)\sin \left(\frac{\mathrm{n\tau \tau }}{\mathrm{w}}\mathrm{x}\right)+\sqrt{\frac{2}{\mathrm{w}}}\left(‐\frac{\mathrm{n\tau \tau }}{{\mathrm{w}}^2}\mathrm{x}\right)\cos \left(\frac{\mathrm{n\tau \tau }}{\mathrm{w}}\mathrm{x}\right)$

Thus,

Substitute the value in the equation (2) to get:

${\mathrm{\gamma }}_{\mathrm{n}}\left(\mathrm{t}\right)0$

Obtain the dynamic phase.

(b)

The dynamic phase change for this process is given by,${\mathrm{\theta }}_{\mathrm{n}}\left(\mathrm{t}\right)=‐\frac{1}{\mathrm{h}}{\int }_0^{\mathrm{t}}{\mathrm{E}}_{\mathrm{n}}\left(\mathrm{t}\text{'}\right)\mathrm{dt}\text{'}$… (3)

But the energy in this case is${\mathrm{E}}_{\mathrm{n}}=\frac{{\left(\mathrm{n\tau \tau h}\right)}^2}{2{\mathrm{mw}}^2}$ . Thus,

$$\begin{gathered} {\theta }_n\left(t\right)=\frac{1}{h}{\int }_0^t\frac{n^2\tau {\tau }^2{h}^2}{2m{w}^2}dt\text{'} \\ =‐\frac{n^2\tau {\tau }^{2}h}{2m}\int \frac{1}{w^2}\frac{dt\text{'}}{dw}dw \\ =‐\frac{n^2\tau {\tau }^{2}h}{2m}{\int }_{w_1}^{w_2}\frac{1}{w^2}\frac{dt\text{'}}{dw}dw \\ =\frac{n^2\tau {\tau }^{2}h}{2m}\left(\frac{1}{w_2}‐\frac{1}{w_1}\right) \\ {\theta }_n\left(t\right)=\frac{n^2\tau {\tau }^{2}h}{2mv}\left(\frac{1}{w_2}‐\frac{1}{w_1}\right) \end{gathered}$$

(c)

The Berry’s phase for the cycle will be zero.