Question

The delta function well (Equation 2.114) supports a single bound state (Equation 2.129). Calculate the geometric phase change whengradually increases from${\mathit{\alpha }}_{\mathbf{1}}\mathbf{}\mathbf{to}\mathbf{}{\mathbf{\alpha }}_{\mathbf{2}}$. If the increase occurs at a constant rate, $\mathbf{(}\mathbf{d\alpha }\mathbf{/}\mathbf{dt}\mathbf{=}\mathbf{c}\mathbf{)}$what is the dynamic phase change for this process?

Short answer

The geometric phase is ${\mathrm{Y}}_{\mathrm{n}}\left(\mathrm{t}\right)=0$.The dynamic phase change for this process is

$\theta \left(t\right)=\frac{m}{6h^{2}c}({\alpha }_2^3-{\alpha }_1^3).$

Step-by-step solution

Define Dynamic 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.Dynamic phases take inspiration from dynamic timers and static phases to generate performance metrics for all functions performed in a single phase invocation.

Obtain the geometric phase.

Let the time independent wave function for the bound state be,$\mathrm{\psi }=\frac{\sqrt{\mathrm{m\alpha }}}{\sout{\mathrm{h}}}{\mathrm{e}}^{-\mathrm{m\alpha }\left|\mathrm{\alpha }\right|\sout{\mathrm{h}}}.....$

(1)

The equation 10.42 is given by,localid="1656050634693" role="math"

Thus,

Obtain the dynamic phase.

The dynamic phase change for this process is given by, ${\theta }_n\left(t\right)=\frac{1}{h}{\int }_0^t{E}_n\left(t^{\text{'}}\right)d{t}^{\text{'}}.....\left(3\right)$

But the energy in this case is $\mathrm{E}=-\frac{{\mathrm{m\alpha }}^2}{2{\mathrm{h}}^2}.\mathrm{Thus},$

$$\begin{gathered} \theta \left(t\right)=-\frac{1}{\sout{h}}{\int }_0^T\left(-\frac{m{\alpha }^2}{2\sout{h^2}}\right)d{t}^{\text{'}} \\ =\frac{m}{2\sout{h^3}}{\int }_{{\alpha }_1}^{{\alpha }_2}{\alpha }^2\frac{d{t}^{\text{'}}}{d\alpha }d\alpha \\ =\frac{m}{2\sout{h^3}C}{\int }_{{\alpha }_1}^{{\alpha }_2}{\alpha }^{2}d\alpha \\ =\frac{m}{6h^{2}c}({\alpha }_2^3-{\alpha }_1^3) \\ \theta \left(t\right)=\frac{m}{6h^{2}c}({\alpha }_2^3-{\alpha }_1^3) \end{gathered}$$