Question

(a) Derive the equation 10.67 from Equation 10.65.

(b) Derive Equation 10.79, starting with Equation 10.78.

Short answer

(a) The Hamiltonian is$H=\frac{1}{1m}\left[-{\sout{h}}^2{\nabla }^2+q^2{A}^2+2iq\sout{h}A.\nabla \right].$.

(b) The required equation is ${\left(\frac{\sout{h}}{i}\nabla -qA\right)}^2\psi =-{\sout{h}}^2{e}^{2g{\nabla }^2\psi }$.

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 Hamiltonian.

(a)

The Hamiltonian in equation 10.65 is given by: $H=\frac{1}{2m}{\left(\frac{h}{i}\nabla -qA\right)}^2+q\phi ...\left(1\right)$

Apply the Hamiltonian equation on an arbitrary function as:

$$\begin{gathered} \mathrm{Hf}=\frac{1}{2\mathrm{m}}\left(\frac{\sout{\mathrm{h}}}{\mathrm{i}}\nabla -\mathrm{qA}\right).\left(\frac{\sout{\mathrm{h}}}{\mathrm{i}}\nabla \mathrm{f}-\mathrm{qAf}\right)+\mathrm{q\phi f} \\ =\frac{1}{2\mathrm{m}}\left[-{\sout{\mathrm{h}}}^2\nabla .(\nabla \mathrm{f})-\frac{\mathrm{q}\sout{\mathrm{h}}}{\mathrm{i}}\mathrm{A}(\nabla \mathrm{f})+{\mathrm{q}}^2\mathrm{A}.\mathrm{Af}\right]+\mathrm{q\phi f} \end{gathered}$$

But, $\nabla .\left(Af\right)=(\nabla .A)f+A.(\nabla f).$Thus,

$Hf=\frac{1}{2m}\left[-{\sout{h}}^2\nabla .(\nabla f)-\frac{q\sout{h}}{i}(\nabla .A)f-\frac{2q\sout{h}}{i}A.(\nabla f)+q^{2}A.Af\right]+q\phi f$ $Hf=\frac{1}{2m}\left[-{\sout{h}}^2\nabla .(\nabla f)-\frac{q\sout{h}}{i}(\nabla .A)f-\frac{2q\sout{h}}{i}A.(\nabla f)+q^{2}A.Af\right]+q\phi f$

After equation 10.66, using the condition $\phi =0$and $\nabla .A=0$, so:

$$\begin{gathered} Hf=\frac{1}{2m}\left[-\sout{h^2}{\nabla }^{2}f+2iq\sout{h}A.\nabla f+q^2{A}^{2}f\right] \\ Hf=\frac{1}{2m}\left[-\sout{h^2}{\nabla }^{2}f+q^2{A}^2+2iq\sout{h}A.\nabla \right] \end{gathered}$$

Obtain the required equation.

(b)

Let the equation 10.78 be,$\left(\frac{\sout{\mathrm{h}}}{\mathrm{i}}\nabla -\mathrm{qA}\right)\mathrm{\psi }=\frac{\sout{\mathrm{h}}}{\mathrm{i}}{\mathrm{e}}^{\mathrm{ig}}\nabla \mathrm{\psi }$

$$\begin{gathered} \mathrm{Apply}\left(\frac{\sout{\mathrm{h}}}{\mathrm{i}}\nabla -\mathrm{qA}\right)\mathrm{to}\mathrm{both}\mathrm{sides}\mathrm{to}\mathrm{get}: \\ {\left(\frac{\sout{\mathrm{h}}}{\mathrm{i}}\nabla -\mathrm{qA}\right)}^2\mathrm{\psi }=\left(\frac{\sout{\mathrm{h}}}{\mathrm{i}}\nabla -\mathrm{qA}\right).\left(\frac{\sout{\mathrm{h}}}{\mathrm{i}}{\mathrm{e}}^{\mathrm{ig}}\mathrm{A}.\nabla \mathrm{\psi }\right) \\ =-{\mathrm{h}}^2\nabla .({\mathrm{e}}^{\mathrm{ig}}\nabla {\mathrm{\psi }}^{\text{'}})-\frac{\mathrm{qh}}{\mathrm{i}}{\mathrm{e}}^{\mathrm{ig}}\mathrm{A}.\nabla {\mathrm{\psi }}^{\text{'}} \\ \mathrm{But},\nabla .({\mathrm{c}}^{\mathrm{ig}}\nabla {\mathrm{\psi }}^{\text{'}})={\mathrm{ic}}^{\mathrm{ig}}(\nabla \mathrm{g}).(\nabla {\mathrm{\psi }}^{\text{'}})+{\mathrm{c}}^{\mathrm{ig}}\nabla .(\nabla {\mathrm{\psi }}^{\text{'}})\mathrm{and}\nabla \mathrm{g}=\frac{\mathrm{q}}{\mathrm{h}}\mathrm{A}.\mathrm{Thus} \\ \left(\frac{\sout{\mathrm{h}}}{\mathrm{i}}\nabla -\mathrm{qA}\right)\mathrm{\psi }=\mathrm{i}{\sout{\mathrm{h}}}^2\frac{\mathrm{q}}{\sout{\mathrm{h}}}{\mathrm{e}}^{\mathrm{ig}}\mathrm{A}.\nabla {\mathrm{\psi }}^{\text{'}}-{\sout{\mathrm{h}}}^2{\mathrm{e}}^{\mathrm{ig}}{\nabla }^2{\mathrm{\psi }}^{\text{'}}+\mathrm{iq}\sout{\mathrm{h}}{\mathrm{e}}^{\mathrm{ig}}\mathrm{A}.\nabla \mathrm{\psi } \\ =-{\mathrm{h}}^2{\mathrm{e}}^{\mathrm{ig}}{\nabla }^2\mathrm{\psi }{\left(\frac{\sout{\mathrm{h}}}{\mathrm{i}}\nabla -\mathrm{qA}\right)}^2\mathrm{\psi } \\ =-{\mathrm{h}}^2{\mathrm{e}}^{\mathrm{ig}}{\nabla }^2\mathrm{\psi } \end{gathered}$$