Question

Check the Equation 10.31 satisfies the time-dependent Schrodinger equation for the Hamiltonian in Equation 10.25. Also confirm Equation 10.33, and show that the sum of the squares of the coefficients is 1, as required for normalization.

Short answer

Hence showed that$i\sout{h}\frac{𝜕x}{𝜕t}=H_x$

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 derivative of the function.

To show that : $i\sout{h}\frac{𝜕x}{𝜕t}=H_x..\left(1\right)$

Here, Equation 10.25....$\mathrm{H}=\frac{\sout{\mathrm{h}}{\mathrm{\omega }}_1}{2}\left(\begin{array}{cc}\mathrm{cos\alpha }& {\mathrm{e}}^{\mathrm{i\omega t}}\sin \mathrm{\alpha }\\ {\mathrm{e}}^{\mathrm{i\omega t}}\sin \mathrm{\alpha }& -\mathrm{cos\alpha }\end{array}\right)....\left(2\right)\mathrm{and}$

$Equation10.31....x\left(t\right)=\left(\begin{array}{c}\left[\cos (\lambda t/2)-i\frac{({\omega }_1-\omega )}{\lambda }\sin (\lambda t/2)\right]\cos (\alpha /2)e^{-i\omega t/2}\\ \left[\cos (\lambda t/2)-i\frac{({\omega }_1+\omega )}{\lambda }\sin (\lambda t/2)\right]\sin (\alpha /2)e^{+i\omega t/2}\end{array}\right).....\left(3\right)$

The derivative of the equation (3) with respect to t is,

role="math" localid="1656062904380" $i\sout{h}\frac{𝜕x}{𝜕t}=i\sout{h}\left(\begin{array}{c}\frac{\lambda }{2}\left[-\sin \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\cos \left(\frac{\lambda t}{2}\right)\right]\cos \left(\frac{\alpha }{2}\right)e^{-i\omega t/2}\\ -\frac{i\omega }{2}\left[\cos \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\cos \left(\frac{\alpha }{2}\right)e^{-i\omega t/2}\\ \frac{\lambda }{2}\left[-\sin \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\cos \left(\frac{\lambda t}{2}\right)\right]\sin \left(\frac{\alpha }{2}\right)e^{i\omega t/2}\\ +\frac{\lambda }{2}\left[\cos \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\sin \left(\frac{\alpha }{2}\right)e^{i\omega t/2}\end{array}\right)$

From (2) and (3), the equation (1) becomes,

$H_x\left(t\right)=\frac{\sout{h}{\omega }_1}{2}\left(\begin{array}{c}\cos \alpha \left[\cos \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\cos \frac{\alpha }{2}{e}^{-i\omega t/2}\\ +e^{-i\omega t}\sin \alpha \left[\cos \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\cos \frac{\alpha }{2}{e}^{i\omega t/2}\\ e^{-i\omega t}\cos \alpha \left[\cos \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\cos \frac{\alpha }{2}{e}^{-i\omega t/2}\\ -\cos \alpha \left[\cos \left(\frac{\lambda t}{2}\right)-\frac{i({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\sin \frac{\alpha }{2}{e}^{i\omega t/2}\end{array}\right)$

Show that the two matrices are identical.

Start the upper element:

$\begin{array}{c}i\sout{h}\frac{\lambda }{2}\left\{\left[-\sin \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\cos \left(\frac{\lambda t}{2}\right)\right]\cos \left(\frac{\alpha }{2}\right)e^{-i\omega t/2}\right.\\ -\frac{i\omega }{2}\left.\left[\cos \left(\frac{\lambda t}{2}\right)-\frac{i({\omega }_1-{\omega }^{\text{'}})}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\cos \left(\frac{\alpha }{2}\right)e^{-i\omega t/2}\right\}\\ =\left\{\frac{\sout{h}{\omega }_1}{2}\cos \alpha \left[\cos \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\cos \frac{\alpha }{2}{e}^{i\omega t/2}\right.\\ \left.+e^{-i\omega t}\left[\cos \left(\frac{\lambda t}{2}\right)-\frac{i({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\sin \frac{\alpha }{2}{e}^{i\omega t/2}\right\}\end{array}$

Cancel the like terms and substitute $\mathrm{sin\alpha }=2\sin \frac{\mathrm{\alpha }}{2}\cos \frac{\mathrm{\alpha }}{2}$in the last term to get:

role="math" localid="1656064057976" $\begin{array}{c}i\left\{\lambda \left[-\sin \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\cos \left(\frac{\lambda t}{2}\right)\right]\right.\\ -\frac{i\omega }{2}\left.\left[\cos \left(\frac{\lambda t}{2}\right)-\frac{i({\omega }_1-{\omega }^{\text{'}})}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\right\}..........\left(4\right)\\ ={\omega }_1\cos \alpha \left[\cos \left(\frac{\lambda l}{2}\right)-i\frac{({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\\ +2{\sin }^2\frac{\alpha }{2}\left[\cos \left(\frac{\lambda t}{2}\right)-\frac{i({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\end{array}$

The sum of sine’s terms is zero, hence,

$$\begin{gathered} \frac{i}{\lambda }\sin \left(\frac{\lambda t}{2}\right)[-{\omega }^2-{\omega }_1^2+2\omega {\omega }_1\cos \alpha -{\omega }_1+{\omega }^2+({\omega }_1^2-\omega {\omega }_1)\cos \alpha \\ +({\omega }_1^2+\omega {\omega }_1\left)\right(1-\cos \alpha \left)\right]=-\frac{i}{\lambda }\sin \left(\frac{\lambda t}{2}\right)[-{\omega }_1^2-2\omega {\omega }_1\cos \alpha -{\omega }_1{\omega }^2 \\ +{\omega }_1^2\cos \alpha -\omega {\omega }_1\cos \alpha +{\omega }_1^2+\omega {\omega }_1-{\omega }_1^2\cos \alpha -\omega {\omega }_1\cos \alpha =0 \\ \mathrm{Similarly},\mathrm{the}\mathrm{same}\mathrm{thin}\mathrm{for}\mathrm{the}\mathrm{lower}\mathrm{term}\mathrm{is}\mathrm{i}\sout{\mathrm{h}\frac{𝜕\mathrm{x}}{𝜕\mathrm{t}}}{\mathrm{H}}_{\mathrm{x}}. \end{gathered}$$

Confirm the equation 10.33.

Let the equation 10.33 be, $\begin{array}{c}x\left(t\right)=\left[\cos \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega \cos \alpha )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]e^{-i\omega t/2}{x}_{+}\left(t\right)\\ +i\left[\frac{\omega }{\lambda }\sin \alpha \sin \left(\frac{\lambda t}{2}\right)\right]e^{-i\omega t/2}{x}_{-}\left(t\right)\end{array}$

Here, $x_{+}\left(t\right)=\left(\begin{array}{c}\cos (\alpha /2)\\ e^{i\omega t}\sin (\alpha /2)\end{array}\right),x_{-}\left(t\right)=\left(\begin{array}{c}{e}^{-i\omega t}\sin (\alpha /2)\\ \cos (\alpha /2)\end{array}\right).\mathrm{Thus},$

$$\begin{gathered} \left[\cos \left(\frac{\lambda t}{2}\right)-i\frac{({\omega }_1-\omega \cos \alpha )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]e^{-i\omega t/2}\left(\begin{array}{c}\cos \frac{\alpha }{2}\\ -e^{i\omega t}\sin \frac{\alpha }{2}\end{array}\right) \\ i\left[\frac{\omega }{\lambda }\sin \alpha \sin \left(\frac{\lambda t}{2}\right)\right]e^{-i\omega t/2}\left(\begin{array}{c}\sin \frac{\alpha }{2}\\ -e^{i\omega t}\cos \frac{\alpha }{2}\end{array}\right)=\left(\begin{array}{c}\alpha \\ \beta \end{array}\right) \end{gathered}$$

Here, localid="1656067456488" $$\begin{gathered} \alpha =\left\{\left[\cos \left(\frac{\lambda t}{2}\right)-\frac{i{\omega }_1}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\cos \frac{\alpha }{2}+\frac{i\omega }{\lambda }\left[\cos \alpha \cos \frac{\alpha }{2}+\sin \alpha \sin \frac{\alpha }{2}\right]\sin \left(\frac{\lambda t}{2}\right)\right\}{e}^{-i\omega t/2} \\ but\cos \alpha \cos \frac{\alpha }{2}+\sin \alpha \sin \frac{\alpha }{2}=\cos \left(\alpha -\frac{\alpha }{2}\right)=\cos \frac{\alpha }{2}. \\ \alpha =\left[\cos \left(\frac{\lambda t}{2}\right)-\frac{i({\omega }_1-\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\cos \frac{\alpha }{2}{e}^{-i\omega t/2} \end{gathered}$$

Hence, the upper term of the equation (3) is confirmed.

$For\beta :\beta =\left\{\left[\cos \left(\frac{\lambda t}{2}\right)-\frac{i{\omega }_1}{\lambda }\left(\frac{\lambda t}{2}\right)\right]\sin \frac{\alpha }{2}+\frac{i\omega }{\lambda }\left[\cos \alpha \sin \frac{\alpha }{2}-\sin \alpha \cos \frac{\alpha }{2}\right]\sin \left(\frac{\lambda t}{2}\right)\right\}{e}^{i\omega t/2}$

role="math" localid="1656068970462" $$\begin{gathered} \mathrm{But},\cos \alpha \cos \frac{\alpha }{2}-\sin \alpha \cos \frac{\alpha }{2}=\sin \left(\frac{\alpha }{2}-\alpha \right)=-\sin \frac{\alpha }{2}.So, \\ \beta =\left[\cos \left(\frac{\lambda t}{2}\right)-\frac{i({\omega }_1+\omega )}{\lambda }\sin \left(\frac{\lambda t}{2}\right)\right]\sin \frac{\alpha }{2}{e}^{i\omega t/2} \end{gathered}$$

Hence, the lower term of the equation (3) is confirmed.

The coefficients are normalized. Hence,

$$\begin{gathered} {\left|c_{+}\right|}^2+{\left|c_{-}\right|}^2={\cos }^2\frac{\lambda t}{2}+\frac{{({\omega }_1-\omega \cos \alpha )}^2}{{\lambda }^2}{\sin }^2\frac{\lambda t}{2}+{\left[\frac{\omega }{\lambda }\sin \alpha \sin \frac{\lambda t}{2}\right]}^2 \\ ={\cos }^2\frac{\lambda t}{2}+{\sin }^2\frac{\lambda t}{2}\left[\frac{{({\omega }_1-\omega \cos \alpha )}^2}{{\lambda }^2}+{\left(\frac{\omega }{\lambda }\sin \alpha \right)}^2\right] \\ ={\cos }^2\frac{\lambda t}{2}+{\sin }^2\frac{\lambda t}{2}\left[\frac{{\omega }_1+{\omega }_1^2-2\omega {\omega }_1\cos \alpha }{{\lambda }^2}\right] \\ ={\cos }^2\frac{\lambda t}{2}+{\sin }^2\frac{\lambda t}{2}\left[\frac{{\omega }^2+{\omega }_1^2-2\omega {\omega }_1\cos \alpha }{{\omega }^2+{\omega }_1^2-2\omega {\omega }_1\cos \alpha }\right] \end{gathered}$$