91影视

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.

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

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

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

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

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

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

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

Start the upper element:

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

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

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

The sum of sine鈥檚 terms is zero, hence,

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

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

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

$$\begin{gathered} \left[\cos \left(\frac{位t}{2}\right)-i\frac{({蝇}_1-蝇\cos 伪)}{位}\sin \left(\frac{位t}{2}\right)\right]e^{-i蝇t/2}\left(\begin{array}{c}\cos \frac{伪}{2}\\ -e^{i蝇t}\sin \frac{伪}{2}\end{array}\right) \\ i\left[\frac{蝇}{位}\sin 伪\sin \left(\frac{位t}{2}\right)\right]e^{-i蝇t/2}\left(\begin{array}{c}\sin \frac{伪}{2}\\ -e^{i蝇t}\cos \frac{伪}{2}\end{array}\right)=\left(\begin{array}{c}伪\\ 尾\end{array}\right) \end{gathered}$$

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

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

$For尾:尾=\left\{\left[\cos \left(\frac{位t}{2}\right)-\frac{i{蝇}_1}{位}\left(\frac{位t}{2}\right)\right]\sin \frac{伪}{2}+\frac{i蝇}{位}\left[\cos 伪\sin \frac{伪}{2}-\sin 伪\cos \frac{伪}{2}\right]\sin \left(\frac{位t}{2}\right)\right\}{e}^{i蝇t/2}$

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

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

The coefficients are normalized. Hence,

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