91影视

Given that

$c_a=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-igt}{C}_b$

And

$c_b=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-igt}{C}_b$

$$\begin{gathered} c_a=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-igt}{C}_b \\ And \\ {c}_b=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-igt}{C}_b \end{gathered}$$

Differentiating with respect to t :

$$\begin{gathered} C_b=-\frac{i}{\sout{h}}{H}_{ab}\left[i{蝇}_0{e}^{i{蝇}_{0}t}{C}_a\right]-\frac{i}{\sout{h}}{H}_{ab}{e}^{i{蝇}_{0}t}\left[-\frac{i}{\sout{h}}{H}_{ab}{e}^{i{蝇}_{0}t}{c}_b\right] \\ or \\ {C}_b=i{蝇}_0{c}_b-\frac{1}{{\sout{h}}^2}{\left|H_{ab}\right|}^{2}cb.Let{伪}^2=\frac{1}{\sout{h^2}}{\left|H_{ab}\right|}^2.Then{c}_b-i{蝇}_0{C}_b+{伪}^2{c}_b=0 \end{gathered}$$

or

This is a linear differential equation with constant coefficients, so it can be solved by a function of the form :

$$\begin{gathered} {位}^2-i{蝇}_0位+{伪}^2=0;位=\frac{1}{2}\left[i{蝇}_0卤-\sqrt{-{蝇}_0^2}-4{伪}^2\right] \\ =\frac{i}{2}\left({蝇}_0卤蝇\right),where蝇=\sqrt{-{蝇}_0^2}-4{伪}^2 \end{gathered}$$

The general solution is therefore

$$\begin{gathered} c_b\left(t\right)=A{e}^{i\left({蝇}_0+蝇\right)t/2}+B{e}^{i\left({蝇}_0+蝇\right)t/2} \\ =e^{i{蝇}_{0}t/2}\left(A{e}^{i{蝇}_{0}t/2}+B{e}^{{蝇}_{0}t/2}\right), \\ Or \end{gathered}$$

$$\begin{gathered} {\mathrm{c}}_{\mathrm{b}}\left(\mathrm{t}\right)={\mathrm{e}}^{{\mathrm{蝇}}_0\mathrm{t}/2}\left[\mathrm{Ccos}\left({\mathrm{蝇}}_0\mathrm{t}/2\right)+\mathrm{Dsin}\left({\mathrm{蝇}}_0\mathrm{t}/2\right)\right].\mathrm{But}{\mathrm{c}}_{\mathrm{b}}\left(0\right)=0,\mathrm{so}\mathrm{C}=0, \\ \mathrm{and}\mathrm{hence} \end{gathered}$$

$c_b=\left(t\right)=D{e}^{{蝇}_{0}t/2}\sin \left({蝇}_{0}t/2\right)$

Then

$$\begin{gathered} c_b=D\left[\frac{i{蝇}_0}{2}{e}^{{蝇}_{0}t/2}\sin \left({蝇}_{0}t/2\right)+\frac{蝇}{2}{e}^{{蝇}_{0}t/2}\cos \left({蝇}_{0}t/2\right)\right] \\ =\frac{蝇}{2}D{e}^{{蝇}_{0}t/2}\left[\cos \left({蝇}_{0}t/2\right)+i\frac{{蝇}_0}{蝇}\sin \left({蝇}_{0}t/2\right)\right] \\ =-\frac{i}{\sout{h}}{H}_{ba}{e}^{{蝇}_{0}t/2}{c}_a \end{gathered}$$

$c_a=\frac{i\sout{h}}{H_{ba}}\frac{蝇}{2}{e}^{-{蝇}_{0}t/2}D\left[\cos \left({蝇}_{0}t/2\right)+i\frac{{蝇}_0}{蝇}\sin \left({蝇}_{0}t/2\right)\right].But{c}_a\left(0\right)=1so\frac{i\sout{h}}{H_{ba}}\frac{蝇}{2}$

Conclusion:

$c_a\left(t\right)=e^{-i{蝇}_{0}t/2}\left[\cos \left(i{蝇}_{0}t/2\right)+i\frac{{蝇}_0}{蝇}\sin \left(i{蝇}_{0}t/2\right)\right],c_b\left(t\right)=\frac{2H_{ab}}{i\sout{h}蝇}{e}^{i{蝇}_{0}t/2}\sin \left(i{蝇}_{0}t/2\right)$

Where

$$\begin{gathered} 蝇=\sqrt{{蝇}_0^2+4\frac{{\left|H_{ab}\right|}^2}{\sout{h^2}}} \\ {\left|c_a\right|}^2+{\left|c_b\right|}^2={\cos }^2\left(蝇t/2\right)+\frac{{蝇}_0^2}{{蝇}^2}{\sin }^2\left(蝇t/2\right)+\frac{4{\left|H_{ab}\right|}^2}{{\sout{h}}^2{蝇}^2}{\sin }^2\left(蝇t/2\right) \\ ={\cos }^2\left(蝇t/2\right)+\frac{1}{{蝇}^2}\left({蝇}_0^2+4\frac{{\left|H_{ab}\right|}^2}{{\sout{h}}^2}\right){\sin }^2\left(蝇t/2\right) \end{gathered}$$

${\cos }^2\left(蝇t/2\right)+{\sin }^2\left(蝇t/2\right)=1$

[In light of the Comment you might question the initial conditions. If the perturbation includes a factor 胃(t) , are we sure this doesn鈥檛 alter and That is, are we sure and are continuous at a step function potential? The answer is 鈥測es鈥�, for if we integrate Eq. 9.13 from to ,

$$\begin{gathered} c_a=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-i{蝇}_{0}t}{c}_{b1,}{c}_a=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-i{蝇}_{0}t}{c}_a \\ {c}_a\left(o\right)-c_a\left(-o\right)=-\frac{i}{\sout{h}}{H}_{ab}{鈭�}_0^o{e}^{-i{蝇}_{0}t}{c}_b\left(t\right)dt \end{gathered}$$

But$\left|c_b\left(t\right)\right|鈮�1,$

So the intergral goes to zero as $o鈫�0,$and hence$c_b\left(-o\right)=c_a\left(o\right).c_b$

The same goes for of course