91影视

Consider a time dependent potential, we can solve the Schrodinger equation for this potential in a two state system if we split the Hamiltonian into a time independent part$H^0$ and time dependent part$H^1$, that is:

$H=H^0+H^1$

This is done in the section 9.1 to get the solution of:

$蠄\left(x,t\right)=c_a\left(t\right)蠄\left(x\right)e^{-Eat/h}+c_b\left(t\right){蠄}_b\left(x\right)e^{-Eat/h}$

For H鈥� independent of t,$鈬�{\mathrm{c}}_{\mathrm{b}}^{\left(2\right)}\left(\mathrm{t}\right)\mathrm{and}{\mathrm{c}}_{\mathrm{b}}^{\left(1\right)}\left(\mathrm{t}\right)=-\frac{\mathrm{i}}{\sout{\mathrm{h}}}{\mathrm{H}}_{\mathrm{ba}}{鈭�}_0^{\mathrm{t}}{\mathrm{e}}^{{\mathrm{i蝇}}_0\mathrm{t}}\mathrm{dt}\text{'}$

localid="1658396453751" $$\begin{gathered} {\mathrm{c}}_{\mathrm{b}}^{\left(2\right)}\left(\mathrm{t}\right)=-\frac{\mathrm{i}}{\sout{\mathrm{h}}}{\mathrm{H}}_{\mathrm{ba}}{\overline{)\frac{{\mathrm{e}}^{{\mathrm{i蝇}}_0\mathrm{t}}}{{\mathrm{i蝇}}_0}}}_0^{\mathrm{t}}=-\frac{{\mathrm{H}}_{\mathrm{ba}}}{\sout{\mathrm{h}}{\mathrm{蝇}}_0}\left({\mathrm{e}}^{{\mathrm{i蝇}}_0\mathrm{t}}-1\right) \\ \frac{d{c}_a^{\left(1\right)}}{dt}=0鈬�c_a^{\left(1\right)}\left(t\right)=1 \end{gathered}$$

$\frac{d{c}_a^{\left(1\right)}}{dt}=-\frac{i}{\sout{h}}{H}_{ba}{e}^{i{蝇}_{0}t}鈬�c_b^{\left(1\right)}=-\frac{i}{\sout{h}}{鈭�}_0^t{H}_{ba}\left(t\text{'}\right)e^{i{蝇}_{0}t}dt$

Meanwhile

localid="1658397606335" ${\mathrm{c}}_{\mathrm{b}}^{\left(2\right)}\left(\mathrm{t}\right)=1-\frac{\mathrm{i}}{\sout{{\mathrm{h}}^2}}{\left|{\mathrm{H}}_{\mathrm{ba}}\right|}^2{鈭�}_0^t{e}^{i{蝇}_{0}t}\left[e^{i{蝇}_{0}t}dt\text{'}\right]dt=1-\frac{\mathrm{i}}{\sout{{\mathrm{h}}^2}}{\left|{\mathrm{H}}_{\mathrm{ba}}\right|}^2\frac{1}{i{蝇}_0}{鈭�}_0^t\left(1-e^{i{蝇}_{0}t}\right)dt$

$=1+\frac{i}{{蝇}_0{\sout{h}}^2}{\left|H_{ab}\right|}^2{\overline{)\left(t=\frac{e^{-i{蝇}_{0}t}}{i{蝇}_0}\right)}}_0^t=1+\frac{i}{{蝇}_0\sout{h^2}}{\left|H_{ab}\right|}^2\left[t+\frac{i}{{蝇}_0}\left(e^{-i{蝇}_{0}t}-1\right)\right]$

$\frac{d{c}_a^{\left(2\right)}}{dt}=-\frac{i}{\sout{h}}{H}_{ab}{e}^{i{蝇}_{0}t}\left(-\frac{i}{\sout{h}}\right){鈭�}_0^t{H}_{ba}\left(t\text{'}\right)e^{i{蝇}_{0}t}dt$

${\mathrm{c}}_a^{\left(2\right)}\left(\mathrm{t}\right)=1-\frac{\mathrm{i}}{\sout{{\mathrm{h}}^2}}{鈭�}_0^t{\mathrm{H}}_{\mathrm{ab}}\left(\mathrm{t}\text{'}\right){\mathrm{e}}^{{\mathrm{i蝇}}_0\mathrm{t}}\left[{鈭�}_0^{\mathrm{t}}{\mathrm{H}}_{\mathrm{ba}}\left(\mathrm{t}\text{'}\right){\mathrm{e}}^{{\mathrm{i蝇}}_0\mathrm{t}}\mathrm{dt}\text{'}\right]\mathrm{dt}$

.

For comparison with the exact answers , note first that $c_b\left(t\right)$is already first order (because of the $H_{ba}$in front), whereas differs from ${蝇}_0$only in second order, so it suffices to replace$蝇鈫�{蝇}_0$in the exact formula to get the second-order result:

localid="1658403529843" $c_b\left(t\right)鈮�\frac{2H_{ba}}{i\sout{h}{蝇}_0}{e}^{i{蝇}_{0}t/2}\sin \left({}^{i{蝇}_{0}t/2}\right)=\frac{2H_{ba}}{i\sout{h}{蝇}_0}{e}^{{蝇}_{0}t/2}\frac{1}{2i}\left(e^{{}^{i{蝇}_{0}t/2}}-e^{{}^{i{蝇}_{0}t/2}}\right)=-\frac{H_{ba}}{\sout{h}{蝇}_0}\left(e^{i{蝇}_{0}t}-1\right)$

in agreement with the result above.

Checkingis more difficult. Note that

$蝇={蝇}_0\sqrt{1+\frac{4{\left|H_{ab}\right|}^2}{{蝇}_0^2{\sout{h}}^2}}鈮�{蝇}_0\left(1+2\frac{{\left|H_{ab}\right|}^2}{{蝇}_0^2{\sout{h}}^2}\right)={蝇}_0+2\frac{{\left|H_{ab}\right|}^2}{{蝇}_0^2{\sout{h}}^2};\frac{{蝇}_0}{蝇}鈮�1-2\frac{{\left|H_{ab}\right|}^2}{{蝇}_0^2{\sout{h}}^2}$

Taylor expansion:

$\left\{\begin{array}{l}\cos \left(x+o\right)=\cos x-o\sin x\cos \left(蝇t/2\right)=\cos \left(\frac{{蝇}_{0}t}{2}+\frac{{\left|H_{ab}\right|}^2}{{蝇}_0\sout{h^2}}\right)鈮�\cos \left({蝇}_{0}t/2\right)-\frac{{\left|H_{ab}\right|}^{2}t}{{蝇}_0\sout{h^2}}\sin \left({蝇}_{0}t/2\right)\\ \begin{array}{l}\sin \left(x+o\right)=\sin x-\sin xo\cos \left(蝇t/2\right)=\sin \left(\frac{{蝇}_{0}t}{2}+\frac{{\left|H_{ab}\right|}^2}{{蝇}_0\sout{h^2}}\right)鈮�\sin \left({蝇}_{0}t/2\right)-\frac{{\left|H_{ab}\right|}^{2}t}{{蝇}_0\sout{h^2}}\cos \left({蝇}_{0}t/2\right)\\ \end{array}\end{array}\right.$

localid="1658402491191" $$\begin{gathered} c_a\left(t\right)鈮�e^{-i{蝇}_{0}t/2}\left\{\cos \left(\frac{{蝇}_{0}t}{2}\right)-\frac{{\left|H_{ab}\right|}^{2}t}{{蝇}_0{\sout{h}}^2}\sin \left(\frac{{蝇}_{0}t}{2}\right)+i\left(1-2\frac{{\left|H_{ab}\right|}^2}{{蝇}_0{\sout{h}}^2}\right)\left[\sin \left(\frac{{蝇}_{0}t}{2}\right)+\frac{{\left|H_{ab}\right|}^{2}t}{{蝇}_0{\sout{h}}^2}\cos \left(\frac{{蝇}_{0}t}{2}\right)\right]\right\} \\ =e^{-i{蝇}_{0}t/2}\left\{e^{-i{蝇}_{0}t/2}-\frac{{\left|H_{ab}\right|}^{2}t}{{蝇}_0{\sout{h}}^2}\left[it{e}^{i{蝇}_{0}t/2}+\frac{2i}{{蝇}_0}\frac{1}{2i}\left(e^{i{蝇}_{0}t/2}-e^{i{蝇}_{0}t/2}\right)\right]\right\} \end{gathered}$$

$=1-\frac{{\left|H_{ab}\right|}^2}{{蝇}_0{\sout{h}}^2}\left[-it+\frac{1}{{蝇}_0}\left(1-e^{i{蝇}_{0}t}\right)\right]=1-\frac{{\left|H_{ab}\right|}^2}{{蝇}_0{\sout{h}}^2}\left[-it+\frac{1}{{蝇}_0}\left(1-e^{i{蝇}_{0}t}\right)\right],asabove$

,asabove.

Thus the formula agree up to second order.

$$\begin{gathered} c_a\left(t\right)鈮�\frac{2H_{ba}}{i\sout{h}{蝇}_0}{e}^{i{蝇}_{0}t/2}\sin \left({}^{i{蝇}_{0}t/2}\right)=\frac{2H_{ba}}{i\sout{h}{蝇}_0}{e}^{{蝇}_{0}t/2}\frac{1}{2i}\left(e^{{}^{i{蝇}_{0}t/2}}-e^{{}^{i{蝇}_{0}t/2}}\right)=-\frac{H_{ba}}{\sout{h}{蝇}_0}\left(e^{i{蝇}_{0}t}-1\right) \\ {c}_b\left(t\right)鈮�\frac{2H_{ba}}{i\sout{h}{蝇}_0}{e}^{i{蝇}_{0}t/2}\sin \left({}^{i{蝇}_{0}t/2}\right)=\frac{2H_{ba}}{i\sout{h}{蝇}_0}{e}^{{蝇}_{0}t/2}\frac{1}{2i}\left(e^{{}^{i{蝇}_{0}t/2}}-e^{{}^{i{蝇}_{0}t/2}}\right)=-\frac{H_{ba}}{\sout{h}{蝇}_0}\left(e^{i{蝇}_{0}t}-1\right) \end{gathered}$$