Question

Calculate ${\mathbf{c}}_{\mathbf{a}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{c}}_{\mathbf{b}}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{}$, to second order, for a time-independent perturbation in Problem 9.2. Compare your answer with the exact result.

Short answer

The formula agree up to second order.

$$\begin{gathered} c_a\left(t\right)=1-\frac{{\left|H_{ab}\right|}^2}{{\omega }_0{\sout{h}}^2}\left[-it+\frac{1}{{\omega }_0}\left(1-e^{i{\omega }_{0}t}\right)\right]=1+\frac{1}{{\omega }_0{\sout{h}}^2}{\left|H_{ab}\right|}^2\left[-it+\frac{1}{{\omega }_0}\left(e^{i{\omega }_{0}t}-1\right)\right] \\ {c}_a\left(t\right)=1-\frac{2H_{ba}}{{\sout{ih}}^2{\omega }_0}{e}^{i{\omega }_{0}t/2}\sin \left({\omega }_{0}t/2\right)=1-\frac{2H_{ba}}{{\sout{ih}}^2{\omega }_0}{e}^{i{\omega }_{0}t/2}\frac{1}{2i}\left(e^{{\omega }_{0}t/2}-e^{{}^{{\omega }_{0}t/2}}\right)=-\frac{H_{ba}}{\sout{h}{\omega }_0}\left(e^{{}^{{\omega }_{0}t/2}}-1\right) \end{gathered}$$

Step-by-step solution

Concept used

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:

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

Calculating  to second order for the perturbation theory

For H′ independent of t,$\Rightarrow {\mathrm{c}}_{\mathrm{b}}^{\left(2\right)}\left(\mathrm{t}\right)\mathrm{and}{\mathrm{c}}_{\mathrm{b}}^{\left(1\right)}\left(\mathrm{t}\right)=-\frac{\mathrm{i}}{\sout{\mathrm{h}}}{\mathrm{H}}_{\mathrm{ba}}{\int }_0^{\mathrm{t}}{\mathrm{e}}^{{\mathrm{i\omega }}_0\mathrm{t}}\mathrm{dt}\text{'}$

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

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

Meanwhile

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

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

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

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

.

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 ${\omega }_0$only in second order, so it suffices to replace$\omega \to {\omega }_0$in the exact formula to get the second-order result:

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

in agreement with the result above.

Checkingis more difficult. Note that

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

Taylor expansion:

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

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

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

,asabove.

Thus the formula agree up to second order.

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