Question

Solve Equation 9.13 to second order in perturbation theory, for the general case $$\begin{gathered} {\mathbf{c}}_{\mathbf{a}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{a}\mathbf{,}{\mathbf{c}}_{\mathbf{b}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{b} \\ {\mathbf{c}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{{\mathbf{H}}^{\mathbf{\text{'}}}}_{\mathbf{ab}}{\mathbf{e}}^{\mathbf{-}{\mathbf{i\omega }}_{\mathbf{0}}\mathbf{t}}\mathbf{}{\mathbf{c}}_{\mathbf{b}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathbf{c}}_{\mathbf{b}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{{\mathbf{H}}^{\mathbf{\text{'}}}}_{\mathbf{ba}}{\mathbf{e}}^{\mathbf{-}{\mathbf{i\omega }}_{\mathbf{0}}\mathbf{t}}\mathbf{}{\mathbf{c}}_{\mathbf{a}} \end{gathered}$$

(9.13).

Short answer

The second order perturbation theory for the equation is

$$\begin{gathered} c_{\alpha }^{\left(2\right)}\left(t\right)=\alpha -\frac{ib}{h}{\int }_0^t{H}_{ab}^{\cancel{c}}\left(t^{\cancel{c}}\right)e^{-i{\omega }_0{t}^{\cancel{c}}}d{t}^{\cancel{c}}-\frac{\alpha }{h^2}{\int }_0^t{H}_{ab}^{\cancel{c}}\left(t^{\cancel{c}}\right)e^{-i{\omega }_0{t}^{\cancel{c}}}\left[{\int }_0^{t^{\cancel{c}}}{H}_{ba}^{\cancel{c}}\left(t^{\cancel{c\cancel{c}}}\right)e^{i{\omega }_0{t}^{\cancel{cc}}}d{t}^{cc}\right]d{t}^{\cancel{c}} \\ {c}_b^{\left(2\right)}\left(t\right)=b-\frac{ia}{h}{\int }_0^t{H}_{ab}^{\cancel{c}}\left(t^{¢}\right)e^{i{\omega }_0{t}^{\cancel{c}}}d{t}^{\cancel{c}}-\frac{b}{h^2}{\int }_0^t{H}_{ab}^{\cancel{c}}\left(t^{¢}\right)e^{i{\omega }_0{t}^{\cancel{c}}}d{t}^{\cancel{c}}\left[{\int }_0^{t^{\cancel{c}}}{H}_{ba}^{\cancel{c}}\left(t^{\cancel{c\cancel{c}}}\right)e^{-i{\omega }_0{t}^{\cancel{cc}}}d{t}^{cc}\right]d{t}^{\cancel{c}} \end{gathered}$$

Step-by-step solution

Concept.

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 $H^{°}$and time dependent part $H^{\text{'}}$, that is:

$H=H^{°}+H^1$

This is done in the section 9.1 to get the solution.

Solving the given equation to second order perturbation theory

$$\begin{gathered} \mathrm{We}\mathrm{have},\mathrm{Zero}\mathrm{order}:c_{\alpha }^{\left(0\right)}\left(t\right)=\alpha ,c_b^{\left(0\right)}\left(t\right)=b \\ \mathrm{The}\mathrm{coefficients}\mathrm{are}\mathrm{solutions}\mathrm{of}\mathrm{the}\mathrm{couples}\mathrm{ODE}’\mathrm{s},\mathrm{that}\mathrm{is}: \\ \mathrm{The}\mathrm{equation}9.13is{\stackrel{.}{c}}_a=-\frac{i}{\hslash }{H}_{ab}^{\text{'}}{e}^{-i{\omega }_{0}t}{c}_b,\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}{\stackrel{.}{c}}_b=-\frac{i}{\hslash }{H}_{ab}^{\text{'}}{e}^{-i{\omega }_{0}t}{c}_{\alpha } \\ \mathrm{First}\mathrm{order};\left\{\begin{array}{l}{\stackrel{.}{\mathrm{c}}}_{\mathrm{a}}=-\frac{\mathrm{i}}{\hslash }{\mathrm{H}}_{\mathrm{ab}}^{\text{'}}{\mathrm{e}}^{-{\mathrm{i\omega }}_0\mathrm{t}}\mathrm{b}\Rightarrow {\mathrm{c}}_{\mathrm{\alpha }}^{\left(1\right)}\left(\mathrm{t}\right)=\mathrm{\alpha }-\frac{\mathrm{ib}}{\hslash }{\int }_0^{\mathrm{t}}{\mathrm{H}}_{\mathrm{ab}}^{\text{'}}\left(\mathrm{t}\right){\mathrm{e}}^{-{\mathrm{i\omega }}_0\mathrm{t}}{\mathrm{dt}}^{\text{'}}\\ {\stackrel{.}{\mathrm{c}}}_{\mathrm{a}}=-\frac{\mathrm{i}}{\hslash }{\mathrm{H}}_{\mathrm{ab}}^{\text{'}}{\mathrm{e}}^{-{\mathrm{i\omega }}_0\mathrm{t}}\mathrm{\alpha }\Rightarrow {\mathrm{c}}_{\mathrm{b}}^{\left(1\right)}\left(\mathrm{t}\right)=\mathrm{b}-\frac{\mathrm{i\alpha }}{\hslash }{\int }_0^{\mathrm{t}}{\mathrm{H}}_{\mathrm{ab}}^{\text{'}}\left(\mathrm{t}\right){\mathrm{e}}^{-{\mathrm{i\omega }}_0\mathrm{t}}{\mathrm{dt}}^{\text{'}}\end{array}\right. \\ \mathrm{Second}\mathrm{order}:{\stackrel{.}{\mathrm{c}}}_{\mathrm{a}}=-\frac{\mathrm{i}}{\hslash }{\mathrm{H}}_{\mathrm{ab}}^{\text{'}}{\mathrm{e}}^{-{\mathrm{i\omega }}_0\mathrm{t}}\left[\mathrm{b}-\frac{\mathrm{i\alpha }}{\hslash }{\int }_0^{\mathrm{t}}{\mathrm{H}}_{\mathrm{ab}}^{\text{'}}\left(\mathrm{t}\right){\mathrm{e}}^{-{\mathrm{i\omega }}_0\mathrm{t}}{\mathrm{dt}}^{\text{'}}\right] \\ {\mathrm{c}}_{\mathrm{\alpha }}^{\left(2\right)}\left(\mathrm{t}\right)=\mathrm{\alpha }-\frac{\mathrm{ib}}{\hslash }{\mathrm{H}}_{\mathrm{ab}}^{\text{'}}\left(\mathrm{t}\right){\mathrm{e}}^{-{\mathrm{i\omega }}_0\mathrm{t}}{\mathrm{dt}}^{\text{'}}-\frac{\mathrm{\alpha }}{{\hslash }^2}{\int }_0^{\mathrm{t}}{\mathrm{H}}_{\mathrm{ab}}^{\text{'}}\left(\mathrm{t}\right){\mathrm{e}}^{-{\mathrm{i\omega }}_0\mathrm{t}}{\mathrm{dt}}^{\text{'}}\left[{\int }_0^{\mathrm{t}}{\mathrm{H}}_{\mathrm{ba}}^{\text{'}}\left(\mathrm{t}\right){\mathrm{e}}^{-{\mathrm{i\omega }}_0\mathrm{t}}{\mathrm{dt}}^{\text{'}}\right]{\mathrm{dt}}^{\text{'}} \\ \mathrm{To}\mathrm{get}{\mathrm{c}}_{\mathrm{b}}\mathrm{just}\mathrm{switch}\mathrm{a}\leftrightarrow \mathrm{b}(\mathrm{which}\mathrm{entails}\mathrm{also}\mathrm{changing}\mathrm{the}\mathrm{sign}\mathrm{of}{\mathrm{\omega }}_0) \\ {\mathrm{c}}_{\mathrm{b}}^{\left(2\right)}\left(\mathrm{t}\right)=\mathrm{b}-\frac{\mathrm{ia}}{\hslash }{\mathrm{H}}_{\mathrm{ba}}^{\text{'}}\left({\mathrm{t}}^{\text{'}}\right){\mathrm{e}}^{{\mathrm{i\omega }}_0\mathrm{t}}{\mathrm{dt}}^{\text{'}}-\frac{\mathrm{b}}{{\hslash }^2}{\int }_0^{\mathrm{t}}{\mathrm{H}}_{\mathrm{ab}}^{\text{'}}\left({\mathrm{t}}^{\text{'}}\right){\mathrm{e}}^{{\mathrm{i\omega }}_0\mathrm{t}}{\mathrm{dt}}^{\text{'}}\left[{\int }_0^{\mathrm{t}}{\mathrm{H}}_{\mathrm{ba}}^{\text{'}}\left({\mathrm{t}}^{\text{'}}\right){\mathrm{e}}^{-{\mathrm{i\omega }}_0\mathrm{t}}{\mathrm{dt}}^{\text{'}}\right]{\mathrm{dt}}^{\text{'}} \end{gathered}$$