Question

Suppose you don’t assume ${\mathit{H}}_{\mathbf{a}\mathbf{a}}\mathbf{=}{\mathit{H}}_{\mathbf{b}\mathbf{b}}\mathbf{=}\mathbf{0}$

(a) Find ${\mathit{c}}_{\mathbf{a}}\left(t\right)$and ${\mathit{c}}_{\mathbf{b}}\left(t\right)$ in first-order perturbation theory, for the case

.show that , to first order in .

(b) There is a nicer way to handle this problem. Let

.

Show that

where

So the equations for are identical in structure to Equation 11.17 (with an extra

(c) Use the method in part (b) to obtain in first-order
perturbation theory, and compare your answer to (a). Comment on any discrepancies.

Short answer

$$\begin{gathered} {\left|c_a\right|}^2=\left[1-\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}\right]\left[1-\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}\right]=1{\left[1-\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}\right]}^2=1 \\ {\left|c_a\right|}^2=\left[1-\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}\right]\left[1-\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}\right]=0 \\ So{\left|c_a\right|}^2+{\left|c_b\right|}^2=1\left(tofirstorder\right) \end{gathered}$$

(b) role="math" localid="1658488366910" $$\begin{gathered} d_a=-\frac{i}{\sout{h}}{e}^{\frac{i}{n}{\int }_0^{t}H\left(t\right)dt}{c}_b{H}_{ab}{e}^{i{\omega }_{0}t} \\ {d}_b=e^{\frac{i}{n}{\int }_0^{t}H\left(t\right)dt}\left(\frac{i}{\sout{h}}{H}_{bb}\right)c_b+e^{\frac{i}{\sout{h}}{\int }_0^{t}H\left(t\right)dt}{c}_b \end{gathered}$$

(c) $d_b=e^{\frac{i}{n}{\int }_0^{t}H\left(t\right)dt}\left(\frac{i}{\sout{h}}{H}_{bb}\right)c_b+e^{\frac{i}{\sout{h}}{\int }_0^{t}H\left(t\right)dt}{c}_b$

Step-by-step solution

(a) Finding  ca(t) and cb(t)

$$\begin{gathered} \left.c_a=-\frac{i}{\sout{h}}\left[c_a{H}_{aa}+c_b{H}_{ab}{e}^{i{\omega }_{0}t}\right] \\ \Rightarrow c_b=-\frac{i}{\sout{h}}\left[c_a{H}_{aa}+c_b{H}_{ab}{e}^{i{\omega }_{0}t}\right]\right\} \end{gathered}$$

$c_b=-\frac{i}{\sout{h}}\left[c_a{H}_{aa}+c_b{H}_{ab}{e}^{-i{\left(E_b-E_b\right)}^{t/h}}\right]$

$c_b=-\frac{i}{\sout{h}}\left[c_a{H}_{aa}+c_b{H}_{ab}{e}^{-i{\left(E_b-E_b\right)}^{t/h}}\right]$

Initial conditions:$c_a\left(t\right)=1,c_b\left(0\right)=0$,

Zero order:.

First order: