Question

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

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

${\mathit{c}}_{\mathbf{e}}\left(0\right)\mathbf{=}\mathbf{1}\mathbf{,}{\mathit{c}}_{\mathbf{b}}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{0}$.show that ${\left|c_a^{\left(1\right)}\left(t\right)\right|}^{\mathbf{2}}\mathbf{+}{\left|c_b^{\left(1\right)}\left(t\right)\right|}^{\mathbf{2}}\mathbf{=}\mathbf{1}$, to first order in$\hat{\mathit{H}}\mathbf{\text{'}}$ .

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

${\mathit{d}}_{\mathbf{a}}{}^{\mathbf{0}}\mathit{e}^{\frac{\mathbf{i}}{\mathbf{h}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}{\mathbf{H}}_{\mathbf{a}\mathbf{a}}^{\cancel{\mathbf{c}}}\mathbf{\left(}{\mathbf{t}}^{\cancel{\mathbf{c}}}\mathbf{\right)}\mathbf{d}{\mathbf{t}}^{\cancel{\mathbf{c}}}}{\mathit{c}}_{\mathbf{a}}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{d}}_{\mathbf{b}}{}^{\mathbf{0}}\mathit{e}^{\frac{\mathbf{i}}{\sqrt{\mathbf{2}}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{t}}{\mathbf{H}}_{\mathbf{b}\mathbf{b}}^{\cancel{\mathbf{c}}}\mathbf{\left(}{\mathbf{t}}^{\cancel{\mathbf{c}}}\mathbf{\right)}\mathbf{d}{\mathbf{t}}^{\cancel{\mathbf{c}}}}{\mathit{c}}_{\mathbf{b}}$.

Show that

role="math" localid="1658561855290" ${\stackrel{\mathbf{-}}{\mathbf{d}}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{\mathit{e}}^{\mathbf{i}\mathbf{\varphi }}\mathit{H}{\mathbf{\text{'}}}_{\mathbf{a}\mathbf{b}}{\mathit{e}}^{\mathbf{-}\mathbf{i}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{t}}{\mathit{d}}_{\mathbf{b}}\mathbf{;}\mathbf{}\mathbf{}\mathbf{}\stackrel{\mathbf{-}}{{\mathbf{d}}_{\mathbf{b}}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{\mathit{e}}^{\mathbf{i}\mathbf{\varphi }}\mathit{H}{\mathbf{\text{'}}}_{\mathbf{b}\mathbf{a}}{\mathit{e}}^{\mathbf{-}\mathbf{i}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{t}}{\mathit{d}}_{\mathbf{a}}$

where

$\mathbf{\varphi }\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{\equiv }\frac{\mathbf{1}}{\sout{\mathbf{h}}}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{l}}\left[H{\text{'}}_{aa}(t\text{'})-H{\text{'}}_{bb}(t\text{'})\right]\mathit{d}\mathit{t}\mathbf{\text{'}}$.

So the equations for${\mathit{d}}_{\mathbf{a}}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{d}}_{\mathbf{b}}$are identical in structure to Equation 11.17 (with an extra role="math" localid="1658562498216" ${\mathit{e}}^{\mathit{i}\mathit{\varphi }}\mathbf{}\mathit{t}\mathit{a}\mathit{c}\mathit{k}\mathit{e}\mathit{d}\mathbf{}\mathit{o}\mathit{n}\mathit{t}\mathit{o}\mathbf{}\hat{\mathit{H}}\mathbf{\text{'}}$)

$\stackrel{\mathbf{-}}{{\mathbf{c}}_{\mathbf{a}}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}\mathit{H}{\mathbf{\text{'}}}_{\mathbf{a}\mathbf{b}}{\mathit{e}}^{\mathbf{-}\mathbf{i}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{t}}{\mathit{c}}_{\mathbf{b}}\mathbf{,}\mathbf{}\mathbf{}\stackrel{\mathbf{-}}{{\mathbf{c}}_{\mathbf{b}}}\mathbf{=}\mathbf{-}\mathbf{}\frac{\mathbf{i}}{\sout{\mathbf{h}}}\mathit{H}{\mathbf{\text{'}}}_{\mathbf{b}\mathbf{a}}{\mathit{e}}^{\mathbf{i}{\mathbf{\omega }}_{\mathbf{0}}\mathbf{t}}{\mathit{c}}_{\mathbf{a}}$

(c) Use the method in part (b) to obtainrole="math" localid="1658562468835" ${\mathit{c}}_{\mathbf{a}}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{}\mathit{a}\mathit{n}\mathit{d}{\mathit{c}}_{\mathbf{}\mathbf{b}}\mathbf{\left(}\mathit{t}\mathbf{\right)}$in first-order
perturbation theory, and compare your answer to (a). Comment on any discrepancies.

Short answer

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

$$\begin{gathered} da=-\frac{i}{\sout{h}}{e}^{\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}}{c}_{b}H{\text{'}}_{ab}{e}^{i{\omega }_{0}t}. \\ {d}_b=-\frac{i}{\sout{h}}{e}^{\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}}\left(\frac{i}{\sout{h}}H{\text{'}}_{bb}\right)c_b+e^{\frac{i}{\sout{h}}{\int }_0^t{H}_{bb}\left(t\text{'}\right)dt\text{'}}{c}_b \end{gathered}$$

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

Step-by-step solution

(a) Finding  ca(t) and cb(t)

$$\begin{gathered} \left.c_a=-\frac{i}{\sout{h}}\left[c_{a}H{\text{'}}_{aa}+c_{a}H{\text{'}}_{ab}{e}^{i{\omega }_{0}t}\right] \\ \Rightarrow c_a=-\frac{i}{\sout{h}}\left[c_{b}H{\text{'}}_{bb}+c_{a}H{\text{'}}_{ba}{e}^{i{\omega }_{0}t}\right]\right\} \end{gathered}$$

role="math" localid="1658555553364" $c_a=-\frac{i}{\sout{h}}\left[c_{a}H{\text{'}}_{aa}+c_{b}H{\text{'}}_{ab}{e}^{-i\left(E_b-E_a\right)t/\sout{h}}\right]$

$c_a=-\frac{i}{\sout{h}}H{\text{'}}_{ab}{e}^{i{\omega }_{0}t}{c}_b,c_b=-\frac{i}{\sout{h}}H{\text{'}}_{ba}{e}^{i{\omega }_{0}t}{c}_a$

role="math" localid="1658555600978" $c_b=-\frac{i}{\sout{h}}\left[c_{b}H{\text{'}}_{bb}+c_{a}H{\text{'}}_{ba}{e}^{-i\left(E_b-E_a\right)t/\sout{h}}\right]$

Initial conditions: role="math" localid="1658555900086" $c_a\left(0\right)=1,c_b\left(0\right)=0,$

Zero order: $c_a\left(0\right)=1,c_b\left(0\right)=0,$

First order:$\left\{\begin{array}{l}{c}_a=-\frac{i}{\sout{h}}H{\text{'}}_{aa}\Rightarrow c_a\left(t\right)=1-\frac{i}{\sout{h}}{\int }_0^{t}H{\text{'}}_{aa}\left(t\text{'}\right)dt\text{'}\\ c_b=-\frac{i}{\sout{h}}H{\text{'}}_{ba}{e}^{i{\omega }_{0}t}\Rightarrow c_b\left(t\right)=1-\frac{i}{\sout{h}}{\int }_0^{t}H{\text{'}}_{ba}\left(t\text{'}\right)e^{i{\omega }_{0}t}dt\text{'}\end{array}\right.$

$$\begin{gathered} {\left|c_a\right|}^2=\left[1-\frac{i}{\sout{\sout{h}}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\right]\left[1-\frac{i}{\sout{\sout{h}}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\right]=1+{\left[\frac{i}{\sout{\sout{h}}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\right]}^2=1\left(tofirstorderinH\text{'}\right) \\ {\left|c_b\right|}^2=\left[-\frac{i}{\sout{\sout{h}}}{\int }_0^t{H}_{ba}\left(t\text{'}\right)e^{i{\omega }_{0}t}dt\text{'}\right]\left[\frac{i}{\sout{\sout{h}}}{\int }_0^t{H}_{ab}\left(t\text{'}\right)e^{i{\omega }_{0}t}dt\text{'}\right]=0\left(tofirstorderinH\text{'}\right) \\ So{\left|c_a\right|}^2+{\left|c_b\right|}^2=1\mathbf{(}\mathbf{to}\mathbf{}\mathbf{first}\mathbf{}\mathbf{order}\mathbf{)} \end{gathered}$$

(b) Showing  da=-iheiϕH'abeiω0tdb;da=-iheiϕH'baeiω0tda

$d_b=-\frac{i}{\sout{h}}{e}^{\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}}\left(\frac{i}{\sout{h}}H{\text{'}}_{bb}\right)c_b+e^{\frac{i}{\sout{h}}{\int }_0^t{H}_{bb}\left(t\text{'}\right)dt\text{'}}{c}_{b}But{c}_a=-\frac{i}{\sout{h}}\left[c_{a}H{\text{'}}_{aa}+c_{b}H{\text{'}}_{ab}{e}^{-i{\omega }_{0}t}\right]$

Two terms cancel, leaving

$$\begin{gathered} d_b=-\frac{i}{\sout{h}}{e}^{\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}}{c}_b{H}_{ab}{e}^{-i{\omega }_{0}t}.But{c}_b=e^{-\frac{i}{\sout{h}}{\int }_0^{t}H{\text{'}}_{bb}\left(t\text{'}\right)dt\text{'}}{d}_b \\ =-\frac{i}{\sout{h}}{e}^{\frac{i}{\sout{h}}{\int }_0^t\left[H{\text{'}}_{aa}\left(t\text{'}\right)-H{\text{'}}_{bb}\left(t\right)\right]dt}{H}_{ab}{e}^{i{\omega }_{0}t}{d}_b,or{d}_a=-\frac{i}{\sout{h}}{e}^{i\varphi }H{\text{'}}_{ab}{e}^{i{\omega }_{0}t}{d}_b \end{gathered}$$

Similarly,$$\begin{gathered} d_b=-\frac{i}{\sout{h}}{e}^{\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}}\left(\frac{i}{\sout{h}}H{\text{'}}_{bb}\right)c_b=e^{-\frac{i}{\sout{h}}{\int }_0^{t}H{\text{'}}_{bb}\left(t\text{'}\right)dt\text{'}}{d}_b.But{c}_b=-\frac{i}{\sout{h}}\left[c_b{H}_{bb}+c_{a}H{\text{'}}_{ba}{e}^{i{\omega }_{0}t}\right] \\ =-\frac{i}{\sout{h}}{e}^{\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}}{c}_{a}H{\text{'}}_{ba}{e}^{i{\omega }_{0}t}.But{c}_a=e^{-\frac{i}{\sout{h}}{\int }_0^t{H}_{aa}\left(t\text{'}\right)dt\text{'}}{d}_a \\ =-\frac{i}{\sout{h}}{e}^{\frac{i}{\sout{h}}{\int }_0^t\left[H{\text{'}}_{bb}\left(t\text{'}\right)-H{\text{'}}_{aa}(t\text{'})\right]dt\text{'}}H{\text{'}}_{ba}{e}^{i{\omega }_{0}t}{d}_a=-\frac{i}{\sout{h}}{e}^{i\varphi }H{\text{'}}_{ba}{e}^{i{\omega }_{0}t}{d}_a \end{gathered}$$

(c) Using the method in part (b) to obtain  ca(t) and cb(t)

Initial conditions:$c_a\left(0\right)=1\Rightarrow d_a\left(0\right)=1;c_b\left(0\right)=0\Rightarrow d_b\left(0\right)=0$.

Zero order:$d_a\left(t\right)=1,d_b\left(t\right)=0$.

First order:${\dot{d}}_a=0\Rightarrow d_a\left(t\right)=1\Rightarrow c_a\left(t\right)=e^{-\frac{i}{\sout{h}}{\int }_0^{t}H{\text{'}}_{aa}(t\text{'})dt\text{'}}$

${\dot{d}}_b=-\frac{i}{\sout{h}}{e}^{-i\varphi }H{\text{'}}_{ba}{e}^{i{\omega }_{0}t}\Rightarrow d_b=-\frac{i}{\sout{h}}{\int }_0^t{e}^{-i\varphi (t\text{'})}H{\text{'}}_{ba}(t\text{'})e^{i{\omega }_{0}t\text{'}}dt\text{'}$

These don’t look much like the results in (a), but remember, we’re only working to first order in $H\text{'}Hso{c}_a\left(t\right)\approx 1-\frac{i}{\sout{h}}{\int }_0^{t}H{\text{'}}_{aa}(t\text{'})dt\text{'}(\mathrm{to}\mathrm{this}\mathrm{order}),\mathrm{while}\mathrm{for}{\mathrm{c}}_{\mathrm{b}},\mathrm{the}\mathrm{factor}\mathrm{H}{\text{'}}_{\mathrm{ba}}$in the integral means it is already first order and hence both the exponential factor in front and $e-i\varphi$should be replaced by 1. Then $c_b\left(t\right)\approx 1-{\int }_0^t\frac{i}{h}H{\text{'}}_{aa}\left(t\text{'}\right)e^{i{\omega }_{0}t\text{'}}dt\text{'}$, and we recover the results in (a).