Question

Solve Equation 9.13 for the case of a time-independent perturbation, assumingthatandcheck that

. Comment: Ostensibly, this system oscillates between “” Doesn’t this contradict my general assertion that no transitions occur for time-independent perturbations? No, but the reason is rather subtle: In this are not, and never were, Eigen states of the Hamiltonian—a measurement of the energy never yields. In time-dependent perturbation theory we typically contemplate turning on the perturbation for a while, and then turning it off again, in order to examine the system. At the beginning, and at the end,are Eigen states of the exact Hamiltonian, and only in this context does it make sense to say that the system underwent a transition from one to the other. For the present problem, then, assume that the perturbation was turned on at time t = 0, and off again at time T —this doesn’t affect the calculations, but it allows for a more sensible interpretation of the result.

${\mathbf{c}}_{\mathbf{a}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{\mathbf{H}}_{\mathbf{ab}}{\mathbf{e}}^{\mathbf{igt}}{\mathbf{c}}_{\mathbf{b}\mathbf{,}}{\mathbf{c}}_{\mathbf{b}}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\sout{\mathbf{h}}}{\mathbf{H}}_{\mathbf{ba}}{\mathbf{e}}^{\mathbf{igt}}{\mathbf{c}}_{\mathbf{a}}$ …(9.13).

Short answer

$c_a\left(o\right)-c_a\left(-o\right)=-\frac{i}{\sout{h}}{H}_{ab}{\int }_0^o{e}^{igt}{c}_b\left(t\right)dt$

But

$\left|c_b\left(t\right)\right|\le 1.$

Step-by-step solution

Given data

Given that

$c_a=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-igt}{C}_b$

And

$c_b=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-igt}{C}_b$

Solving the case of a time independent perturbation

$$\begin{gathered} c_a=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-igt}{C}_b \\ And \\ {c}_b=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-igt}{C}_b \end{gathered}$$

Differentiating with respect to t :

$$\begin{gathered} C_b=-\frac{i}{\sout{h}}{H}_{ab}\left[i{\omega }_0{e}^{i{\omega }_{0}t}{C}_a\right]-\frac{i}{\sout{h}}{H}_{ab}{e}^{i{\omega }_{0}t}\left[-\frac{i}{\sout{h}}{H}_{ab}{e}^{i{\omega }_{0}t}{c}_b\right] \\ or \\ {C}_b=i{\omega }_0{c}_b-\frac{1}{{\sout{h}}^2}{\left|H_{ab}\right|}^{2}cb.Let{\alpha }^2=\frac{1}{\sout{h^2}}{\left|H_{ab}\right|}^2.Then{c}_b-i{\omega }_0{C}_b+{\alpha }^2{c}_b=0 \end{gathered}$$

or

This is a linear differential equation with constant coefficients, so it can be solved by a function of the form :

$$\begin{gathered} {\lambda }^2-i{\omega }_0\lambda +{\alpha }^2=0;\lambda =\frac{1}{2}\left[i{\omega }_0\pm -\sqrt{-{\omega }_0^2}-4{\alpha }^2\right] \\ =\frac{i}{2}\left({\omega }_0\pm \omega \right),where\omega =\sqrt{-{\omega }_0^2}-4{\alpha }^2 \end{gathered}$$

The general solution is therefore

$$\begin{gathered} c_b\left(t\right)=A{e}^{i\left({\omega }_0+\omega \right)t/2}+B{e}^{i\left({\omega }_0+\omega \right)t/2} \\ =e^{i{\omega }_{0}t/2}\left(A{e}^{i{\omega }_{0}t/2}+B{e}^{{\omega }_{0}t/2}\right), \\ Or \end{gathered}$$

$$\begin{gathered} {\mathrm{c}}_{\mathrm{b}}\left(\mathrm{t}\right)={\mathrm{e}}^{{\mathrm{\omega }}_0\mathrm{t}/2}\left[\mathrm{Ccos}\left({\mathrm{\omega }}_0\mathrm{t}/2\right)+\mathrm{Dsin}\left({\mathrm{\omega }}_0\mathrm{t}/2\right)\right].\mathrm{But}{\mathrm{c}}_{\mathrm{b}}\left(0\right)=0,\mathrm{so}\mathrm{C}=0, \\ \mathrm{and}\mathrm{hence} \end{gathered}$$

$c_b=\left(t\right)=D{e}^{{\omega }_{0}t/2}\sin \left({\omega }_{0}t/2\right)$

Then

$$\begin{gathered} c_b=D\left[\frac{i{\omega }_0}{2}{e}^{{\omega }_{0}t/2}\sin \left({\omega }_{0}t/2\right)+\frac{\omega }{2}{e}^{{\omega }_{0}t/2}\cos \left({\omega }_{0}t/2\right)\right] \\ =\frac{\omega }{2}D{e}^{{\omega }_{0}t/2}\left[\cos \left({\omega }_{0}t/2\right)+i\frac{{\omega }_0}{\omega }\sin \left({\omega }_{0}t/2\right)\right] \\ =-\frac{i}{\sout{h}}{H}_{ba}{e}^{{\omega }_{0}t/2}{c}_a \end{gathered}$$

$c_a=\frac{i\sout{h}}{H_{ba}}\frac{\omega }{2}{e}^{-{\omega }_{0}t/2}D\left[\cos \left({\omega }_{0}t/2\right)+i\frac{{\omega }_0}{\omega }\sin \left({\omega }_{0}t/2\right)\right].But{c}_a\left(0\right)=1so\frac{i\sout{h}}{H_{ba}}\frac{\omega }{2}$

Conclusion:

$c_a\left(t\right)=e^{-i{\omega }_{0}t/2}\left[\cos \left(i{\omega }_{0}t/2\right)+i\frac{{\omega }_0}{\omega }\sin \left(i{\omega }_{0}t/2\right)\right],c_b\left(t\right)=\frac{2H_{ab}}{i\sout{h}\omega }{e}^{i{\omega }_{0}t/2}\sin \left(i{\omega }_{0}t/2\right)$

Where

$$\begin{gathered} \omega =\sqrt{{\omega }_0^2+4\frac{{\left|H_{ab}\right|}^2}{\sout{h^2}}} \\ {\left|c_a\right|}^2+{\left|c_b\right|}^2={\cos }^2\left(\omega t/2\right)+\frac{{\omega }_0^2}{{\omega }^2}{\sin }^2\left(\omega t/2\right)+\frac{4{\left|H_{ab}\right|}^2}{{\sout{h}}^2{\omega }^2}{\sin }^2\left(\omega t/2\right) \\ ={\cos }^2\left(\omega t/2\right)+\frac{1}{{\omega }^2}\left({\omega }_0^2+4\frac{{\left|H_{ab}\right|}^2}{{\sout{h}}^2}\right){\sin }^2\left(\omega t/2\right) \end{gathered}$$

${\cos }^2\left(\omega t/2\right)+{\sin }^2\left(\omega t/2\right)=1$

[In light of the Comment you might question the initial conditions. If the perturbation includes a factor θ(t) , are we sure this doesn’t alter and That is, are we sure and are continuous at a step function potential? The answer is “yes”, for if we integrate Eq. 9.13 from to ,

$$\begin{gathered} c_a=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-i{\omega }_{0}t}{c}_{b1,}{c}_a=-\frac{i}{\sout{h}}{H}_{ab}{e}^{-i{\omega }_{0}t}{c}_a \\ {c}_a\left(o\right)-c_a\left(-o\right)=-\frac{i}{\sout{h}}{H}_{ab}{\int }_0^o{e}^{-i{\omega }_{0}t}{c}_b\left(t\right)dt \end{gathered}$$

But$\left|c_b\left(t\right)\right|\le 1,$

So the intergral goes to zero as $o\to 0,$and hence$c_b\left(-o\right)=c_a\left(o\right).c_b$

The same goes for of course