3
$\begingroup$

I was going thru Ch7 of Foot and trying to fill in the gaps. However I got stuck on (7.14). So Foot was working with a two level system with a small perturbation in the Hamiltonian resulted from an oscillating electric field $$H_I(t)=e\mathbf{r}\cdot \mathbf{E_0}cos(\omega t)$$.

Now we assume that the new eigenstate for the perturbed Hamiltonian is $$|\Psi(\mathbf{r},t)\rangle= c_1 |1\rangle e^{-i\omega_1 t}+c_2 |2\rangle e^{-i\omega_2 t}$$

And we work out relationships for $c_1$ and $c_2$ and try to solve them. It's not hard to get (7.9) and (7.10) in his book.

$$i\dot{c_1}=\Omega cos(\omega t)e^{-i\omega_0t}c_2$$ $$i\dot{c_2}=\Omega^* cos(\omega t)e^{i\omega_0t}c_1$$

where $\Omega$ is the Rabi frequency and $\omega_0=\omega_2-\omega_1$.

Now Foot says "when all the population starts in the lower level, $c_1(0)=1$ and $c_2(0)=0$. Integration of eqns (7.9) and (7.10) leads to

$$c_1(t)=1$$ $$c_2(t)=\frac{\Omega^*}{2}\{\frac{1-e^{i(\omega_0+\omega)t}}{\omega_0+\omega}+\frac{1-e^{i(\omega_0-\omega)t}}{\omega_0-\omega}\}"$$

I was wonder how we could integrate them and get those two eqns at the bottom? I tried two different ways but I still can't quite get what I want. Also I tried to simplify (7.9) and (7.10) a bit

$$i\dot{c_1}=c_2(e^{i(\omega-\omega_0)t}+e^{-i(\omega+\omega_0)t})\frac{\Omega}{2}$$

But not sure if this helps with the integration.

Any thoughts?

My apologies for previous ambiguity. In particular, I apologise to @neutrino for asking the wrong question. The question is: We all know that it would make our lives much easier if we apply rotating wave approximation before solving the ODEs. However Foot solved the ODEs first to get this result $$c_1(t)=1$$ $$c_2(t)=\frac{\Omega^*}{2}\{\frac{1-e^{i(\omega_0+\omega)t}}{\omega_0+\omega}+\frac{1-e^{i(\omega_0-\omega)t}}{\omega_0-\omega}\}"$$ He then applied the rotating wave approximation. I was wondering if anyone could derive the above equation in the way that Foot did?

$\endgroup$
5
  • $\begingroup$ It may be useful if you comment which ways have you tried. $\endgroup$
    – J L
    Jul 30, 2013 at 7:59
  • 1
    $\begingroup$ I think your las equation should be: $$i\dot{c_1}=c_2(e^{i(\omega-\omega_0)t}+e^{-i(\omega+\omega_0)t})\frac{\Omega}{2}$$ $\endgroup$
    – neutrino
    Jul 30, 2013 at 10:30
  • $\begingroup$ @Nivalth Thanks for the comment. I tried the way neutrino did it. I also tried to do it by brute force. But the former does not give me the answer in the book. The later was a bit too complicated to work out. $\endgroup$
    – Evariste
    Jul 30, 2013 at 14:49
  • $\begingroup$ @Evariste Are you sure that the book states that $c_1(t)=1$? In that case, (7.9) implies that $c_2(t)=0$, or $\Omega = 0$... $\endgroup$
    – neutrino
    Jul 31, 2013 at 6:18
  • $\begingroup$ @Evariste I've confirmed that (7.9) is correct, I've been able to get a copy of Foot's book. But I don't know how to arrive to that solution, with $c_1(t)=1$ $\endgroup$
    – neutrino
    Jul 31, 2013 at 6:29

2 Answers 2

2
$\begingroup$

The Schrödinger equation for a periodic electric potential is: $$-\frac{\hbar}{i}\dot\psi=[H+E_{0}cos(\omega t)]\psi$$

Now, as you said, let's assume that the new eigenstate for the perturbed Hamiltonian is $$|\Psi(\mathbf{r},t)\rangle= c_1 |1\rangle e^{-i\omega_1 t}+c_2 |2\rangle e^{-i\omega_2 t} $$

Here $|1\rangle$ and $|2\rangle$ are solutions for the stationary state: $$H|1\rangle = \hbar \omega _{1}$$ $$H|2\rangle = \hbar \omega _{2}$$

If you substitute $|\Psi(\mathbf{r},t)\rangle$ in the Schrödinger equation, and then, perform a scalar product with $\langle1|$ and $\langle2|$, you arrive at two differential equations, for $c_{1}(t)$ and $c_{2}(t)$:

$$i \hbar \dot{c_1}=cos(\omega t) (\langle1|E_{0}|1\rangle c_1 + \langle1|E_{0}|2\rangle c_2e^{-i(\omega_2-\omega_1)t})$$ $$i \hbar \dot{c_2}=cos(\omega t) (\langle2|E_{0}|1\rangle c_1e^{-i(\omega_1-\omega_2)t} + \langle2|E_{0}|2\rangle c_2)$$

Now, let's call $\omega_0 = \omega_2 - \omega _1$, and $\Delta \omega = \omega - \omega_0$, and assume that $\Delta \omega << \omega_0$, then the previous two equations can be written as:

$$i\dot{c_1}=\frac{1}{2\hbar} (\langle1|E_{0}|1\rangle c_1 (e^{i\omega t}+e^{-i\omega t})+ \langle1|E_{0}|2\rangle c_2(e^{i\Delta\omega t}+e^{-i(\omega+\omega_0) t}))$$ $$i\dot{c_2}=\frac{1}{2\hbar} (\langle2|E_{0}|1\rangle c_1 (e^{-i\Delta\omega t}+e^{i(\omega+\omega_0) t})+ \langle2|E_{0}|2\rangle c_2(e^{i\omega t}+e^{-i\omega t}))$$

Now, we can apply Rotating Wave Approximation to discard the fast rotating terms with frequency $\omega$: $$i\dot{c_1}=\frac{1}{2\hbar} (\langle1|E_{0}|2\rangle c_2(e^{i\Delta\omega t}+e^{-i(\omega+\omega_0) t}))$$ $$i\dot{c_2}=\frac{1}{2\hbar} (\langle2|E_{0}|1\rangle c_1 (e^{-i\Delta\omega t}+e^{i(\omega+\omega_0) t}))$$

Let's do the subsitution $\Omega = \frac {\langle1|E_{0}|2\rangle}{\hbar}$ to obtain:

$$i\dot{c_1}=c_2(e^{i(\omega-\omega_0)t}+e^{-i(\omega+\omega_0)t})\frac{\Omega}{2}$$ $$i\dot{c_2}=c_1(e^{-i(\omega-\omega_0)t}+e^{i(\omega+\omega_0)t})\frac{\Omega^{\star}}{2}$$

Then, apply again the Rotating Wave Approximation to discard the fast rotating terms $\omega + \omega_0$, which are of order $2\omega$:

$$i\dot{c_1}=c_2e^{i(\omega-\omega_0)t}\frac{\Omega}{2}$$ $$i\dot{c_2}=c_1e^{-i(\omega-\omega_0)t}\frac{\Omega^{\star}}{2}$$

If you derive the first of the last two equations with respect to time: $$i\ddot{c_1}=\dot{c_2}e^{i(\omega-\omega_0)t}\frac{\Omega}{2}+c_2i(\omega-\omega_0)e^{i(\omega-\omega_0)t}\frac{\Omega}{2}$$ You can eliminate $c_{2}$ from the second equation, obtaining a second order ODE. Then, try as an ansatz: $$c_{1}=Ae^{\lambda t}+Be^{-\lambda t}$$

And apply the initial conditions.

Hope that helps!

$\endgroup$
11
  • $\begingroup$ Thanks a lot for the help! It was a good answer. And personally I would do it this way(i.e. applying the rotating-wave approximation first). However I would really like to know how Foot got to this equation $$i\dot{c_1}=c_2(d^{i(\omega-\omega_0)t}+e^{-i(\omega+\omega_0)t})\frac{\Omega}{2}$$. He claims that he integrated the system of ODEs first before applying the rotating-wave approximation, which sounds a bit sus. $\endgroup$
    – Evariste
    Jul 30, 2013 at 14:50
  • $\begingroup$ By the way, I voted your answer up. But forgive me for not accepting your answer for now(simply because I was looking for something else, i.e. the way Foot did this question). $\endgroup$
    – Evariste
    Jul 30, 2013 at 14:52
  • 1
    $\begingroup$ @Evariste I've edited my answer to give you a more detailed explanation, including the derivation for the "problematic" Foot's equation. $\endgroup$
    – neutrino
    Jul 30, 2013 at 17:24
  • 1
    $\begingroup$ I don't understand your notation, particularly the difference between $\omega$ and $\omega_0$. If $\omega$ is the difference in frequency between the two levels, then $\omega$ should be independent from the electric field frequency, which you called $\omega_0$. Yet, you seem to be using $\omega$ to refer to the electric field frequency also. This choice of notation makes it difficult to see how you went from the first set of equations for the $\dot{c}$'s to the second. Am I not understanding something? $\endgroup$ Jul 30, 2013 at 18:34
  • 1
    $\begingroup$ @NowIGetToLearnWhatAHeadIs You're right, I made a mistake! Corrected! $\endgroup$
    – neutrino
    Jul 30, 2013 at 18:36
0
$\begingroup$

He used time-dependent perturbation theory to get the result without RWA. You can see Griffiths, introduction to quantum mechanics.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.