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


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?

  • $\begingroup$ It may be useful if you comment which ways have you tried. $\endgroup$
    – J L
    Jul 30, 2013 at 7:59
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    $\begingroup$ 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


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!

  • $\begingroup$ 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
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    $\begingroup$ @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
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    $\begingroup$ 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
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    $\begingroup$ @NowIGetToLearnWhatAHeadIs You're right, I made a mistake! Corrected! $\endgroup$
    – neutrino
    Jul 30, 2013 at 18:36

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


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