# Understanding the Derivation of Rabi-Oscillations

In his scriptum, Jan Krieger proves on page 56 the probability of finding a system in a state $$\vert2\rangle$$ if it was at time $$t = 0$$ in the state $$\vert1\rangle$$, where both $$\vert1\rangle$$ and $$\vert2\rangle$$ denote the eigenstates of the unperturbed Hamiltonian $$\hat{H}^{0}$$:

Now, I do not quite follow the following two transitions: $$\langle2|\psi(t)\rangle=e^{i\varphi/2}\left(\cos\frac\theta2\cdot e^{-iE_+t/\hbar}\langle2|+\rangle-\sin\frac\theta2\cdot e^{-iE_-t/\hbar}\langle2|-\rangle\right)\tag{1}$$ $$= e^{i\varphi}\cdot\cos\frac\theta2\cdot \sin\frac\theta2 \cdot (e^{-iE_+t/\hbar}-e^{-iE_-t/\hbar})\tag{2}$$ $$\mathbb{P}_{12}(t)=|\langle2|\psi(t)\rangle|^2=\frac14\sin^2\theta\cdot(e^{-iE_+t/\hbar}-e^{-iE_-t/\hbar})^2\tag{3}$$ $$= \sin^2\theta\cdot\sin^2\left(\frac{E_+-E_-}{2\hbar}\cdot t\right)\tag{4}$$

(i). From equation $$(2)$$ to $$(3)$$, if we have $$\langle2\vert\psi(t)\rangle$$ and we now want to calculate $$\left|\langle2\vert\psi(t)\rangle\right|^2$$, then I get: $$\left|\langle2\vert\psi(t)\rangle\right|^2 \propto \left| \exp\left( -\frac{iE_{+}t}{\hbar}\right) - \exp\left( -\frac{iE_{-}t}{\hbar} \right) \right|^2 = \left[\exp\left( -\frac{iE_{+}t}{\hbar}\right) - \exp\left( -\frac{iE_{-}t}{\hbar} \right) \right] \cdot \left[ \exp\left( \frac{iE_{+}t}{\hbar}\right) - \exp\left( \frac{iE_{-}t}{\hbar} \right) \right],$$ which is not $$\left( e^{-iE_+t/\hbar} - e^{-iE_{-}t/\hbar} \right)^2$$.

(ii). I also do not yet obtain the equation $$(4)$$ from $$(3)$$, but maybe this gets clearer once I understand (i).

• Have you tried to derive the final expression on your own? Jan 22, 2021 at 9:47
• Hi Jakob, so you mean whether I tried to do this with my own calculation that I wrote down under (i)? No, let me give it a try then.
– user248824
Jan 22, 2021 at 9:55
• If you need further help, let me know. But I think it boils down to the fact that the author has used $(\ldots)^2$ instead of $|\ldots|^2$, as you did (which is correct). Jan 22, 2021 at 19:09
• Dear Jakob, I will take a look at what you did tonight, promised, I just haven't managed yet ... :/
– user248824
Jan 22, 2021 at 19:32

I will give it a try, but of course there could be some errors. Please double check everything. First, we will write $$x\equiv \frac{\theta}{2}$$, $$A\equiv E_+\,t$$ and $$B\equiv E_-\,t$$. We will further set $$\hbar=1$$. Additionally to the equations you included in the post, we need that $$\langle 2|+\rangle = c_1\, \sin(x)$$ and $$\langle 2|-\rangle = c_1\, \cos(x)$$, with $$c_1 \in \mathbb{C}$$ and $$|c_1|^2=1$$. We then find $$\langle 2|\Psi(t)\rangle = c\,\sin(x)\,\cos(x)\, \left(e^{-iA}-e^{-iB}\right) \quad,$$ with $$c\in\mathbb{C}$$ and $$|c|^2=1$$.
Hence, we obtain $$|\langle 2|\Psi(t)\rangle|^2 = \mathbb{P}_{12}(t) = \frac{\sin^2(2x)}{4} \,\left|\left(e^{-iA}-e^{-iB}\right)\right|^2 \quad .$$
We evaluate $$\left|\left(e^{-iA}-e^{-iB}\right)\right|^2 = 2-2\,\cos(A-B)$$. Moreover, it holds that $$2-2\,\cos(A-B) = 4\,\sin^2\left(\frac{A-B}{2}\right)\quad .$$ Finally, if we re-substitute our quantities, then we obtain the desired expression: $$\mathbb{P}_{12} (t) =\sin^2(\theta)\, \sin^2\left(\frac{(E_+-E_-)t}{2}\right) \quad .$$
Edit: I think that it is indeed wrong to write $$(\ldots)^2$$, because we deal with complex numbers. It should read $$|(\ldots)|^2$$.
• I like this answer, as it focuses itself on the essentials and eliminates the unnecessary burdens of notation by defining $x$, $A$ and $B$. I think it would have been great to leave the $\exp(i\varphi/2)$, as it took me some time to understand that we absorbed this into $c_1$, but the proof is in my opinion correct.