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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 $\hat{H}^{0}$:

http://www.jkrieger.de/download/quantenmechanik.pdf

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).

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  • $\begingroup$ Have you tried to derive the final expression on your own? $\endgroup$ Jan 22, 2021 at 9:47
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    $\begingroup$ 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. $\endgroup$
    – user248824
    Jan 22, 2021 at 9:55
  • $\begingroup$ 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). $\endgroup$ Jan 22, 2021 at 19:09
  • $\begingroup$ Dear Jakob, I will take a look at what you did tonight, promised, I just haven't managed yet ... :/ $\endgroup$
    – user248824
    Jan 22, 2021 at 19:32

1 Answer 1

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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 $|(\ldots)|^2$.

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    $\begingroup$ 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. $\endgroup$
    – user248824
    Jan 22, 2021 at 21:54

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