TITLE: Can there be an interference term in a two-state quantum system?
QUESTION [0 upvotes]: Say I have a two-state system $|\psi \rangle=c_1|0\rangle+c_2|1\rangle$. If the states $|0\rangle$ and $|1\rangle$ are orthogonal, then the probability is:
$$
\begin{align}
\langle \psi |\psi \rangle&=(\langle 0|c_1^* + \langle 1|c_2^* )(c_1|0\rangle+c_2|1\rangle)\\
&=c_1^*c_1\langle 0|0\rangle+c_2c_2^*\langle 1|1\rangle + c_1^*c_2\langle 0|1\rangle+c_1c_2^*\langle 1|0\rangle\\
&=c_1^*c_1+c_2c_2^*=1
\end{align}
$$
Unlike the double split experiment, there are no interference terms:
$$
P=|\psi_1|^2+|\psi_2|^2+2|\psi_1||\psi_2|\cos(\theta_1-\theta_2)
$$
Can a two-state system produce a similar $\cos$-based interference term?

REPLY [0 votes]: Take
\begin{align}
\vert\psi_1\rangle&=e^{-\frac{i \gamma }{2}} \cos \left(\frac{\beta }{2}\right)
\vert 0\rangle+ e^{-\frac{i \gamma }{2}} \sin \left(\frac{\beta }{2}\right)\vert 1\rangle\, ,\\
\vert\psi_2\rangle&=
e^{\frac{i \gamma }{2}} \sin \left(\frac{\beta }{2}\right)\vert 0\rangle + e^{\frac{i \gamma }{2}} \cos \left(\frac{\beta }{2}\right)\vert 1\rangle
\end{align}
for which $\langle \psi_1\vert \psi_2\rangle=e^{-i\gamma}\sin\beta$.
{"set_name": "stack_exchange", "score": 0, "question_id": 566198}