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?

  • $\begingroup$ Look at what happens if you calculate $\langle \psi | \varphi\rangle$ where the 2 states are not the same $\endgroup$ Jul 16, 2020 at 20:16
  • $\begingroup$ @BySymmetry So $|\psi \rangle=c_1|0\rangle+c_2|1\rangle$ and $\phi\rangle=a_1 |0\rangle+a_2 |1\rangle$. Then $\langle \psi | \phi \rangle = c_1^*a_1 + c_2^*a_2$. That is equal to $1$, right? Where is the cos interference term? $\endgroup$
    – Anon21
    Jul 16, 2020 at 20:51
  • $\begingroup$ Not is it not $=1$ in general. Take $c_1=1/\sqrt{3}, c_2=\sqrt{\frac{2}{3}}$ but $a_1=a_2=\frac{1}{\sqrt{2}}$. $\endgroup$ Jul 16, 2020 at 21:00
  • $\begingroup$ @BySymmetry Is this what you have in mind: $|\psi \rangle =c_1|0\rangle +c_2|1\rangle$ and $|πœ™\rangle =π‘Ž_1|0\rangle+π‘Ž_2|1\rangle$? If so, then $\langle πœ“|πœ™\rangle=𝑐_1^*π‘Ž_1+𝑐_2^*π‘Ž_2$, which contains no interference terms. $\endgroup$
    – Anon21
    Jul 16, 2020 at 21:53
  • $\begingroup$ Your previous statement $\uparrow$ is incorrect, as demonstrated by a full calculation. In fact this is in general a complex number. $\endgroup$ Jul 16, 2020 at 21:57

1 Answer 1


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 $\langle \psi_1\vert \psi_2\rangle=e^{-i\gamma}\sin\beta$.

  • $\begingroup$ Can you explain why $\langle 1 | 0 \rangle$ is not orthogonal? $\endgroup$
    – Anon21
    Jul 16, 2020 at 20:52
  • $\begingroup$ @AlexandreH.Tremblay no I cannot as this was a typesetting error on my part. I just corrected it. $\endgroup$ Jul 16, 2020 at 20:56
  • $\begingroup$ So the key is to use 2 parameters $\beta$ and $\gamma$ in both wavefunctions , instead of 4 unique parameters? $\endgroup$
    – Anon21
    Jul 16, 2020 at 21:36
  • $\begingroup$ there are still 4 coefficients. you could write $c_1$ etc as you did before. Your error is in assuming things like $c_1^*a_1+ c_2^* a_2=1$, which is not true in general. If you want take the angle $\beta$ in $\vert \psi_2\rangle$ to be different from the one in $\vert\psi_1\rangle$ to get an even more general overlap. $\endgroup$ Jul 16, 2020 at 21:54
  • $\begingroup$ I think I might have it, but it appears that the interference term of P is the interference of a mixture, not a pure state. If I have $|\psi_1 \rangle = z_1 e^{i\theta_1} | 0\rangle$ and $|\psi_2 \rangle = z_2 e^{i\theta_2} |0\rangle$, then $\langle \psi_1+\psi_2 | \psi_1+\psi_2 \rangle=|z_1|^2+ |z_2|^2 + 2 |z_1||z_2|\cos (\theta_2-\theta_1)$ $\endgroup$
    – Anon21
    Jul 17, 2020 at 0:06

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