I have read that if you have a Mach–Zehnder interferometer (doing a single-photon experiment) and put a non-destructive detector in only one of the two arms (connected to the first beam splitter), you lose interference when the wave packets of the photon are recombined at the second beam splitter.

I am trying to model this as a wave function, but am having trouble.

Suppose the interferometer consists of a first beam splitter, which the photon enters from its source, followed by arm $A_x$, heading along the cartesian $x$ direction, and arm $A_y$, heading along the cartesian $y$ direction. (so, the two arms are spatially orthogonal to each other). At the end of each arm is a mirror which directs the wave packets to a second beam splitter, which re-combines them. Also, suppose that somewhere in arm $A_x$ is a non-destructive detector. There is no detector, or anything else, in arm $A_y$

Neglecting the detector in $A_x$, right after the photon passes through the first beam splitter (time $t_1$), we would have

$|\Psi_{t_1}\rangle = (1/\sqrt 2) (|A_x\rangle + |A_y\rangle)$

That is, there is a 50% chance for the photon to be in either arm.

Taking the detector in $A_x$ into account, right after the photon passes through the first beam splitter (but before reaching the detector), we would have:

$|\Psi_{t_1}\rangle = |D_{off}\rangle (1/\sqrt 2) (|A_x\rangle + |A_y\rangle)$ where is the detector in its "off" state.

Next, at $t_2$, if the non-destructive detector registers the presence of the photon, then this should evolve to

$|\Psi_{t_2}\rangle = (1/\sqrt 2)(|D_{on}\rangle |A_x\rangle + |D_{off}\rangle |A_y\rangle $ where is the detector in its "on" state and the particle is in $|A_x\rangle)$.

Or, at $t_2$, if the non-destructive detector does not register the presence of the photon, and the particle is hence in $|A_y\rangle$, then this should evolve to

$|\Psi_{t_2}\rangle = (1/\sqrt 2)(|D_{off}\rangle |A_x\rangle + |D_{off}\rangle |A_y\rangle) = |\Psi_{t_1}\rangle$ where is the detector remains in its "off" state.

If $\langle \Psi_{t_1}|\Psi_{t_2}\rangle = 0$, then after the second beam splitter, we would expect to see no interference.

However, I do not see how in both $t_2$ scenarios, described above, we end up with $\langle \Psi_{t_1}|\Psi_{t_2}\rangle = 0$

For instance, let $|A_y\rangle = \left(\begin{array}{c}0\\1\end{array}\right) \quad |A_x\rangle = \left(\begin{array}{c}1\\0\end{array}\right) \quad |D_{off}\rangle = \left(\begin{array}{c}0\\1\end{array}\right) \quad |D_{on}\rangle = \left(\begin{array}{c}1\\0\end{array}\right)$

Then $|\Psi_{t_1}\rangle = |D_{off}\rangle (1/\sqrt 2) (|A_x\rangle + |A_y\rangle) = \left(\begin{array}{c}0\\0\\1/\sqrt 2\\1/\sqrt 2\end{array}\right)$

$|\Psi_{t_2}\rangle = (1/\sqrt 2)(|D_{on}\rangle |A_x\rangle + |D_{off}\rangle |A_y\rangle = \left(\begin{array}{c}1/\sqrt 2\\0\\0\\1/\sqrt 2\end{array}\right) $

In that case, I do not get $\langle \Psi_{t_1}|\Psi_{t_2}\rangle = 0$.

Could you please tell me, where is my mistake? Thanks.

  • $\begingroup$ Interference is completely independent of the number of photons in the interferometer. A simple Mach-Zehnder interferometer shows interference in both arms even with detection, as the light leaves the interferometer anyway, whether detectors are present at the outputs or not. Putting a detector in one of the arms destroys the interference because there is no interferometer left... the technically interesting case that you are trying to model is not a simple Mach-Zehnder but a delayed-choice quantum eraser experiment, I suppose? $\endgroup$ – CuriousOne Feb 9 '16 at 8:06
  • $\begingroup$ @ CuriousOne: Actually, I neglected to say that the detector is non-destructive. Also, maybe the single photon part is not relevant, but I am not sure what is wrong with my model of this, knowing that there should be no interference but not being able to get that result from my model. I would be very interested in seeing the correct model $\endgroup$ – David Feb 9 '16 at 15:45
  • $\begingroup$ The state of the system at $t_2$ is $|\Psi_{t_2}\rangle = (1/\sqrt 2)(|D_{on}\rangle |A_x\rangle + |D_{off}\rangle |A_y\rangle. $ Your second expression for $|\Psi_{t_2}\rangle $ is only true if the detector doesn't work (i.e. if its state is unchanged by the presence of the photon). The interference visibility, as defined in my other answer (physics.stackexchange.com/questions/233190/…), is $\langle D_{on}|D_{off}\rangle = 0$. $\endgroup$ – Mark Mitchison Feb 9 '16 at 16:21
  • $\begingroup$ @ Mark Mitchison: In that answer (to the other post), you wrote that the amount of interference depends on $$\langle C_a \lvert \rho_C(t) \rvert C_b\rangle = \langle \Psi_{P_i}\rvert \langle \Psi_{A_i}\rvert T \exp \left ( \int_0^t\mathrm{d}s \,H_{AP}(s) \right)\lvert \Psi_{A_i}\rangle \lvert \Psi_{P_i}\rangle $$ which seems to me to be equivalent to $\langle\Psi_{t_1}|\Psi_{t_2}\rangle$ in this question. (The detector is non-destructive, if that matters here.) $\endgroup$ – David Feb 9 '16 at 16:39
  • $\begingroup$ @David No, in your other question you assumed the existence of an auxiliary system $A$ which does not appear here. The correct expression is $\langle \Psi_{P_i}\rvert T \exp \left (\int_0^t\mathrm{d}s\, H(s)\right) \lvert \Psi_{P_i}\rangle$, with $P$ the detector system and $H$ a Hamiltonian that switches the detector on. Given that here $\lvert \Psi_{P_i}\rangle = \lvert D_{off}\rangle$, in the end this reduces to $\langle D_{off}\lvert D_{on}\rangle = 0$. $\endgroup$ – Mark Mitchison Feb 9 '16 at 16:52

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