Are all emitted photons accounted for at the detector(s) in the single-photon Mach-Zehnder interferometer? Layman alert: I am sorry if I am asking stupid questions. I'm just trying to grasp what is happening with a single photon in MZ interferometer.
Say we have a single-photon source, and let the second detector be the one with the probability zero.
Do we know how many photons were emitted and do we then detect the same number of them at the first detector?
In other words, does the photon which should end up in the second detector get annihilated somehow (and therefore is missing) after passing the second splitter (either straight or reflecting), or does it always somehow choose to go to the first detector?
I say "choose" because if it is not missing then it seems to me that the behaviour at the second splitter is not random. If the photon always ends up in the first detector, then it must choose the different behaviour in two consecutive splitters. If it went straight through the first splitter then it must reflect of the second, and vice versa.
 A: In the usual setup, the probabilities of detection are related to the transmitivity and reflectivity of the beam splitter (ignoring phases for this discussion) so you cannot get 0 probability at one detector unless you remove the beam splitter or place a full mirror in your interferometer (reflectivity $0$ or $1$).
In general, if you detect photons one a the time, yes indeed you will detect all photons provided there is no loss and your detectors have 100% efficiency: photons will appear individually at one or the other detector.  If you measure a classical beam, you can check via energy conservation (or more accurately power) that indeed there is no loss.
Here are some details for the single photon input case:
A 2-channel MZ device effects the unitary transformation
$$
U=\left(
\begin{array}{cc}
 e^{-\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right) &
   -e^{-\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta
   }{2}\right) \\
 e^{\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta }{2}\right) &
   e^{\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right)
   \\
\end{array}
\right)
$$
between modes.  Thus, acting on an input state with one photon in the first mode:
$$
a_1^\dagger\vert 0\rangle \equiv \vert 1\rangle \mapsto \left(\begin{array}{c} 1\\0\end{array}\right)
$$
we get the superposition
$$
\left(
\begin{array}{c}
 e^{-\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right)
   \\
 e^{\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta }{2}\right) \\
\end{array}
\right)  = e^{-\frac{1}{2} i (\alpha +\gamma )} \cos \left(\frac{\beta }{2}\right)\vert 1\rangle +  e^{\frac{1}{2} i (\alpha -\gamma )} \sin \left(\frac{\beta }{2}\right) \vert 2\rangle
$$
where $a_2^\dagger\vert 0\rangle\equiv \vert 2\rangle$ is the state with one photon in the second mode.
Thus, the probability of finding the input state in the 2nd mode is $\sin^2(\beta/2)$, which is $0$
only if the transmittivity $\cos^2(\beta/2)=1$.  In this case, the interferometer is then equivalent to a relative phase shifter:
$$
U\vert_{\beta=0}=\left(
\begin{array}{cc}
 e^{-\frac{1}{2} i (\alpha +\gamma )} & 0 \\
 0 & e^{\frac{1}{2} i (\alpha +\gamma )} \\
\end{array}
\right)
$$
and there is no splitting.
A similar analysis shows that the probability of getting $0$ in the first mode corresponds to $\beta=\pi$, in which case the input state is completely reflected to mode $2$ but accumulates an extra $\hbox{e}^{\frac{1}{2} i (\alpha -\gamma )}$ phase.
One often-used configuration is one where a 50-50 beamslitter is followed by a phase shifter and followed by a "reverse" 50-50 beam splitter.  In matrix form this is:
$$
\left(
\begin{array}{cc}
 \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
 -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)\cdot 
\left(
\begin{array}{cc}
 e^{-\frac{i \alpha }{2}} & 0 \\
 0 & e^{\frac{i \alpha }{2}} \\
\end{array}
\right)\cdot \left(
\begin{array}{cc}
 \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\
 \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\
\end{array}
\right)=\left(
\begin{array}{cc}
 \cos \left(\frac{\alpha }{2}\right) & -i \sin
   \left(\frac{\alpha }{2}\right) \\
 -i \sin \left(\frac{\alpha }{2}\right) & \cos
   \left(\frac{\alpha }{2}\right) \\
\end{array}
\right)
$$
which is an adjustable beamsplitter.  One can vary the phase $\alpha$ by varying the optical path length in one arm so that $\cos(\alpha/2)=1$ (nothing reflected) or $\cos(\alpha/2)=0$ (nothing transmitted).
(A sketch by the OP would be useful to clarify if this is what they have in mind.)
Either way the previous argument runs the same: for detectors with 100% efficiency, the individual photons entering via input channel $1$ will be detected in one or the other detector, depending on $\alpha$.
A: Consider Feynman path integral ... every photon travels its own path, if the path is not ideal (per Feynman path integral) it is not travelled. Areas of non-travel are dark, all the photons go to the bright areas. The word 'interference" is out of date and historical and it violates conservation of energy! Photons never cancel, they are only emitted and eventually absorbed.
