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Consider a single photon in a Mach-Zehnder interferometer. Considering the photon only, the output state is the sum over both paths

$$\vert 1 \rangle + \vert 2 \rangle=\vert \psi \rangle + e^{i\varphi}\vert \psi \rangle,$$

where $\psi$ is an arbitrary initial shape of the wave-function. So we get an interference pattern depending on the relative phase $\varphi$.

Now, take into account the apparatus as a quantum mechanical system. At the mirrors A and B (and also at the half-transparent ones not considered) the photon interacts with one or more electrons of the mirror, transferring momentum. The output state is now rather

$$\vert \psi \rangle\vert A \rangle + e^{i\varphi}\vert \psi \rangle\vert B \rangle.$$

Of course, if we measured the momentum transferred to A or B, we'd have which-way information, thus no interference pattern. But even if we don't measure, the fact that an entanglement has been created between photon and apparatus seems to destroy the pattern, because the $\vert \psi \rangle$s cannot cancel out.

So why does the interferometer work at all? I can't find a standard argument for this, apart from the Copenhagen-like 'Thou shalt not treat the apparatus quantum mechanically'. The only way out seems to suppose that the states $A$ and $B$ of the apparatus are not really orthogonal...

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This hand waving assumption you are making is the crux:

At the mirrors A and B (and also at the half-transparent ones not considered) the photon interacts with one or more electrons of the mirror, transferring momentum

The photon is not interacting with one or more electron on its way, it is interacting with the lattice of atoms. This means that the masses involved in momentum exchanges are orders of magnitude larger than the mass of an electron, and because of this practically no energy is transferred to the lattice. ( similar to a ball bouncing off a wall).

Because of the large mass of the lattice, the center of mass "photon + lattice" and the laboratory "lattice at rest" coincide within measurement errors. Thus the geometry of the lattice acts as a simple boundary condition to the quantum mechanical problem;it defines the effective potential well whose solutions give the wavefunctions. If the boundary did not exist the wavefunctions would be those of a free photon.

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  • $\begingroup$ Thanks for the quick answer, but I was looking for a more in-depth theoretical explanation, rather than a hand waving one. $\endgroup$ – NUU May 8 '16 at 13:47
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A full quantum mechanical description of the interferometer is complicated because it's not an isolated system. But we can do a thought experiment where we imagine it to be made out of mirrors that are floating in free space. Then as Anna has explained, the photon will interact with the entire system, it will not excite vibrational modes of the lattice. This means that we only need to consider the center of mass part of the wavefunction of the mirror. The overlap determines the visibility of the interference pattern and this is determined by the width of the wavefunction of the center of mass of the mirror. How can we estimate this?

Let's first consider a particle of mass m in a thermal bath at temperature T. The particle is then in a mixed state. The density matrix in the position representation will have off diagonal components that decay to zero over the distance of the order of the thermal de-Broglie wavelength $\lambda = \frac{h}{2\pi m k T}$. You can picture this as the thermal bath localizing the particle somewhere to within a distance of order $\lambda$. The particle can then be considered to be in a pure state described by a wavefunction that has this width.

We can then apply this reasoning to the mirror. If the mirror is at some temperature $T$ in the sense that the atoms the mirror consists of form their own heath bath, then we can factor out the center of mass degree of freedom. The wavefunction of the center of mass can then be taken to be a pure state of a width of order $\lambda$. The width of the momentum space wavefunction will then be of the order of

$$\lambda_p = \sqrt{\frac{m k T}{8\pi}}$$

If you insert realistic numbers in here, you find that this width is enormously larger than the momentum of the photon. The overlap between |A> and |B> will thus be very close to 1.

A more elementary way to argue would be to note that the surface of the mirror must be localized to within a fraction of the wavelength of the photon which automatically guarantees that the uncertainty in the momentum will be larger than that of the photon.

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  • $\begingroup$ Pretty solid explanation, thanks. I would have expected, however, that decoherence would "make things worse", i.e. lead to less overlap of $A$ and $B$, which now have to be considered as combined states of apparatus and environment. Is there an explanation (or readable paper) why this is not the case? $\endgroup$ – NUU May 8 '16 at 13:43
  • $\begingroup$ @NUU Note that decoherence leads to localization in the position basis. The dynamics of a system is described by a local Hamiltonian which means that the environment will de-facto end up "measuring the position" of a system. This leads to a certain spread in momentum space. $\endgroup$ – Count Iblis May 8 '16 at 20:42

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