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A classic setup (I suppose it's classic) in introductory quantum mechanics has a single photon passing through a double slitted grating. Directly across each slit of the grating is a device that can identify which particular slit a photon passed through. In this setup, the uptake is that, if you can distinguish which slit a photon passes through (i.e. localize the photon), you collapse the wavefunction and prevent any interference pattern from forming on the screen past the grating.

A practice question -- which forms the basis for this conceptual question (see below) -- localizes a photon but, this time, has it sent through an interferomter. It asks what would happen to the probability of such a particle if we also moved the right arm slightly outwards (please read the question too!). From what I understand, however, localization shouldn't bear any single consequence in interferometry. This is what I would argue:

a) Despite the fact that a single photon enters an interferometer, the photon's path can be thought to be a superposition of all possible ones. Thus, it is possible to think of the single photon entering the setup as if two photons entered different arms of an interferometer.

b) If we localize the particle, we confine its wavefunction to a certain range.

c) Consequently, unable to take on its original, full range (given by $\psi$), the photon doesn't superimpose to produce all possible intereference constructions of amplitude (the phase angle, $\phi$, has a limited set of values).

d) This could mean anything! Without knowledge of where $\psi$ is limited, many possible, albeit limited, wavefunctions result.

How does localizing the particle, in interferomtery, prevent an inteference pattern from forming? enter image description here

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  • $\begingroup$ Here's a paper that is related: Phys. Rev. Lett. 75 3783 1995 also available without paywall as chapmanlabs.gatech.edu/papers/scattering_ifm_prl95.pdf (thanks anna). $\endgroup$ – dmckee♦ Oct 11 '15 at 3:31
  • $\begingroup$ @dmckee "Our experiment has the interesting property that the loss of coherence cannot be attributed to smearing of the interference pattern caused by momentum transferred in the scattering process, but, rather, is the result of random phase shifts between the two interfering paths." I don't think their results are any less applicable here just because they localized photons by varying slit separation(i.e by making the slit separation larger than $\lambda$). Is this a fair assessment? $\endgroup$ – Muno Oct 11 '15 at 15:51
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The localization referred to in the actual problem has to do with the length of the pulse (duration $\rightarrow$ length), whereas localization via which-path measurement in the double-slit setup concerns the transversal extent of the wave function ("around one slit" vs. "over two slits"). Since the pulse entering the interferometer is generally assumed to be rather localized transversally, and propagation in the interferometer does not affect transversal localization, the analogy with the double-slit doesn't provide much insight in this case.

As for how the longitudinal localization of the pulse determines whether or not there is an interference pattern, take a careful look at the geometry of the interferometer. You have a 0.3m long wave packet propagating at light speed that is split in two by the beam-splitter at first pass. Let $L=2m$ be the length of the long arm of the interferometer. In a time $\Delta t = L/c$, one branch of the split wave packet goes all the way around the shorter arm and returns to the beam splitter for a second pass. During the same time, the other branch of the wave packet just about reaches the mirror at the end of the longer arm. It will take it $\Delta t$ more time to return to the beam splitter, and by now the first branch has long cleared the interferometer. The second wave packet has nothing to interfere with. The situation doesn't change if the right mirror is moved to the right, since it only adds to the length $L$ and the time $\Delta t$.

To get a probability explicitly, simply notice that after the second pass through the beam splitter the wave function is split in 4 mutually orthogonal components with identical weight ( $\frac{1}{2}$ ) and extract the probability from the corresponding amplitudes.

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  • $\begingroup$ This process -- that, by the time the wave packet from the longer arm returns, the shorter arm's wave packet is long gone -- depends on the fact that a single photon entering the interferometer does travel through both paths. I thought the same way when I first gave my reasoning. Now that you reference the delocalization of the photon in the double slit setup as being caused by the ability of the photon to, in essence, "wrap around" both slits, I must ask: what causes delocalization in the interferometer's case? $\endgroup$ – Muno Oct 11 '15 at 19:45
  • $\begingroup$ You wrote that 'localization via which-path measurement in the double-slit setup concerns the transversal extent of the wave function ("around one slit" vs. "over two slits").' I took this to mean that delocalization would occur if the photon could traverse at least two slits. $\endgroup$ – Muno Oct 11 '15 at 19:47
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    $\begingroup$ The splitter is basically a partially transparent mirror. Pulse delocalization, as in split along two distinct paths, comes from partial reflection and partial transmission. The relative weight of the two, that is probability of transmission vs. probability of reflection, can vary, but unless otherwise stated we usually assume a perfect 50/50 splitter. $\endgroup$ – udrv Oct 11 '15 at 19:54
  • $\begingroup$ Oh wow, even a single photon can be split. Makes sense. Thanks $\endgroup$ – Muno Oct 11 '15 at 20:04
  • $\begingroup$ Re: delocalization in the double-slit, yes this is what I meant. After passing the slits the wave function "spreads laterally", propagating in all directions from the slits to points on the screen. In the interferometer, interference occurs between wave packets propagating in a single direction along a common path (output path). The transverse size of the wave packet is not affected and/or is not important to interference. At least not if the splitter is perfectly flat. $\endgroup$ – udrv Oct 11 '15 at 20:19

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