Gravitational signature of a photon in a double slit experiment

[PeterDonis]

Here just have polarization filters in front of the slits at 90deg relative orientation.

If they are polarization filters, as opposed to quarter-wave plates, then we have a different setup, because each individual photon now passes through only one slit or the other, whereas with QWPs, each photon still passes through both slits, the QWP just shifts its phase.

The two setups are treated differently in terms of the basic math of QM, as follows:

(1) With QWPs, the wave function of the photon after the slits but before the detector screen is a coherent superposition of the two one-slit wave functions, with an appropriate phase applied to each according to how the QWP at the corresponding slit is aligned. The amount of interference seen at the detector screen is then a function of the relative phases at the two slits. This is the kind of setup I took @DrChinese to be describing, since he talked about being able to continuously adjust how much interference there is.

(2) With polarization filters oriented at 90 degrees relative, there is no single wave function after the slits but before the detector screen, because we have to apply the projection postulate: we can only describe the state after the slits as a density matrix showing equal probability of the photon coming from each slit. Of course this means it is impossible to have interference.

We can talk about either setup, but we need to be clear about which one we are talking about.

 

[vanhees71]

If this were true, it would be impossible to obtain partial interference with such a setup. But it isn't; by adjusting the relative directions of the polarizers at the two slits, one can adjust the amount of interference continuously from 100% to zero. But which-way information is binary: either it's there (and completely destroys interference) or it isn't (and interference is unaffected).

Of course if you don't measure the polarization you also don't have the wwi. If you instead register the photon at a screen you get an interference pattern with a contrast determined by the angles of the qgps. It's also a good example for the important difference between state preparation, here realized by the photon source, the double slit and the qgps, and a measurement, which here is realized by the far-distant screen behind the slit. By the statd preparation you get the entanglement between wwi and the polarization but not the corresponding information of these properties of each individual photon. If you measurse a different observable (in this case the position of photon detection through the screen) of course you never make use of the entanglement to gain wwi by measuring the polarization of the photons. The characteristic feature of entanglement is that you have 100% correlations between observables which themselves have no certain values.

 

[vanhees71]

If they are polarization filters, as opposed to quarter-wave plates, then we have a different setup, because each individual photon now passes through only one slit or the other, whereas with QWPs, each photon still passes through both slits, the QWP just shifts its phase.

The two setups are treated differently in terms of the basic math of QM, as follows:

(1) With QWPs, the wave function of the photon after the slits but before the detector screen is a coherent superposition of the two one-slit wave functions, with an appropriate phase applied to each according to how the QWP at the corresponding slit is aligned. The amount of interference seen at the detector screen is then a function of the relative phases at the two slits. This is the kind of setup I took @DrChinese to be describing, since he talked about being able to continuously adjust how much interference there is.

(2) With polarization filters oriented at 90 degrees relative, there is no single wave function after the slits but before the detector screen, because we have to apply the projection postulate: we can only describe the state after the slits as a density matrix showing equal probability of the photon coming from each slit. Of course this means it is impossible to have interference.

We can talk about either setup, but we need to be clear about which one we are talking about.

That's of course right, though I don't understand which phase shift you are referring to in the qgp setup. Of course there's a relative phaseshift between the partial waves from the two slitx which cause the two-slit interference fringes if the qgps are not in 90deg relative orientation. If they are, there's no interference between these partial waves because the polarization states are perpendicular to each other and that's why jn thix case the interference pattern is gone completely and partially present in other orientations of the qgps. Is that what you mean?

 

[DrChinese]

If they are polarization filters, as opposed to quarter-wave plates, then we have a different setup, because each individual photon now passes through only one slit or the other, whereas with QWPs, each photon still passes through both slits, the QWP just shifts its phase.

The two setups are treated differently in terms of the basic math of QM, as follows:

(1) With QWPs, the wave function of the photon after the slits but before the detector screen is a coherent superposition of the two one-slit wave functions, with an appropriate phase applied to each according to how the QWP at the corresponding slit is aligned. The amount of interference seen at the detector screen is then a function of the relative phases at the two slits. This is the kind of setup I took @DrChinese to be describing, since he talked about being able to continuously adjust how much interference there is.

(2) With polarization filters oriented at 90 degrees relative, there is no single wave function after the slits but before the detector screen, because we have to apply the projection postulate: we can only describe the state after the slits as a density matrix showing equal probability of the photon coming from each slit. Of course this means it is impossible to have interference.

We can talk about either setup, but we need to be clear about which one we are talking about.

Just making sure we are discussing the same setup. A. In my citation, there are polarizers (not wave plates). B. The detector I am referring to is something that would identify which-slit (if that information is available). I am not referring to the screen that displays the pattern.

You can continuously adjust the amount of interference from 0 to 100% as you adjust the relative angle between the slits.

 

[PeterDonis]

In my citation, there are polarizers (not wave plates).

Yes, I was mixing up terminology from the two papers, the one you referenced and the one @vanhees71 referenced. The former paper deals with linear polarizations while the latter deals with circular polarizations.

 

[vanhees71]

The two setups are of course significantly different. With the polarizers you really measure the polarization through a filter measurement, i.e., you absorb part of the photons (according to the corresponding probability given their polarization state) and let through only photons with a certain polarization (say H for horizontally polarized) when they go through slit 1 and only photons with polarization V when going through slit 2. Now the information about the WWI is realized by determined values for the polarization and there's of course no interference pattern, because if a photon comes through then it must have gone through either one of the slits due to the polarizers.

Defining the single-photon Fock states for a photon going through slit 1 and the one going through slit 2 as $|\psi_1 \rangle=\hat{A}_1^{(-)}(\vec{x},\text{H}), \quad |\psi_2 \rangle=\hat{A}_2^{(-)}(\vec{x},\text{V})$, where $\hat{A}_1^{(-)}$ creates a state described by a spherical-wave with center in slit 1, and $\hat{A}_2^{(-)}$ a spherical wave with center in slit 2, the state for photons after the slit is the mixture
$$\hat{\rho}=P_H |\psi_1 \rangle \langle \psi_2| + P_L |\psi_2 \rangle \langle \psi_2|.$$
Here $P_H$ and $P_V$ are the probabilities that the incoming photon goes through a H-polarization filter and a V-polarization filter, respectively. Of course $P_H+P_V=1$.

I gave the example with quarter-wave plates, because this is not a filter setup but all photons which would go through the slits without the qwps goes also through if the qwps are in place. The (ideal) qwps represent unitary operators making a left-circular polarized photon when going through slit 1 and a right-circular polarized photon when going through slit 2 if the incoming photons are H polarized (in the setup where the qwps are oriented in $\pm \pi/4$ orientation). So in this case the wwi is only gained if you measure the polarization state in the LR ($h=\pm 1$) basis. The single photon has not a determined polarization after the slits but the path the photon took and the polarization state are entangle. In this case the single-photon state after the slit are something like
$$|\Psi \rangle = \frac{1}{\sqrt{2}} [\hat{A}_1^{(-)}(t,\vec{x},h=1)+ \hat{A}_2^{(-)}(t,\vec{x},h=-1)]|\Omega \rangle,$$
where $\hat{A}_1$ is the operator creating a spherical wave with center at slit 1 and $\hat{A}_2$ is one that creates a spherical wave with center at slit 2. Note that there's no wave function but a single-photon Fock state, i.e., in accordance with relativistic QFT (there are no wave functions for massless particles because there's no position operator). The field operators look like the negative-frequency part of the corresponding solution of the classical diffraction problem in the Kirchhoff approximation, i.e., using Huygen's principle. Thus, if you don't measure the polarization but just register the photons at a far-distant screen, such that you'd find an two-slit interference pattern (of course modulated by the single-slit interference pattern as in the experiment with classical em. waves) without the qwps in position (then both creation operators would create H-linear polarizations), in the setup with qwps in position you don't get double-slit interference fringes, because the two parts don't interfere, because they are perpendicular single-photon state vectors and thus you only see the incoherent superposition of two single-slit interference patterns as in the setup with the polarizers.

These are two examples that the two-slit interference pattern is completely gone if there is the possibility to gain WWI for any photon. You don't need to really get the WWI by measuring the polarization state in the setup with the qwps. It's enough that you have this 100% entanglement between through which slit the photon came such that you can get with 100% certain the information through which slit each photon came by measuring the polarization in the $h=\pm 1$ basis. It's characteristic for entanglement that the entangled observables are indetermined when the quantum object is prepared in the entangled state but that you have still the (in this case 100%) correlation for the outcomes of the corresponding measurements.

This enables also the realization of postselection as in the Walborn et al quantum eraser measurement. Depending on which partial ensemble from a totally measured ensemble of entangled photon pairs you choose from the established measurement protocols you can either keep the potential WWI encoded in the polarization state of each photon (no interference fringes) or you select a partial ensemble, for which the WWI is not encoded in the corresponding polarization state (by filtering the signal photon from the entangled photon pairs by a 45 deg linear polarization filter of the idler photon) and then get the two-slit interference pattern back for this subensemble. Taking the other half with the idler going through a -45 deg linear polarization filter you also get a two-slit interference pattern, but shifted against the former one. Both together give the full ensemble, not showing the two-slit interference pattern.