B Gravitational signature of a photon in a double slit experiment

[vanhees71]

In the setup with quarter-wave plates at 45 and -45 degree orientation and linearly H-polarized photons you have 100% entanglement between polarization states (left and right circular polarized) photons anc which-way information, i.e. you can know by measuring these polarizations with 100% certainty through which slit the photon came. These photons don't make a double-slit interference pattern if you observe them on a far distant screen. Without the quarter-wave plates at the slits you don't have any possibility to know the which-way information through any measurement, because in this setup it's completely indetermined through which slit each photon came and you get a two-slit interference pattern. This has been empirically proven, even in the fascinating extension of this experiment enabling "quantum erasing" the which-way info. See the above quoted paper by Walborn et al.

 

[PeterDonis]

you can know by measuring these polarizations

Yes, but that requires adding a polarization measuring device to the experiment. The quarter-wave plates themselves don't measure the polarizations.

 

[vanhees71]

Yes, of course. But the point is that you have the certain which-way information available for the photons, and this destroys the two-slit-interference pattern no matter whether you really make use of this by measuring the polarization or not. So @DrChinese 's statements were right.

 

[PeterDonis]

you have the certain which-way information available for the photons, and this destroys the two-slit-interference pattern

In the absence of a polarization measurement, no, you do not have "which-way information" available. In the absence of a polarization measurement device of some kind, there is nothing to cause decoherence at each slit, and it is decoherence that provides "which-way information", even if that information is not usable by humans. To put it another way, with just the quarter wave plates, the operation performed on the photon at each slit is reversible, and you need an irreversible change, even if its details are not detectable by humans, to have "which-way information" that destroys interference.

With just quarter wave plates and no polarization measurement, the two-slit experiment is like a Mach-Zehnder interferometer with a phase shifter in each of the arms. What determines how much signal comes out of each side of the final beam splitter in the MZI is the relative phase in the two arms coming into the beam splitter, not the amount of "which-way information" available, and the phase shifters simply change that relative phase. Similarly, what determines the pattern on the detector in a two-slit experiment is the relative phase of the wave functions coming from each slit, and the quarter wave plates simply change that relative phase.

 

[DrChinese]

In the case of the polarizers, you can't get which-slit information just from the polarizers alone. The polarizers just induce a phase shift in photons passing through by a simple unitary operation, and no entanglement is produced between the polarizers and the photons passing through them. That means nothing takes place at the polarizers that could be subject to decoherence and could produce an irreversible record (even one that is not recoverable by humans, such as a decoherent interaction with an environment) of which slit each photon passed through.

Again, the case of the polarizers is simply different from the case in which there is a which-slit measurement, or even the in principle possibility of a which-slit measurement. The polarizers themselves aren't measuring the polarization of the photons; they are just shifting their phase. (Perhaps "polarizers" is the wrong term to use to describe these devices; I think @vanhees71 used the term "quarter-wave plates", which might be better.)

Not sure what you are saying here. Perhaps @vanhees71 has convinced you that the slit differential IS needed, but that the actual detector is NOT needed. So I will be specific. You can induce which-slit information with polarizers alone (and note that polarizers are always present, just their relative alignment changes). See reference.

"When either the vertical or the horizontal filter covers both slits, the double-slit interference pattern is preserved, albeit at a reduced intensity compared to no filter. When the vertical filter covers one slit and the horizontal filter covers the other, the double-slit pattern disappears completely. Two superimposed single-slit patterns are all that remain."

My point, as I have said repeatedly, is that the shifting of the particle phase so there is a DIFFERENCE between the slits is what causes interference to disappear. By analogy: gravity would need to accomplish a similar feat (so extreme I believe it is impossible to create such conditions) to cause interference to disappear.

Further, no detector is necessary for the cited experimental result.

"This new arrangement changes the setup into a which-path experiment in the sense that it is now (in principle) possible to know which slit the photon passed through; this destroys the quantum interference."

The presence or absence of a sensitive "gravity detector" would make no difference. We already know this, since such a detector is not needed in the cited experiment.

https://sciencedemonstrations.fas.h...-demonstrations/files/single_photon_paper.pdf

 

[PeterDonis]

You can induce which-slit information with polarizers alone

No, you can affect how much interference is observed at the detector with polarizers alone. But that is not equivalent to inducing which-slit information. Affecting the amount of interference at the detector screen only requires a phase shift; it does not require decoherence. Inducing which-slit information requires decoherence.

the shifting of the particle phase so there is a DIFFERENCE between the slits is what causes interference to disappear

I agree that this is the case in the particular experimental setup you are using, with quarter-wave plates at each slit. I am only saying, as above, that shifting the phase is not the same thing as inducing which-slit information. And that means there are two different ways of making inteference disappear at the detector screen: (1) create the appropriate amount of relative phase shift; (2) induce which-slit information. These are not the same thing.

gravity would need to accomplish a similar feat (so extreme I believe it is impossible to create such conditions) to cause interference to disappear.

To do it by method (1) above, yes. But I have been talking all along about doing it by method (2) above. The two methods are not the same.

 

[PeterDonis]

there are two different ways of making inteference disappear at the detector screen: (1) create the appropriate amount of relative phase shift; (2) induce which-slit information. These are not the same thing.

Note also that these two methods do not have identical capabilities. Method (1) allows a continuous variation in the amount of interference, by adjusting the amount of the relative phase shift. Method (2) is discrete: the only effect it can produce is zero interference.

Considering the MZI analogy I suggested in post #64 may be helpful here. Method (1) means putting a phase shifter in each arm of the MZI: this allows a continuous adjustment of the output at the detectors in each output arm of the final beam splitter, from 100% signal in one arm and 0% in the other, all the way to 0% in the first arm and 100% in the second. Method (2) means putting a photon detector (some device that registers a blip when a photon passes through it) in each arm of the MZI; the only effect this can produce is a 50-50 split in output signal in the two output arms of the final beam splitter.

 

[vanhees71]

In the absence of a polarization measurement, no, you do not have "which-way information" available. In the absence of a polarization measurement device of some kind, there is nothing to cause decoherence at each slit, and it is decoherence that provides "which-way information", even if that information is not usable by humans. To put it another way, with just the quarter wave plates, the operation performed on the photon at each slit is reversible, and you need an irreversible change, even if its details are not detectable by humans, to have "which-way information" that destroys interference.

With just quarter wave plates and no polarization measurement, the two-slit experiment is like a Mach-Zehnder interferometer with a phase shifter in each of the arms. What determines how much signal comes out of each side of the final beam splitter in the MZI is the relative phase in the two arms coming into the beam splitter, not the amount of "which-way information" available, and the phase shifters simply change that relative phase. Similarly, what determines the pattern on the detector in a two-slit experiment is the relative phase of the wave functions coming from each slit, and the quarter wave plates simply change that relative phase.

Again: With the qgps as described the which-way information is contained in the polarization state of the photon. You don't need to measure which polarization the individual photon has to destroy the interference pattern. For that it's sufficient that it is possible to gain which-way info when measured. All needed is the correlation between wwi and polarization through the prepared state through which wwi and polarization state are entangled.

 

[PeterDonis]

With the qgps as described the which-way information is contained in the polarization state of the photon. You don't need to measure which polarization the individual photon has to destroy the interference pattern.

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).

 

[vanhees71]

Not sure what you are saying here. Perhaps @vanhees71 has convinced you that the slit differential IS needed, but that the actual detector is NOT needed. So I will be specific. You can induce which-slit information with polarizers alone (and note that polarizers are always present, just their relative alignment changes). See reference.

"When either the vertical or the horizontal filter covers both slits, the double-slit interference pattern is preserved, albeit at a reduced intensity compared to no filter. When the vertical filter covers one slit and the horizontal filter covers the other, the double-slit pattern disappears completely. Two superimposed single-slit patterns are all that remain."

My point, as I have said repeatedly, is that the shifting of the particle phase so there is a DIFFERENCE between the slits is what causes interference to disappear. By analogy: gravity would need to accomplish a similar feat (so extreme I believe it is impossible to create such conditions) to cause interference to disappear.

Further, no detector is necessary for the cited experimental result.

"This new arrangement changes the setup into a which-path experiment in the sense that it is now (in principle) possible to know which slit the photon passed through; this destroys the quantum interference."

The presence or absence of a sensitive "gravity detector" would make no difference. We already know this, since such a detector is not needed in the cited experiment.

https://sciencedemonstrations.fas.h...-demonstrations/files/single_photon_paper.pdf

Which phaseshift are you talking about? Here just have polarization filters in front of the slits at 90deg relative orientation. These do not create additional phase-shift differences between the photons running through the slits. The wwi is encoded in the polarisarion state of the photon. Since the photons running through slit 1 have perpendicular polsrization states to those runnung through slit 2 there's no interference, i.e. no two-slit interference fringes. The only difference to my qwp example is that you don't need polarized photons as source and that you absorb a certain number of photons at the polarization filters.

 

[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.