2-photon interference question

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Consider a simple setup with two independent but coherent light beams of identical intensity that converge on a 50:50 half mirror. One beam has a phase shift to the other such that they interferer and only one beam leaves the mirror in direction C while destructive interference causes no light to leave in direction D.

This setup is simply meant to compare two beams for coherence and ensure that their intensities are equal, since otherwise a detector positioned at D will measure something. In the following however I will assume perfect coherence, so only the sensitivity to intensity deviations remains a factor.

Now what happens when the both beam intensities are tuned down until there are only individual photons? Does the behavior at some point fundamentally change from the classical case – apart from the obvious that detectors will only measure individual photons instead of a continuous beam? [EDIT: of course the two incoming photons need to be assumed to be perfectly synced and arrive at the same time at the mirror] This is a case of two photon interference but unlike the setup of the HOM-effect there is phase shift in the incoming beams. If my quick mental calculation applying a displacement operator before the mirror matrix to the initial state is correct, there should still be no photons detected at D while C should measure both each time.

Now the thing I am interested in, is whether this setup still remains sensitive to the exact intensities, for example what happens if the initial beams are reduced even further e.g. by using additional beam splitters with adjustable splitting ratios ahead of the interference mirror? Would it be still possible from the detection frequencies at C and D to calculate the intensity ratios? Given that the calculus is completely linear I would assume that an interference of two half-photons should behave just the same as in the full case.

To give some background, I am trying to understand one of the central axioms of quantum mechanics that gives me the greatest headaches. And unlike the axioms defining the time evolution of a quantum system I know no experiments that would motivate it nor have I found any other convincing explanation so far. It usually just falls down from heaven without a comment. So i am trying to finding my own explanation by understanding how the above setup behaves.
 
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  • #2
PeterDonis
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This is a case of two photon interference but unlike the setup of the HOM-effect there is phase shift in the incoming beams.
The phase shift just switches which output beam is the "constructive interference" beam and which is the "destructive interference" beam. It doesn't change the fact that, assuming coincidence in time of the two incoming photons, only the "constructive interference" beam will show an output photon detection. So I think your quick calculation answer is correct.

Note that the "coincidence in time" requirement is critical, though, just as it is for the ordinary HOM effect. Normally, two independent sources of photons, even if they are assumed to have identical frequency, phase, and intensity, will not emit photons at the same time when their intensity is turned down enough. So at low enough intensity, most of the time a photon will come into the beam splitter from just one arm, not both at the same time, and you will get a 50-50 chance of detection in each output beam. Only in the rare case that both sources emit a photon at the same time (or close enough in time, given the frequency) will photons come into the beam splitter from both arms at once and allow interference to take place.
 
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Note that the "coincidence in time" requirement is critical, though, just as it is for the ordinary HOM effect. Normally, two independent sources of photons, even if they are assumed to have identical frequency, phase, and intensity, will not emit photons at the same time when their intensity is turned down enough. So at low enough intensity, most of the time a photon will come into the beam splitter from just one arm, not both at the same time, and you will get a 50-50 chance of detection in each output beam. Only in the rare case that both sources emit a photon at the same time (or close enough in time, given the frequency) will photons come into the beam splitter from both arms at once and allow interference to take place.
oh, you are right, i should have specified time-coincidence as a prerequisite rather then just coherence. I am actually quite amazed that people have managed to time things just right to make such experiments even possible. But experimental challenges aside, let's take this as a given for the consideration here as the question how the system behaves is a theoretical one - so ideal conditions can be assumed :).
 
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PeterDonis
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I am actually quite amazed that people have managed to time things just right to make such experiments even possible.
The timing can't be controlled. The best you can do is to try to measure the timing using coincidence counters or something like that, and correlate the timing to the measured output signals. But there's no way to avoid having a lot of runs where you don't have coincidence in time.
 
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The timing can't be controlled. The best you can do is to try to measure the timing using coincidence counters or something like that, and correlate the timing to the measured output signals. But there's no way to avoid having a lot of runs where you don't have coincidence in time.
Doesn't have to be as long as I can deduct it retroactively from the measured samples and pick out only those where that was the case. Then i can do all interesting analysis on only that subset. EDIT: or might not even need that. Maybe it could suffice to know the ratio/fraction of coincided events - by somehow measuring that first and then i only need to check event frequencies at C and D for singe and double photon detection.
 
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PeterDonis
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Maybe it could suffice to know the ratio/fraction of coincided events
I don't think that's sufficient. You need to know which individual events had a coincidence, so you can pick out just that subset. Or, alternatively, instead of just a binary "coincidence" vs. "no coincidence" measurement for each event, you could have an actual numerical measure of, roughly speaking, how far apart in time the two photons were for each event. IIRC that has been done for some fairly recent runs measuring the HOU effect.
 
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I don't think that's sufficient. You need to know which individual events had a coincidence, so you can pick out just that subset. Or, alternatively, instead of just a binary "coincidence" vs. "no coincidence" measurement for each event, you could have an actual numerical measure of, roughly speaking, how far apart in time the two photons were for each event. IIRC that has been done for some fairly recent runs measuring the HOU effect.
For the start it is sufficient. What i want to look at first is the term ##r_{CD} = \frac {n_C - n_D} {n_C + n_D}## where ##n_C## is either the beam intensity for the coherent beam at C or number of detection events at C for individual photons. With coherent light it shouldn't be difficult to achieve ##r_{CD}=1## and for individual particles it's pretty straight forward to deduct the uncoiciding photons since they won't experience any interference and will evenly distribute between C and D so they yield ##r_{CD} = 0## on average while the two photon case should yield ##r_{CD} = 1## due to destructive interference preventing anything reaching D in theory, no?

and just to clarify by HOU you meant HOM effect?

The phase shift just switches which output beam is the "constructive interference" beam and which is the "destructive interference" beam. It doesn't change the fact that, assuming coincidence in time of the two incoming photons, only the "constructive interference" beam will show an output photon detection. So I think your quick calculation answer is correct.
I had to contemplate a little on this to realize that all samples of a half-mirror use kind of the same value for the reflection phase shift of ##\pi## and I assumed the same. However in the HOM effect the incoming beams have no phase shift towards each other so the resulting interference is a different one that i want - i.e. C and D will detect evenly many events, however the two particles will always leave together in the same direction. So the HOM interference only eliminates the case where two particles leave in different beams. Therefore I need the additional phase shift or I end up just observing HOM. It is supposed to turn one creation operator for the D beam into an annihilation to make it vanish when combined with the operators from the other beam.

Okay I have figured that even if I filter perfectly coinciding photon pairs I won't entirely reproduce the coherent light beam results. Because using simple beams I should be able get get ##r_{CD}## to ##1## easily and adding any beam splitters reducing the intensity ahead won't change that (unless they introduce a not intended relative phase shift that is). However with individual photons it gets a little more complicated it seems. Without the beam splitters and filtering only the coindicing pairs I should still be able to reproduce ##r_{CD} = 1## but once the prefixed beam splitters are added this changes and my result seems to become a linear combination of the classical cases of two same beams in + only one beam + only the other beam in (the case of no beams is also part of the linear combo but doesn't affect ##r_{CD}## obviously). So assuming the ahead beam splitters are 50:50 for both incoming beams then that would reduce ##r_{CD}## to ##\frac 1 3##? Ist that correct?
 
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  • #8
PeterDonis
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individual particles it's pretty straight forward to deduct the uncoiciding photons since they won't experience any interference and will evenly distribute between C and D
If you don't have a way to separate the "no coincidence" events from the "coincidence" events, you don't have a way to separate the "no interference" results from the "interference" results. If you just lump them all together, you will compute an effective ##r_{CD}## that is somewhere between ##0## and ##1##, and you won't have any way of telling how much of that is due to lack of interference from the "no coincidence" events, which is predicted by the theory, and how much is due to lack of interference from the "coincidence" events, which is not predicted by the theory. You have to be able to distinguish the two types of events in order to test the theory at all.

by HOU you meant HOM effect?
Yes, sorry, that was a typo.

the HOM interference only eliminates the case where two particles leave in different beams
By itself, yes. But, as you note, by introducing an appropriate phase shift you can make all the "interference" events go to either C or D only, instead of being 50-50 split between them as in the basic HOM effect. At least, that's my understanding; I haven't done the detailed math to check. If I have time to soon I will.

Ist that correct?
I have no idea because I can't follow your jumble of words. If you have actually done the math, showing the math would help.
 
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If you don't have a way to separate the "no coincidence" events from the "coincidence" events, you don't have a way to separate the "no interference" results from the "interference" results.
I'm aware, but for now I not sure if I really need that for my purpose. Let's put that aside until it get's relevant.

I have no idea because I can't follow your jumble of words. If you have actually done the math, showing the math would help.
Okay so let's start with the classical optics first. neglecting leading amplitude factors if I have a 50:50 half mirror that has a matrix form ##\begin{pmatrix}i & 1\\1 & i\end{pmatrix}## such that multiplying a beam vector ##\begin{pmatrix}1\\ i\end{pmatrix}## gives ##\begin{pmatrix}2i\\ 0\end{pmatrix}## such that the outgoing beam C takes it all and leaves nothing for D. Let's make sure this general configuration is correct first before going any further. After that I want to check how the very same setup treats two coinciding photons.

... also really great that the Latex guide link is now right there next to the reply text field!
 
  • #10
Cthugha
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Now what happens when the both beam intensities are tuned down until there are only individual photons? Does the behavior at some point fundamentally change from the classical case – apart from the obvious that detectors will only measure individual photons instead of a continuous beam? [EDIT: of course the two incoming photons need to be assumed to be perfectly synced and arrive at the same time at the mirror] This is a case of two photon interference but unlike the setup of the HOM-effect there is phase shift in the incoming beams. If my quick mental calculation applying a displacement operator before the mirror matrix to the initial state is correct, there should still be no photons detected at D while C should measure both each time.
Just to make sure some common misunderstandings are avoided: If one reduces the intensity of a coherent beam to the point that there is on average only one photon per time range of interest changes absolutely nothing with respect to the expected results. Everything is still perfectly classical. Most importantly, it is fundamentally impossible to realize your scenario "the two incoming photons need to be assumed to be perfectly synced and arrive at the same time at the mirror". This is something that cannot be achieved by starting from a coherent beam. The waiting time distribution between two photon emission events is given by the waiting time distribution for emission events that are Poisson distributed and therefore cannot be synced in principle.

If one wants to do that syncing, one needs to use light sources that are intrinsically non-classical (single atoms, single ions, single quantum dots, heralded SPDC, ...). The difference is that for coherent beams you get one photon on average per time range of interest (buut there is no upper limit on the photon number present), while for non-classical light sources you get one photon at most. This is a huge difference. Most importantly, the latter choice of light source will give show the HOM effect, while a coherent beam tuned down in intensity will never do that and just show the same interference pattern you would also see at higher intensities.
 
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Just to make sure some common misunderstandings are avoided: If one reduces the intensity of a coherent beam to the point that there is on average only one photon per time range of interest changes absolutely nothing with respect to the expected results. Everything is still perfectly classical. Most importantly, it is fundamentally impossible to realize your scenario "the two incoming photons need to be assumed to be perfectly synced and arrive at the same time at the mirror". This is something that cannot be achieved by starting from a coherent beam. The waiting time distribution between two photon emission events is given by the waiting time distribution for emission events that are Poisson distributed and therefore cannot be synced in principle.
You are right when you are talking about experiments with a single coherent beam, but because in this setup there are two independent ones, tuning down the intensity down will eventually start to change the results quite massively. the interference relies on the presence of both beams but once we are down to individual photons the chances that a photon will find a coinciding partner to make the interference happen will steadily go to near zero thus the interference should disappear - since classically this just corresponds to one beam being switched off.

Therefore indeed a different photon production mechanism will be required - i.e. a different but analoge experiment. The classical reference experiment is just there for reference to make clear which type of mulit-particle interference I want to achieve/understand with individual particles. Anyhow, I am not so much interested in the practical implementation and challenges of such an experiment but rather the theoretical description. For now I just want to understand the mechanism how to achieve that kind of destructive interference with two photons.
 
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Cthugha
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You are right when you are talking about experiments with a single coherent beam, but because in this setup there are two independent ones, tuning down the intensity down will eventually start to change the results quite massively. the interference relies on the presence of both beams but once we are down to individual photons the chances that a photon will find a coinciding partner to make the interference happen will steadily go to near zero thus the interference should disappear - since classically this just corresponds to one beam being switched off.
No, it absolutely does not and this is a very common misunderstanding. Photons are not point particles, but all photons within one coherence volume are completely indistinguishable. Assuming infinite coherence length or time therefore is fully equivalent to assuming that both beams are derived from the same laser. There is absolutely no difference. Also note that this is not about "practical implementation". This is fundamental quantum optics. For the same reason, you can do boson sampling only with Fock states and not with weak coherent states. As long as you use coherent beams, there is no two-photon-interference at all and all that is taking place is single photon interference. Otherwise you could do all the boson sampling stuff simply with lasers.
 
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No, it absolutely does not and this is a very common misunderstanding. Photons are not point particles, but all photons within one coherence volume are completely indistinguishable. Assuming infinite coherence length or time therefore is fully equivalent to assuming that both beams are derived from the same laser. There is absolutely no difference. Also note that this is not about "practical implementation". This is fundamental quantum optics. For the same reason, you can do boson sampling only with Fock states and not with weak coherent states. As long as you use coherent beams, there is no two-photon-interference at all and all that is taking place is single photon interference. Otherwise you could do all the boson sampling stuff simply with lasers.
There is one significant difference: a photon coming from one beam is always perfectly time synced with itself, photons from different beams won't be. You can emulate the problem with a single beam like that: take an Mach-Zehnder interferometer, where the obstacle in one leg of the experiment is just an glass-fiber loop of random length to increase the photon travel time. This way the photon desyncs with itself and both beams miss each other in time when they arrive at the final half mirror so their interference will be gone. indistinguishably isn't the issue here, time sync is.
 
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Cthugha
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There is one significant difference: a photon coming from one beam is always perfectly time synced with itself, photons from different beams won't be. You can emulate the problem with a single beam like that: take an Mach-Zehnder interferometer, where the obstacle in one leg of the experiment is just an glass-fiber loop of random length to increase the photon travel time. This way the photon desyncs with itself and both beams miss each other in time when they arrive at the final half mirror so their interference will be gone. indistinguishably isn't the issue here, time sync is.
That is not correct. In order to have the beams miss each other, the delay introduced by the fiber loop needs to be longer than the longitudinal coherence length of the light field. The "desync" you mean occurs due to the delay being longer than the coherence length which results in the relative phase of the two light fields becoming random with respect to each other.

For using two different beams, you now have three options: you can perfectly synchronize their phase. Then they will also always show interference if you interfere them, so you essentially get physics equivalent to a single beam with infinite coherence time. You can leave them completely unsynchronized. Then you will not see any interference for experiments with integration times longer than the coherence time of the individual beams. However, if you repeat the experiment several times with integration times shorter than the mutual coherence time, you will see different interference patterns for each single integration. Averaging over all these measurements will then yield no interference at all. In all realistic cases, you will be able to maintain the phase synchronization for some finite duration. Then you will get absolutely the same scenario as when using the fiber loop and the two beams are fully comparable to a single beam with finite coherence time.

What one can do, is to consider the beamsplitter as a simple device with two input ports and two output ports and apply the beamsplitter transform to the input ports to see what comes out. When doing so, you will find that it makes a huge difference whether you take Fock states as the input (which do not have any well defined phase) or whether you use weak coherent states, which have some photon number uncertainty and which require to take the off-diagonal terms into account. The essential physics of the beamsplitter in quantum optics is well described in the "bible" ("optical coherence and quantum optics" by Mandel and Wolf).
 
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[...]
So you are saying that if i tune down my classical beam intensities down to 1 photon per minute, I will still see a interference, where my detector at D detects nothing and at C I only see a single photon at once every now and then?

I guess in your words at such low intensities the beams will lose coherence. I guess in practice this will be the dominant issue. Theoretically however you can still assume they are perfectly in phase but it won't affect the outcome in any way. I mean if one beam doesn't send out a photon for a minute then how is that distinguishable from the beam not being present at all?

Oh, and by "sync" I so far meant time coincidence of photon pairs, not phase coherence.

So in any case I want to reconstruct that interference with a minimal amount of photons. Yet it must be done with at least two separate photons for a specific reason: measurement kind of distinguishes between not entangled particles up to a higher level and messes with the probability amplitudes of their superpositions. The amplitudes of an independent photon however cannot be impacted by that - or if it is partially entangled then only up to a certain degree. I however want to know what part these amplitudes play in interference. Also interference it is not known to cause "measurment/wave function callapse" so far which is why I picked that example to study. For that I cannot just take the two beams as originating from a single one.
 
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Hmm, then again, you can reduce beam intensity using beam splitters. each one taking it down by half - so losing intensity exponentially this therefore won't take many splitters to reach a region where the chance of detecting a single photon in the remaining beam per set time interval gets very low/quite close to zero. In that region interference patterns between separate beams should start to break down, at least that's what I would assume - roughly proportional to the chance of at least one photon of each beams being present in the same time interval.
 
  • #17
Cthugha
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So you are saying that if i tune down my classical beam intensities down to 1 photon per minute, I will still see a interference, where my detector at D detects nothing and at C I only see a single photon at once every now and then?
Assuming that the relative phase of the two beams is stable and there is no additional noise: yes. The interference pattern depends only on the fields and not on the intensity at all

I guess in your words at such low intensities the beams will lose coherence. I guess in practice this will be the dominant issue. Theoretically however you can still assume they are perfectly in phase but it won't affect the outcome in any way. I mean if one beam doesn't send out a photon for a minute then how is that distinguishable from the beam not being present at all?
No, the beams will not lose coherence. They do not lose anything. Absolutely nothing changes but the intensity of the signal. The interference pattern is completely unchanged. Only the photon count rates will go down. The relative photon distribution at the output is unchanged, irrespective of whether you put a million photons per second there or one per year.

The whole notion of "sending out a photon" is potentially complicated because things do not work that way. The light field has some coherence time and that is the typical time scale during which there is a superposition of the photon being emitted or the photon still being in the source. Consider a simple laser with a gain medium and two mirrors, where the light goes back and forth. When using an unbalanced interferometer with different arm lengths, you will still see interference if the path difference is shorter than the coherence time of the light. In practice this means that you cannot distinguish whether the photon left the laser cavity earlier and took the long path or whether the photon made several more round trips, left the cavity later and took the shorter path. Things work in a similar manner for emission from individual atoms. The uncertainty of the exact moment of emission is given by the coherence time of the light field. At the low light levels you are interested in, you calculate probability amplitudes for the photon detection events. For the experiment you are interested in, these behave like the classical fields.

Oh, and by "sync" I so far meant time coincidence of photon pairs, not phase coherence.
These are not independent. You cannot localize a photon better than its coherence volume. Accordingly, you cannot force photons to be detected within a time interval shorter than their coherence time (unless you shorten the coherence time), which in turn influences phase coherence.

So in any case I want to reconstruct that interference with a minimal amount of photons. Yet it must be done with at least two separate photons for a specific reason: measurement kind of distinguishes between not entangled particles up to a higher level and messes with the probability amplitudes of their superpositions. The amplitudes of an independent photon however cannot be impacted by that - or if it is partially entangled then only up to a certain degree. I however want to know what part these amplitudes play in interference. Also interference it is not known to cause "measurment/wave function callapse" so far which is why I picked that example to study. For that I cannot just take the two beams as coming from a single source.
This sounds odd on many different levels. As planned, it will not work. In the standard interference pattern, there simply is no two-photon interference contribution for coherent light fields. Also, as emphasized several times, coherent light never yields single photons. The statistics is always Poissonian and one usually counts the number of photons per coherence time. Then it does not matter whether you choose this time scale as nanoseconds, minutes or years. I assume you want to have a photon number much smaller than 1 per coherence time. In that case, in each arm the probability to have 0 photons present will be huge. The probability to have 1 photon present will be small and the probability to have more photons present will be tiny. Now the probability distribution of where these photons will end up is given only by the probability amplitudes.

A different scenario would be to actively modulate the signal, so that you actively change the mean value of the photon number for different intervals of time. This is a very different animal and cannot in principle be achieved by attenuating light beams. This is what constitutes a non-classical light source. Many people mistakenly think that the mean intensity is what makes a light source quantum. This is not the case. What makes it quantum is its non-linearity. If a single photon source emits a photon, the probability to emit another one directly afterwards drops to zero. For a laser, no matter how high or low the probability to emit some photon at some instant is, the probability to emit another one directly afterwards stays unchanged. This is the defining property of these two light fields and this property is also crucial in determining whether one will find standard single photon interference or two-photon interference. Therefore, what you call syncing in principle cannot be realized by attenuating laser beams. You need non-classical light sources for that.
 
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  • #18
Cthugha
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Hmm, then again, you can reduce beam intensity using beam splitters. each one taking it down by half - so losing intensity exponentially this therefore won't take many splitters to reach a region where the chance of detecting a single photon in the remaining beam per set time interval gets very low/quite close to zero. In that region interference patterns between separate beams should start to break down, at least that's what I would assume - roughly proportional to the chance of at least one photon of each beams being present in the same time interval.
Sorry, I noted this part after posting my reply.

No, this does not change anything. Interference pattern between different beams only take place anyway, if they are phase-stabilized with respect to each other. Otherwise there is no interference at all. The interference pattern then is a standard single photon interference pattern caused by the interference of fields, which is completely independent of the mean photon number - however large or small that may be. For coherent beams, it is completely irrelevant whether the photons are present in the same interval or not. That is different for thermal or non-classical light fields, but for coherent fields there is no difference.
 
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No, the beams will not lose coherence. They do not lose anything. Absolutely nothing changes but the intensity of the signal. [...]
Hmm okay. I am not so sure, but if you say so, let's then take a more complex setup to test it out. for the nomenclature i'd call a beam merger a half mirror in a setup where it is supposed to take two coherent beams in and merge them via interference into one with a detector placed at the destructive leg to check. So that makes the Mach-Zehnder interferometer (MZI) setup the first mirror a splitter the second a merger.

Now let's start with a classical beam that is evenly splitt into n beams with each having an intensity about below a photon per specified time interval. let each single one of them run into an empty MZI setup. Assuming perfect conditions each beam fraction should pass its MZI without any loss, in particular no merger should measure anything at its destructive D leg.

Now take a single photon source producing a photon every now and then with identical frequency to our beam. analogously let that photon run through a cascade of splitters to fraction it into n-beams and than feed them each to a MZI like this ##><> \rightarrow ##. Even if the photon is out of phase with the main beam it will have the same displacement for each MZI. Besides any disturbance will do to make the destructive interference imperfect and since the amplitudes can be tuned to have similar magnitudes this should be detectable somehow - which the merger D-detectors are there to notice. And having n-MZIs yields a n-measurements to probe a single photon per time frame.

From such data you could derive information about a single probed photon, possibly stuff that cannot be linearly extracted (i.e. using a linear operator = observable) - and also with some time resolution. One quantity in particular, the probability amplitude of each of the probed photons n fraction beams, should be not measurable in any way. So such a setup shouldn't work like that simply.

Well, yeah sorry if this draft idea contains any major flaws. trying to come up with an example to test your assumptions isn't that trivial.
 
  • #20
Cthugha
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I am not sure I get your experiment correctly. Let me try.

Hmm okay. I am not so sure, but if you say so, let's then take a more complex setup to test it out. for the nomenclature i'd call a beam merger a half mirror in a setup where it is supposed to take two coherent beams in and merge them via interference into one with a detector placed at the destructive leg to check. So that makes the Mach-Zehnder interferometer (MZI) setup the first mirror a splitter the second a merger.

Now let's start with a classical beam that is evenly splitt into n beams with each having an intensity about below a photon per specified time interval. let each single one of them run into an empty MZI setup. Assuming perfect conditions each beam fraction should pass its MZI without any loss, in particular no merger should measure anything at its destructive D leg.
Do you mean that these beams enter a MZI and the other input port is empty? In this case: yes, assuming that one ideal stability is given and there is no stray light and there are no further perturbations, such a scenario is possible.

Now take a single photon source producing a photon every now and then with identical frequency to our beam. analogously let that photon run through a cascade of splitters to fraction it into n-beams and than feed them each to a MZI like this ##><> \rightarrow ##. Even if the photon is out of phase with the main beam it will have the same displacement for each MZI. Besides any disturbance will do to make the destructive interference imperfect and since the amplitudes can be tuned to have similar magnitudes this should be detectable somehow - which the merger D-detectors are there to notice. And having n-MZIs yields a n-measurements to probe a single photon per time frame.
Here, I am not exactly sure what you mean. Do you intend to have the single photon arrive at the MZI in addition to the coherent beam at the input port, which has not been used so far? If so: yes, this perturbation will be detectable.
As a little side note: a single photon has no phase in the standard sense. Their phase is completely undefined.

From such data you could derive information about a single probed photon, possibly stuff that cannot be linearly extracted (i.e. using a linear operator = observable) - and also with some time resolution. One quantity in particular, the probability amplitude of each of the probed photons n fraction beams, should be not measurable in any way. So such a setup shouldn't work like that simply.
Well, usually one would do things a bit differently. One would still mix the single photon and the coherent light field at the beam splitter, but one would rather place detectors at the output ports and measure the difference signal between the two outputs. This is called balanced homodyne detection and is one of the most sensitive detection schemes for light fields. It has been used to measure the quantum state of single photons (https://arxiv.org/abs/quant-ph/0101051) or squeezed light (https://link.springer.com/content/pdf/10.1038/387471a0.pdf).
 
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Here, I am not exactly sure what you mean. Do you intend to have the single photon arrive at the MZI in addition to the coherent beam at the input port, which has not been used so far? If so: yes, this perturbation will be detectable.
uhm... yeah, that was exactly the idea. Didn't expect that to work though :O.

Well, usually one would do things a bit differently. One would still mix the single photon and the coherent light field at the beam splitter, but one would rather place detectors at the output ports and measure the difference signal between the two outputs. This is called balanced homodyne detection and is one of the most sensitive detection schemes for light fields. It has been used to measure the quantum state of single photons (https://arxiv.org/abs/quant-ph/0101051) or squeezed light (https://link.springer.com/content/pdf/10.1038/387471a0.pdf).
...great, now I blame you for giving me a sleepless night of reading!

I mean this sounds quite fantastic but also quite weird. Because now I am not sure by what magic some measurement axioms of QM are supposed to hold up. Okay the problem I have with this setup is that I would expect the disturbance introduced by the fractioned probed photon to depend on the probability amplitude in the corresponding term in its superposition. but if that would be the case it would make that amplitude itself measurable which creates problems.

The idea with a single photon in a superposition like ##|a>+|b>+|c>## when ##|a>## hits a detector with a negative result the amplitude of the remaining terms is proportionally raised, so the remaining probability gets back to one (i.e. the photon has to be somewhere and its state needs to be normalized). For a single photon this has no impact on its further behavior because the ratios between amplitudes of the remaining superposition terms stay the same. But for interactions with another photon or a weak coherent beams this isn't exactly obvious. If it were the case and they were sensitive to such changes however you would be able to check when a remote part of a photons superposition was measured - even if that was on another side of the world. Bell experiments then suggest this information would travel super luminously fast - and you could measure it. So yeah, that's sounds unlikely.

Alternatively if measurement of any fraction of the probed photons superposition has no measurable effect on the disturbance of interference in the MZIs it would require the probability amplitude and its real EM-field accompanying it to be separate entities - the one collapsing would not necessarily trigger the other to do the same. But's that's also sacrilege as such behavior would introduce new information to the model, i.e. a hidden variable.

And for now I am too tiered to come up with a trick/mechanism to take the exact value of probability amplitude out of detectable physical results. Then again, forums are there to help people with questions :D
 
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  • #22
Cthugha
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...great, now I blame you for giving me a sleepless night of reading!
Always glad to be a bad influence! ;)

I mean this sounds quite fantastic but also quite weird. Because now I am not sure by what magic some measurement axioms of QM are supposed to hold up. Okay the problem I have with this setup is that I would expect the disturbance introduced by the fractioned probed photon to depend on the probability amplitude in the corresponding term in its superposition. but if that would be the case it would make that amplitude itself measurable which creates problems.
Actually, what is happening is quite boring in comparison. The disturbance or perturbation is quite small. It will only show up in a probabilistic manner. I will explain below.

The idea with a single photon in a superposition like ##|a>+|b>+|c>## when ##|a>## hits a detector with a negative result the amplitude of the remaining terms is proportionally raised, so the remaining probability gets back to one (i.e. the photon has to be somewhere and its state needs to be normalized). For a single photon this has no impact on its further behavior because the ratios between amplitudes of the remaining superposition terms stay the same. But for interactions with another photon or a weak coherent beams this isn't exactly obvious. If it were the case and they were sensitive to such changes however you would be able to check when a remote part of a photons superposition was measured - even if that was on another side of the world. Bell experiments then suggest this information would travel super luminously fast - and you could measure it. So yeah, that's sounds unlikely.
Well, what you get is a state, where you will find the photon at one of the detectors, while you find the vacuum state (this is an important thing in quantum optics) at all the other detectors. Now you get a superposition of the all of these possible states. You already mentioned the single photon case in the absence of any other states (usually called single rail qubit). Now for adding different beams at the other input port one has to consider how strong the perturbation introduced by the single photon will be. If you have a look on how the interference in a Mach-Zehnder-type interferometer changes with phase shifts, the non-intuitive thing is that operating it at full destructive interference makes it pretty insensitive to perturbations. If you have a look on how the intensity distribution changes with phase, you will find that it goes as the squared sine of the phase. Close to a minimum this is a pretty weak and sublinear dependence on the phase. For that very reason people using sensitive interferometers (think about gravitational wave detection and similar stuff) aim to avoid working at the minimum because it is pretty insensitive. Translating that to the MZI-scenario, this means that the perturbation caused by the single photon is also sublinear. This means that probabilistically speaking, it is not guaranteed that a photon will go to the side where destructive interference has been observed so far. It is rather something like "this will happen in 1 out of 100 cases". In practice, it is more probable that the single photon itself will go to the side where destructive interference took place so far as compared to a photon from the weak coherent beam going there due to the imbalance caused by the single photon.

Alternatively if measurement of any fraction of the probed photons superposition has no measurable effect on the disturbance of interference in the MZIs it would require the probability amplitude and its real EM-field accompanying it to be separate entities - the one collapsing would not necessarily trigger the other to do the same. But's that's also sacrilege as such behavior would introduce new information to the model, i.e. a hidden variable.
"No measurable effect" is terminology that gives rise to a lot of pit traps in quantum stuff. The possibility of "having no measurable effect in a single run of the experiment" and "having no measurable effect when the experiment is repeated a million times" are very different things. My personal advice on making sense of QM would be to take probabilities seriously.
 
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  • #23
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"No measurable effect" is terminology that gives rise to a lot of pit traps in quantum stuff. The possibility of "having no measurable effect in a single run of the experiment" and "having no measurable effect when the experiment is repeated a million times" are very different things. My personal advice on making sense of QM would be to take probabilities seriously.
Thanks for your patient answer. Yes, I know my way with probabilities quite well, at that is the mathematical field i come from. I know that the pitfalls to words like "measurable effect" in this regard.

The general idea of overcoming the problem of having to deal with probabilities is to apply the law of large numbers. Of course one way to do that is to repeat an experiment a lot of times. My hope however was to splitting the beam into n-fractions and measuring each fraction independently could also do the trick. But if i get right what you are saying is that this would only lead to one of the detectors measuring a disturbance (with some probability) while all the others would remain silent. In that scenario one would have to assume that the superposition would have to collapse upon the very first interference with the main coherent beam.

The balanced homodyne detection seems like a very powerful detection schema but I'll have to read and think a little more about it. The reason I wanted to work with complete destructive interference however is due to a rough idea: it allows to measure 0-detections, and is great because you can repeat it a million times without your original particle getting consumed, so it stays in the game. In that scenario reducing the chance of disturbing the interference is actually not unintentional as it allows for more repeats. And of course a lot of repeated measurements for a single entity allow again to call the law of large number, hence in the right configuration this could lead to an increased measurement accuracy. I just haven't figured out the proper configuration yet to make use of it. With the enlightment you have granted me a parallel MZI setup won't do me any good. Perhaps a serial setup however might do the trick?

One could try to serialize the MZIs like this: ##\rightarrow<>\rightarrow<>\rightarrow## where only the destructive D-outputs are measured while the C beams goes on into the next MZI. Then at each splitter one could inject another n-th fraction of the probed photon. If the superposition is indeed collapsed at each interference one would expect a uniform distribution at which MZI the disturbance is detected. However if the collapse isn't complete in some way the disturbance would become more likely to be detected towards the later positioned MZIs. Oh, and if one D-detector measures a photon - because it could be a photon from the main beam it leaves a chance that another disturbance will happen further down the road.
 
  • #24
Cthugha
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The general idea of overcoming the problem of having to deal with probabilities is to apply the law of large numbers. Of course one way to do that is to repeat an experiment a lot of times. My hope however was to splitting the beam into n-fractions and measuring each fraction independently could also do the trick. But if i get right what you are saying is that this would only lead to one of the detectors measuring a disturbance (with some probability) while all the others would remain silent. In that scenario one would have to assume that the superposition would have to collapse upon the very first interference with the main coherent beam.
Essentially yes. Whatever causes the first detectable event will destroy the superposition.

The balanced homodyne detection seems like a very powerful detection schema but I'll have to read and think a little more about it. The reason I wanted to work with complete destructive interference however is due to a rough idea: it allows to measure 0-detections, and is great because you can repeat it a million times without your original particle getting consumed, so it stays in the game. In that scenario reducing the chance of disturbing the interference is actually not unintentional as it allows for more repeats. And of course a lot of repeated measurements for a single entity allow again to call the law of large number, hence in the right configuration this could lead to an increased measurement accuracy. I just haven't figured out the proper configuration yet to make use of it. With the enlightment you have granted me a parallel MZI setup won't do me any good. Perhaps a serial setup however might do the trick?
Ah, I see where you try to go. Indeed there are approaches to measurements that rely on non-projective measurements. These yield less information per measurement, but also result in less perturbation per measurement. Such quantum-nondemolition experimets might be what you are after. A good example of what is possible in this direction has been presented by Nobel prize winner Serge Haroche a while ago: https://arxiv.org/abs/1107.4027
 
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Essentially yes. Whatever causes the first detectable event will destroy the superposition.

Ah, I see where you try to go. Indeed there are approaches to measurements that rely on non-projective measurements. These yield less information per measurement, but also result in less perturbation per measurement. Such quantum-nondemolition experimets might be what you are after. A good example of what is possible in this direction has been presented by Nobel prize winner Serge Haroche a while ago: https://arxiv.org/abs/1107.4027
Yes, I guess what I have in mind is a kind of a weak measurement.

And even though it makes a lot of sense that the multi-photon interference disruption at the half-mirror itself causes the destruction of the superposition and not the later measurement at an actual detector I do have to wonder if there are experiments confirming this behavior. From a data analyst perspective I've learned never to trust any model predictions in a region where it had no training data to learn from.

...and that supposed collapse at the interference was also a my initial reason to assume that when two coherent beams which are in phase meet at a beam splitter will stop showing an interference once they are reduced to a extremely low intensities (i.e. individual photon level). I mean it's a guesswork to what basis interference might collapse/project a superposition to but assuming a somewhat localized state would make it possible for photons to miss each other - that was my thought.

So maybe there is a chance it's not the interference but rather indeed the detectors that are responsible for the collapse (well, as in the first collapse that impacts the superpositions). In this case however a series of MZIs configuration like described in my earlier post would allow the probed photon to pass multiple MZIs before it's chance of hitting a detector becomes significant enough. That way it's disturbance of the original beam interferences would be allowed to grow at each iteration/MZI. Counting the iteration at which a disturbance is first detected plus the degree of disturbance in all following MZIs would thus allow for a rather accurate measurement of the amplitude of the photons probability amplitude term in its superposition - especially as this setup is in theory scaleable thus allowing to minimize the error to arbitrary level.
 

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