2-photon interference question

[vanhees71]

The great thing are effective theories, however. In quantum optics you get very far with describing the optical elements like lenses, mirrors, beam-splitters, polarizers, gratings, etc. in an experiment classically and just quantize the em. field. For a very good introduction to quantum optics, see

J. Garrison and R. Chiao, Quantum optics, Oxford University Press, New York (2008),
https://doi.org/10.1093/acprof:oso/9780198508861.001.0001.

 

[Cthugha]

Every interference experiment you have read about in any physics textbook is always interference of a single photon with itself, and the "classical" results you hear about are just histograms of a large number of single particle interferences.

From Nobel prize winner Roy Glauber, "Quantum Optics and Heavy Ion Physics" (https://arxiv.org/pdf/nucl-th/0604021.pdf):

" When you read the first chapter of Dirac’s famous textbook in quantum mechanics [8],however, you are confronted with a very clear statement that rings in everyone’s memory. Dirac is talking about the intensity fringes in the Michelson interferometer, and he says, Every photon then interferes only with itself. Interference between two different photons never occurs.
Now that simple statement, which has been treated as scripture, is absolute nonsense. First of all, the things that interfere are not the photons themselves, they are the probability amplitudes associated with different possible histories. You can obviously have different histories that involve more than one photon at a time. "

I have heard arguments that due to Bose-Einstein nature of photons one can in principle have two photons in exact same state, and because they are indistinguishable from each other, the two can interfere. That may or may not be true. In practice as far as I know nobody has ever managed to perform an experiment where they observe interference patterns which are clearly two particle interference and not a single particle interference (occuring twice)...

Your arguments were outdated already several decades ago. In fact you are postulating some non-mainstream personal theories here, which is not the purpose of these forums. With respect to two-photon interference, there are lots of papers on multi-photon interference.
For example, there is the standard Hong-Ou-Mandel experiment:
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.59.2044
There are detailed studies in the time domain:
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.93.070503
There is two-photon interference of two independently created photons
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.96.240502
Two-photon interference in the emission from two quantum dots:
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.104.137401
Two-photon interference between lasers and quantum dots:
https://www.nature.com/articles/nphys1373
Two photon interference between sunlight and quantum dots (although the title is somewhat overselling in my opinion):
https://journals.aps.org/prl/abstract/10.1103/PhysRevLett.123.080401
The purely classical Hanbury Brown-Twiss experiment is a good example of two-photon interference
https://www.nature.com/articles/1781046a0
as explained e.g. by Fano:
https://aapt.scitation.org/doi/abs/10.1119/1.1937827
Also the higher-order version of photon bunching of course is generally accepted. Back in the days of my diploma thesis I used to play around with it:
https://science.sciencemag.org/content/325/5938/297

Feel free to write rebuttals to all of these papers. This would be something to talk about here, but this is not a place for personal theories.

 

[Killtech]

Ah, here comes the cavalry... and brings a lot of firepower via quotes of so many two-photon experiments :). that should settle the argument I guess - unless someone is too sure of himself to ever challenge his believes unable to learn anything new.

Anyhow, back to my original questions. I am starting to doubt whether it might be in fact possible to measure the exact wave-function of a single photon itself to arbitrarily accuracy using such multi-photon interference setups - but a lot more tricky ones. given that unlike position or impulse the wave function represent actually the native information information to the quantum object it would be less destructive for the state of extract. But of course one has to use detectors at some point so that will ultimately impact and destroy it the original state.

The thing is that I do not see any principal theoretical or experimental aspect left that unambiguously prohibits one from doing that. I originally though that restricting observables to linear operators instead of general functionals would be that theoretical limitation but opening the state space into new dimensions (Fock space) really removes that problem. In the end one can linearize any problem at the expense of adding a lot of dimensions.

But that would open up a door to measure the effects of measurement - i.e. when it happens what and how it affects interactions. The thing is that interaction with beam splitters is unitary but it is not linear in the probability state space - i.e. it without breaking the state (no collapse) exposes the physical nature/interaction of probability amplitudes (talking from a probability theory perspective here; as in pure probabilities cannot do interference). But mathematically this should make it possible to precisely observe the impact of projection operations (measurement, perfectly linear in both the Fock space and in the probability space) when combined with probability-non-linear operations (interference) in setups that iteratively use both.

 

[Cthugha]

Sorry, I have been offline for some time. Some short first answers.

Right unitarity. The interference between a coherent beam and a single photon in a weird state is not a simple interaction but I guess it should be reversible still. So that's more leaning towards the interference not causing the probed photons superposition to break.

Interference is indeed not supposed to break anything. Usually anything, which qualifies as a von-Neumann (projective) measurement is supposed to "break" things. In optics, things are usually not that simple due to experimental limitations and sometimes one instead used postive-operator-valued measures (POVMs). Consider e.g. a detector that has finite detection probability. For an ideal detector, a non-detection event means that there really is only the vacuum present. For a detector with finite detection probability things become complicated.

that said, I have a dumb question... what happens if i capture a photon between perfect mirrors running in a loop. Let one mirror be a little imperfect with a marginal transition rate and a detector placed directly behind it. So each loop the photon has a marginal chance to escape and get detected. ... assuming i know when the photon entered i can measure the time it took it to get out and thus deduct how many loops it took to pass through the mirror. Now let's assume the photon was run through a beam splitter ahread so it already starts in a superposition like $|a>+r|rest>$ and only one output $|a>$ entered the loop - would it now take for the photon longer/more loops to get detected on average as it's probability amplitude was lower to begin with (if it gets detected at all, since that isn't guaranteed in that case)?? I mean sure guessing the p-parameter of a geometric probability distribution from merely a single sample is prone to yield a huge error (though it can be offset by sticking to a very small p parameter region ~ transmission coefficient) but still.

I am not sure I get what you mean. Say you use a long fiber loop for this purpose. Now the probability amplitude essentially just tells you in which percentage of repeated experiments the photon will enter the fiber loop. If it does, the dwelling time distribution inside the fiber loop will be the same, no matter how large or small the probability amplitude was.

Anyhow, back to my original questions. I am starting to doubt whether it might be in fact possible to measure the exact wave-function of a single photon itself to arbitrarily accuracy using such multi-photon interference setups - but a lot more tricky ones. given that unlike position or impulse the wave function represent actually the native information information to the quantum object it would be less destructive for the state of extract. But of course one has to use detectors at some point so that will ultimately impact and destroy it the original state.

Just be careful with the term wave-function, when it comes to photons. There are no wave-functions in the traditional sense due to photons being relativistic particles without rest frames. Howver, I guess I know what you mean.

But that would open up a door to measure the effects of measurement - i.e. when it happens what and how it affects interactions. The thing is that interaction with beam splitters is unitary but it is not linear in the probability state space - i.e. it without breaking the state (no collapse) exposes the physical nature/interaction of probability amplitudes (talking from a probability theory perspective here; as in pure probabilities cannot do interference). But mathematically this should make it possible to precisely observe the impact of projection operations (measurement, perfectly linear in both the Fock space and in the probability space) when combined with probability-non-linear operations (interference) in setups that iteratively use both.

I am not sure I exactly get what you mean. But yes, interference may give a lot of information. There is a reason for Boson sampling being a problem that cannot be calculated well classically.

 

[Killtech]

I am not sure I get what you mean. Say you use a long fiber loop for this purpose. Now the probability amplitude essentially just tells you in which percentage of repeated experiments the photon will enter the fiber loop. If it does, the dwelling time distribution inside the fiber loop will be the same, no matter how large or small the probability amplitude was.

Yes, I noticed the error in my question and figured it out myself. As no one answered and I was not able to edit my original post anymore so I left it as it was. I must admit that because QM isn't properly modeled via probability theory terms translating it myself for each experimental setup and making the correct probability model is sometimes a little tricky and I slip up every now and end up doing a stupid mistake.

Just be careful with the term wave-function, when it comes to photons. There are no wave-functions in the traditional sense due to photons being relativistic particles without rest frames. Howver, I guess I know what you mean.

Well, I would think one can try to describe the photon by the actual spatial distribution of the electro magnetic E B fields accompanying it. If I get the quantization of the EM field right you can actually think of a single photon as a pure classical wave solution to Maxwell that is normalized to the photons energy. I would furthermore guess its "probability distribution" $<\Psi|\Psi>$ would then have to correspond directly to the physical energy distribution of the fields.

I am not sure I exactly get what you mean. But yes, interference may give a lot of information. There is a reason for Boson sampling being a problem that cannot be calculated well classically.

Well, probability amplitude plays a role in interference which is not probabilistic in nature i.e. physical. On the other hand measurement is supposed to impact these amplitudes or more specifically the renormalization of the state. If that renormalization really happens it should have a measurable effect in the physical interactions it is involved in.

As in it actually transports energy and impulse. Of course when a photon goes through a beam splitter we only say the half its energy went one way and half the other by its expectation and if everything was probabilistic in nature this would only mean detection just updates our information and does not really transfer anything - as in it actually went one way or the other but we didn't know until we measured. but if those "probabilities" gets involved into phyisical interactions that isn't holding up anymore as the energy has to go indeed both ways until measured.

 

[Killtech]

Hmm, I have perhaps a simpler question to improve my understanding and I think I can put it into a simple experimental setup: Take a simple laser beam and tune it down to the level of a photon per minute. Now let that beam go through a beam splitter at which one end $B$ is a detector whiles the other outgoing beams ($A$) state is what i wonder about.

Now going through a 50:50 beam splitter I would expect the intensity of the beam to drop by half. But now I wonder how to treat the measurement at B: The measurement is supposed to change a superposition state $|A> + |B>$ into a probability distribution of those two states i.e. an ensemble of those two states. In particular a non-detection at $B$ and using a very classical wave function collapse assumption the superpositions fraction at $|B>$ would vanish and renormalizing the state back to $1$ would then restore the beams intensity as if it did not pass the beam splitter. In the other case of a detection the beams intensity should be severely reduced down to the probability of having another photon in the beam at the same time. So it would seem the $A$ beam would now have a stochastically distributed intensity rather then a deterministic one as it was just before the measurement??

But that would seem to make a huge impact in all followup interactions of that beam, no? Meaning that if it were to be interfered with another beam of matching phase and matching the expected intensity of $|A>$, it would normally be possible to achieve a perfect destructive interference at another beam splitter. But the measurement at $B$ now seems to introduce a massive perturbation to $A$ and with a jumping intensity that will never match that of the other beam a perfect interference would seem to become impossible to achieve. So in the measurement at $B$ would therefore result in a quite measurable remote interaction.

So where is my logical mistake?

EDIT: I just noted that in Quantummechanics a hidden variable interpretation like De Broglie-Bohm should yield a radically different prediction here. In particular the measurement at $B$ having no measurable effect as the particle and its carrying field are different entities. Thus detecting a particle would not impact its underlying field meaning that it would still be visible in the interference interaction independently to what happened to its particle. In this setup it would seem that this and the collapse interpretation yield quite different results - which would make them distinguishable and falsifiable.
... or it's just me that has a lot more to learn about QM :).

Well, in a single particle setup I understand now the mathematical reason why the interpretations would become equivalent in their predictions but with multiple particle on stage this mechanisms is much harder to make work so it opens up some questions.

 

[vanhees71]

A dimmed laser light never produces a pure one-photon Fock state. It's always (well approximated by) a coherent state with low intensity, i.e., it consists mostly of vacuum, a pretty small one-photon component but also (though even smaller) two- and more-photon parts.

I don't understand the rest of your setup, because a quantum state of the em. field always describes detection probabilities and nothing else and as such "ensembles".

 

[Killtech]

A dimmed laser light never produces a pure one-photon Fock state. It's always (well approximated by) a coherent state with low intensity, i.e., it consists mostly of vacuum, a pretty small one-photon component but also (though even smaller) two- and more-photon parts.

The laser doesn't have to produce a pure one-photon state - In that regard take $|A>$ as a quite complex multi-photon state. But whatever state it produces and the even more complex state after the beam splitter will have to go through a measurement which changes the state. All elements of the superposition within B will have to vanish in case of a non-detection. The state will be changed differently in case of a detection. The thing is that I don't know really know which from to assume for a dimmed laser state $|A>$ to calculate what effect measurement might do to it.

I don't understand the rest of your setup, because a quantum state of the em. field always describes detection probabilities and nothing else and as such "ensembles".

Here this the thing though: detection probabilities come in two flavors: those that do interference (superpositions) and those that don't (ensembles, real probability distributions). Given both their linear time evolution they are mostly indistinguishable other then during interference. Indeed however raw QM does not have that probabilistic layer which merely represents the state of our classical knowledge. Yet in modelling experiments where there are random interactions (detection) involved during the experiments this makes things easier.

EDIT: Thein again I guess I figured out my error. If the laser state is something of a superposition of many individual photons of the form $\Pi_i^n |\psi_i>$ with each $|\psi_i>$ having individually an extremely low detection probability - and respectively $\Pi_i^n (|\psi_{A,i}> + |\psi_{B,i}>)$ after the beam splitter. This allows to calculate the impact of measurement on the state and it becomes clear that with rising $n$ those states become very much alike for the case of a non-detection and a detection. Or in simpler words I just found out that detecting a photon at $B$ is an independent (probability-)event from detecting one in $A$ for such a beam, so whatever I measure at $B$ doesn't change detection chances or anything really about $A$. Thus you are right $|A>$ doesn't really end up being an ensemble unless there is an experimental way to limit the maximum photon number $n$ in a laser beam. Meh, that means such an experiment can only be done with individual photons which I guess will make it practically a lot more messy. Even so, the question still stands in that case.

 

[vanhees71]

There is no such distinction of states. I'm still not clear about the concrete scenario you have in mind. Are you referring to the difference between the superposition of two state vectors $|A \rangle + | B\rangle$ (the corresponding pure state is represented not by the vector but by the corresponding statistical operator, which in this case is a projection operator $|\psi \rangle \langle \psi|$ with a normalized vector $|\psi \rangle$) and a mixed state of the form $p_{A} | A \rangle \langle A| + p_B |B \rangle \langle B|$?

An ideal beam split is described by a unitary operator, applied to the vectors/statistical operators.

To say something concrete, I still lack the information about which specific experiment you have in mind.

 

[PeterDonis]

going through a 50:50 beam splitter I would expect the intensity of the beam to drop by half

No, you wouldn't. You would expect roughly half of the beam to go to A and half to go to B. That is not the same as the intensity of the beam dropping by half.

renormalizing the state back to would then restore the beams intensity

No, it wouldn't. If you are just looking at one photon detection event, there is no such thing as "beam intensity".

To measure beam intensity at all, you need to measure a large number of photon detection events over a period of time; the intensity is then the number of events divided by the time it took for them to occur. You would then find that roughly half of the events were at A and half were at B. The total beam intensity would not change.

that would seem to make a huge impact in all followup interactions of that beam

No. Once a particular photon has been detected at B, say, then obviously you need to use the wave function describing a photon detection at B for future predictions about that photon. But that says nothing at all about the wave function you would use for predictions about future photons to be emitted from the source.

detection probabilities come in two flavors: those that do interference (superpositions) and those that don't (ensembles, real probability distributions).

No. Detection probabilities are just detection probabilities. The variation in detection probability with other parameters (such as position on the detector screen) and the correlations between probabilities for different detection events are what tell you whether there is interference or not.

Also, you are conflating "interference" and "superposition". They are not the same thing. Whether or not a particular quantum state is a superposition is basis dependent. Whether or not interference occurs in a particular measurement setup is not basis dependent.

that means such an experiment can only be done with individual photons which I guess will make it practically a lot more messy. Even so, the question still stands in that case.

No, your question does not stand for the "individual photons" case (which would require a source producing Fock states or something close to them). See above.

 

[Killtech]

There is no such distinction of states. I'm still not clear about the concrete scenario you have in mind. Are you referring to the difference between the superposition of two state vectors $|A \rangle + | B\rangle$ (the corresponding pure state is represented not by the vector but by the corresponding statistical operator, which in this case is a projection operator $|\psi \rangle \langle \psi|$ with a normalized vector $|\psi \rangle$) and a mixed state of the form $p_{A} | A \rangle \langle A| + p_B |B \rangle \langle B|$?

An ideal beam split is described by a unitary operator, applied to the vectors/statistical operators.

To say something concrete, I still lack the information about which specific experiment you have in mind.

Ah, \rangle does the magic trick!

Anyhow beware my superior MS Paint skillz!

By $|A\rangle$ the state within a somewhat localization basis i.e. the state that passes through $A$ and in particular i expect the state vector to be $\frac 1 {\sqrt 2}(|A \rangle + | B\rangle)$ after it leaves the first beam splitter. The thing I ask is what happens to that state after it reaches B and what interaction follows in the following beam splitter which then is finally measured.

I want to know how the calculus handles the measurement during the experiment and explicitly want to see its effects calculated before it reaches the second beam splitter and its detectors - and compare it to the case where the detector at B is placed further back to measure only after the other detectors. In particular what happens to $p_A$ after measurement at $B$ since this number is actively taking part in the physical interaction at the second beam splitter.

And yes, Beam2 must carry at a photon from another source then Beam1. It might have been run through its own beam splitter (if its a single particle source rather then a laser) to ensure the intensities match $A$.

 

[Killtech]

No, you wouldn't. You would expect roughly half of the beam to go to A and half to go to B. That is not the same as the intensity of the beam dropping by half.

Well, that is not that easy and after a lengthy discussion with the optics expect... I kind would go with his detailed explanations backed up with various quotes of experimental papers, sorry.

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. [...]

 

[PeterDonis]

that is not that easy

Sure it is: you said a 50-50 beam splitter. That's what a 50-50 beam splitter means.

I kind would go with his detailed explanations

Nothing in what you quoted from @Cthugha contradicts anything I said.

 

[Killtech]

Sure it is: you said a 50-50 beam splitter. That's what a 50-50 beam splitter means.

No, you wouldn't. You would expect roughly half of the beam to go to A and half to go to B. That is not the same as the intensity of the beam dropping by half.

No, it wouldn't. If you are just looking at one photon detection event, there is no such thing as "beam intensity".

I was starting with a laser beams with very weak intensities but also considering the individual photon case.
In the way the calculus depicts a beam splitter by a unitary matrix the probability amplitude takes effectively a similar role of intensity by the effect it has in the interference. Changes in that quantity therefore impact the outgoing distribution. In a pure single photon scenario you cannot do that much with it but in a two photon experiment it's less clear.

No. Once a particular photon has been detected at B, say, then obviously you need to use the wave function describing a photon detection at B for future predictions about that photon. But that says nothing at all about the wave function you would use for predictions about future photons to be emitted from the source.

Huh? The problem is about that particular photon, an no future one! It is supposed to meet another photon at a follow up beam and impact its behavior. If that particular photon however vanishes at the detector in B the 2nd photon has no partner to interact with so will behave differently! Check the picture i posted, maybe it'll clarify.

No. Detection probabilities are just detection probabilities. The variation in detection probability with other parameters (such as position on the detector screen) and the correlations between probabilities for different detection events are what tell you whether there is interference or not.

Yeah, and there is a theory where all those words come from which gives a good framework to calculate all the probabilities and correlations in particular. now say you run a beam through a 50:50 beam splitter you get the regular result. but if you run it though another type of a 50:50 beam splitter that is in fact a 100% mirror that randomly changes its orientation between two options. At first the resulting detection probabilities and correlation will look the same for both types. But once you allow those beams to progress to a further experimental setup (like an MZ interferometer) they will produce vastly different results. If you have more complex random interactions in your experiment it just makes it a lot easier handle the time evolution of that system by adding another classic probability layer on top of the Fock state to account for these sources of classical randomness. That said for the second beam splitter type will yield probability distribution between the two states $|A\rangle$ and $|B\rangle$ with zero probability for every other state including something like $a|A\rangle + b|B\rangle$.

Now measurement, if one takes it by the axioms describing it, is supposed to introduce such a source of randomness. But yeah, you are right the joint probability distribution between the detection events are what tells whether there is interference or not (correlations alone may not always be sufficient). And obviously you can calculate that from a probability distribution over the Fock space taken which is then your probability space.

 

[PeterDonis]

I was starting with a laser beams with very weak intensities

Which will still result in individual photon detection events, and if you are just looking at one such event, that tells you nothing about intensities.

The problem is about that particular photon, an no future one!

There is no problem as far as the math and predictions of QM are concerned. The only problem is interpretation, and different QM interpretations will say different things about what happens to the state in one arm of the beam splitter (in your case, A) if a detection is made in the other arm (in your case, B). Any discussion of that problem belongs in a separate thread in the QM interpretations forum, not here.

say you run a beam through a 50:50 beam splitter you get the regular result. but if you run it though another type of a 50:50 beam splitter that is in fact a 100% mirror that randomly changes its orientation between two options.

Your second option isn't a "beam splitter", it's a mirror that is either there or not there. But if you are using a laser as your source, then you have no way of linking whether or not the mirror is there to the emission of a photon, since a laser does not record when a photon is emitted--it can't, since in general the laser will be in a superposition of having emitted or not emitted photons (this is an obvious consequence of the fact that the EM field state emitted by the laser is a coherent state, not a Fock state). So to make your second version of the experiment meaningful at all, you would need to use a photon source that (a) emits Fock states, and (b) "heralds" each emission so you can link the random choice of whether or not the mirror will be there to the heralding event.

Experiments involving photons are routinely done using "heralding" sources of this type, so the technology certainly exists. And since such an experiment produces extra data that is not produced in the first type of experiment--the "heralding" data and the random choice of which way the mirror was on each run--it is obviously going to allow you to look at statistics and correlations that you could not look at in the first type of experiment, where there is no such data.

The real question for this comparison is: if all you have is the output of the detectors downstream of the second beam splitter in your setup, would you be able to tell, from that data alone, which of the two types of setup were being used at the first beam splitter location? You appear to believe the answer is yes. Can you elaborate? What difference do you think you would see in the data from the detectors downstream of the second beam splitter, that would tell you which type of setup (50-50 beam splitter, or mirror there/not there by random choice) is being used at the first beam splitter location?

 

[vanhees71]

No, you wouldn't. You would expect roughly half of the beam to go to A and half to go to B. That is not the same as the intensity of the beam dropping by half.

I'm not so sure about this, though I'd need to do the calculation or look it up somewhere: If you have a coherent state entering a beam 50:50 splitter I'd expect to detect coherent states of half the intensity at the two "exits" of the beam splitter. That's the difference between a "dimmed laser" (low-intensity coherent state) and a true one-photon Fock state: In the former case there's some (though very small) probability to detect two photons (one at each exit of the beam splitter), while in the latter exactly this can never happen.

 

[vanhees71]

There is no problem as far as the math and predictions of QM are concerned. The only problem is interpretation, and different QM interpretations will say different things about what happens to the state in one arm of the beam splitter (in your case, A) if a detection is made in the other arm (in your case, B). Any discussion of that problem belongs in a separate thread in the QM interpretations forum, not here.

As long as you stay with standard QED (and the standard effective theory describing optical equipment like beam splitters in quantum optics, which is a very well established accurate approximation of a fully microscopic description, which fortunately is not needed here) the outcome of measurements (i.e., detection probabilities/photon statistics) are independent of the interpretation. So we don't need to deal with interpretation here, and if so, indeed one should do so in a separate thread in the interpretations forum (it'll lead to nothing anyway ;-)).

 

[Cthugha]

The laser doesn't have to produce a pure one-photon state - In that regard take |A> as a quite complex multi-photon state. But whatever state it produces and the even more complex state after the beam splitter will have to go through a measurement which changes the state. All elements of the superposition within B will have to vanish in case of a non-detection. The state will be changed differently in case of a detection. The thing is that I don't know really know which from to assume for a dimmed laser state |A> to calculate what effect measurement might do to it.

In addition to all the correct responses that have been given already, just a short additional one: Splitting a coherent light beam and performing measurements on one of the outputs will do absolutely nothing to the state of the light field in the other output beam. This is the evry definition of a coherent beam. A coherent beam follows a Poissonian phootn number distribution, which is the distribution for independent events. Accordingly, if you have a photon detection event, also the conditional mean photon number of the remaining beam does not change at all. Even for a beam that contains only one photon per minute on average, there is still a finite probability for having two photons. For a coherent beam, the probability for that is exactly so large that the conditional photon number for the remaining beam after one detection event stays exactly at one photon per minute. All detection events are statistically completely independent foor coherent beams.

 

[PeterDonis]

If you have a coherent state entering a beam 50:50 splitter I'd expect to detect coherent states of half the intensity at the two "exits" of the beam splitter.

Yes, for that particular case, it makes sense to talk about the intensity being split in half, half in each output beam of the beam splitter.

But the OP was not limiting his claim to that case.

 

[vanhees71]

The problem is that the OP has not defined the setup in clear terms. On the other hand the idea with the "dimmed laser" came up, and this has to be distinguished very clearly from the case of single-photon states (see @Cthugha 's very clear explanation in posting #53 too).

 

[Killtech]

As long as you stay with standard QED the outcome of measurements (i.e., detection probabilities/photon statistics) are independent of the interpretation. So we don't need to deal with interpretation here, and if so, indeed one should do so in a separate thread in the interpretations forum (it'll lead to nothing anyway ;-)).

The real question for this comparison is: if all you have is the output of the detectors downstream of the second beam splitter in your setup, would you be able to tell, from that data alone, which of the two types of setup were being used at the first beam splitter location? You appear to believe the answer is yes. Can you elaborate? What difference do you think you would see in the data from the detectors downstream of the second beam splitter, that would tell you which type of setup (50-50 beam splitter, or mirror there/not there by random choice) is being used at the first beam splitter location?

Well, this is not exactly what I believe but rather what I am struggling to understand how it is supposed mathematically avoid that - admittedly because I probably either don't understand the calculus of optical interference or measurement well enough.

Speaking a litte more general besides the my specific experiment, the underlying issue i struggle with interference is the following: For a single particle an operation representing a half mirror acts like this on a beams i call here a and b: $\begin{pmatrix}a \\b\end{pmatrix} \mapsto \frac 1 {\sqrt 2} \begin{pmatrix}1 & i\\i & 1\end{pmatrix} \begin{pmatrix}d \\e\end{pmatrix}$. Now written as an operator acting on the entirety of the state space and assuming the particle was already beam split into multiple beams like $a|A\rangle + b|B\rangle + c|C\rangle>$ (where B and C were are the result of another beam split) the operator should look like this:
$\begin{pmatrix}0 & 0 & 0 & 2^{-0.5} & i 2^{-0.5}\\0 & 0 & 0 & i 2^{-0.5} & 2^{-0.5}\\0 & 0 & 1 & 0 & 0\\0 & 0 & 0 & 1 & 0\\0 & 0 & 0 & 0 & 1\end{pmatrix}$ with dimensions being $\begin{pmatrix}A\\B\\C\\D\\E\end{pmatrix}$ and my initial state $\begin{pmatrix}a\\b\\c\\0\\0\end{pmatrix}$.
Now as far as I understand in order to achieve perfect destructive interference at D, one condition would have to be $|\frac {a} {b}|=1$, otherwise one beam will be stronger then the other allowing a few particles to cross into D. But that would make the ratio of $a$ and $b$ act physically with measurably different results. As for the "fake mirror beam splitter" I mentioned, it will always preserve the input amplitude value so anything sensitive to an amplitude ratio should immediately notice i.e. the amplitude of the quantum state is not actually halved since that is only done within the pure probability space above.

Now I understand that for a single particle measurement calculus will keep all ratios like $|\frac {a} {b}|$ constant, so nothing can be done with it. For a laser beam i learned that the detection outputs behind a beam splitter will be independent (thus no correlation) so this cannot be used either. For a two particle interference however the amplitudes of different particles can be independent of each other with respect to measurement which would seem to open a way to modify their ratio via measurement alone. I don't understand the 2-photon calculus well enough but if one can achieve destructive interference with two particles and the probability amplitudes are similarly involved in determining the output distribution then I don't understand how the measurement calculus is capable of handling this without producing detectable disturbances - or without the use additional variables.

 

[PeterDonis]

I probably either don't understand the calculus of optical interference or measurement well enough.

You seem to be using an incorrect representation of both the degrees of freedom in the state space and the operation of the beam splitter.

Take just a single beam splitter. It has two input arms and two output arms, and the operator describing it has to tell us how the amplitude in each input arm gets split between the two output arms. However, the amplitude in each input arm is not just a single amplitude, because the particles we are testing have spin, which we are considering to be spin 1/2 (photon polarization can be treated this way at least for this discussion), so each particle has two spin degrees of freedom. That means the total state space that the beam splitter acts on has four degrees of freedom (spin up/down for each input arm, mapped to spin up/down for each output arm), so the operator describing the beam splitter has to be a 4 x 4 matrix, not a 2 x 2 matrix.

If we want to look at the state in a particular arm of a particular beam splitter, we then have to take the appropriate two degrees of freedom from the four that describe the total state coming into or going out of that beam splitter. In a multiple beam splitter experiement, we can then take those two degrees of freedom and combine them with two others to get the full input state for the next beam splitter.

Now let's apply the above to your setup. At the first beam splitter, we only have nonzero amplitude coming in on one arm. So the 4-component input vector will look something like this:

$$
\begin{pmatrix} a \\ b \\ 0 \\ 0 \end{pmatrix}
$$

It's worth taking a moment to unpack what this vector means. It means we have amplitude $a$ for spin up in the first input arm, amplitude $b$ for spin down in the first input arm, and amplitude $0$ for spin up and spin down in the second input arm.

The beam splitter operator matrix will look like this:

$$
\frac{1}{\sqrt{2}} \begin{pmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 0 \\
0 & 1 & 1 & 0 \\
1 & 0 & 0 & 1
\end{pmatrix}
$$

As you can easily verify, the result of applying this operator to the input vector is:

$$
\frac{1}{\sqrt{2}} \begin{pmatrix} a \\ b \\ b \\ a \end{pmatrix}
$$

Can you see how this describes what you already intuitively know that the beam splitter does?

Now, your setup has a second beam splitter downstream of the "A" output arm of the first. The full input state of that beam splitter will be the first two components of the output vector above (can you see why?), combined with two new amplitudes coming in on the other arm, so it will look like this:

$$
\begin{pmatrix} \frac{1}{\sqrt{2}} a \\ \frac{1}{\sqrt{2}} b \\ c \\ d \end{pmatrix}
$$

Now you can just apply the same beam splitter operator as above to this input state to get the output state, and the two pairs of components of that output state will tell you what to predict for each detector.

 

[Killtech]

You seem to be using an incorrect representation of both the degrees of freedom in the state space and the operation of the beam splitter.

Take just a single beam splitter. It has two input arms and two output arms, and the operator describing it has to tell us how the amplitude in each input arm gets split between the two output arms. However, the amplitude in each input arm is not just a single amplitude, because the particles we are testing have spin, which we are considering to be spin 1/2 (photon polarization can be treated this way at least for this discussion), so each particle has two spin degrees of freedom.

There were no additional degrees of freedom in the operator over the entire space I wrote. I was just trying to represent the total time evolution operator for the entire system in discrete time, which is why it had to include all possible modes. but I'm fine handling each beam splitter individually as you did - the math stays the same anyway. So let's leave it at that.

Fair point about the photon polarization, though. I didn't clarify that and frankly I forgot about it but since I am looking for interference having a preset linear polarization to deal with makes sense. And in my defense that's what most sources in on the internet do too, so eliminating the modes $b=0$ and $c=0$ in your post we can reduce the matrix to 2x2 for the only present polarization. Anyhow your beam splitter matrix is missing two $-1$ values somewhere i think otherwise it allows total absorption with all beams vanishing.

And your example displays the problem I am struggling with quite well: At the second beam splitter the amplitudes going in is $a'=\frac 1 {\sqrt 2} a$ and $d$. Now a perfect destructive interference will show if $\frac 1 {\sqrt 2} (a'+d)=0$, right? So let's pick $d=-\frac 1 {\sqrt 2} a$ to have that. Now not a single photon will ever be able to reach the detector at the destructive leg. However, if you replace the first beam splitter by a "fake mirror" one, then for a single photon $a'$ is going to be either $a$ or $0$ with a 50:50 chance. So in neither case a perfect interference can show: in the first case the supposedly destructive leg will still have an amplitude of $\frac 1 {\sqrt 2} (a'+d) = \frac 1 {\sqrt 2} (1- \frac 1 {\sqrt 2}) a$ $= \frac {\sqrt 2 - 1} 2 a$ and in the other case it's $\frac 1 {\sqrt 2} (a'+d) = - \frac 1 2 a$. Averaging the two resulting detection probabilities over the two possibilities will always yield a detection rate larger then zero, since both probabilities are positive definite and the detected particle numer is also a positive integer. Or in expressed in numbers that's $(0.5 \frac {(\sqrt 2 -1)^2} 4 + 0.5 \frac 1 4)|a|^2 \neq 0$ (if $a\neq 0$) chance of detecting a photon in the fake splitter setup. Therefore using the different types of beam splitters should result in different detection statistics. The basic idea of this setup is to detect deviances in an amplitude from a reference one.

However the fake mirror beam splitter acts differently then measurement does: it will only affect the amplitude $a'$ whilst leaving $d$ unchanged, while measurement will scale both $a'$ and $d$ proportionally so no difference will be observable still (i.e. the destructive interference will be uphold in each of the two possibilities). Yeah, my setup cannot work if the reference amplitude isn't kept constant. This is why a two photon interference becomes of interest: detection of one photon should not affect the amplitude of the other photon.

 

[PeterDonis]

I'm fine handling each beam splitter individually as you did - the math stays the same anyway

I don't see how what you wrote is the same mathematically as what I wrote. What you wrote doesn't make sense to me.

your beam splitter matrix is missing two $-1$ values somewhere i think otherwise it allows total absorption with all beams vanishing

I think I got the signs right but it's possible I flipped some. I'll have to check when I have time.

 

[vanhees71]

I'm not familiar with your $4 \times 4$ matrix representation. Here's an old posting of mine, using the standard approach with $2 \times 2$ matrices:

https://www.physicsforums.com/threa...nce-in-an-interferometer.975742/#post-6216820

 

[Killtech]

I don't see how what you wrote is the same mathematically as what I wrote. What you wrote doesn't make sense to me.

I think I got the signs right but it's possible I flipped some. I'll have to check when I have time.

Your matrix has a zero eigenvalue with eigenvectors $\begin{pmatrix}1 & 0 & 0 & -1\end{pmatrix}^T$ and $\begin{pmatrix}0 & 1 & -1 & 0\end{pmatrix}^T$ i.e. that's a total absorption for both polarizations. Therefore I assume you meant:
$$
\frac{1}{\sqrt{2}} \begin{pmatrix}
1 & 0 & 0 & 1 \\
0 & 1 & 1 & 0 \\
0 & 1 & -1 & 0 \\
1 & 0 & 0 & -1
\end{pmatrix}
$$
which is what I used in the calculations in my previous post. So that matrix would corresponds to a $\frac \pi 2$ phase shift on reflection. Indeed my mirror used a $\frac \pi 4$ phase shift, so sorry, you are right it's not mathematically the same. However projecting it down to a single polarization subspace doesn't change anything mathematically - because the spaces are independent and can be cleanly separated.

 

[vanhees71]

I've no clue what your $4 \times 4$ matrices mean. There are two incoming photon modes and two outgoing ones. So you need only a $2 \times 2$ matrix as explained in any textbook of quantum optics, or are you referring to something else?

 

[Killtech]

I've no clue what your $4 \times 4$ matrices mean. There are two incoming photon modes and two outgoing ones. So you need only a $2 \times 2$ matrix as explained in any textbook of quantum optics, or are you referring to something else?

each photon beam is composed of two independent components, that makes 4 degrees of freedom for 2 beams. See photon polarization. I haven't seen a 4x4 matrix for beam splitters that often on the internet but it makes sense to me when one has to deal with different incoming polarization states in an experiment. In the 2x2 matrix form I suppose you would have trouble explaining why two orthogonal polarizations don't interfere even though their phases are right.

 

[vanhees71]

Hm, in the usual formalism that's all included in the annihilation operators used to describe the beam splitter for arbitrary modes. That's why in any description of a beam splitter usually there's only a $2 \times 2$ matrix. See my summary of the standard treatment here:

https://www.physicsforums.com/threa...nce-in-an-interferometer.975742/#post-6216820

for more details see

https://journals.aps.org/pra/abstract/10.1103/PhysRevA.40.1371

 

[Killtech]

Hm, in the usual formalism that's all included in the annihilation operators used to describe the beam splitter for arbitrary modes. That's why in any description of a beam splitter usually there's only a $2 \times 2$ matrix. See my summary of the standard treatment here:

Yes, well, this is just because I am great at confusing people. The 4x4 matrix probably came up because I tried to write the total time evolution operator for my setup - just so the presence of other beams outside the beam splitter is not forgotten... and I guess this confused PerterDonis what those additional dimensions were supposed to mean - and the degrees of freedom that come with it. Hence I guess he wrote it down in a way to point out which degrees of freedom he sees there. Technically though the additional ones in my matrix are those of the total experiment setup which has a few more then just those needed for the interaction at the beam splitter. But I also made an honest mistake: I (intuitively) chose an identity matrix to say that the beam splitter does affect anything about beams that aren't its input... which would however mean the photons amplitude in those beams would be trapped in place and could "interfere" with anything coming up in the next time step.

Anyhow let's skip this and get back to discussing the experiment.

 

[PeterDonis]

I've no clue what your $4 \times 4$ matrices mean.

As @Killtech said, I was attempting to capture two polarization degrees of freedom for each input/output arm of the beam splitter. However, I agree that if you don't measure polarizations in the experiment you can ignore that degree of freedom and just use a $2 \times 2$ matrix to represent the beam splitter operator.

 

[vanhees71]

I still don't get it. Do you have a paper/book, where this convention is used? The formalism, I summarized in

https://www.physicsforums.com/threa...nce-in-an-interferometer.975742/#post-6216820

includes the polarization in the symbol of the annihilation operators, i.e., $\hat{a}_1$ stands for $\hat{a}(\vec{k}_1,\lambda_1)$, where $\lambda_1 \in \{1,2 \}$. It refers to a mode with wave number (momentum) $\vec{k}_1$ and a polarization state labelled by $\lambda_1$ and $\lambda_2$ (e.g., you can take some linear polarization states or the circular or elliptical ones). Of course the transfer matrix depends on the specific polarization states, determined by the specifically used beam splitter.

 

[PeterDonis]

Do you have a paper/book, where this convention is used?

Who are you asking? If you are asking me, I'm no longer sure that the $4 \times 4$ matrix I gave correctly captures the polarization degrees of freedom, but anyway I've already agreed we can just ignore them for this discussion.

Of course the transfer matrix depends on the specific polarization states, determined by the specifically used beam splitter.

Can you give an example of a transfer matrix that changes the polarization states? For example, that takes some of the amplitude coming in on an input arm in the "right circular polarization" degree of freedom and transfers it to the "left circular polarization" degree of freedom in an output arm?

That kind of transfer matrix was what I was attempting to capture (quite possibly incorrectly). I don't see how you can capture that in a $2 \times 2$ matrix, since each input arm now has to have two degrees of freedom, not one (the momentum is fixed on each input arm, but now you have two polarization basis states).

 

[Killtech]

In addition to all the correct responses that have been given already, just a short additional one: Splitting a coherent light beam and performing measurements on one of the outputs will do absolutely nothing to the state of the light field in the other output beam. This is the evry definition of a coherent beam. A coherent beam follows a Poissonian phootn number distribution, which is the distribution for independent events. Accordingly, if you have a photon detection event, also the conditional mean photon number of the remaining beam does not change at all. Even for a beam that contains only one photon per minute on average, there is still a finite probability for having two photons. For a coherent beam, the probability for that is exactly so large that the conditional photon number for the remaining beam after one detection event stays exactly at one photon per minute. All detection events are statistically completely independent foor coherent beams.

I really want to know the mathematical model for this. So how do you calculate that it is Poisson distributed?

With my own understanding I would say it should work like this: now a single photon (partial) beam with probability amplitude $a= \sqrt p$ i.e. $\sqrt p |1\rangle$ will be simply Bernoulli distributed with parameter $p$ when it runs into a detector. Now if i have a beam with multiple independent identically distributed photons I get a Binomial distribution and I would write the beam down as $a |n\rangle$ $=\sqrt {\frac p n} \prod_{i=0}^n |A_i>$ with each $|A_i\rangle>$ still Bernoulli distributed with parameter $\frac p n$ and independent of the other $|A_i\rangle$. The beam is tuned down to an amplitude of $\sqrt {\frac p n}$ such that the expected photon number is the same as for the single photon. With this using the Poisson limit theorem it follows that $$\lim_{n \rightarrow +\infty} \sqrt {\frac p n} \prod_{i=0}^n |A_i>$$
is Poisson distributed with $\lambda = lim_{n \rightarrow +\infty} n \frac p n$ $=p$ and with this it is easy to verify that it yields exactly the behavior you said with a beam splitter. That is: detection events with such beams will be always statistically independent due to $n$ being infinite (since taking one particle out of an infinite pool via measurement leaves the pool unchanged) and therefore conditional particle numbers stay unaffected by detections. This will however only hold for as long $n$ is a large number though I guess this is probably always true for a laser beam?

This makes a lot of sense to write down a coherent laser beam in such a way as it enables me to properly calculate its behavior at beam splitters and other devices but I wager this isn't a standard way to denote and calculate such things. So how do you do it normally?

 

[vanhees71]

It's sufficient to consider this for a single mode. Then you have basically just a harmonic oscillator with one annihilation and one creation operator $\hat{a}$ and $\hat{a}^{\dagger}$, obeying the commutation relation
$$[\hat{a},\hat{a}^{\dagger}]=\hat{1}.$$
One complete orthonomal set are the eigenvectors of the number operator
$$\hat{N}=\hat{a}^{\dagger} \hat{a}.$$
These Fock states are $|n \rangle$ with the eigenvalues $n \in \mathbb{N}_0$ and
$$\langle n|n' \rangle=\delta_{nn'}.$$
The annihilation operator acts as
$$\hat{a} |n \rangle=\sqrt{n} |n-1 \rangle.$$
Now a coherent state is an eigenstate of the (NOT self-adjoint!) annihilation operator with complex eigenvalues $\alpha \in \mathbb{C}$. To evaluate it we set
$$|C_{\alpha} \rangle=\sum_{n=0}^{\infty} c_n |n \rangle.$$
Then we have
$$\hat{a} |C_{\alpha} \rangle=\alpha |C_{\alpha} \rangle=\sum_{n=1}^{\infty} \sqrt{n} c_n |n-1 \rangle=\sum_{n=0}^{\infty} \sqrt{n+1} c_{n+1} |n \rangle = \sum_{n=0}^{\infty} \alpha c_n |n \rangle.$$
This leads to the recursion
$$c_{n+1} = \frac{\alpha}{\sqrt{n+1}} c_n \; \Rightarrow \; c_{n} = \frac{c_0}{\sqrt{n!}} \alpha^{n}.$$
So we have
$$|C_{\alpha} \rangle=c_0 \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle.$$
For the norm we get
$$\langle C_{\alpha}|C_{\alpha} \rangle=|c_0|^2 \sum_{n=0}^{\infty} \frac{|\alpha|^{2n}}{n!}=|c_0|^2 \exp(|\alpha|^2) \; \Rightarrow\; c_0=\exp(-|\alpha|^2/2).$$
So finally
$$|C_{\alpha} \rangle=\exp(-|\alpha|^2/2) \sum_{n=0}^{\infty} \frac{\alpha^n}{\sqrt{n!}} |n \rangle.$$
What's now the probability to find exactly $n$ photons? Just use Born's rule:
$$P(n)=|\langle n|C_{\alpha} \rangle|^2 = \exp(-|\alpha|^2) \frac{|\alpha|^{2n}}{n!},$$
which is a Poisson distribution with mean $\langle N \rangle=|\alpha|^2$.