So we have two more cases to consider:

Case 3: we pulse the laser with very short pulses, emitting only one photon at a time. This is actually not possible with a laser, but it is possible with something like this single-photon-emitting light source (which was actually case 5 when I first made up the list).

Before analyzing this case I have to hedge: I'm pretty sure I know the right answer, but not 100%. I have actually asked one of the authors of that paper to confirm my suspicions but he's busy and so it will be a while before I expect to get an answer. If his answer turns out to be substantially different from what I say here I will definitely let you all know.

So with that disclaimer in mind, I'm quite confident that this case will turn out to have the same behavior as the case where we have a laser dimmed by a filter,

*and*the photon emissions post-filter detected using a parametric down-converter, i.e. there will be no (first-order) interference (with one exception, which will arrive at in case 4). The reason I'm confident is that this case is structurally the same as that one: we have made a modification to the experiment that allows us to know when the photon was emitted, which allows us to determine which path the photon took by comparing departure and arrival times, and so there can be no interference. Please note that I have taken pains

*not*to say that our (potential) knowledge of the emission time

*causes*the interference to go away. It doesn't. What

*causes*the loss of interference is the entanglement of the photon with something else. Entanglement is a pre-requisite for measurement, which is a pre-requisite for knowledge, so it is true that if there is (potential) knowledge that there can be no interference, but the knowledge bit is a red herring. The only reason I'm using that phraseology is that it is, sadly, ubiquitous in QM pedagogy.

So, with that out of the way, let us proceed to the interesting case: We turn the laser on and off with a duty cycle of 50% and a period that is long enough that the pulses actually do produce interference. The lengths of the arms of the interferometer are adjusted so that the Nth pulse coming from the long arm exactly coincides at the detector with the N+Kth pulse coming from the short arm for some integer K>=1.

Before I go into detail on this I want to say a word about how practical such an experiment might be. To make this work, the period of the pulses has to be long enough that the individual pulses are coherent, but short enough that we can "store" at least one them "in flight" while waiting for the next one without losing coherence. Can this actually be done? Yes, it can, at least if you believe Wikipedia. There is says that fiber lasers can be built with coherence lengths exceeding 100km. That's about 300 microseconds, which is essentially forever by the standards of quantum optics. You could actually do this experiment with a regular semiconductor laser with a coherence length of "merely" 100m. It's pretty straightforward to power-cycle a laser, produce time delays, and measure the results with events happening at nanosecond/meter scales. So this would not even push the boundaries of the state of the art.

There is no doubt what would happen in this case: you would see interference, but (and this is really important)

*only after the Kth cycle*. Before that we know that all of the light at the detector arrived via the short arm, so there is no interference. After the Kth cycle, light is arriving from both arms, so we do get interference. There would be some transient effects at the beginning and end of each pulse, but at steady-state the interference would be easily detectable. This is predicted both by quantum theory and classical E-M theory. There is absolutely nothing interesting going on here, until, that is, you make one more little modification: in addition to a 50% duty cycle, you

*also*dim the laser with a filter.

The outcome predicted by QM is clear: dimming the laser with a filter makes no difference. Whatever we saw when the laser was bright we should also see when the laser is dim, namely, for the first K cycles of the laser we get no interference.

How is this possible? The dramatic narrative of interference usually goes something like this: interference happens when a photon (or some other particle, it doesn't really matter) is placed in a quantum superposition, usually a quantum superposition of physical locations. The usual slogan is "the photon goes both ways". The two paths are then brought together in such a way that no which-way information is available in the final state. The result is interference.

But can this possibly be happening in this case? The two paths that the photon can take are separated by an enormous amount of

*time*, big enough that we are able to turn the laser off and back on while we are waiting for it to travel the long way 'round through our interferometer. It's already enough of a mind bender to say that we cannot know when a particular photon was emitted when the laser is on continuously, but now we seem to be going a step further: in order to have interference, we cannot even know

*which power cycle*a detected photon was produced by! Is it really possible for a laser to produce a photon that is in a quantum superposition

*across power cycles*? That seems extremely improbable. Surely once you turn the laser off, the universe is committed: there's a batch of photons flying through space at the speed of light in some particular quantum mechanical configuration. Surely that configuration can't be changed by something that happens (or not -- we could choose at any time not to turn the laser back on!)

*in the future*?

There is another possibility. Maybe the interference we see is not created by one photon interfering with itself, but rather interfering with another photon produced in a different cycle. This seems a lot more plausible, but is it actually possible? Paul Dirac, one of the founders of quantum theory, once famously wrote “...each photon then interferes only with itself. Interference between different photons never occurs.” Interestingly, while I can find this quote all over the internet, and it is invariably attributed to Dirac, I cannot locate its original source. So it's possible that this quote is apocryphal. But it doesn't matter. What matters is that this sentiment was taken seriously for decades until an experiment by Magyar and Mandel debunked it in 1963. There have been books written about, and even entitled, multi-photon interference. So it's definitely a thing.

The experiment as we have described it to this point is an interesting variant on the Magyar and Mandel experiment. There they used two different lasers (actually they were masers, but it doesn't matter) to generate their photons, whereas we are using one laser and separating the production of the two photons by time using power cycles and bringing them back together using delay lines. But it amounts to the same thing. The key is that we're bring the photons back together

*at the same time*. That's the reason that the delay time in the interferometer is an integer multiple of the cycle time on the laser, otherwise it doesn't work.

So this might be a mildly interesting but not earth-shattering result. Maybe someone has even done it, I don't know.

But there is one thing that should make us a little queasy about this line of thought, and that is that

*we cannot actually control*how many photons enter the interferometer on any given cycle. We can attenuate the beam so that

*on average*we get one photon per cycle, but that will only be an average. Some cycles we will get more than one, some cycles we will get none. If we're depending on photons to interfere

*with each other*then we need the

*same number*of photons each cycle, otherwise some photons won't have partners to interfere with.

In fact, we can actually

*completely eliminate*the possibility that what we see is multi-photon interference simply by making the laser

*even dimmer*. Let's attenuate the laser to the point where, rather than of one photon every cycle, we instead get one photon

*every 2K cycles*(or more). In other words, most of the time the interferometer is totally dark. Every now and then we will get a photon. The temporal separation between photons is now much more than the coherence time of the laser, much more than the cycle time of the laser, much more than the travel time through the long arm of the interferometer. We can make it an hour or more between photons. Theory predicts that we will still see interference! Not only that, but it will be the

*exact same*interference pattern that we saw when the interferometer was fully illuminated.

*that*? In particular: remember how above we noted that we only saw interference after the Kth cycle? How would any given photon know whether the index of the cycle that produced it was more or less than K?

Note well that this is in no way intended to highlight a problem with QM. The outcome predicted by QM is very clear, and I would give you very long odds that this prediction is correct. The problem is only in trying to tell a

*story*about what the fleep is going on here that involves

*photons being emitted by the laser*. I don't see any way to do it.

In fact, I'll go one step further: AFAICT, this is a strong argument for the following remarkable conclusion (and if this holds up I think it really could be earth-shattering):

*the quantum wave function must be physically real*because that is the

*only thing*that could account for the K-cycle delay in the onset of interference. If anyone can see a flaw in my reasoning I would really appreciate it if you would point it out.

Interesting thought experiments that you've constructed!

ReplyDeleteThanks! :-)

ReplyDeleteMy question is, how many languages are we going to mine for number words before this is through?

ReplyDeleteSeriously, I second Don's comment. I agree with you about the QM prediction. I'll need to think over the implications: the combination of the laser cycling and the time delay is an ingenious way to interpose something that is normally thought of as "macroscopic" (turning the laser on and off) with a process that normally produces interference.

ReplyDelete@me:

ReplyDeleteI agree with you about the QM prediction.Actually, as I start to think this over, I'm no longer sure I agree about the QM prediction. I'm not sure it's possible for the quantum state being emitted to maintain coherence across a stop/restart cycle.

(For the single photon emitting light source, as opposed to a laser, I'm not sure the quantum state it emits is a coherent state to begin with; but I think in order to get a real understanding of that I would have to read a number of the other papers referenced by that one, since it looks like a considerable body of theoretical experimental work is being drawn on. I have not had time to do that and I don't expect it to be a short process.)

> I'm not sure it's possible for the quantum state being emitted to maintain coherence across a stop/restart cycle.

ReplyDeleteThen leave the laser on and use a shutter to cycle it.

@Ron:

ReplyDeleteThen leave the laser on and use a shutter to cycle it.I agree this would resolve the coherence question, but I think it also removes the problem you refer to here:

The problem is only in trying to tell a story about what the fleep is going on here that involves photons being emitted by the laser.Using a shutter seems to me to be equivalent to having a filter with a time-dependent probability for passing a photon: it's 0% for the "off" portion of the cycle, and some positive probability for the "on" portion. But the presence of a filter does not prevent telling a story that involves photons being emitted by the laser; at least, that seems to be established by your previous articles and our discussions of them. So if a filter doesn't raise that problem, a shutter shouldn't either.

@Peter:

ReplyDelete> Using a shutter seems to me to be equivalent to having a filter with a time-dependent probability for passing a photon: it's 0% for the "off" portion of the cycle, and some positive probability for the "on" portion.

Yes, thats true, but remember that the duty cycle of the shutter is quite long. If we're trying to rescue the photon story then we have to concede that a LOT of them get through the shutter on each cycle. Most of them are then blocked by the filter (but this is irrelevant -- I mention it only to complete the mental picture). What matters is that the filter is illuminated with the exact same timing and duty cycle as the shutter, so it's not too much of a stretch to say that any photon that reaches the filter must have gone through the shutter on the corresponding cycle. If this were not the case, you would have to have photons "hanging around" somehow to delay themselves by a full shutter cycle. (And in fact I think it's even worse than that: you would need a single photon that splits into a superposition, one part of which "hangs around" while the other does not.)

But this is an interesting line of thought. The idea that laser needs to stay on to maintain coherence seems to me to be not only correct, but insightful. My gut tells me there's another profound truth lurking there.

@Ron:

ReplyDeleteIf we're trying to rescue the photon story then we have to concede that a LOT of them get through the shutter on each cycle.You seem to be implicitly assuming that a photon must go through the shutter and the filter during a particular cycle. But in the previous version (filter but no shutter) we showed that the photon must be in a superposition of going through the filter at different times (because it's in a superposition of being emitted by the laser at different times). It seems like a straightforward extension of that to say that when the shutter is present but its duty cycle is arranged appropriately, the photon state that predicts the probability of detection at the detector is a superposition of the photon going through the shutter during different cycles.

In fact, the most straightforward way of telling the "photon" story for any of these cases is to tell it backwards: you start with the detection of a photon at a specific time and a specific place on the detector, and you trace back the possible histories of the photon that could have ended up there. These histories will end up at the source (the laser) at different times, as long as the appropriate pathways to those emission events are open. The fact that other pathways between the detector and the laser that involve different times and places on the detector might be closed (because the shutter was closed) is irrelevant to explaining that particular detection; more precisely, they are irrelevant to computing the probability amplitude for a detection at that particular time and place.

Of course, the issue with telling the photon story backwards this way is that it seems to be inconsistent with causality. Our intuitions about causality tell us (or at least they seem to be telling you) that there should be a feasible forward version of the story, but there doesn't seem to be one that just involves a single photon state evolving in time. But I think that particular version of the intuition about causality is wrong. A "photon" is an approximation anyway; there is no requirement that all possible states (or better, histories) of the underlying entity, the quantum electromagnetic field, must have an interpretation as photons being emitted, traveling through space, and then being detected. Properly defining causality for a quantum field does not involve any notion of "particles" or any particular interpretation of the field; it's just a statement of the commutation relations that the field operators at different events must satisfy.

In short, while I think I understand the intuition that lies behind wanting to tell a "photon story", I'm not sure that not being able to tell one (or at least one that goes forward in time) has the implications you are saying it does. At the very least, I think this scenario indicates that all of the philosophical machinery that has been harnessed to the topic of interpretations of QM needs to be looking at quantum field theory, not non-relativistic QM; and I don't see much of that in the literature I've read.

"

ReplyDeletetell it backwards: you start with the detection of a photon at a specific time and a specific place on the detector"And yet I might have a problem with the "backwards" telling too. Since that implicitly seems to assume that "photon detected here/now" is a necessary singular event in the universe's block history, whereas the MWI interpretation says that even if "you" detected a particular event, that was only one version of you. The universe as a whole includes parallel histories where that particular photon detection event did not occur.

Causality does seem to be only "forwards". It seems like a causal theory

mustinvolve a forward explanation. And attempted backwards explanations have additional problems of their own.@Don:

ReplyDeletethat implicitly seems to assume that "photon detected here/now" is a necessary singular event in the universe's block historyFor purposes of calculating the probability amplitude for a detection at that particular time and place, it is. Different "worlds" from the MWI standpoint would, as you say, correspond to detections at different times and places, which would involve sums over different sets of histories, with different macroscopically distinguishable endpoints, in order to compute the probability amplitudes for those different detections. I didn't mean to rule out the possibility of doing this; sorry if that wasn't clear.

The "sum over histories" viewpoint in itself is agnostic about whether all of the different possible history sums, corresponding to different macroscopically distinguishable detections (i.e., detections at different times and places) all "really exist" or are just calculational devices to compute probabilities. I'm not trying to take a position here in favor of, or against, any particular interpretation. I'm just saying that to properly consider the question of QM interpretations, the most fundamental version of QM we have is not non-relativistic QM, but relativistic quantum field theory. (And the "sum over histories" version of QFT is not even the only possible version of QFT; it's just the one that I find easiest to handle conceptually. But any version of QFT is similarly agnostic about QM interpretations, if you just look at the math and the predictions.)

Causality does seem to be only "forwards".Yes, but in QFT "forwards" does not have to mean there is a single "story" you can tell in the forwards direction, at least not one that involves a single "state" evolving in time. In fact, from a QFT standpoint, "forwards" is more a matter of how we label things than how the actual math works; the commutation relations between field operators are symmetric with respect to future and past. What picks out the "cause" vs. the "effect" is which one is in the past light cone vs. the future light cone of the other.

> Of course, the issue with telling the photon story backwards this way is that it seems to be inconsistent with causality.

ReplyDeleteI think it's worse than that. I think it violates the principle of continuity of identity. To make sense of this experiment you'd have to say something like: a photon is emitted by the laser, and then some fairly long time later, the *same photon* is emitted by the laser *again*.

I'm also pretty sure I can come up with a variation that would require the "same photon" to be emitted by a *different* laser.

@Ron:

ReplyDeletethe principle of continuity of identityWhat do you think this principle requires? Bear in mind that, from the standpoint of quantum field theory, there is no such thing as a "photon" as an "object" with continuous identity. There is a source configured a certain way, devices such as beam splitters and mirrors, and a detector that shows detection events with particular spacetime coordinates (time and location). The term "photon" is just a shorthand for describing a particular state of the quantum field, and that shorthand does not entail that there has to be any valid interpretation of the quantum field state that corresponds to a single massless particle traveling through spacetime. (Which means the shorthand is misleading if it leads you to believe that there must be such an interpretation.)

To make sense of this experiment you'd have to say something like: a photon is emitted by the laser, and then some fairly long time later, the *same photon* is emitted by the laser *again*.If this is indeed a problem, it's a problem for any version of the time delay interferometer experiment that shows interference, including the simplest version where there's just the laser and the interferometer with different path lengths, nothing else. Any story you tell that involves photons is going to require that "the same photon" is emitted by the laser at two different times, since those are the two alternatives that have to interfere to obtain a correct prediction of the probability of a single photon detection at a particular time and location on the detector.

I start to hesitate when people begin asking questions about the "same" photon. Elementary particles do not have unique identity. It isn't a meaningful question to ask whether two potential particles (that happen to share the same characteristics) are "really" the "same", or "different". (Can You Prove Two Particles Are Identical?)

ReplyDelete> from the standpoint of quantum field theory, there is no such thing as a "photon" as an "object" with continuous identity.

ReplyDeleteBe that as it may, some notable people have insisted that photons really do exist. Feynman, for example, was quite adamant about it:

“This is a particle in every sense… I don’t know how much I can emphasize this, especially to young students who have learned it’s waves. It IS particles [emphasis in original] in every way...”

Furthermore, there is a whole field of study called multi-photon interference. e.g.

https://www.springer.com/us/book/9780387255323

If the phrase "multi-photon interference" is to have any meaning at all, we have to be able to count photons, even if we cannot distinguish among them. If photons don't exist, what are we counting?

> I start to hesitate when people begin asking questions about the "same" photon. Elementary particles do not have unique identity.

Sure, but that doesn't mean that we can't distinguish between them under some circumstances, e.g. if I watch a movie twice, the photons absorbed by my retina at the first showing are not the same photons as those in the second showing, even if it's the same movie.

> > the principle of continuity of identity

> What do you think this principle requires?

If an object exists at time T1 and the same object exists at time T2 then that object exists at all times between T1 and T2.

From this it follows logically that if an object does not exist at time T1 but does exist at time T2 > T1 then it must have been created at some unique time Tc, where T1< Tc < T2.

Note that there is an analogous rule for an object's destruction: if an object exists at time T1 but not at T2>T1, then it must have been destroyed at some unique time Td where T1 < Td < T2, and that this is entirely uncontroversial. You can't destroy the same thing twice.

@Ron:

ReplyDeletesome notable people have insisted that photons really do exist.What they mean is that we always

detectsufficiently faint EM radiation as discrete detection events. For example, in the experiments we have been discussing, if the laser intensity (post filter) is sufficiently low, we will always see individual dots on the detector screen. We never see a pattern of light and dark that gets fainter and fainter without ever resolving into individual dots observed one at a time.What they do not mean is that the entire process that leads up to those discrete detection events must be ultimately "made" of little billiard balls moving at the speed of light, with exactly one billiard ball corresponding to each discrete detection event.

If photons don't exist, what are we counting?Sources and detection events. "Multi-photon interference" means you have, for example, two sources, both aimed at the same beam splitter, and two detectors, each of which shows discrete detection events, and you find interference patterns in the detection events.

If an object exists at time T1 and the same object exists at time T2 then that object exists at all times between T1 and T2.Then this principle seems to me to be obviously false in QM, so I don't see why you would want to use it. The time delayed interferometer experiment is simply a way to make the falsity of this principle in QM obvious.

"

ReplyDeleteare not the same photons"The suggestion is that this is not a meaningful distinction. Photons (assuming the same properties: wavelength, etc.) only have a count, not a unique identity. A better analogy is the dollars in your bank account. Your bank account only has a total quantity. It isn't meaningful to ask whether one particular dollar is "the same" or "different" than some other dollar. (If you deposit a dollar, and then later write a check to spend a dollar, are you spending "the same" dollar that you deposited?) There is only a (conserved) count, not identity.

"

If an object exists at time T1 and the same object exists at time T2 then that object exists at all times between T1 and T2."I think you're confusing conservation with identity. Energy doesn't get created or destroyed, sure. That just means the counts have to remain the same. If there was one photon at T1, then there will still have to be one photon at T2 (unless there was an event to transform -- but not destroy -- the energy in that photon).

But that's not quite the same as saying it is "the same" photon at T2. It's only saying that "there is" a photon at T2.

With only a single photon, the distinction doesn't seem important. But when you have multiple photons coming on multiple paths, where you can't distinguish which photon went which way, it's important to realize that there is no "fact of the matter" about whether some particular photon is "the same" as some previous photon.

The only thing conserved is the count (in a local region of space-time), not the identity.

@me:

ReplyDeletethis principle seems to me to be obviously false in QMActually, on consideration, I should modify this. The principle seems to me to be obviously false in quantum field theory. It's not false in non-relativistic QM since one can adopt interpretations in which it is true, at least for some definition of what an "object" is (but that definition still might not be one that is natural to our pre-quantum intuitions). But we know non-relativistic QM is just an approximation; and as I've said before, this approximation is not valid for photons--or more precisely for the quantum EM field--anyway.

@Don:

ReplyDeletePhotons (assuming the same properties: wavelength, etc.) only have a count, not a unique identityI agree with what you say about photon identity. In addition, unless the quantum EM field is in an eigenstate of the photon number operator, there is not even a well-defined count; trying to count the photons in the same non-eigenstate will give a random result with probabilities defined by the amplitudes of the different number eigenstates in the overall state.

The state emitted by a laser is a coherent state, which is not a photon number eigenstate, so there is no well-defined "count" of photons in such a state. When we talk about states emitted by a very low intensity laser, we are talking about states where the amplitude of any photon number eigenstate with a photon number of more than 1 is so low that there is negligible probability of detecting more than one. (And most of the time we don't detect any photon at all; that corresponds to the state having a much larger amplitude for the vacuum state--the state of photon number zero--than for any other photon number eigenstate.)

@Peter:

ReplyDelete> this approximation is not valid for photons

Yes, we agree on all this. This is about pedagogy, not physics.

> > If photons don't exist, what are we counting?

> Sources

OK, then take the time-delay interferometer experiment and instead of a single laser, use two lasers and a mirror mounted on a pivot to select one of the two beams to illuminate the experiment. Align the lasers so that they are in phase with one another. Rotate the mirror to swap one laser for the other while the shutter is closed. Do you see that setup as fundamentally different from the single-laser case? Why?

@Don:

> Your bank account only has a total quantity.

Of course, but there is still a meaningful distinction to be made between my money and your money. In an analogous manner one can in many situations draw a meaningful distinction between this photon over here and that photon over there -- or, more to the point, between the photon that was emitted by this object over here versus the one emitted by that object over there. If this were not so, we couldn't see.

@Ron:

ReplyDeleteDo you see that setup as fundamentally different from the single-laser case?Not if the two lasers can truly be kept coherent with each other. But in both cases (single laser and this new dual laser case), there is no source event: we are not detecting individual photon emissions. If we were, there would be no interference. And if we're not, then we have no discrete emission events to count; we only have discrete detection events to count, so that is the only count of photons that we have. (I see that the word "sources" is ambiguous: I should have said "discrete emission events" to make my meaning clear, and for better correspondence with "discrete detection events".)

@Peter:

ReplyDelete> we have no discrete emission events to count; we only have discrete detection events to count

But for every detection there must be a corresponding emission because conservation laws. We might not be able to pin down exactly which atoms they came from or when they happened (which indeed we cannot) but we can still know how many there were despite the fact that we can't detect them directly.

Take N atoms of U238. Put them in a geiger counter and and wait for it to register one click. I can't tell you which atom will decay, or exactly when it will happen, or which particular protons and neutrons will be emitted as an alpha particle, or even if it makes sense to talk about "a particular neutron" (probably not). But I can tell you with absolute certainty that the end of this process you will have exactly N-1 atoms of U238, one Th234 atom, and one alpha particle.

@Ron:

ReplyDeletefor every detection there must be a corresponding emission because conservation lawsThe energy that gets deposited in a detection does not have to be carried by single particle from a single emission. The time delay interferometer, again, is a good illustration of why that cannot be true.

I can tell you with absolute certainty that the end of this process you will have exactly N-1 atoms of U238, one Th234 atom, and one alpha particle.The count you are giving here is not energy conservation, it's charge and baryon number conservation. Photon number is not a conserved quantity, so there is no analogous conservation law in the photon processes we have been discussing.

Energy conservation in the alpha decay example means that the total energy (including rest energy) of the Th234 atom and the alpha particle must equal the original energy (including rest energy) of the U238 atom. In a frame in which the U238 atom starts out at rest, that means the Th234 atom and the alpha particle, between them, must have 4.38 MeV of kinetic energy in addition to their rest energy. And we can detect and measure all of these quantities directly; we can measure the U238 atom before the decay, and we can measure the products after the decay.

In the photon processes we have been discussing, we measure the detection (and in principle we could measure the energy deposited at the detection event, though in practice experiments of this sort usually don't), but we don't measure the emission. So all we can say is that the energy deposited at the detection came from the source (the laser), since that's the only place it could have come from. We can't say exactly how that energy was transported from the source to the detection. To be able to say that with precision, we would have to measure it, by measuring individual photon emissions. And doing that would remove the interference, so you would no longer need to account for the detection events by using a superposition of photons emitted at different times.

@Peter:

ReplyDeletePlease keep in mind again that this is not about physics but rather about interpretation and pedagogy. We agree on the physics.

> The energy that gets deposited in a detection does not have to be carried by single particle from a single emission.

That's kind of the point of this thought experiment, to create a situation where this must be the case. See below.

> The count you are giving here is not energy conservation, it's charge and baryon number conservation.

That was just supposed to be an illustration of how conservation laws allow you to infer that there must have been a single emission corresponding to a single detection in a situation where you could actually go back after the fact and verify that the prediction was correct. Yes, the quantity conserved here is different, but a conserved quantity is a conserved quantity. Baryon number is just easier to measure before and after than total system energy.

> in principle we could measure the energy deposited at the detection event, though in practice experiments of this sort usually don't

We don't bother measuring the energy because we already know that it is hc/λ.

Let me try to make this as bullet-proof as I can:

1. Lasers emit monochromatic light. That means that all of the photons coming out of the laser have the same wavelength λ, and the same energy, hc/λ.

2. When we detect a photon, the energy in that photon (i.e. hc/λ) is deposited into the detector. That energy was originally resident in the power supply for the laser. By some process, that energy was transferred from the power supply, into the laser, through space, and to the detector.

3. The canonical story of how that happened is: the energy in the power supply was transferred into an atom in the laser to excite an electron to a higher energy level. Then that electron dropped back down to a lower energy level, emitting a photon. In a laser, this process happens multiple times through stimulated emission, so that all the photons are emitted in phase, but that's an irrelevant detail if all that concerns us is figuring out where the energy goes. We can tell this story equally well using an incoherent monochromatic source.

4. A photon is not a billiard ball. It travels through space as a wave, potentially occupying a very large volume. Multiple "photon waves" can overlap and occupy the same space at the same time. Individual photon waves are indistinguishable from each other, but they can be sufficiently separated in time and space so that it makes sense to treat the spatially and/or temporally separated components as if they were distinct entities. Furthermore, the total energy contained in these waves (the field) is quantized: at any given time it must be an integer multiple of hc/λ. Let's call this multiple N.

5. When a photon is detected, N is decreased by 1. We can know when this happens.

6. Therefore, there must be a corresponding process that increases N by 1. We may not know when it happens. The whole concept of "when it happens" may not even be meaningful. But it must happen.

Do you disagree with any of that?

Note that the thought experiment is designed to produce a well-defined region of space where the only possible values of N are 0 and 1. The purpose of the shutter is to create well-separated periods of time where N cannot increase, and thus to constrain the times during which N can increase.

As an aside, it just occurred to me that saying that we detected a photon with a particular energy at a particular time should have exactly the same problem as saying we produced a photon with a particular energy at a particular time! And yet we can obviously say that we detected a photon at a particular time. If we know that photon came from a laser then we can know what its energy must have been even if we don't measure it. How is this not a counterexample to the uncertainty principle? I actually don't know the answer to that. But a cursory investigation reveals that "energy and time (unlike position q and momentum p, for example) do not satisfy a canonical commutation relation", so the answer might not be straightforward.

ReplyDelete@Ron:

ReplyDelete1. Lasers emit monochromatic light. That means that all of the photons coming out of the laser have the same wavelength λ, and the same energy, hc/λ.I disagree. Lasers emit light in a coherent state. A coherent state is not an eigenstate of the Hamiltonian or the photon number operator, so it does not have a definite energy, and it does not contain a definite number of photons. It does, however, have a definite phase at each point in spacetime; that is what makes it coherent. (Because of the definite phase, I think it can be thought of as having a definite frequency and wavelength; however, because it is not an eigenstate of photon number, you can't relate that definite frequency and wavelength to the energy of a photon.)

2. When we detect a photon, the energy in that photon (i.e. hc/λ) is deposited into the detector. That energy was originally resident in the power supply for the laser. By some process, that energy was transferred from the power supply, into the laser, through space, and to the detector.I would agree with "when we detect a photon, energy is deposited in the detector, and that energy was transferred from the power supply to the detector via the laser". But I do not agree that we can say the energy deposited in the detector was "the energy of the photon", because the laser state does not contain a definite number of photons, and there is no way to say that the energy transported was carried by a single photon.

3. The canonical story of how that happened is: ... We can tell this story equally well using an incoherent monochromatic source.If we're just looking at a source and a detector, and talking about energy transport, then I agree that the source doesn't have to be coherent. However, I'm not sure what quantum state would model an incoherent monochromatic source.

4. A photon is not a billiard ball. It travels through space as a waveI agree with the first part, but not the second. If we aren't measuring individual photon emissions, then we don't know how a photon travels through space.

Individual photon waves are indistinguishable from each other, but they can be sufficiently separated in time and space so that it makes sense to treat the spatially and/or temporally separated components as if they were distinct entities.I agree there are cases in which you can do this, but those cases do not include the state emitted by a laser.

the total energy contained in these waves (the field) is quantized: at any given time it must be an integer multiple of hc/λ. Let's call this multiple N.Not if the state is not an eigenstate of the Hamiltonian and the photon number operator, which it isn't. See above.

The correct statement of energy quantization in this context is that when a photon is *detected*, it's always detected as a discrete detection event. But that does not mean the overall state of the quantum EM field is a photon number eigenstate.

Further comments in a follow-up post.

Continuing from my previous post:

ReplyDelete@Ron:

5. When a photon is detected, N is decreased by 1. We can know when this happens.

6. Therefore, there must be a corresponding process that increases N by 1.

I disagree, for the reasons given in my previous post.

the thought experiment is designed to produce a well-defined region of space where the only possible values of N are 0 and 1.As I've already said, I think what it actually does is show why thinking of the state emitted by the laser as containing a definite number of photons, and the energy deposited in the detector as carried by a single photon, cannot be right. The only way to explain the observed probabilities of photon detections at all of the possible times and places on the detector in the time delay interferometer is, as we've already agreed, to think of the detected photon as being in a superposition of having been emitted at two different times. That obviously invalidates any interpretation in which the state emitted by the laser has to have a definite photon number at all times. As I said in my previous post, the state emitted by the laser is a coherent state, which doesn't have a definite photon number, so there's no problem with the actual quantum field theory involved.

saying that we detected a photon with a particular energy at a particular time should have exactly the same problem as saying we produced a photon with a particular energy at a particular timeNo, it doesn't. The point is that in the laser, we are *not* saying we produced a photon with a particular energy at a particular time. If we want to be able to say that, we need a different kind of photon source, one that explicitly measures each individual photon and its energy as it's emitted. You can make the same sorts of statements about that kind of source as you can make about a photon detector that explicitly measures individual discrete photon impacts. But a laser is not that kind of source.

If we know that photon came from a laser then we can know what its energy must have been even if we don't measure it.No, we know what the photon's energy was when we detected it, because we *did* measure it. But, as above, that does not mean we can trace that single photon back to a single emission event at which that same definite quantity of energy was removed from the laser.

How is this not a counterexample to the uncertainty principle? ... a cursory investigation reveals that "energy and time (unlike position q and momentum p, for example) do not satisfy a canonical commutation relation"Yes, that is why it's possible to detect a photon with a definite energy at a definite time, whereas it's not possible to detect a particle at a definite position with a definite momentum (you can measure one or the other with a single measurement, but not both). But the issue under discussion is a different one: whether it's possible to link a given discrete detection event to a single emission event. In the case we're discussing, it's not.

@me:

ReplyDeleteThe only way to explain the observed probabilities of photon detections at all of the possible times and places on the detector in the time delay interferometer is, as we've already agreed, to think of the detected photon as being in a superposition of having been emitted at two different times.I'll go ahead and throw out a further hint here. The statement quoted above is true for *any* photon that we detect that was emitted by a laser. The time delay interferometer just makes that fact explicit in our explanation of what's happening.

But if the quantum field is in a superposition of the same photon ("the same" meaning "the discrete detection event at a particular time and place, whose amplitude we are trying to calculate") being emitted at different times, that must mean that the *laser* itself is in a superposition of emitting the same photon at different times. But the laser photons are emitted by atoms dropping from excited states to ground states; so that means the state of the quantum EM field emitted by the laser must be *entangled* with the states of the atoms in the laser, so that the atoms themselves are in superpositions of emitting photons at different times.

It might be interesting to consider the implications of that.

@Peter:

ReplyDelete> > Lasers emit monochromatic light.

> I disagree. Lasers emit light in a coherent state.

Huh??? Isn't "coherent" a superset of "monochromatic"?

> I'm not sure what quantum state would model an incoherent monochromatic source.

Monochromaticity is a kind of coherence, so "incoherent monochromatic source" is at best an underspecified concept and at worst an oxymoron. You can also have spatial coherence and phase coherence. Lasers have all three. Sodium vapor lamps, for example, are monochromatic (more or less, they actually have two closely spaced spectral lines in the visible range), but are not spatially coherent (their output is not collimated like a laser) nor are they phase coherent.

> the laser state does not contain a definite number of photons

Why not? If I turn on a laser for some time T, and during that time the laser emits a total amount of energy E, and the energy is emitted in the form of light at a wavelength λ, then the total number of photons produced must be Eλ/hc.

Either that or all the physics textbooks are wrong.

Note that my argument for there being a definite number of photons relies *only* on monochromaticity, conservation of energy, and E=hc/λ. I don't need collimation or phase coherence (even though I get those for free if my source is a laser).

On a totally separate topic:

> the state of the quantum EM field emitted by the laser must be *entangled* with the states of the atoms in the laser, so that the atoms themselves are in superpositions of emitting photons at different times.

> It might be interesting to consider the implications of that.

Indeed. I've wondered about that. Since we know that entangled photons do not produce first-order interference, and we agree that photons emitted by lasers are entangled with the laser, why do lasers produce first-order interference?

I don't know the answer to that.

@Peter:

ReplyDeleteI think I may have just had an "aha!" moment:

> I disagree. Lasers emit light in a coherent state.

Do you mean that it is the additional coherence (phase and spatial coherence) that *causes* my argument for counting photons to fall apart? All the photons emitted by the laser the entire time it's on (or at least during the coherence time) are all part of a single non-separable quantum state, and it doesn't make sense to talk about them as individual entities for the same reason it doesn't make sense to talk about the two members of an EPR pair as individual entities?

Something still seems wrong here, but at least it seems kinda sorta plausible. And it would explain why you're drawing the distinction between monochromatic and coherent, which was a real head-scratcher for me.

@Ron:

ReplyDeleteIsn't "coherent" a superset of "monochromatic"?Not if by "monochromatic" you mean what you described in your previous post. When I say a laser emits light in a coherent state, I am referring to a specific mathematical description of the state of the quantum EM field, and that state does not have a definite energy or a definite number of photons, as I said. If not having a definite energy or a definite number of photons is compatible with your definition of "monochromatic", then yes, a coherent state is monochromatic. But then you can't draw the inferences from "monochromatic" that you are trying to draw.

Wikipedia actually gives a good summary of the mathematical description of coherent states here:

https://en.wikipedia.org/wiki/Coherent_states

Either that or all the physics textbooks are wrong.Can you show me a physics textbook that says the state emitted by a laser is an eigenstate of the photon number operator? The coherent states described on the Wikipedia page, which as I understand it are the states emitted by a laser, are obviously not eigenstates of the photon number operator or the Hamiltonian, so they do not have a definite energy or a definite number of photons.

Do you mean that it is the additional coherence (phase and spatial coherence) that *causes* my argument for counting photons to fall apart?I mean just what I've said several times now: a coherent state of the quantum EM field is not an eigenstate of the photon number operator or the Hamiltonian, so it does not have either a definite number of photons or a definite energy. No argument for "counting photons" can be applied to such a state since any such argument relies on the number of photons in the state being definite.

All the photons emitted by the laser the entire time it's on (or at least during the coherence time) are all part of a single non-separable quantum state, and it doesn't make sense to talk about them as individual entities for the same reason it doesn't make sense to talk about the two members of an EPR pair as individual entities?I think this is a reasonable way of looking at it, yes. But I think you also will have to throw in the atoms in the laser being entangled with the quantum EM field, so in a sense it doesn't even make sense to talk about the laser and the field as being separate entities. I agree that makes things even more counterintuitive, but that's just how we roll in quantum land. :-)

@me:

ReplyDeleteNo argument for "counting photons" can be applied to such a state since any such argument relies on the number of photons in the state being definite.Just to be clear, this statement does not just apply to the coherent state emitted by a laser. It applies to any state of the quantum EM field that is not an eigenstate of the photon number operator. That's why I have been emphasizing it: because it gives a very general test for whether arguments that involve "counting photons" are applicable.

@Peter:

ReplyDelete[Reposting to correct a minor but crucial typo]

> I mean just what I've said several times now: a coherent state of the quantum EM field is not an eigenstate of the photon number operator or the Hamiltonian

I believe you. But you are still missing the point.

> Can you show me a physics textbook that says the state emitted by a laser is an eigenstate of the photon number operator?

Obviously not. But I can show you physics textbooks that say:

1. Light is quantized

2. The energy of a light quantum is hc/λ

3. Laser light is monochromatic, which is to say that all [1] of the light quanta emitted by a laser have the same wavelength and hence the same energy (that's just what "monochromatic" *means*).

4. Energy is conserved.

From these premises, all of which can be found in physics textbooks, one can not only conclude *that* there is a definite number of photons emitted by a laser in a given time period, one can even compute exactly what that number is: Eλ/hc. This is not rocket science. (It's barely even quantum mechanics!)

The question is not: is this reasoning wrong? I believe you when you tell me that it is. The question is *where* is the flaw in the reasoning? The answer to *that* question is not going to involve the word "eigenstates".

> any such argument relies on the number of photons in the state being definite.

No, it doesn't. That's the *conclusion* of the argument, not one of the premises.

---

[1] The word "all" was missing in my original comment

"

ReplyDeleteThe question is *where* is the flaw in the reasoning?"Oooh! Oooh! Let me guess. You guys are way beyond my understanding here, but I'm enjoying following the discussion. That said, I can guess an answer to this question:

The laser itself is in a superposition, of different energy states and different numbers of emitted photons. Saying that "there are"

exactly10 photons emitted, is the same kind of "wavefunction collapse" error as the Copenhagen interpretation.And (I'm going to guess) that Peter is going to say: you can see this, with a clever experimental setup, once you get a detection event. Because that event will demonstrate interference from one version of the laser that emitted a photon at T1, along with another version of the laser that emitted a photon at a later time T2, even though any measurement of the energy in the system will show that "only one" photon got emitted, and so the energy remaining in the laser and/or power supply will only show a drop of a single photon's worth of energy. But the only reasonable explanation of the detection event (because of interference), necessarily involves a story where photon(s) were emitted at different times.

So you have to resolve the fact that there is only one photon's worth of energy, with the simultaneous fact that photons will need to have been emitted at different times (in order to cause the interference). The question, "how many photons were flying through the air" isn't well specified and doesn't have a definite answer. It remains a superposition of multiple possibilities.

(I eagerly await my grade on my attempted answer.)

@Don:

ReplyDelete> (I eagerly await my grade on my attempted answer.)

I give you an A for effort. But a proper answer to my question has to have one of the following two forms:

1. Your Nth premise is actually false, notwithstanding that you can find it in physics textbooks. The physics textbooks are wrong.

2. All your premises are correct, but your conclusion is nonetheless faulty because X.

I can't even begin to imagine what X might be. Note that X cannot be "the state emitted by a laser is not an eigenstate of the photon number operator". That's just telling me how you know that my conclusion is wrong (which, I take pains to stress once again, I totally accept). It doesn't tell me where my reasoning ran off the rails.

Your answer, that the laser is in a quantum superposition, likewise doesn't help because my reasoning never assumed that the laser was not in a quantum superposition. *All* light is quantized (or so the physics books tell me). *All* photons have energy equal to hc/λ (or so the physics books tell me). Energy is conserved always, period, end of story (or so the physics books tell me) [1]. Entanglement is not even in play here.

---

[1] It's actually the momentum-energy 4-vector that is conserved, but that's a pedantic quibble.

@Ron:

ReplyDeleteI can show you physics textbooks that say:

1. Light is quantized

2. The energy of a light quantum is hc/λ

3. Laser light is monochromatic, which is to say that all [1] of the light quanta emitted by a laser have the same wavelength and hence the same energy (that's just what "monochromatic" *means*).

4. Energy is conserved.

I don't doubt that you can find those ordinary language statements in physics textbooks, yes. But physics is not done in ordinary language. It's done in math. I've already linked you to the math of coherent states. I don't know whether that math makes statements 1. through 4. above true or false; ordinary language is vague. What I do know is that the math invalidates any reasoning about the state emitted by a laser that depends on counting individual photons traveling through space.

The question is not: is this reasoning wrong? I believe you when you tell me that it is. The question is *where* is the flaw in the reasoning? The answer to *that* question is not going to involve the word "eigenstates".Um, yes, it does. You said your reasoning was "barely even quantum mechanics". Here's some basic quantum mechanics:

A. If a quantum system is not in an eigenstate of an observable, it does not have a definite value for that observable.

The coherent state of the EM field emitted by a laser is not an eigenstate of the photon number operator. That's trivial to verify by looking at the math on the Wikipedia page I linked you to. Therefore, the coherent state of the EM field emitted by a laser does not have a definite number of photons.

That's the *conclusion* of the argument, not one of the premises.I don't care whether you call it a premise, conclusion, or whatever. The point is that, if your argument says that there is a definite number of photons in the state of the quantum EM field emitted by the laser, it's wrong, because it contradicts the basic quantum mechanics I just described.

Btw, about all those physics textbooks: it should not be surprising that they use sloppy ordinary language in describing experiments like those with lasers. It's extremely hard *not* to use sloppy ordinary language when talking about QM, because we have no ordinary language to use that is *not* sloppy. That's why it's crucial to look at the actual math: what is the specific mathematical description of the state the textbook is talking about? What are the specific operators that describe the observables? What happens when you act on the state with those operators? Those are questions that you have to answer with math, not ordinary language; and if the math tells you one thing (for example, that there isn't a definite number of photons in the state emitted by the laser), and the ordinary language seems to be telling you something different, then the math is right and the ordinary language is wrong. Even if the ordinary language is in a textbook.

@Don:

ReplyDeleteThe laser itself is in a superposition, of different energy states and different numbers of emitted photons.Yes.

And (I'm going to guess) that Peter is going to say: you can see this, with a clever experimental setup, once you get a detection event. Because that event will demonstrate interference from one version of the laser that emitted a photon at T1, along with another version of the laser that emitted a photon at a later time T2, even though any measurement of the energy in the system will show that "only one" photon got emitted, and so the energy remaining in the laser and/or power supply will only show a drop of a single photon's worth of energy. But the only reasonable explanation of the detection event (because of interference), necessarily involves a story where photon(s) were emitted at different times.I don't know that you needed to guess, since I've already said exactly this; the time delay interferometer is precisely such an experiment. :-)

@Ron:

ReplyDeleteX cannot be "the state emitted by a laser is not an eigenstate of the photon number operator". That's just telling me how you know that my conclusion is wrong (which, I take pains to stress once again, I totally accept). It doesn't tell me where my reasoning ran off the rails.Ok, fair enough. See below.

I can't even begin to imagine what X might be.I can. You are arguing that your four premises, taken together, necessarily entail the conclusion: "There is a definite number of photons in the state of the quantum EM field emitted by a laser".

In order for such an argument to be valid, your premises must all be about states of the quantum EM field. But they're not. They're about photon detection events. Here's how I would rephrase them to make that clear:

1. Light is quantized -- Whenever we detect sufficiently faint light, it is as individual discrete events.

2. The energy of a light quantum is hc/λ -- When we detect faint light as individual discrete events, each of them transfers a definite energy, which we can relate to the wavelength of the light using the formula hc/λ. (Note that this assumes the light *has* a definite wavelength, which is not always the case; but I think it's valid for a laser.)

3. Laser light is monochromatic, which is to say that all of the light quanta emitted by a laser have the same wavelength and hence the same energy (that's just what "monochromatic" *means*). -- Whenever we detect laser light as individual discrete events, they all transfer the same energy.

4. Energy is conserved. -- The energy received in any discrete detection event of laser light must have originally come from the laser.

With the premises rephrased this way, it's clear that they do not entail that the state of the quantum EM field emitted by a laser has a definite number of photons. (And if you cornered the authors of the textbooks that make such statements, I think they would admit that statements like the premises above are really about detection events, not the field itself.)

@Peter [1 of 2]:

ReplyDelete> physics is not done in ordinary language. It's done in math

It is *done* in math, but it is *explained* in language. You can't teach physics merely by writing down equations (notwithstanding that some of my old physics professors tried). You can't even do *math* merely by writing down equations. Look at any math book or math paper or physics book or physics paper: they are all full of words.

In order to do *physics* you need *words* to provide the math with *semantics*, to relate the math to actual, you know, *physical situations*. Knowing the Schroedinger equation and how to solve it is not enough. You also need to know (at the very least) how to translate a description of a physical situation into a Hamiltonian.

> the math invalidates any reasoning about the state emitted by a laser that depends on counting individual photons traveling through space

Once again: my math does not *depend* on this, it *predicts* that you can do this.

You keep repeating the same thing: my math must be wrong because the conclusion it reaches is wrong. And I keep giving you the same response: I believe you. But that doesn't tell me what I've done wrong, only that I've done *something* wrong. Divided by zero somewhere. Made some unwarranted assumption. I have no idea. That is what I'm *asking*.

> The coherent state of the EM field emitted by a laser is not an eigenstate of the photon number operator. That's trivial to verify by looking at the math on the Wikipedia page I linked you to.

It might be trivial for you. It's not for me, nor, I suspect, for the typical reader of this blog. (Don, Luke, you want to back me up on this?)

BTW, I started digging in to your reference and found this:

"[T]he average photon number in a coherent state is..."

and then it gives a formula which I cannot reproduce here because typography.

Furthermore, in figure 1 there are examples shown of coherent states and their corresponding photon numbers.

That seems to me to call into question your claim that:

> the coherent state of the EM field emitted by a laser does not have a definite number of photons.

It also turns out (as I suspected) that there are lots of different kinds of coherent states. That page has a ton of math, and zero occurrences of the word "laser".

Also, as an aside, it turns out Wikipedia has a good page on single-photon sources.

@Peter [2 of 2]:

ReplyDelete> Ok, fair enough. See below.

Cool, now we're getting somewhere!

> 1. Light is quantized -- Whenever we detect sufficiently faint light, it is as individual discrete events.

Surely you are not saying that quantization is only true in the faint limit? Do you not accept the correspondence principle?

Other than this I don't see how your rephrasing is in any substantive way different from mine. And it is certainly not "clear that they do not entail that the state of the quantum EM field emitted by a laser has a definite number of photons."

Forget photons for a moment. Would you agree that the EM field emitted by a laser contains a definite amount of energy? And that at any given time this energy must be an integer multiple of hc/λ?

@Ron:

ReplyDeleteIt might be trivial for you. It's not for me, nor, I suspect, for the typical reader of this blog.I admit there is a lot on that Wikipedia page; let me narrow it down.

In the section of the Wikipedia page on "Quantum mechanical definition", there is an expression for the coherent state in the basis of Fock states. The Fock states are the eigenstates of photon number, and form an orthonormal basis of the Hilbert space for the quantum EM field; they are denoted by the kets |n>, where n is the number of photons in the Fock state. The expression for the coherent state in terms of the Fock states is a sum containing more than one term (actually an infinite number of them); therefore a coherent state is not an eigenstate of photon number, since an eigenstate would be a single Fock state, not a superposition of multiple Fock states. Note that this can be seen by simple inspection of the state; no calculation is required.

That seems to me to call into question your claimNo, it doesn't. "Average photon number" means "expectation value of photon number". That's not the same as the state being an eigenstate of photon number and having a definite photon number. An expectation value can be computed for any state for a given observable; it's a probability weighted average of all the possible measurement results for that observable with the quantum system in the given state. So for a state of the EM field, the average photon number is a probability weighted average of all the possible photon numbers that could be obtained by measuring the photon number of the state.

It also turns out (as I suspected) that there are lots of different kinds of coherent states. That page has a ton of math, and zero occurrences of the word "laser".Yes, a laser is not the only kind of source that emits coherent states, and coherent states emitted by different sources can have different properties. But all of them have the key property I've been emphasizing: they aren't eigenstates of particle number (where "particle" means whatever particle corresponds to discrete detection events of the quantum field in question), so they don't have a definite particle number.

it turns out Wikipedia has a good page on single-photon sources.Yes, it does. Note that this page explicitly gives lasers as its example of coherent light sources, which are contrasted with single photon sources.

@Ron:

ReplyDeleteSurely you are not saying that quantization is only true in the faint limit?No, but if the light is not sufficiently faint the quantized nature of the detections is not directly observable because so many detections are happening so fast that we cannot resolve what's happening at the detector into individual discrete detection events. That's why it took so long to figure out that light is quantized in this way.

Would you agree that the EM field emitted by a laser contains a definite amount of energy?No, for the same reason it doesn't have a definite number of photons; the state is not an eigenstate of the Hamiltonian.

Remember that the quantum EM field is entangled with the laser, so not only does the EM field not have a definite energy, the laser doesn't either. Both are in superpositions of states with different energies, because the overall state of the field-laser system is in a superposition of photons being emitted at different times.

@Peter:

ReplyDelete> if the light is not sufficiently faint the quantized nature of the detections is not directly observable because so many detections are happening so fast that we cannot resolve what's happening at the detector into individual discrete detection events

That sentence turned into kind of an "aha" moment for me, and I've been ruminating on it all day. It explains why there has to be a lasing threshold. It's not just an engineering constraint, it's really a fundamental requirement, like a "phase change" for light.

Also, in my reading I learned that the complementary observable for photon number is phase angle, and that is kind of blowing my mind.

Still... Take a 1 milliwatt 633 nm laser and run it for one millisecond. If you crunch the numbers on that, it works out to about 2^40 photons. (Yes, I know this is not legitimate, but bear with me.) Run the output of the laser through a binary tree of beam splitters with a depth of 40. At the leaves of the tree put 2^40 photon counters. (Thought experiments are not subject to economic constraints!) Assume 100% efficiency (and a spherical chicken).

The field strength at each counter is now low enough that individual photons should be countable. If the counts don't sum to 2^40, where is the extra energy coming from / going to?

@Ron:

ReplyDeleteAlso, in my reading I learned that the complementary observable for photon number is phase angle, and that is kind of blowing my mind.I've seen this said too, and I've never quite been able to grok it. I've also seen it used specifically to explain the physics of Josephson junctions, which doesn't really help me any. :-)

Take a 1 milliwatt 633 nm laser and run it for one millisecond. If you crunch the numbers on that, it works out to about 2^40 photons. (Yes, I know this is not legitimate, but bear with me.)As an expectation value I think it's fine. But you have to be clear that it's an expectation value. It's not a definite count of anything.

The field strength at each counter is now low enough that individual photons should be countable. If the counts don't sum to 2^40, where is the extra energy coming from / going to?What extra energy? The 2^40 was an expectation value, not an exact count. Your measurement gives an exact count (assuming that the intuitive reasoning we're both using to analyze it will stand up to an actual rigorous mathematical analysis). There is no reason why the exact count must equal the expectation value; in general it won't, for the same reason the exact count of heads in 40 flips of a fair coin won't in general be exactly 20, even though 20 is the expectation value of the number of heads.

What will be true, however, is this: multiply the actual number of counts measured by the energy hc/lambda of 633 nm photons. Measure the actual energy output by the laser during the 1 millisecond pulse. (Note that this will not be exactly 1 milliwatt-millisecond; that's an expectation value, just like the 2^40 photon counts. The laser power output is subject to some unavoidable variation, and you have to actually measure how much energy it outputs in a given time.) The two energies will be equal.

Another way of stating what I just said is that the laser state and the EM field state are entangled, and hence measurements on them are correlated. The equality of energies that I just described is one of the correlations. When you measure a particular number of 633 nm photons in the detectors, you are also implicitly measuring the number of atoms in the laser that emitted 633 nm photons, because of the entanglement of the two.

@Peter:

ReplyDelete> What will be true, however, is this: multiply the actual number of counts measured by the energy hc/lambda of 633 nm photons. Measure the actual energy output by the laser during the 1 millisecond pulse. (Note that this will not be exactly 1 milliwatt-millisecond; that's an expectation value, just like the 2^40 photon counts. The laser power output is subject to some unavoidable variation, and you have to actually measure how much energy it outputs in a given time.) The two energies will be equal.

OK, so that all makes sense. I believe I can either measure or control the actual energy input to the laser, and so I can calculate/control the photon number from that. But let's leave that aside for the moment because I have a more immediate concern. It's a question I asked before but I guess you missed it. This has actually bothered me for a very long time:

If the photons coming out of a laser are entangled with the laser (and of course they are -- how could a photon not be entangled with its source?) how can they exhibit first-order interference? If they are entangled with the laser, then their state is not |α> but rather |α>|L> where L is the state of the laser.

(Note that this question actually applies to any photon, not just coherent ones.)

@Ron:

ReplyDeleteIf the photons coming out of a laser are entangled with the laser (and of course they are -- how could a photon not be entangled with its source?) how can they exhibit first-order interference?Ah, I see. My initial guess is that the rule that "entangled photons don't produce interference" only applies to photons entangled with each other (for example, in the parallel polarization state that typically comes out of parametric down conversion), not photons entangled with atoms. After all, any photon emission process has an element of quantum indeterminacy to it; AFAIK there is no photon source where we can control the exact time of emission or which atom the photon gets emitted from. That means any emitted photon is going to be entangled with an atom (or something) inside the source. So if the rule "entangled photons don't produce interference" applied to entanglement with anything, we would never observe photon interference at all.

But I need to think about this some more. I also want to go back and look at the underlying math behind the "entangled photons don't interfere" rule.

@Peter:

ReplyDelete[Oops, blogger ate my angle brackets]

[And again to fix another typo. Damn, I wish it were possible to edit comments! Or at least preview them. Luke, where's your new discussion system?]

> entangled photons don't interfere

Entangled photons do interfere, they just don't exhibit *first order* interference. To observe (second-order) interference in entangled photons you have to perform a measurement on both members of the pair, and then combine the information from both measurements. (See e.g. https://arxiv.org/pdf/quant-ph/9903047.pdf) But that's clearly not the case for the output of a laser.

The first-order math is pretty straightforward, arguably the most elementary QM calculation there is (it's so simple, even I can do it!) For a pure superposition state |x> + |y> the square of the amplitude is |x|^2 + |y|^2 + x*y + y*x, where a*b denotes the inner product of a and the complex conjugate of b. Those terms can be non-zero (and in particular they can be negative) so interference.

For an entangled state |xq> + |yr> the square of the amplitude for the x-y subsystem is:

|x|^2 + |y|^2 + x*y <q|r> + y*x <r|q>

Because photons are massless, <q|r> and <r|q> are both zero (massless particles have a zero amplitude for spontaneously changing state), so the last two terms disappear, so no (first-order) interference.

@Ron:

ReplyDeleteEntangled photons do interfere, they just don't exhibit *first order* interference.Yes, you're right. I meant "inteference" to just mean first-order interference, since that's the kind that's relevant to the particular question we're discussing, but I should have said that explicitly.

Because photons are massless, and are both zero (massless particles have a zero amplitude for spontaneously changing state), so the last two terms disappear, so no (first-order) interference.I don't think this is because photons are massless. The same logic holds for a pair of entangled qubits stored as, e.g., electron spins, and electrons aren't massless. The key point is that the states q and r of the q-r subsystem are orthogonal, i.e., that they represent mutually exclusive measurement results. That's what makes their inner products zero. And similar logic applies if we consider the square of the amplitude for the q-r subsystem, if the x and y states of the x-y subsystem are orthogonal.

I think the difference in the photon entangled with laser case is that, for any particular photon detection at one of the 2^40 detectors, we don't know which atom in the laser emitted the photon. So, if we model the photon as the x-y subsystem and the laser as the q-r subsystem in the math you give, there is no way to pick out a pair of orthogonal "q" and "r" states; i.e., there is no way to write down the entanglement of the photon with the laser such that the inner product of the q and r states is zero. In a different experiment, where the source is a single atom, and we can measure whether or not that atom emits a photon independently of detecting the photon itself, then I think the "no first order interference" rule would hold, because the q and r states would be "this particular atom did not emit a photon" vs. "this particular atom emitted a photon", and those are orthogonal.

> I don't think this is because photons are massless. The same logic holds for a pair of entangled qubits stored as, e.g., electron spins, and electrons aren't massless.

ReplyDeleteI learned this calculation from the Feynman lectures, where q and r are presented as states of a (macroscopic) detector, and <q|r> and <r|q> are described as "the amplitude for the detector to spontaneously change from one state to the other", i.e. the amplitude for it to give the wrong result. So I presumed that it was possible for some systems to have a non-zero amplitude for the even if q and r are orthogonal, but that this was not possible for photons because massless particles can't change state.

Anyway, this is a detail. We agree on what matters here.

> there is no way to write down the entanglement of the photon with the laser such that the inner product of the q and r states is zero

> In a different experiment, where the source is a single atom, and we can measure whether or not that atom emits a photon independently of detecting the photon itself, then I think the "no first order interference" rule would hold

I don't think so. In practice, entangled states are produced by aiming lasers at macroscopic chunks of BBO. We don't know precisely which atom of BBO emitted which pair. So the mere presence of a macroscopic source can't be what makes the difference.

@Ron:

ReplyDeleteIn practice, entangled states are produced by aiming lasers at macroscopic chunks of BBO. We don't know precisely which atom of BBO emitted which pair. So the mere presence of a macroscopic source can't be what makes the difference.I agree that for any macroscopic source you can't tell which atom emitted a given photon. And so for any macroscopic source, I don't think entanglement of the emitted photons with the source makes first-order interference impossible. But entanglement between a pair of photons emitted by the same source does.

What I was hypothesizing in what you quoted was a source (which AFAIK nobody presently knows how to build) in which you could tell exactly which atom emitted a given photon, either because there was only one atom present, or because you had some way of measuring which specific atom changed state. If such a source existed, I think the "entanglement makes first-order interference impossible" rule would apply to the emitted photon. But that's just a hypothesis, and doesn't apply to the sources we are discussing here.

> What I was hypothesizing in what you quoted was a source (which AFAIK nobody presently knows how to build) in which you could tell exactly which atom emitted a given photon, either because there was only one atom present

ReplyDeleteActually, I think that is exactly what is described in this paper that I cited earlier:

https://arxiv.org/pdf/1302.2897.pdf

> I think the "entanglement makes first-order interference impossible" rule would apply to the emitted photon.

This is far from clear to me. I'm still waiting to hear back from the author of that paper.

> But entanglement between a pair of photons emitted by the same source does.

That is clearly true. But why?

This illustrates the point I was making earlier. The hard part is not doing the math. The hard part is translating a physical description into math, and in particular in QM, what parts of a system it is safe to ignore.

@Ron:

ReplyDeleteI think that is exactly what is described in this paper that I cited earlierYes, you're right. It's not clear to me from the paper whether they are observing first order interference with these photons; I see a second-order correlation function described, but that's all.

@Ron:

ReplyDeleteThat is clearly true. But why?Wasn't this question (why don't entangled pairs of photons produce interference) already answered by your previous articles in this series?

@Peter:

ReplyDeleteYeah, sorry, I lost track of my antecedents. Yes, we know that entangled photons don't produce (first order) interference. The question is: why do *any* photons produce first-order interference, because every photon must be entangled with its source.

I think I may be making some progress. I went back to look at how entangled photons are produced in parametric down converters. The light that comes out of one of those consists of two cones, one polarized horizontally, the other vertically. The photons always come out in pairs, but they are only entangled at the points where the two cones intersect, otherwise they are merely (classically) correlated. That led me to the realization that an entangled state is the quantum superposition of two classically correlated states, namely, |HV> and |VH>. Which is completely obvious in retrospect (the entangled state is |HV>+|VH> duh!), but it felt like an "aha" moment to me. Likewise, a coherent state is a superposition of Fock states, but the situation there is more complicated because there are (vastly!) more than two states. I strongly suspect that the answer is hiding in there somewhere.

ReplyDelete@Ron:

ReplyDeletean entangled state is the quantum superposition of two classically correlated states, namely, |HV> and |VH>. Which is completely obvious in retrospect (the entangled state is |HV>+|VH> duh!), but it felt like an "aha" moment to me. Likewise, a coherent state is a superposition of Fock states, but the situation there is more complicated because there are (vastly!) more than two states. I strongly suspect that the answer is hiding in there somewhere.I've been pondering how to express the difference my intuition sees between these two cases, and I think you've hit on a key point. I would view the comparison between the PDC case and the laser case like this:

(1) In the PDC case, as you say, the two-photon system is in a superposition of classically correlated states; but not just classically correlated, but 100% classically correlated, orthogonal states. In other words, if we were to measure the polarizations of both photons in the same direction (H or V), the results would be 100% anticorrelated. And in this case, each photon can display no interference.

(2) In the laser case, the photon-laser system is in a superposition of states, which we have described as the photon being emitted at two different times, and consequently the laser emitting a photon at two different times. But the states of the laser emitting a photon at different times are not classically distinguishable; as the laser is set up, there is no measurement that tells us when the laser emits a photon. So there is no measurement that can be made on the laser that tells us when to expect the photon to hit the detector. And in this case, the photon can display interference.

The laser case can be clarified somewhat by observing that, as we have seen, altering the setup to provide time-of-emission information from the laser (for example, by turning the laser on and off at controlled times) removes the interference. In other words, it's not so much that it's impossible to measure the time at which the laser emits a photon: it's that doing so in a way that removes interference necessarily changes the behavior of the laser as a source. Whereas, in the PDC case, the photons as they come from the source already are set up in a way that measuring either one's polarization gives information about the other's polarization; there's no need to change the behavior of anything in the experiment.

The above is still heuristic; it's hard to express all this in ordinary language. But it might help to briefly expand on coherent states. Coherent states are superpositions of multiple Fock states (the latter are the states of definite photon number); in fact, strictly speaking, any coherent state is a superposition of an infinite number of Fock states (i.e., there is a nonzero amplitude for any photon number, including zero, in the coherent state). What that means is that, if you detect a photon when the EM field is in a coherent state, that detection does not tell you that the field is in any particular Fock state; by contrast, when measure the H/V polarization of a PDC photon, that does tell you that that photon is in either the H or V state.