## Tuesday, May 08, 2018

### A quantum mechanics puzzle, part deux

This post is (part of) the answer to a puzzle I posed here.  Read that first if you haven't already.

To make this discussion concrete, let's call the time it takes for light to traverse the short (or Small) arm of the interferometer Ts, the long (or Big) arm Tb (because Tl looks too much like T1).

So there are five interesting cases here.  Let's start with the easy one: we illuminate the interferometer with a laser which we turn on and leave on.  In that case it's a no-brainer: the photons arrive at the business end of the interferometer first from the short path Ts seconds after turning the laser on.  At this point we know (because timing) which way the photons went so there is no interference.  Then at time Tb the photons arrive from the long path.  All the photons are identical, so we no longer know which path they took.  So interference.

Case 2 (the one the original puzzle was about): the laser is turned on and stays on, but the power is modulated (or filters are put in place) so that the light level is very low, so low that the average time of arrival between individual photons at the detector is much larger than Tb.

There are two plausible-sounding answers in this case.  Plausible-sounding answer #1 is that the result is exactly the same as before: after Tb we still get interference.  The equations of quantum mechanics are independent of brightness, so wherever we get interference when the light is bright, we still get interference when the light is dim.

Plausible-sounding answer #2 is that when the light is dim we can tell which way the photon went by looking at the timing.  Whenever we get a detection there are two possibilities: either the photon was emitted Ts seconds earlier and took the short path, or the photon was emitted Tb seconds earlier and took the long path.  So if we can tell when the photon was emitted, then there will be no interference.

But can we tell?  Well (and this is the answer I was originally looking for when I composed the puzzle) it depends on exactly how we make the laser dim!  There are at least two ways, and they produce different results.

The first is to put some sort of shutter in front of the laser that only lets through one photon through at a time. This is equivalent to turning the laser on only for very short periods of time.  If we do it this way we will get no interference.

The other way is to do nothing to the laser itself, but rather to put a filter between the laser and the interferometer that blocks (or reflects) most of the photons.  The photons are blocked by the filter at random, so there is no way to tell when a particular photon got through the filter.  Hence: interference.

But suppose we tweak our setup slightly so that we can tell when a photon was transmitted by the filter.  How can we do this?  It is this exploration that leads to (IMHO) a profound insight.

Think about it: how do you detect a photon without destroying it?  You can't!  The only way to detect a photon is to have some atom absorb it, and that process destroys the photon.  But there is a sneaky trick we can do: we can run the photons through a parametric down-converter (PDC).  A PDC is a crystal made of some material (typically some stuff called beta-barium borate or BBO) whose atoms absorb photons at one wavelength and then re-emit them as two photons at different wavelengths.  The key is that these two emissions happen at more or less the same time.  So we can send one of these photons into the interferometer and use the other one to tell us when this event happened.  Experimental physicists actually do experiments like this routinely.  To distinguish between the two photons, the one that goes into the apparatus is called the signal photon, while the other that is measured to figure out the timing is called the idler.  By measuring (and hence destroying) the idler photon we can tell when the signal photon entered the interferometer, and so we can tell which way the signal photon went (by comparing the timing of entry and exit).  So we cannot have interference.

Here is the profound insight: this setup will not produce interference even if we don't actually measure the idler photon!  Why?  (You might want to think about that for a moment before proceeding.)  Because if it did, then we could use that fact to transmit information faster than light!

Here is how we would do it: instead of measuring the idlers, we send them off (via mirrors) to some distant location (let's call it L1)  At the same time, we take our interferometer and move it away from the PDC by the same distance but in the opposite direction to a location we will call L2.  The distance between L1 and L2 is much greater than the length of the long arm of the interferometer.

If we could produce interference by choosing not to perform any measurements on the idlers then we could use this setup to communicate faster than light by selectively measuring the idlers or not.  When we measured the idlers at L1, the interference would be destroyed at L2.  And this effect would have to happen instantaneously because if it didn't then we could measure idlers at L1 and still have interference at L2, and that is impossible.

The profound conclusion is that photons emitted by a parametric down-converter do not produce interference!  [1]

Those of you who have read my paper on the EPRG paradox will find this all to be familiar territory.  In fact, this is the exact same conclusion that was reached in that paper, and for the exact same reason: the photons emitted by a PDC are entangled, and entangled photons do not self-interfere.  The reason they don't self-interfere is that entanglement is the first step of the measurement process, and it, not measurement per se, is what destroys (first-order) interference.

This is all old news (at least 17 years old).  So why is this (IMHO) cool?  Because we could reach this conclusion without knowing anything about entanglement!  We didn't need to invoke EPR or Bell's theorem or polarization or anything like that.  All we needed was the principle that which-way information destroys interference to reach the conclusion that if there is any physical process that reliably produces multiple photons at the same time, then those photons cannot self-interfere.  We have shown, without doing any math, only from elementary first principles, that entangled particles are different in some deep and profound way from non-entangled ones.

I think that's cool.

That's probably enough for one post.  I'll finish up the other three cases later.

---

[1]  This is not quite true.  The strictly correct statement is that entangled photons do not produce first-order interference.  They can and do produce second-order interference, which can only be detected by transmitting classical information from L1 to L2.

Don Geddis said...

Not disagreeing with any of your conclusions, but ... it makes me uncomfortable to use language like "the principle that which-way information destroys interference". That may be true, but there is a question of causation vs. correlation here. I fear that many laypeople may interpret that sentence as something preciously close to magic, or something about consciousness ("if we can know which way the photon went, then that knowledge itself causally destroys interference"), etc.

The causation is much more plausibly the reverse: the same fundamental cause that allows you to determine "which-way", is also the cause that stops interference. It is not that which-way knowledge itself directly destroys interference.

I prefer to use the multiple worlds interpretation, so I would describe it along the lines that: if you were ever going to be able to determine which-way, then a previous event would have already entangled (aka "split") you (the observer himself) into the component of the photon that is already on only one side of the interference (and thus that slice of you will no longer be able to observe the full interference effect).

Don Geddis said...

P.S. I do feel a little vindicated, that on your original post I guessed an answer of theory #2, which was: "There is interference after T2, but it goes away if the laser is dim enough". On this post, you say: "Then at time Tb the photons arrive from the long path. All the photons are identical, so we no longer know which path they took. So interference." and also "The first is to put some sort of shutter in front of the laser that only lets through one photon through at a time. This is equivalent to turning the laser on only for very short periods of time. If we do it this way we will get no interference."

I know you have other variations and much more complexity, but I'm going to call that close enough to my guess, to make me happy.

Ron said...

@Don:

> I fear that many laypeople may interpret that sentence as something preciously close to magic, or something about consciousness

Hm, I thought I made it clear that this was not the case when I wrote "this setup will not produce interference even if we don't actually measure the idler photon". It's entanglement, not measurement per se, that destroys interference.

That was really intended to be the whole point, so I'm a little dismayed that this didn't come through.

Peter Donis said...

@Ron:
It's entanglement, not measurement per se, that destroys interference.

I think if you had said "the principle that entanglement destroys interference" instead of "the principle that which-way information destroys interference", it would have made your point with much less risk of the kind of misinterpretation that Don is describing.

Peter Donis said...

@Don:
if you were ever going to be able to determine which-way, then a previous event would have already entangled (aka "split") you (the observer himself) into the component of the photon that is already on only one side of the interference (and thus that slice of you will no longer be able to observe the full interference effect).

But this isn't actually true. What prevents observation of the interference is not entanglement of an observer (or measuring device) with either photon. It's the entanglement of the two photons coming out of the PDC. By the time the observer observes the signal photon, the entanglement that destroyed the interference is already done. It isn't that worlds have "split" so that no individual world can see the interference effect; there is no interference effect at all, not even if you look at all of the "worlds" taken together, i.e., the overall pure state of the system including all branches.

Don Geddis said...

"there is no interference effect at all, not even if you look at all of the "worlds" taken together"

I may well still be confused, but I don't yet understand your claim.

Assuming we agree on Ron's post, he starts the setup with this description: "The photons are blocked by the filter at random, so there is no way to tell when a particular photon got through the filter. Hence: interference." There is the interference that I'm claiming. The (MWI) "actual truth" is that individual photons both go through the filter, and also get blocked. Simultaneously, for a single photon. (Hence, interference, even for single photons.)

And how do you "get rid" of interference, in that setup? By running the photon through a PDC to get a pair. But this pair only gets generated, in the "split" world where the photon made it through the filter. In the "full" description of the world state, every photon both does and doesn't make it through the filter. So pairs of photons both do and don't get created, simultaneously.

But once you restrict yourself to the slice where the pair of photons really did get created ... then you're in the slice where interference can no longer be observed.

But I would still assert that interference is still happening ... in the union of worlds that includes slices where that same photon is absorbed by the filter instead, and no pair gets created.

Do you think I'm still mistaken? How would you describe the "overall pure state of the system including all branches"? What do you think that encompasses? And why isn't that simply Ron's world of "Hence: interference"? (Or do you disagree with the description in Ron's post?)

Peter Donis said...

@Don:
Assuming we agree on Ron's post

I agree with Ron's analysis of all the cases he presents, yes.

The (MWI) "actual truth" is that individual photons both go through the filter, and also get blocked. Simultaneously, for a single photon. (Hence, interference, even for single photons.)

No, these possibilities can't interfere, because they're macroscopically distinguishable. (In decoherence language, these possibilities will be decohered.)

In any case, interference between these possibilities is not what's being detected when you put the photons that get through the filter into the interferometer setup. That interference, which is what's being detected, is only produced for photons that get through the filter, and is produced by the interferometer. And the two possibilities for the photon in the interferometer (i.e., the pieces of its wave function in the two arms) are not decohered and are not macroscopically
And how do you "get rid" of interference, in that setup? By running the photon through a PDC to get a pair. But this pair only gets generated, in the "split" world where the photon made it through the filter. In the "full" description of the world state, every photon both does and doesn't make it through the filter. So pairs of photons both do and don't get created, simultaneously.

I agree with this, yes. But as above, the alternatives "photon makes it through filter" and "photon doesn't make it through filter" are macroscopically distinguishable, so they can't interfere.

I would still assert that interference is still happening ... in the union of worlds that includes slices where that same photon is absorbed by the filter instead, and no pair gets created.

Do you think I'm still mistaken?

Yes. See above.

How would you describe the "overall pure state of the system including all branches"? What do you think that encompasses?

See follow-up post to come right after this one.

Peter Donis said...

How would you describe the "overall pure state of the system including all branches"? What do you think that encompasses?

It depends on how much you want to include. If you want to include the filter itself, then the overall pure state would have to include the two states of the filter: "absorbed photon" and "did not absorb photon". And, as above, these states are macroscopically distinguishable, so they are decohered, so there is an initial branching of worlds before we ever get to the point where the PDC comes into play. So the overall state would look something like this:

a |photon absorbed>|filter absorbed photon>|PDC did not down convert anything>|nothing in interferometer>

+

b |photon not absorbed>|filter did not absorb photon>|PDC down converted photon>|signal + idler photon entangled state>

Here a and b are complex amplitudes whose squared moduli sum to 1 and give the probabilities for the filter to absorb and not absorb the photon, respectively.

But these two terms cannot interfere, because they are macroscopically distinguishable. And the first term just says nothing happens inside the apparatus because there is no photon there. So if we want to analyze what happens inside the apparatus, we can ignore the first term and just use the second term, and we can remove the kets from that term that describe the filter and the PDC since those will not add anything to the analysis. (And of course we would need to write down the signal + idler photon entangled state in more detail, but I am assuming there is no dispute about how that would be done.) Doing that will not ignore any interference that would otherwise be included.

Peter Donis said...

@me:
That interference, which is what's being detected, is only produced for photons that get through the filter, and is produced by the interferometer. And the two possibilities for the photon in the interferometer (i.e., the pieces of its wave function in the two arms) are not decohered and are not macroscopically

Agh, this got garbled, should have ended with "macroscopically distinguishable" (and then a nice line break before the next quote in italics).

Also, to clarify, when I say the interference is "produced by the interferometer", I mean that, for cases where there is interference, the interferometer is what is operating on the photon wave function to produce it. I did not mean to imply that the interferometer can do that for cases where there is no interference; the interferometer's operation is the same, but the photon's wave function in these cases is different.

Don Geddis said...

"So the overall state would look something like this ... And the first term just says nothing happens inside the apparatus because there is no photon there."

Thanks for the detail, and I understand what you're trying to say ... but I'm still having a hard time incorporating it with the rest of what I think I understand. Can you help me by doing this same analysis of Ron's original case? "The photons are blocked by the filter at random, so there is no way to tell when a particular photon got through the filter. Hence: interference."

You say that you agree (with Ron) that there is interference. This should be a case where there is no PDC, no pair of photons, a filter in front of the laser, photons generated with much greater (average) time between than Tb, and also Tb>>Ts. There ought to (in general) only be one photon at a time in the interferometer. And if it takes path s, it should hit the detector long, long before it can get through path p.

Why is there interference? What does the overall state look like?

Oh, and maybe you could contrast it with the "overall state" for this description of Ron's: "some sort of shutter in front of the laser that only lets through one photon through at a time. This is equivalent to turning the laser on only for very short periods of time. If we do it this way we will get no interference."

You seem to be suggesting that photons which don't get into the interferometer, can be ignored and play no role. Which would suggest that it doesn't matter whether you use a filter, or a shutter (or turn the laser on & off). Yet you also (apparently) agreed with Ron that you get interference with a filter, but not with a shutter.

This set of claims (by you) seems inconsistent to me. Can you help clarify it so I can understand?

Peter Donis said...

@Don:
if it takes path s, it should hit the detector long, long before it can get through path p.

Why is there interference? What does the overall state look like?

In this case, we are not measuring when the photon is emitted from the laser and passed through the filter. So when a photon is detected at some time t at the detector, the quantum state that predicts the probability of that detection is a superposition of

|photon emitted at time t - Ts>|photon went through short arm>

and

|photon emitted at time t - Tb>|photon went through long arm>

In other words, it's a superposition of the photon being emitted at two different times.

Ron discusses this in the article, but the two hints are separated by several paragraphs. First is this, in the paragraph discussing plausible sounding answer #2:

Whenever we get a detection there are two possibilities: either the photon was emitted Ts seconds earlier and took the short path, or the photon was emitted Tb seconds earlier and took the long path. So if we can tell when the photon was emitted, then there will be no interference.

And then, three paragraphs later, discussing the filter option for making the laser dim:

The photons are blocked by the filter at random, so there is no way to tell when a particular photon got through the filter. Hence: interference.

In other words, with the laser left on and the filter present (but no PDC), a photon can be in a superposition of being emitted at different times, because the time of the photon's emission is not entangled with anything else. And the interference is between the two possible times the photon could have been emitted and passed through the filter; not between the photon being passed through the filter and not passed through the filter.

Peter Donis said...

@Don:
maybe you could contrast it with the "overall state" for this description of Ron's: [shutter/laser only on for short periods]

In this case (with a caveat, see below), the photons arriving at the detector can't be in a superposition of having been emitted at two different times, because the laser was not on at both of those times. In other words, for a photon detected at time t, only one of the two times (t - Ts) and (t - Tb) corresponds to a time that the laser was on. And the laser is not in a superposition of being on and being off; it's either on or off. (By contrast, with the laser always on but the filter present, photons are in a superposition of passing the filter and not passing the filter.)

The caveat (which Ron hinted at in the comment thread of the article prior to this one) is that *if* we time the laser's on/off switching just right, we can make it so that the laser *is* on at both times (t - Ts) and (t - Tb). And if we do that, then we *will* see interference at the detector.

Peter Donis said...

it's a superposition of the photon being emitted at two different times.

To clarify a little bit, this sort of superposition is not always possible. Roughly speaking, the requirements for it to work are:

(1) The photon state has to be a coherent state (which is true for a laser);

(2) The state has to have valid "support" at both times (i.e., the source--laser in this case--has to be on at both times);

(3) The longest path length traveled by the photon has to be less than the coherence length of the state.

Ron hasn't talked about the last of these three, but it does come into play if you make the path length of the long arm of the interferometer too long. IIRC typical coherence lengths for lasers used in these types of experiments can be as large as 100 meters or so, so it's certainly feasible to have a wide variation in the path length of the long arm while still meeting condition #3 above. But if, for example, you wanted to have the mirror of the long arm on the Moon and the rest of the apparatus on Earth, the laser photons would not stay coherent over that path length.

Ron said...

@Peter:

Hey, stop steppin' on my lines! ;-)

Yes, coherence length, blah blah. "Singlemode fiber lasers with linewidths of a few kHz can have coherence lengths exceeding 100 km."

But the really interesting thing is [spoiler alert!] that, for a laser, it actually doesn't make sense to say that a particular photon was emitted by the laser at a particular time or by a particular atom, and this is *essential* to the operation of the laser. This is well known (it's the Heisenberg uncertainty principle applied to energy/time) but often overlooked in QM pedagogy.

Peter got this right:

> a photon can be in a superposition of being emitted at different times

I think that's a profound and often overlooked fact. I myself didn't really realize it until I started thinking about the time-delay interferometer.

Peter Donis said...

@Ron:
"Singlemode fiber lasers with linewidths of a few kHz can have coherence lengths exceeding 100 km."

Ah, my memory was faulty.

for a laser, it actually doesn't make sense to say that a particular photon was emitted by the laser at a particular time or by a particular atom, and this is *essential* to the operation of the laser.

Yes, you're right. I'm not sure how to more accurately describe the properties of the coherent state emitted by the laser in ordinary language, though.

Ron said...

> I'm not sure how to more accurately describe the properties of the coherent state emitted by the laser in ordinary language, though.

Well, there's a lot more to say about laser light than just that. But it's a surprising property (at least it was surprising to me) of laser light that its photons are in superpositions of emission times. The fact that light travels at a constant speed in all reference frames is a bedrock mantra of physics. If light is made of photons then it would seem to follow logically that if a photon is detected at time T, then it must have been emitted at time T-D/c where D is the distance from the source to the detector. But that is not the case.

Luke said...

> The profound conclusion is that photons emitted by a parametric down-converter do not produce interference! [1]
>
> … the photons emitted by a PDC are entangled, and entangled photons do not self-interfere. …
>
> [1] This is not quite true. The strictly correct statement is that entangled photons do not produce first-order interference. They can and do produce second-order interference, which can only be detected by transmitting classical information from L1 to L2.

Is this consistent with Two-photon interference of polarization-entangled photons in a Franson interferometer? There is also Time-resolved double-slit interference pattern measurement with entangled photons.