Does a beam of entangled photons create interference fringes?

[bruce2g]

Does a single beam of entangled photons create interference?

Hi,
This question has come up in an FTL thread going on right now. I've seen two different answers to this question, and I'd be interested if anyone knows the real answer.

Here's the setip: shine a laser beam on a crystal which performs SPDC (spontaneous parametric down conversion). This produces two beams, A and B. The photons in beam A are entangled with the photons in beam B.

Here's the question: If you run beam A through a double slit to a detector, will you see an interference pattern?

I know that SPDC is not perfect, and beams A and B can probably never be perfectly entangled (i.e., some unentangled optical 'junk' gets through), so perhaps this is why there is no definitive experimental evidence. I guess you would need to filter the frequency and do coincidence counting on the two beams to insure that you were only looking at photons that were members of an entangled monochromatic pair.

Still, this seems like a simple question that should have a clear anwer. I hope someone can provide it.

Bruce

 

[vanesch]

Here's the setip: shine a laser beam on a crystal which performs SPDC (spontaneous parametric down conversion). This produces two beams, A and B. The photons in beam A are entangled with the photons in beam B.
Here's the question: If you run beam A through a double slit to a detector, will you see an interference pattern?

Short answer: no.
The quantum-mechanical answer is of course the entanglement, and the fact that the interference term <psi1 | psi2> includes the in product of the "local" photon and the "remote" photon states, and the remote photon states are orthogonal.
But the "classical" answer is that the light coming out of a PDC xtal is a "rainbow". So for two different slits, they will receive slightly different components of the rainbow (slightly different colors), which will hence not interfere.
Now, of course, you can select out a very tiny part of the rainbow, and have it interfere ; but then it will turn out that the entanglement is essentially gone (you've essentially selected out ONE term of the (|a>|b> + |c>|d>) state) and you end up with a separated product state.

 

[dlgoff]

What if you only produced two photons at a time from the PDC and let one of the photons go through the slits i.e. one at a time. Do you get interference then?

Thanks

 

[DrChinese]

What if you only produced two photons at a time from the PDC and let one of the photons go through the slits i.e. one at a time. Do you get interference then?

Thanks

The result Vanesch mentions is independent of the number of photons going through. I.e. he is already assuming you are considering a pair at a time.

Anytime you are talking about PDC, you are pretty much guaranteed to see only a lone pair at a time and the data is built from that. You will only see 2 entangled pairs within a specified time window on the order of twice a minute.

 

[RandallB]

Here's the question: If you run beam A through a double slit to a detector, will you see an interference pattern?

I know that SPDC is not perfect

I think the answer yes of course it would.

Based on the question you asked the SPDC would not be important at all.
In your set up it’s only a part of creating a beam of photons into an experiment that has nothing to do with entanglement.
You are throwing away the B side and not using it, right.
So A is just a beam of photons that will behave as any other in a double slit. When you move the detector and count hits as needed you will accumulate an inference pattern, no reason it shouldn’t.

But if your intent was to compare some part of that area that the pattern builds up on in some polarization test of correlation with beam B. I sure going though the double slit would collapse & destroy any entanglement.

(This wasn't a trick question was it?)

 

[bruce2g]

Short answer: no.
The quantum-mechanical answer is of course the entanglement, and the fact that the interference term <psi1 | psi2> includes the in product of the "local" photon and the "remote" photon states, and the remote photon states are orthogonal.
But the "classical" answer is that the light coming out of a PDC xtal is a "rainbow". So for two different slits, they will receive slightly different components of the rainbow (slightly different colors), which will hence not interfere.
Now, of course, you can select out a very tiny part of the rainbow, and have it interfere ; but then it will turn out that the entanglement is essentially gone (you've essentially selected out ONE term of the (|a>|b> + |c>|d>) state) and you end up with a separated product state.

I have read that if you do coincidence counting between the A and B photon beams (see for example http://grad.physics.sunysb.edu/~amarch/ ) then the interference pattern is detected. Do you know why the coincidence counting would make a difference in this simple setup (in the delayed choice erasure experiments, the coincidence counting separates out the fringes from the anti-fringes; but in this simple case, I don't think there are any anti-fringes).

Bruce

 

[bruce2g]

I think the answer yes of course it would.

Based on the question you asked the SPDC would not be important at all.
In your set up it’s only a part of creating a beam of photons into an experiment that has nothing to do with entanglement.
You are throwing away the B side and not using it, right.
So A is just a beam of photons that will behave as any other in a double slit. When you move the detector and count hits as needed you will accumulate an inference pattern, no reason it shouldn’t.

But if your intent was to compare some part of that area that the pattern builds up on in some polarization test of correlation with beam B. I sure going though the double slit would collapse & destroy any entanglement.

(This wasn't a trick question was it?)

No, it's not a trick. Apparently, the existence of the entanglement changes the behavior of the two photons. I read the book Entanglement : the Greatest Mystery In Physics by Amir D. Aczel, and he says that somehow the two photons are only 'half' photons and that there is no interference unless they are coincidence counted. Unfortunately, Aczel's book is written for the general public (no equations, but lots of anecdotes about Schrödinger's sex life), and I am hoping to get a deeper understanding of the interference phenomenon here.

 

[Edgardo]

Hi,
Here's the setip: shine a laser beam on a crystal which performs SPDC (spontaneous parametric down conversion). This produces two beams, A and B. The photons in beam A are entangled with the photons in beam B.

Here's the question: If you run beam A through a double slit to a detector, will you see an interference pattern?
Bruce

I think yes, there will be an interference pattern. I don't see why there shouldn't be one due to entanglement. Indeed, in Fig. 2 of the Walborn paper it is shown that interference occurs also if you detect the photons in coincidence.

The interference pattern vanishes in the Walborn experiment because you mark the photons by their polarization and thus you get which-way-information. In Walborn's experiment it's important to have coindicent counts otherwise you can't say which of the photon's which-path-information has been erased.
It's all about which-path information!

I have read that if you do coincidence counting between the A and B photon beams (see for example http://grad.physics.sunysb.edu/~amarch/ ) then the interference pattern is detected. Do you know why the coincidence counting would make a difference in this simple setup (in the delayed choice erasure experiments, the coincidence counting separates out the fringes from the anti-fringes; but in this simple case, I don't think there are any anti-fringes).
Bruce

The reason why anti-fringes and fringes occur can be maybe explained the following way: In Fig. 7 you see a non-interference pattern (bell shaped curve). You will notice that it is somehow the sum of Fig. 4 (fringe) and Fig. 5 (anti-fringe) (I don't know why the counts in Fig. 7 are lower). So actually what you see in Fig. 7 are both the photons from Fig. 5 and 4.
Or differently explained: Imagine you have the bell-shaped curve, someone tells you that she doesn't see any interference pattern. But if you take the coincidence counts, you will see interference (fringe). Next step, you substract the fringe pattern from the bell-shaped curve, that is Fig. 4 from Fig. 7. Then, in my opinion, you should get the anti-fringe.

In short: fringe + anti-fringe = bell shaped curve.

See also here:
http://www.mat.ufmg.br/~tcunha/2003-07WalbornF.pdf
Look at the last page of the above pdf at the right bottom picture showing the orange fringe and the red anti-fringe pattern under the black bell shaped curve.

 

[bruce2g]

I think yes, there will be an interference pattern. I don't see why there shouldn't be one due to entanglement. Indeed, in Fig. 2 of the Walborn paper it is shown that interference occurs also if you detect the photons in coincidence.

As near as I can tell, in the very simple setup I described, you will not get interference from a single beam, but you will get an interference pattern if you use coincidence counting.

So, I'm wondering, 'why is it that coincidence counting between the two entangled beams produces interference, but a single entangled beam alone will not produce interference?'

Bruce

(PS - I'm sorry I mentioned the anti-fringes, they don't really apply to the simple setup I described.)

 

[vanesch]

I think the answer yes of course it would.

Based on the question you asked the SPDC would not be important at all.
In your set up it’s only a part of creating a beam of photons into an experiment that has nothing to do with entanglement.

Well, the way the question was posted, it was assumed that no preselection was going to take place such that the entanglement was gone ; and that is exactly what you have to do with the light coming out of a SPDC crystal if you want to have a beam of which you can obtain interference

 

[zekise]

No fringes (unless you do selection)

A few threads below, on Quantum Erasure 9903047, this issue has been discussed. Also a search on entanglement will produce similar discussions.

The answer is that you will NOT observe an interference pattern on pA.

pA not being the whole object, cannot simply interfere with itself. You will have to include the effect of photon pB in the analysis. Photon pB will act to obscure the interference pattern, so that without the help of pB (through coincidence selection), you will not see the interference pattern for pA on the double slit. Instead what you will see is a gaussian distribution.

I understand what happens is conceptually as follows: The wave function governing pA also contains equal components from pB. If pA collapses on its own components, you will get a fringe. But there is equal probability that pA will collapse on the wavefunction for pB. In this case pA and pB being asymmetric, such a collapse will generate an anti-fringe.

You will never know just by looking at pA whether it has collapsed on the fringe or anti-fringe. So you cannot separate out (without the help of pB) the two fringes. The end result is that the anti-fringe will exactly normalize the fringe resulting in a gaussian (normal) distribution, that contains NO information (of the wavelength - momentum).

However, if you measure pB, transmit the info locally (classically) to pA, you will be able to tell when pA falls on the fringe vs. the anti-fringe, and hence select out the fringe. Note that the frequency you will get is the sum of the frequencies of pA and pB, which shows that pA is acting non-locally.

In entangled pairs, each element contains no information. It is the sum of the two entangled parts that contain information.

Yes, the biography in Aczal's book is quite amusing.

 

[RandallB]

, it was assumed that no preselection was going to take place such that the entanglement was gone ; and that is exactly what you have to do with the light coming out of a SPDC crystal if you want to have a beam of which you can obtain interference

Producing an interference pattern, ala young’s double slit, has nothing to do with “coincidence counting” it uses statistical accumulation of many problematic events to build up an interference pattern. So I have no idea why Bruce things it does.

The idea that with no “pre-selection”, thus no loss of entanglement; somehow leaves a beam of light, or individual photon in it, abnormal or missing something as compared to another beam able to produce interference when sent through a double slit.

What exactly do you expect QM says would be shown on a negative given appropriate exposure time with both slits open for a single SPDC created beam?
If not interference, what is predicted? Seems like very simple experiment to conduct – IF interference is produced would you say QM or some part of QM is falsified?

I’m positive interference will be seen, but I can see no part of QM such a demonstration would falsify.

 

[vanesch]

What exactly do you expect QM says would be shown on a negative given appropriate exposure time with both slits open for a single SPDC created beam?
If not interference, what is predicted? Seems like very simple experiment to conduct – IF interference is produced would you say QM or some part of QM is falsified?

Imagine an entangled pair:
|red>|blue> + |blue>|red>

Note, that in order for the photons to be entangled in the first place, we need them to come in two DIFFERENT flavors (here, red and blue). Indeed, if all the photons in the first beam were in the pure "red" state, we would not have entanglement, because that state could be factored out.

Ok, this means, if you are only going to look at the first photon, that they will appear to come in a statistical mixture, which is 50% red, and 50% blue. If you send them through a thing to have them interfere, then you will have 50% a "red" interference pattern, and 50% a "blue" interference pattern, which comes down to no pattern at all. That is not surprising, and you don't need to look at the second beam for this, because the first beam simply appears to be "white" (a stochastical mixture of red and blue).

But you can now put a filter in there. You can filter out only the red photons. But then, you're left (Copenhagen view) only with the |red> |blue> state. Entanglement is gone. Interference is possible.

So there is nothing "magical" about photons from an entangled pair, which would suddenly make them "avoid to interfere". It is only that, when you view them from one side, they COME IN A STATISTICAL MIXTURE of orthogonal states (here, color). So it is not amazing that the beam mixture doesn't seem to give interference. From the moment you purify the mixture, you select only ONE term of the entanglement (which means that you end up with a product state). Then of course you can now have interference.

ADDED: that's why the beams coming out of a PDC Xtal come in a *rainbow*.

 

[Sherlock]

Imagine an entangled pair:
|red>|blue> + |blue>|red>

Note, that in order for the photons to be entangled in the first place, we need them to come in two DIFFERENT flavors (here, red and blue). Indeed, if all the photons in the first beam were in the pure "red" state, we would not have entanglement, because that state could be factored out.

Ok, this means, if you are only going to look at the first photon, that they will appear to come in a statistical mixture, which is 50% red, and 50% blue. If you send them through a thing to have them interfere, then you will have 50% a "red" interference pattern, and 50% a "blue" interference pattern, which comes down to no pattern at all. That is not surprising, and you don't need to look at the second beam for this, because the first beam simply appears to be "white" (a stochastical mixture of red and blue).

But you can now put a filter in there. You can filter out only the red photons. But then, you're left (Copenhagen view) only with the |red> |blue> state. Entanglement is gone. Interference is possible.

So there is nothing "magical" about photons from an entangled pair, which would suddenly make them "avoid to interfere". It is only that, when you view them from one side, they COME IN A STATISTICAL MIXTURE of orthogonal states (here, color). So it is not amazing that the beam mixture doesn't seem to give interference. From the moment you purify the mixture, you select only ONE term of the entanglement (which means that you end up with a product state). Then of course you can now have interference.

ADDED: that's why the beams coming out of a PDC Xtal come in a *rainbow*.

This is interesting, especially since I don't understand SPDC yet. Whether an interference pattern emerges when a beam of photons is sent through a double-slit device depends on the original photon source (eg., the crystals used in SPDC)?

So, the answer to RandallB's question:What exactly do you expect QM says would be shown on a negative given appropriate exposure time with both slits open for a single SPDC created beam? -- is that no interference pattern will emerge? So, exactly what would you see on the negative after running beam A through the double-slit device -- just a single-slit distribution?

Sorry to be redundant, but I'm having a difficult time understanding this. These red and blue states of the photons in beam A --- is the beam a combination of photons having different wavelengths?

 

[vanesch]

Sorry to be redundant, but I'm having a difficult time understanding this. These red and blue states of the photons in beam A --- is the beam a combination of photons having different wavelengths?

Yes, and/or different polarizations. That's the entire idea! Classically, you can understand the different wavelengths (at different angles) as the different solutions to the pump -> idler + signal conservation of energy and momentum conditions, when the angles are slightly different. That's why the output is a *rainbow*. Let us forget for the moment, polarization, it makes the issue a bit more complicated. Let us assume that we are going to consider the entangled states based upon wavelength (and hence, outgoing angle). In fact, because of the relationship between angle and color, the outgoing photons will indeed be in an entangled state |red>|blue> + |blue> |red>. One could do EPR-like experiments with that (usually it is not the wavelength, but the polarization that is used as discriminating property, but wavelength is simpler to explain).

This means you have to accept, of course, photons of different wavelength in your experiment! Otherwise, you have already selected out ONE term of the entanglement. But "accepting photons of different wavelength" means, having essentially an incoherent light beam. So if you try your interference experiment on it, you will not see interference. And you won't find that strange, because the beam, to you, looks incoherent. You're not surprised.
Now, you can then purify your beam, to have a coherent beam. And then of course you will have interference ; but you destroyed the entanglement by purifying the beam! You projected out ONE of the terms, in quantum speak.

It is of course important that the entangled degree of freedom is the one causing, or not, interference. If the interference is independent of the quantity that is entangled (meaning, if that kind of interference is not going to change wrt the measured property of the entangled partner, and as thus, is not the right experiment to do an EPR demonstration), then of course you can still obtain the interference, with entanglement.

This is thinkable, for instance, with a pure wavelength entangled pair, the entanglement being the polarization. If we have the state:

|red+>|blue-> + |red->|blue+> and we do a double-slit interference experiment on the red beam which is INSENSTIVE TO POLARIZATION, then of course we will get an interference pattern. This is because the |red+> state and the |red-> state will give rise to identical interference patterns, and so the statistical mixture of 50% |red+> and 50% |red-> will also yield that interference pattern. But with such an interference experiment, you cannot do any EPR like demonstration, because no matter what you do, and no matter what you subselect on the "blue" side, you'll ALWAYS get the same red interference pattern. That is because the red interference pattern is INDEPENDENT of the entangled degree of freedom (which is, here, polarization) - in fact, it can be factored out, and one could write:

|red>|blue> (|+>|-> + |->|+>)

So the "red" and the "blue" degrees of freedom are NOT entangled, only the + and - polarization degrees of freedom are. As such, interference of the "red" degree of freedom is possible, and of the "blue" degree of freedom too.

But not a polarization-dependent interference pattern. That will appear to be an unpolarized mixture at both sides.

 

[bruce2g]

That's why the output is a *rainbow*...
This means you have to accept, of course, photons of different wavelength in your experiment! Otherwise, you have already selected out ONE term of the entanglement. But "accepting photons of different wavelength" means, having essentially an incoherent light beam. So if you try your interference experiment on it, you will not see interference. And you won't find that strange, because the beam, to you, looks incoherent. You're not surprised.
...

Just by way of background, when you do a PDC, you shine in a monochromatic laser light and the non-linear crystal produces two photons for each one input. Conservation of energy requires that the frequencies of the new photons sum to the original laser frequency. However, the new photons do not each have 1/2 of the original frequency -- they just need to sum to the original frequency. So, even though you started with a monochromatic laser, your new beams will be incoherent.

Moving right along, thanks for the nice explanation, Vanesch. I'm still slightly confused by a few things I see in the published experiments, so let me see if I've got this right:

DetectorA <-- Double-slit <-- FilterA <-- SPDC --> FilterB --> DetectorB

1. If you have no filters and no coincidence counting, then there will not be any interference at DetectorA.

2. If you have no filters but you do coincidence counts between DetectorA and DetectorB, there I think there will be an interference pattern revealed by the coincidence counters.

3. If you have FilterA, and no coincidence counts, then you will see interference (monochromatic light, entanglement lost).

4. If you have FilterB but not FilterA, and you perform coincidence counts, then you will see interference (since monochormatic B implies monochromatic A).

The reason I'm not sure about (2) is that in the Walborn et al paper ([PLAIN]http://grad.physics.sunysb.e...//www.mat.ufmg.br/~tcunha/2003-07WalbornF.pdf

 

[Sherlock]

Yes, and/or different polarizations. That's the entire idea! Classically, you can understand the different wavelengths (at different angles) as the different solutions to the pump -> idler + signal conservation of energy and momentum conditions, when the angles are slightly different. That's why the output is a *rainbow*. Let us forget for the moment, polarization, it makes the issue a bit more complicated. Let us assume that we are going to consider the entangled states based upon wavelength (and hence, outgoing angle). In fact, because of the relationship between angle and color, the outgoing photons will indeed be in an entangled state |red>|blue> + |blue> |red>. One could do EPR-like experiments with that (usually it is not the wavelength, but the polarization that is used as discriminating property, but wavelength is simpler to explain).

This means you have to accept, of course, photons of different wavelength in your experiment! Otherwise, you have already selected out ONE term of the entanglement. But "accepting photons of different wavelength" means, having essentially an incoherent light beam. So if you try your interference experiment on it, you will not see interference. And you won't find that strange, because the beam, to you, looks incoherent. You're not surprised.

The entangled photons produced by PDC crystals are entangled in both wavelength and polarization. So, if beam A or B (which are each superpositions of two different wavelengths) is directed through a double-slit device (with no other filtering), then the banding characteristic of a coherent light source directed through a double-slit device won't be seen at screenA or screenB.

Now, you can then purify your beam, to have a coherent beam. And then of course you will have interference ; but you destroyed the entanglement by purifying the beam! You projected out ONE of the terms, in quantum speak.

It is of course important that the entangled degree of freedom is the one causing, or not, interference. If the interference is independent of the quantity that is entangled (meaning, if that kind of interference is not going to change wrt the measured property of the entangled partner, and as thus, is not the right experiment to do an EPR demonstration), then of course you can still obtain the interference, with entanglement.

So it isn't entanglement, per se, that disallows an interference pattern sans filtering.

This is thinkable, for instance, with a pure wavelength entangled pair, the entanglement being the polarization.
If we have the state:

|red+>|blue-> + |red->|blue+> and we do a double-slit interference experiment on the red beam which is INSENSTIVE TO POLARIZATION, then of course we will get an interference pattern. This is because the |red+> state and the |red-> state will give rise to identical interference patterns, and so the statistical mixture of 50% |red+> and 50% |red-> will also yield that interference pattern.

The PDC created state is:
|red+>|blue-> + |blue-> |red+>

... and I see the difference between the PDC state and:
|red+>|blue-> + |red->|blue+>

What sources produce the latter state (substituting whatever colors or wavelengths are characteristic of the emissions)?

But with such an interference experiment, you cannot do any EPR like demonstration, because no matter what you do, and no matter what you subselect on the "blue" side, you'll ALWAYS get the same red interference pattern. That is because the red interference pattern is INDEPENDENT of the entangled degree of freedom (which is, here, polarization) - in fact, it can be factored out, and one could write:

|red>|blue> (|+>|-> + |->|+>)

So the "red" and the "blue" degrees of freedom are NOT entangled, only the + and - polarization degrees of freedom are. As such, interference of the "red" degree of freedom is possible, and of the "blue" degree of freedom too.

But not a polarization-dependent interference pattern. That will appear to be an unpolarized mixture at both sides.

Ok, thanks vanesch. A clear enough explanation for laymen like me (though I did spend about 4 hours yesterday refreshing my memory of the classical optics math involved in interference and diffraction).

 

[vanesch]

The PDC created state is:
|red+>|blue-> + |blue-> |red+>

... and I see the difference between the PDC state and:
|red+>|blue-> + |red->|blue+>

The PDC Xtal generates a lot of states. The degrees of freedom of a photon are wavevector and polarization ; the wavevector itself consists of direction and wavelength. So you could think of a "photon state" as:
|direction>|color>|spin>

Now, a PDC xtal generates a lot of terms of the kind:

|direction1>|color1>|spin1> |direction2> |color2> |spin2>:

we have:

|2-photon state> = weighted sum over
|direction1_a>|color1_a>|spin1_a> |direction2_a> |color2_a> |spin2_a>
|direction1_b>|color1_b>|spin1_b> |direction2_b> |color2_b> |spin2_b> +
|direction1_c>|color1_c>|spin1_c> |direction2_c> |color2_c> |spin2_c>
+...

where a,b,c... are different possible cases, which respect the physics of the PDC, that is: conservation of energy, angular momentum and momentum in the |pump> => |c> transition.

As such, the PDC generates a very complex entangled 2-photon state.

But the experimental setup selects out usually a single direction for idler and signal beam, and uses a wavelength filter. As such, only ONE set of direction1 and direction2, and color1 and color2 are selected ; what remains are the different possibilities of spin. Which gives us:

|direction1>|color1>|spin1a>|direction2>|color2>|spin2a>
+ |direction1>|color1>|spin1b>|direction2>|color2>|spin2b>

which can be rewritten as:

|direction1>|color1>|direction2>|color2> (|spin1a>|spin2a> +
|spin1b>|spin2b>)

As the direction and colors of the two beams are now factored out, this is a non-interesting degree of freedom, and one usually doesn't even mention it.

cheers,
Patrick.

 

[RandallB]

But "accepting photons of different wavelength" means, having essentially an incoherent light beam. So if you try your interference experiment on it, you will not see interference. And you won't find that strange, because the beam, to you, looks incoherent. You're not surprised.

Ridiculous – I’d be absolutely shocked not to get an interference pattern!
You can make an interference paten with sunlight how much more incoherent do you need. Also, the wavelength diff between the two “different colors” is such a tiny fraction it’s not worthy of being called a “rainbow”.

“If not interference, what is predicted?”
Your saying – a single-slit distribution through double slits!

You quoted my other question but didn’t answer:
“Seems like very simple experiment to conduct – IF (and IMO when) interference is produced would you say QM or some part of QM is falsified?”

What part of QM stands to be falsified if your claim fails??

Before you dig this any deeper or common sense kicks in, PM another mentor to take a look at this.

This is simple double slit - entanglement has no effect.

 

[QuantumUnbound]

Ridiculous – I’d be absolutely shocked not to get an interference pattern!

I agree. You would see an interference pattern.

If you just viewed a screen behind the double slit you would "see" an interference pattern. Of course, the way I understand it you are viewing much more than just entangled photons. PDC (parametric down conversion) only produces a few hundred entangled pairs per second. So the entangled pairs are pretty much washed out by the millions of non-entangled photons (Isn't that correct?)
But, even if you use a coincidence counter to only record entangled pairs then you will still see an interference pattern, because you have done nothing to determine which-way informaton.
Now, if you start adding quarter-wave plates and polarizers things get a little crazy. The best explanation I have seen is a link earlier in this thread: http://www.mat.ufmg.br/~tcunha/2003-07WalbornF.pdf

"Anytime you send photons, of any kind or flavor, through a double slit you will see an interference pattern unless you provide a method for which-way information." I believe that statement to be incontrovertible. (I said that, it wasn't in a paper somewhere.)

Coherence doesn't matter. Talk of colors, frequency, and other stuff is just confusing and not needed. Polarization techniques are the easiest to explain and actually perform.

Of course, I could be wrong

 

[JesseM]

Ridiculous – I’d be absolutely shocked not to get an interference pattern!

...

This is simple double slit - entanglement has no effect.

I agree. You would see an interference pattern.

Did you guys look over the earlier thread abou the delayed choice quantum eraser experiment, or the followup thread about the same experiment? In this experiment we have a pair of entangled photons generated at two possible locations (exactly analogous to the slits in the double-slit experiment), one of which goes to a screen, another of which goes through a measurement apparatus which can either measure the photon in such a way that you can retroactively determine which location both photons come from, or can measure it in such a way that the which-path information is "erased". By your argument, since there were no detectors to determine which location the photon that went to the screen came from, does this automatically mean you would expect interference? But if you were to see interference and then measure the other photon in such a way that you figure out which location they both came from, this would seem to be a violation of complementarity, since you aren't supposed to get interference when you have the which-path information. On the other hand, if you saw interference on the screen when you recovered the which-path information but lost the interference when you erased it, this would imply a possibility of FTL (or even backwards-in-time) communication. So it seems like the only acceptable conclusion is that you won't see interference on the screen here, and indeed that's what is found experimentally, and is also what is predicted by the mathematics of QM. However, if you do some coincidence-counting, looking at only the subset of photons on the screen whose twins were detected at a detector which erased their which-path information, then you do see interference in this subset, even though the total pattern of photon hits on the screen shows no interference.

Whenever you're dealing with entanglement, you have to worry about the possibility that by measuring one member of the entangled pair, you can recover the which-path information of the other. Vanesch pointed out that there are some situations involving entanglement where there is no way of doing this, but there are others where there is--Dr. RMC pointed one out on this thread, where you have two photons with entangled momentum aimed at two different double-slits on the left and the right, such that conservation of momentum insures that if you detect the photon on the left at the bottom slit, this tells you the photon on the right must have gone through the top slit, even if there were no detectors at the double-slit on the right. Again, if we want to both preserve complementarity and avoid the possibility of FTL communication, this means that we cannot see any interference in the total pattern of photons on the screen, because there's always the possibility that you could retroactively determine which slit it went through by performing a measurement on its entangled twin.

 

[QuantumUnbound]

JesseM:
I guess I was wrong. I wonder if Feynman had this much trouble keeping it all straight? And I only capitulate because the experiments seem to say what your saying. But I still don't understand why.
I mean, if only about 200 photons a second are actually entangled then why would the rest of them act differently? 99.9% of the photons are not entangled. I think I will read some more.

 

[vanesch]

Also, the wavelength diff between the two “different colors” is such a tiny fraction it’s not worthy of being called a “rainbow”.

You seem not to have understood what I wrote. You CAN have, in certain circumstances, an interference pattern, if the relevant degree of freedom that is responsible for the interference pattern, is not entangled. It depends on how the interference experiment is set up, and I initially assumed that the interference experiment was set up to be relevant to the entanglement situation (in the sense that the interference pattern would somehow depend on how the other state was analysed, so as to provide an EPR paradox because you'd have both the interference pattern and the which-way information).
The point was that there is no magic to this, because from the viewpoint of one beam, you have a statistical mixture which will, in such a case, wash out any interference. But if the statistical mixture you obtain DOES NOT INFLUENCE the particular interference experiment you're performing, then of course you WILL see interference. But that interference is insensitive to the entanglement (the relevant degree of freedom is in a product state) and as such useless as a tool to look at the entanglement.

“Seems like very simple experiment to conduct – IF (and IMO when) interference is produced would you say QM or some part of QM is falsified?”

What part of QM stands to be falsified if your claim fails??

My claim will not fail, that if your interference experiment is sensitive to the entangled degrees of freedom (wether it be spin, wavelength, direction,...), then you will not see interference. I didn't state explicitly the condition, my fault (I thought that it was obvious that the interference was going to be relevant to the entangled degree of freedom). The reason why you will not see interference is that, locally, your beam is seen as in a MIXTURE of two complementary interference-generating states (in the reduced density matrix).

 

[Edgardo]

...
By your argument, since there were no detectors to determine which location the photon that went to the screen came from, does this automatically mean you would expect interference? But if you were to see interference and then measure the other photon in such a way that you figure out which location they both came from, this would seem to be a violation of complementarity, since you aren't supposed to get interference when you have the which-path information. On the other hand, if you saw interference on the screen when you recovered the which-path information but lost the interference when you erased it, this would imply a possibility of FTL (or even backwards-in-time) communication.
...
complementarity and avoid the possibility of FTL communication, this means that we cannot see any interference in the total pattern of photons on the screen, because there's always the possibility that you could retroactively determine which slit it went through by performing a measurement on its entangled twin.

I find this post by JesseM very interesting, since it describes what the problem is really about.

Question to Jesse:
You say that no interference occurs because of the mere possibility gaining which-path information.

But my question is: Why do we get an interference pattern when watching at the data with coincidence counting? I do not see why coincidence counting erases which-path information in the setup by Walborn, Fig. 2.
http://grad.physics.sunysb.edu/~amarch/Walborn.pdf

http://grad.physics.sunysb.edu/~amarch/PHY5655.gif
http://grad.physics.sunysb.edu/~amarch/PHY5651.gif

Or is it erased because in coincidence counts we have made sure that the upper photon can't be used anymore to gain which-path information?

 

[vanesch]

Or is it erased because in coincidence counts we have made sure that the upper photon can't be used anymore to gain which-path information?

Yes. See the neighbouring thread on the same subject...

 

[Edgardo]

Yes. See the neighbouring thread on the same subject...

1) So, if I took a metal plate and blocked the upper photon right behind the crystal such that nobody would be able to gain which-path information, interference would definitely occur?

2) In the setup of Fig. 2 in the Walborn paper:
What if I replaced the upper detector by a better one, namely one that can measure the position of the upper photon when it is detected (left or right)?

Imagine that this detector has two channels, left and right and repeat the original poster's experiment.
Then, if I took coincidence again without differing left and right channel,
I get interference because which path-information is erased. So far so good, that is what we expect, we get a Sinc-shaped distribution
(Sinc-function is the typical interference pattern)

But then, if I detected with coincidence for example only on left channel,
the interference should vanish because I have which-path information.
Coincidence with only right channel would also yield no interference pattern. Result for this thought-experiment: By measuring coincidence with one "position"-channel, you destroy the interference pattern. That is, choosing a subset of the above photons yields a Gaussian-shaped distribution.

My problem with this result:
Total photons = Subset(left channel) + Subset(right channel)
=> Sinc = Gaussian + Gaussian

with "Total photons = Sinc" and
"Subset=Gaussian"

How can this be?

I hope my thought-experiment makes sense and the setup is clear.
Note: This is only a thought experiment and I don't know whether the use of such "position"-channels and coincidence would destroy interference.

 

[vanesch]

1) So, if I took a metal plate and blocked the upper photon right behind the crystal such that nobody would be able to gain which-path information, interference would definitely occur?

No ! Because you've lost the essential subselection information. Nature is not going "to check whether you could eventually...". It wants you to prove that you cannot restore the which-path information, by USING the complementary information. (or vice versa). If you can't show anything, you get nothing.

2) In the setup of Fig. 2 in the Walborn paper:
What if I replaced the upper detector by a better one, namely one that can measure the position of the upper photon when it is detected (left or right)?

because of the entanglement between the two directions ? If the entanglement between the two directions would be sufficient, then you would need another subset selection, which would go as follows: you'd need a double slit in the upper beam, and detect the interference pattern (and as such, loose the directional information) at the upper detector as subselection information. The reason is that you will now have created a beam (on the lower side) which will need to have two clearly separated modes (one going to the upper slit, and the other going to the lower slit) which will look like statistically mixed (incoherent modes).

Imagine that this detector has two channels, left and right and repeat the original poster's experiment.
Then, if I took coincidence again without differing left and right channel,
I get interference because which path-information is erased.

No, you wouldn't get interference in this case. In order for you to have interference, you need to make sure that the source is common to both channels (for instance, by making a - diffracting - pinhole or so), which will screw up the link between the direction of the first and the second photon (you kill direction entanglement). If you allow for direction entanglement, then there must be a clear "left path" and a clear "right path", and it will seem to you that you get independently a photon in the "left path" and then in the "right path" from the xtal (in other words, that the two clearly separated beams are incoherent).

But then, if I detected with coincidence for example only on left channel,
the interference should vanish because I have which-path information.
Coincidence with only right channel would also yield no interference pattern. Result for this thought-experiment: By measuring coincidence with one "position"-channel, you destroy the interference pattern. That is, choosing a subset of the above photons yields a Gaussian-shaped distribution.

No, I don't think that you can ever obtain NON-interference out of a subselection of an interference pattern (due to the zeros!).

Note: This is only a thought experiment and I don't know whether the use of such "position"-channels and coincidence would destroy interference.

I think it should, from a QM perspective:

You have: |left>|right> + |right>|left>

(were left and right are two clearly separated modes of emission from the xtal, otherwise there would be no link between the direction of the left and the right beam).

When looking only at the first beam, you then get, through the reduced density matrix, that this looks like a STATISTICAL mixture of:
50% |left> and 50% |right>
as such, the two modes coming out of the xtal seem to be completely incoherent (a photon in the left one, then a photon in the right one...).

What you usually do to do a 2-slit experiment, is to use a pin hole, or a lens or so, to GET EQUAL ILLUMINATION FROM THE SAME BEAM of the two slits. (and if you do that, you DESTROY the link left-right between signal and idler beam!). If you let two clearly separated modes emanent from the Xtal shine on the two slits, it is as if you had two different light bulbs, each illuminating ONE slit: that doesn't give an interference pattern.

At least, that's how I interpret it from a purely QM point of view.

 

[RandallB]

My claim will not fail, that if your interference experiment is sensitive to the entangled degrees of freedom (wether it be spin, wavelength, direction,...), then you will not see interference. I didn't state explicitly the condition, my fault (I thought that it was obvious that the interference was going to be relevant to the entangled degree of freedom). The reason why you will not see interference is that, locally, your beam is seen as in a MIXTURE of two complementary interference-generating states (in the reduced density matrix).

‘QuantumUnbound’ may be easily swayed by some mumbo jumbo about quantum erasure, but that requires the actual use of the B beam, so I am not.
I understand you think your CLAIM WILL NOT FAIL.
But, what you are still not stating is what part of QM does your claim stand on.
IF (just in case IF) it was to fail. Would it be QM in general or some part QM that you’re using, would be falsified by that failure, in your opinion; can you at least be clear on this?
[For me this will just be your interpretation of QM that fails, because the pattern will be seen.]

As I said easy enough experiment, as per the OP; Parametric Down Conversion is used to create two beams A & B.
First all needed tests are done to ensure each beam is entangled with the other including “a MIXTURE of two complementary interference-generating states” and “rainbow” in each beam is being generated.

Then without changing the beams just change the test on beam A to a double slit test, rotating that test to any or all polar alignments as you like. And made as sensitive to “entangled degrees of freedom” (whatever that means) as you like.
Just as long as no part of beam B is brought into the experiment to convert it into some kind of entanglement or erasure test. (beam B is discard, redirect it to empty space or a blank wall whatever).

If you don’t get interference I’ll eat my hat, or whatever stakes you care to put on it (and I do have a hat).

 

[JesseM]

‘QuantumUnbound’ may be easily swayed by some mumbo jumbo about quantum erasure, but that requires the actual use of the B beam, so I am not.

Did you actually bother looking over the delayed choice quantum eraser experiment? This experiment has actually been performed, and no interference is observed. Also, if you would bother to try to follow my argument, you would see why I say that this is what we must expect if we want to preserve complementarity and avoid FTL signalling, and if you disagreed you would have some kind of actual counterargument beyond calling it "mumbo jumbo".

IF (just in case IF) it was to fail. Would it be QM in general or some part QM that you’re using, would be falsified by that failure, in your opinion; can you at least be clear on this?
[For me this will just be your interpretation of QM that fails, because the pattern will be seen.]

It has nothing to do with interpretational issues; the mathematics of QM that is used to predict what probability distribution to expect in any given experimental setup is not disputed by any interpretation, and presumably vanesch is basing his statements on such mathematical predictions.

If you don’t get interference I’ll eat my hat, or whatever stakes you care to put on it (and I do have a hat).

Looks like you'll have to start eating then, because the delayed choice quantum erasure experiment already shows that photons coming from one of two possible locations (just like the slits) will not create interference on the screen if they are members of entangled pairs, with the other member of the pair going in the opposite direction and capable of being measured in such a way that tells you which of the two locations both members of the pair came from.

 

[JesseM]

I find this post by JesseM very interesting, since it describes what the problem is really about.

Question to Jesse:
You say that no interference occurs because of the mere possibility gaining which-path information.

But my question is: Why do we get an interference pattern when watching at the data with coincidence counting?

Well, you can only do coincidence counting after the idler photon has been detected at one of the detectors. And of the 4 possible detectors the idler can be found at in the setup given in Scully's paper, 2 will "erase" the which-path information so that there is no longer any possibility of doing a measurement which would retroactively tell you where the two entangled photons; it is only when you do coincidence-counting between hits at one of those detectors and the screen that you see an interference pattern.

I do not see why coincidence counting erases which-path information in the setup by Walborn, Fig. 2.
http://grad.physics.sunysb.edu/~amarch/Walborn.pdf

I don't understand the exact details of this setup, but it apparently has something to do with placing different polarizers over the two slits which the signal photon (which they label 's') can go through, and also measuring the polarization of the idler (which they label 'p'). Look at fig. 1, where they say that "quarter-wave plates" are placed over the slits that s goes through and a linear polarizer is placed over the detector that p goes to. Section A, 'Obtaining the which path information', suggests that given certain polarization measurements on p, you can tell something about the polarization of s which allows you to tell which slit it must have gone through (because polarizers block light with certain polarizations, and presumably the orientation of the polarizers over the two slits is different):

If photon p is detected with polarization x

 

[JesseM]

To add to my last post, I just came across http://grad.physics.sunysb.edu/~amarch/ which describes the experiment that Edgardo linked to in terms that are more accessible to a laymen.

diagram: http://grad.physics.sunysb.edu/~amarch/PHY5656.gif

To make the "which-way" detector, a quarter wave plate (QWP) is put in front of each slit. This device is a special crystal that can change linearly polarized light into circularly polarized light. The two wave plates are set so that given a photon with a particular linear polarization, one wave plate would change it to right circular polarization while the other would change it to left circular polarization.

With this configuration, it is possible to figure out which slit the s photon went through, without disturbing the s photon in any way. Because the s and p photons are an entangled pair, if we measure the polarization of p to be x we can be sure that the polarization of s before the quarter wave plates was y. QWP 1, which precedes slit 1, will change a y polarized photon to a right circularly polarized photon while QWP 2 will change it to a left circularly polarized photon. Therefore, by measuring the polarization of the s photon at the detector, we could determine which slit it went through. The same reasoning holds for the case where the p photon is measured to be y. The following table provides a summary.

[I couldn't reproduce the table here, but go to the page to see it]

The presence of the two quarter wave plates creates the possibility for an observer to gain which-way information about photon s. When which-way information is available, the interference behavior disappears. It is not necessary to actually measure the polarization of p and figure out what slit s passed through. Once the quarter wave plates are there, the s photons are marked, so to speak.

So, if you know the linear polarization of the s photon before it went through the quarter-wave plates, then by measuring its circular polarization at the detector Ds (which functions as the 'screen' in the double-slit setup), you can retroactively tell which slit it went through. But circular polarization and linear polarization are noncommuting variables like position and momentum, so once you measure the circular polarization of s, its linear polarization should be scrambled; however, you can tell what the linear polarization of the s photon was before it hit the quarter-wave plates by measuring the linear polarization of the p photon that it was entangled with. But by putting a polarizer in the p photon's path before you measure its linear polarization, you can lose the information about what its original polarization was, and thus also lose the information about the linear polarization of the s photon:

diagram: http://grad.physics.sunysb.edu/~amarch/PHY5657.gif

Increasing the strangeness of this scenario, the next step is to bring back the interference without doing anything to the s beam. A polarizer is placed in the p beam, oriented so that it will pass light that is a combination of x and y. It is no longer possible to determine with certainty the polarization of s before the quarter wave plates and therefore we cannot know which slit an s photon has passed through. The s photons are no longer marked. The potential to gain which-way information has been erased.

In this case you regain interference if you do a coincidence count between the two detectors. But because the linear polarizer is in the path of the p photons, I'm guessing that only half as many p photons would make it to the detector with the polarizer in place, so that many of the s-photon detections do not have a corresponding p-photon detection to include in the coincidence count; presumably the total pattern of s photons would still show no interference, because if they did in this case, then as always this would lead to the possibility of FTL or backwards-in-time signalling by changing the total pattern of s photons by removing or replacing the polarizer in the path of the p photons.

 

[DrChinese]

If you don’t get interference I’ll eat my hat...

You already know you won't get interference, because then you could perform FTL signalling if there were. So the real question is: why no interference?

The answer, in case it wasn't clear, is that entangled photons will act differently than photons which are not entangled. There are different constraints - i.e. different degrees of freedom.

Or seen a different way: for there to be interference, a photon needs to be able to able to traverse all possible paths so it can interfere with itself. As Mandel has stated in his writings, "On the other hand, if there is some way, even in principle, of distinguishing between the possible photon paths, then the corresponding probabilities have to be added and there is no interference."

This behavior occurs in the quantum world, but has no classical counterpart. In the classical world, there would always be interference.

 

[vanesch]

Then without changing the beams just change the test on beam A to a double slit test, rotating that test to any or all polar alignments as you like. And made as sensitive to “entangled degrees of freedom” (whatever that means) as you like.
Just as long as no part of beam B is brought into the experiment to convert it into some kind of entanglement or erasure test. (beam B is discard, redirect it to empty space or a blank wall whatever).

First of all, I'd like to point out that interpretations of QM have nothing to do with this, it is all standard quantum theory (different interpretations all agree on extracted probabilities of observation ; if they don't, they are not *interpretatons* but different theories). So, or I'm making elementary errors in using the quantum formalism, or the result is independent of any interpretational scheme.

If you have an entangled system, and you dump the partner (as you suggest), then you can only do measurements on the first beam. So all observables we're going to discuss are determined by the *reduced density matrix* of the first system.

Now, what is this "interference, sensitive to the entangled degrees of freedom" ?

First of all, what is interference ? You have interference when you have a 2 states |a> and |b> and you have (at least) 2 observables A+ and A-, which measure (|a> + |b>) and (|a> - |b>). If the incoming state is |a>+|b>, then A+ clicks 100% sure, and A- doesn't click, 100% sure. You can have more than 2 observables, for instance, A(x), where x is the position on a screen. A(x) clicks 100% sure if the state is |a> + exp(i x) |b> and doesn't click if the state is |a> - exp(i x)|b>. A+ then corresponds to x = 0 and A- corresponds to x = 180 degrees (pi).

In a setup such as the double slit experiment, |a> = |through-the-left-slit> and |b> = |through-the-right-slit>

We say that we have interference on an incoming state if A+ clicks always, and if A- doesn't click at all, or vice versa. So in our example, if the incoming state is |a>+|b>, we have interference, and if the incoming state is |a>-|b>, we have (complementary) interference. However, if the state is |a>, we don't, and if the state is |b>, we don't: A+ and A- both have 50% chance to click, in both cases.

It should be clear that interference occurs when the incoming state is |a>+|b>, or |a>-|b>. However, when the incoming state is a STATISTICAL MIXTURE of 50% |a> and 50% |b>, then you DON'T have interference.

Now, what did I mean when I said that we cannot have interference if the *relevant* degrees of freedom are entangled ? It means that we have an entangled state |a>|u> + |b>|v>. That the degree of freedom (in this case, the degree of freedom, or hilbert space, in which |a> and |b> live) of which we are going to take "sum" and "difference" for A+ and A-, occur in an entangled way.

Note that |a>|u> + |b>|v> = 1/2 [(|a>+|b>)(|u>+|v>) + (|a>-|b>)(|u>-|v>)]

Clearly, the |u> and the |v> states in the second beam contain the "which slit" information.

If we are going to perform an interference experiment on the second beam, and we are going to trigger only on (|u> + |v>) (using a U+ detection of interference on the second beam), then the selected states of the first system are |a> + |b>, and, as we know, they interfere, because we now have selected the states |a>+|b> for the first beam.

If we trigger on the |u>-|v> state (using U-), then we will trigger on the events with |a>-|b>, having the complementary interference pattern. Clearly, we cannot perform AT THE SAME TIME a measurement for, say, |u> AND a measurement for U+ or U-, which makes it impossible to have at the same time the 'which path' information (|a> or |b>, or: |u> or |v>) AND the "which interference pattern" (U+ or U-, hence A+ or A-) information.

But now you also see why, when only looking at ONE beam, you cannot get this interference pattern: the reduced density matrix of the one beam gives us an equivalence of the beam with a statistical mixture of 50% |a> and 50% |b>. A+ and A- will have 50% chance to trigger for this mixture, so no interference.

You can of course also consider this mixture as 50% (|a>+|b>) (which will have A+ trigger, and not A-), and 50% (|a> - |b>), which will have A- trigger, and not A+. So also in this view, we get 50% A+ triggering, and 50% A- triggering.

How can you have "interference" then from an entangled pair ? Well, simply by using a degree of freedom that is NOT entangled. For instance, we could have as initial state:

(|a+> + |b+>) |u> + (|a-> + |b->)|v>

We have added here an extra degree of freedom, which is the +/- (say, the spin of the particle. It is the SPIN of the particle that is now entangled with |u> and |v>, NOT the |a>+|b> quality. So if |a> is still "the left slit" and |b> is still "the right slit", then the "which path" degree of freedom (a/b) is NOT entangled. So we CAN get interference now, if that interference is NOT depending on the entangled degree of freedom (+ or -). But the entanglement is then also useless to determine "which path" information - because the SAME combination is present in both terms (and can hence be factored out).

If you don’t get interference I’ll eat my hat, or whatever stakes you care to put on it (and I do have a hat).

Bon appetit, as they say around here

 

[RandallB]

or I'm making elementary errors in using the quantum formalism,

I assume this is the case

If you have an entangled system, and you dump the partner (as you suggest), then you can only do measurements on the first beam. So all observables we're going to discuss are determined by the *reduced density matrix* of the first system.

Not my suggestion that was the point of the opening post.

Now, what is this "interference, sensitive to the entangled degrees of freedom" ?
... |a> = |through-the-left-slit> and |b> = |through-the-right-slit>
...
But the entanglement is then also useless to determine "which path" information - because the SAME combination is present in both terms (and can hence be factored out).

Bon appetit, as they say around here

What is this nonsense? No one is trying to figure out “which way” here. Are you just making this up to bail out of a gross error? Don’t be a wimp own up.

I notice you still have not detailed what it will mean to you if the experiment described provides interference patterns. On that ‘wild chance’ what will it mean to you?
Nor have you put up anything, what no got hat? – I’ll send you some shorts.

If you have contacts to a research facility or at a decent school lab this is a reasonably simple experiment to have run. Let us know when you find someone to give you real results.

 

[RandallB]

Did you actually bother looking over the delayed choice quantum eraser experiment?

I know how a delayed choice quantum eraser experiment works. What’s that got to do with any experiment that only uses one side of the PDC output?
DID YOU read this problem??
We are not sending the A side to one slit and the B side of the PDC to the other slit.
We are sending the A side to the double slit, then to the film to record the pattern - that’s it. Doesn’t get much simpler.

What are you putting up for your lunch?