Explaining DCQE - via coherence in layman terms

In summary: When the polarizer is moved to one of those angles, only the photons that are polarized in that particular direction will be detected, thus "filling-in" the blanks between the fringes.
  • #1
San K
911
1
Does the below sound ok?

1. Interference is caused between coherent waves

2. When we try to find which-way we break the coherence
a) When we try to "partially" find which way, we partially break the coherence and hence a "partial" interference pattern

3. DCQE is explained by the fact that the sub-sample is taken either of coherent or in-coherent bunch of photons (via pairing in the co-incidence counter) depending upon what we did to p-photon ...and thus matches with (what we did to) p-photon...for the sub-sample of s-photon as well.
 
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  • #2
I thought the interference patters was simply because of each photon interfering with itself through the double slit. It looked to me like the polarizers at the slits caused a change in the interference patterns on the detector, with each the different polarization interference patterns adding together to look like there was no pattern. Adding another polarizer in the other beam simply took one set of photons out of the results, resulting in the return of the interference pattern, which had always been there but the other set of photons polarized differently added to the results making it look like the interference was gone.
 
  • #3
Drakkith said:
I thought the interference patters was simply because of each photon interfering with itself through the double slit. It looked to me like the polarizers at the slits caused a change in the interference patterns on the detector, with each the different polarization interference patterns adding together to look like there was no pattern. Adding another polarizer in the other beam simply took one set of photons out of the results, resulting in the return of the interference pattern, which had always been there but the other set of photons polarized differently added to the results making it look like the interference was gone.

ya...but how do we explain going from an interference pattern to a non-interference pattern?...because an interference pattern would have blank spaces between the fringes...how do they get "filled-in" by photon ...when we are getting a sub-set (sub-sample) of the interference pattern...

i mean how can a sub-sample of an interference pattern be a non-interference pattern?...how do the blank spaces (between the fringes) get filled-in...


what am i missing here?

i.e. in the DCQE we do no-which-way for the s-photons and (a few nano seconds) later do which-way for p-photons...
 
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  • #4
When you add the polarizer in the P beam, ONLY the photons that pass through will be recorded. All of these photons will be polarized in the same way. Remember that the coincidence counter only counts when the input is from BOTH detectors. All those photons getting to the S detector but having their entangled P partners blocked by the polarizer at P will NOT be counted. So you have removed whatever pattern would be caused by them by only counting events where both S and P are detected.
 
  • #5
San K said:
ya...but how do we explain going from an interference pattern to a non-interference pattern?...because an interference pattern would have blank spaces between the fringes...how do they get "filled-in" by photon ...when we are getting a sub-set (sub-sample) of the interference pattern...

i mean how can a sub-sample of an interference pattern be a non-interference pattern?...how do the blank spaces (between the fringes) get filled-in...


what am i missing here?

i.e. in the DCQE we do no-which-way for the s-photons and (a few nano seconds) later do which-way for p-photons...

No, that is backwards ... at least it is for the Walborn version of the DCQE, which is what I think you are referring to. In that case, the s-photon path is associated with the which-way measurement due to the QWP's that induce right-hand circular polarization on photons passing through the right hand slit and left-hand circular polarization on photons passing through the left hand slit. Thus the two sets of photons are distinguishable, and there is no interference in the single-photon measurements on Ds. If there is no polarizer in the p-photon path, then the photons arriving at the p-detector are also distinguishable, so there is ALSO no interference pattern observed in the coincident two-photon measurements. It is only when the polarizer is in place, and set to the angle of one of the QWP's (i.e. 45 or 135), that the interference fringes are observed in the coincident two-photon measurements.

I *think* that this is only the choice where the interference is strongest, and that progressively weaker fringes would be observed as the detector is rotated off those values, disappearing entirely for angles of 0 or 90. (My justification for that statement is equation [14] from the http://arxiv.org/abs/quant-ph/0106078" .)

As to how the "blanks" in the interference pattern are filled in .. Drakkith had it right when he said that the two-photon count in the absence of the polarizer is a superposition of the interference patterns that are observed at polarizer angles of 45 and 135 degrees. Those patterns are perfectly out of phase, and so the fringes cancel when they are superposed.
 
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  • #6
SpectraCat said:
No, that is backwards ... at least it is for the Walborn version of the DCQE, which is what I think you are referring to. In that case, the s-photon path is associated with the which-way measurement due to the QWP's that induce right-hand circular polarization on photons passing through the right hand slit and left-hand circular polarization on photons passing through the left hand slit. Thus the two sets of photons are distinguishable, and there is no interference in the single-photon measurements on Ds. If there is no polarizer in the p-photon path, then the photons arriving at the p-detector are also distinguishable, so there is ALSO no interference pattern observed in the coincident two-photon measurements. It is only when the polarizer is in place, and set to the angle of one of the QWP's (i.e. 45 or 135), that the interference fringes are observed in the coincident two-photon measurements.

I *think* that this is only the choice where the interference is strongest, and that progressively weaker fringes would be observed as the detector is rotated off those values, disappearing entirely for angles of 0 or 90. (My justification for that statement is equation [14] from the http://arxiv.org/abs/quant-ph/0106078" .)

As to how the "blanks" in the interference pattern are filled in .. Drakkith had it right when he said that the two-photon count in the absence of the polarizer is a superposition of the interference patterns that are observed at polarizer angles of 45 and 135 degrees. Those patterns are perfectly out of phase, and so the fringes cancel when they are superposed.


Great answers Drakkith and SpectraCat. I think you are right but I don't fully grasp it yet.

SpectraCat, Yes i am talking about the Walborn experiment and I get it.

However I still not get it if we do the Walborn experiment "backwards". I am missing something about the photons and the patterns.

I am still trying to understand both your post (to resolve the confusion, in my mind, I understand your posts one-way) in the meantime I will post what if confusing me.

There is only one fringe. So I don't understand where the superposition would come from.

The one fringe is caused by s-photons going through no-which-way. i.e. a fringe caused by a simple bare double slit.

Now when we do which-way on the p-photon there would be a scatter of p-photons around a band.

Now when we do the co-incidence counting...there would be no pairing because

all the s-photons went no-which-way and all the p-photons went which-way?

So if there is no pairing, would the sheet look blank?
 
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  • #7
Drakkith said:
When you add the polarizer in the P beam, ONLY the photons that pass through will be recorded. All of these photons will be polarized in the same way. Remember that the coincidence counter only counts when the input is from BOTH detectors. All those photons getting to the S detector but having their entangled P partners blocked by the polarizer at P will NOT be counted. So you have removed whatever pattern would be caused by them by only counting events where both S and P are detected.


Maybe I am confusing polarization with which-way (or no-which-way) by considering them to be corelated/synonymous.
 
  • #8
SpectraCat said:
No, that is backwards ...

As to how the "blanks" in the interference pattern are filled in .. Drakkith had it right when he said that the two-photon count in the absence of the polarizer is a superposition of the interference patterns that are observed at polarizer angles of 45 and 135 degrees. Those patterns are perfectly out of phase, and so the fringes cancel when they are superposed.

In the "backwards" experiment (of Walborn's), there would not be absence of a polarizer but a presence of a polarizer.

Let's start with a new simple experiment below:

we send s-photons through a double slit...and get an interference pattern

now we send p-photons through a polarizer to find out which-way (i.e. mark the photon)

then we do co-incidence count-

what pattern would s-photons have on the co-incidence counts?
what pattern would p-photons have on the co-incidence counts?
 
  • #9
You realize you have to have both P and S photons being detected at the same time for coincidence counts right?
 
  • #10
Drakkith said:
You realize you have to have both P and S photons being detected at the same time for coincidence counts right?

yes...(..not same time though...but...correlated to same emission time, i guess that is what you mean to say..)...but see my reply to zonde's post where...


you have both interference pattern and which-way...its the same issue...and ...zonde has expressed what I am talking about….

its the last post on this link -->

https://www.physicsforums.com/showthread.php?t=503667&page=6
 
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  • #11
San K said:
we send s-photons through a double slit...and get an interference pattern

now we send p-photons through a polarizer to find out which-way (i.e. mark the photon)

then we do co-incidence count-

what pattern would s-photons have on the co-incidence counts?
what pattern would p-photons have on the co-incidence counts?

I'm very much a layman but this is the way I see it (apologies if this is wrong, but I'm sure someone will correct me): -
You can't get which way info simply by looking at the p-photon. You can only get which path information by comparing the polarization of the p-photon with the polarization of the s-photon.
e.g.
if p-photon is vertical and s-photon is rotated left then s-photon went through slit 1
if p-photon is vertical and s-photon is rotated right then s-photon went through slit 2
if p-photon is horizontal and s-photon is rotated right then s-photon went through slit 1
if p-photon is horizontal and s-photon is rotated left then s-photon went through slit 2

So, without QWPs in place I would expect there to be an interference pattern, as we don't have which way information.
 
  • #12
Joncon said:
You can't get which way info simply by looking at the p-photon. You can only get which path information by comparing the polarization of the p-photon with the polarization of the s-photon.
e.g.
if p-photon is vertical and s-photon is rotated left then s-photon went through slit 1
if p-photon is vertical and s-photon is rotated right then s-photon went through slit 2
if p-photon is horizontal and s-photon is rotated right then s-photon went through slit 1
if p-photon is horizontal and s-photon is rotated left then s-photon went through slit 2

So, without QWPs in place I would expect there to be an interference pattern, as we don't have which way information.


If the above is correct then you have answered my question. ..:-) thanks Joncon, that's what I was missing

and

this ties in with what zonde said...in the link below

https://www.physicsforums.com/showthread.php?t=503667&page=6


Does this also say/mean that (in a limited sense) for the DCQE:

a) you can go from which-way to no-which way but you cannot go from

b) no-which-way to which-way...

i.e. a) you can get an interferene pattern from a no-interference pattern

b) but you cannot get a no-interfernce pattern from an interference pattern

because an interference pattern is sub-set of a non-interference pattern

and that is why walborn does the experiement in the sequence a) above and not b) above
 
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  • #13
SpectraCat said:
No, that is backwards ... at least it is for the Walborn version of the DCQE, which is what I think you are referring to. In that case, the s-photon path is associated with the which-way measurement due to the QWP's that induce right-hand circular polarization on photons passing through the right hand slit and left-hand circular polarization on photons passing through the left hand slit. Thus the two sets of photons are distinguishable, and there is no interference in the single-photon measurements on Ds. If there is no polarizer in the p-photon path, then the photons arriving at the p-detector are also distinguishable, so there is ALSO no interference pattern observed in the coincident two-photon measurements. It is only when the polarizer is in place, and set to the angle of one of the QWP's (i.e. 45 or 135), that the interference fringes are observed in the coincident two-photon measurements.

I *think* that this is only the choice where the interference is strongest, and that progressively weaker fringes would be observed as the detector is rotated off those values, disappearing entirely for angles of 0 or 90. (My justification for that statement is equation [14] from the http://arxiv.org/abs/quant-ph/0106078" .)

As to how the "blanks" in the interference pattern are filled in .. Drakkith had it right when he said that the two-photon count in the absence of the polarizer is a superposition of the interference patterns that are observed at polarizer angles of 45 and 135 degrees. Those patterns are perfectly out of phase, and so the fringes cancel when they are superposed.

SpectraCat: my question was about doing the Walborn "backwards".

Jocon has answered it in post number 11 & 12 above.
 
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  • #14
Walborn wanted to demonstrate delayed choice eraser as simply as possible, so he avoided the multiple paths of previous experiments by the inspired experimental setup of SPDC entangled pairs, quarter wave plates (to enable which way info) and a (distant) polarizer (to destroy which way)

The fact that everyone has over-analyzed the experiment makes it appear much more complex than it really is, it's like analysing tests of General Relativity by appealing to statistical issues in the apparatus, yeah they might be valid if the experiment hasn't been devised sufficiently cleanly.

Subsequent experiments have shown QM to be non-classical in even more stunning ways, so arguments about its implications are kinda silly
 
  • #15
unusualname said:
Walborn wanted to demonstrate delayed choice eraser as simply as possible, so he avoided the multiple paths of previous experiments by the inspired experimental setup of SPDC entangled pairs, quarter wave plates (to enable which way info) and a (distant) polarizer (to destroy which way)

The fact that everyone has over-analyzed the experiment makes it appear much more complex than it really is, it's like analysing tests of General Relativity by appealing to statistical issues in the apparatus, yeah they might be valid if the experiment hasn't been devised sufficiently cleanly.

Subsequent experiments have shown QM to be non-classical in even more stunning ways, so arguments about its implications are kinda silly

the DCQE is explainable by sub-samples, SpectraCat/Cthuga are right however their answers (to my questions) were incomplete.
 
  • #16
San K said:
the DCQE is explainable by sub-samples, SpectraCat/Cthuga are right however their answers (to my questions) were incomplete.

The sub-samples of course correlate afterwards, but so what? How are the sub-samples created? Even if you have a polarization beam splitter so that ALL photons are measured, it is still a PROBABILSITISIC law that determines which detector they go to.

How do the s-photons know which p-photons will go where if there is a delay after the s-photons are measured?

SpectraCat/Cthuga are WRONG if they think a simple subsampling argument explains this. You HAVE to invoke the nonlocal and/or non-separable nature of QM to explain it.
 
  • #17
Joncon said:
I'm very much a layman but this is the way I see it (apologies if this is wrong, but I'm sure someone will correct me): -
You can't get which way info simply by looking at the p-photon. You can only get which path information by comparing the polarization of the p-photon with the polarization of the s-photon.
e.g.
if p-photon is vertical and s-photon is rotated left then s-photon went through slit 1
if p-photon is vertical and s-photon is rotated right then s-photon went through slit 2
if p-photon is horizontal and s-photon is rotated right then s-photon went through slit 1
if p-photon is horizontal and s-photon is rotated left then s-photon went through slit 2

So, without QWPs in place I would expect there to be an interference pattern, as we don't have which way information.

Joncon, what do you think would happen if we placed QWPs (not polarizer) in the path of the p-photon (and none in the path of s-photons) after s-photon has stuck detector Ds (by going through the double slit without any QWPs)?
 
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  • #18
unusualname said:
The sub-samples of course correlate afterwards, but so what? How are the sub-samples created? Even if you have a polarization beam splitter so that ALL photons are measured, it is still a PROBABILSITISIC law that determines which detector they go to.

How do the s-photons know which p-photons will go where if there is a delay after the s-photons are measured?

SpectraCat/Cthuga are WRONG if they think a simple subsampling argument explains this. You HAVE to invoke the nonlocal and/or non-separable nature of QM to explain it.

the s-photons don't know which way p-photons will go. However the probabilities of the p-photons path are frozen at the time s-photons strikes the detector. The entanglement is broken (when s-photon strikes the detector) and the p-photon's behavior/path is "somewhat/probabilistically" determinable.

i.e. you can tell what is the probability of p-photon taking a particular path, (out of the choices available)

but this again cannot be used to transmit (volitional) information faster than the speed of light...

when we do co-incidence count, we are simply picking only those photons (from both s and p) marks/positions that match the pattern. the co-incidence counter acts like a filter.

the peaks of one interference pattern coincide with the troughs of the other...causing a no-interference pattern.

and Yoon in his paper also explains it this way...the sub-sampling way...
 
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  • #19
unusualname said:
The sub-samples of course correlate afterwards, but so what? How are the sub-samples created? Even if you have a polarization beam splitter so that ALL photons are measured, it is still a PROBABILSITISIC law that determines which detector they go to.

How do the s-photons know which p-photons will go where if there is a delay after the s-photons are measured?

SpectraCat/Cthuga are WRONG if they think a simple subsampling argument explains this. You HAVE to invoke the nonlocal and/or non-separable nature of QM to explain it.

unusualname, how does probability explain the interference patterns in this version of the experiment? http://arxiv.org/abs/quant-ph/9903047

The thing I don't understand is that the D1 and D2 detectors both show interference fringes and anti-fringes when the sub-samples are examined. What I don't get is that the idler photons encounter a beam splitter before going to either of the detectors. As I understand it, the chance of passing through this BS or reflecting off it is 50/50. So I would expect no interference patterns in these sub-samples.

To put it another way - how do idler photons, of signal photons which contribute to an interference pattern, always end up at the same detector?
 
  • #20
San K said:
Joncon, what do you think would happen if we placed QWPs (not polarizer) in the path of the p-photon (and none in the path of s-photons) after s-photon has stuck detector Ds (by going through the double slit without any QWPs)?

I don't see how this can change anything. You still have no information about the s-photon.
 
  • #21
San K said:
the s-photons don't know which way p-photons will go. however the probabilities of the p-photons path are frozen at the time s-photons strikes the detector. the entanglement is broken (when s-photon strikes the detector) and the p-photon's behavior/path is "somewhat/probabilistically" determinable.

however the probabilities of the p-photons path are frozen at the time s-photons strikes the detector

but they aren't, that's what Bell tests show us, that's why QM is so nonintuitive. In fact you could do a bell test inbetween the p-photon arm delay to prove this. But much more sophisticated tests of the non-realism of QM have been done, google GHZ states, Hardy, Zeillinger.

EDIt I actually agree with you, but in much more subtle/amazing way.
 
  • #22
Joncon said:
unusualname, how does probability explain the interference patterns in this version of the experiment? http://arxiv.org/abs/quant-ph/9903047

The thing I don't understand is that the D1 and D2 detectors both show interference fringes and anti-fringes when the sub-samples are examined. What I don't get is that the idler photons encounter a beam splitter before going to either of the detectors. As I understand it, the chance of passing through this BS or reflecting off it is 50/50. So I would expect no interference patterns in these sub-samples.

To put it another way - how do idler photons, of signal photons which contribute to an interference pattern, always end up at the same detector?

the don't end up on the same detector, idler goes to Do and the signal ones (that contribute to an int pattern) go to either D1 or D2

not sure what you are asking...
 
  • #23
Joncon said:
unusualname, how does probability explain the interference patterns in this version of the experiment? http://arxiv.org/abs/quant-ph/9903047

The thing I don't understand is that the D1 and D2 detectors both show interference fringes and anti-fringes when the sub-samples are examined. What I don't get is that the idler photons encounter a beam splitter before going to either of the detectors. As I understand it, the chance of passing through this BS or reflecting off it is 50/50. So I would expect no interference patterns in these sub-samples.

To put it another way - how do idler photons, of signal photons which contribute to an interference pattern, always end up at the same detector?

Hi Joncon,

I don't want to analyse every type of experiment, but San K accidentially gave the correct answer above, if the experiment is static, then the probabilities are fixed once one side of the entangled pair is measured. Of course this requires non-locality/non-separablity if one arm of the entangled pair is longer than the other.
 
  • #24
San K said:
the don't end up on the same detector, idler goes to Do and the signal ones (that contribute to an int pattern) go to either D1 or D2

not sure what you are asking...

No, the signal photons which make up the pattern go to D0. The idlers, which are used to determine the path, go to D1 or D2.

From the document: -
"The signal photon (photon 1, either from A or B) passes a lens LS to meet detector D0"

"The idler photon (photon 2) is sent to an interferometer with equalpath optical arms."
 
  • #25
unusualname said:
however the probabilities of the p-photons path are frozen at the time s-photons strikes the detector

but they aren't, that's what Bell tests show us, that's why QM is so nonintuitive. In fact you could do a bell test inbetween the p-photon arm delay to prove this. But much more sophisticated tests of the non-realism of QM have been done, google GHZ states, Hardy, Zeillinger.

EDIt I actually agree with you, but in much more subtle/amazing way.

Bell tests talk about stuff during entanglement, not after entanglement is broken.

the entanglement (or whatever we later discover the phenomena to be) is broken when one of the pair is measured, ...and whichever is measured first/earlier...


so bell test don’t come into the picture once entanglement is broken...

in what (amazing) way do you agree with my post unusual name?


the probabilities are frozen once entanglement is broken ...and this can be verified by the experiment itself...you can (probabilistically) predict the p-photon's (or whichever of the pair is to be detected later) path...because you know what pattern to expect...

just found an actual example...a commonly used experiment/instrument...where the probabilities are predicted/known/calculable

http://en.wikipedia.org/wiki/Mach-Zehnder_interferometer
 
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  • #26
San K said:
Bell tests talk about stuff during entanglement, not after entanglement is broken.

the entanglement (or whatever we later discover the phenomena to be) is broken when one of the pair is measure, ...whichever is measured first/earlier...

so bell test don’t come into the picture once entanglement is broken...

in what (amazing) way do you agree with my post unusual name?
No. If you had apparatus in place in the p-photon arm at the time the s-photons are measured you could show the p-photons had RANDOM polarization before they are measured (before they pass through a polarizer)

Sorry if this is confusing, but it's correct.
 
  • #27
Joncon said:
No, the signal photons which make up the pattern go to D0. The idlers, which are used to determine the path, go to D1 or D2.

From the document: -
"The signal photon (photon 1, either from A or B) passes a lens LS to meet detector D0"

"The idler photon (photon 2) is sent to an interferometer with equalpath optical arms."

ok fine...i swapped the idlers and signals by mistake...but still they are going to different detectors...not sure what you are asking...

are you asking why they are matching (results/patterns)?...well they are matching because they are entangled...
 
  • #28
unusualname said:
No. If you had apparatus in place in the p-photon arm at the time the s-photons are measured you could show the p-photons had RANDOM polarization before they are measured (before they pass through a polarizer)

Sorry if this is confusing, but it's correct.

you can call that random if you wish however:

you can predict that with X probability p-photons will be this
you can predict that with Y probability p-photons will be that
you can predict that with Z probability p-photons will be this-that
etc

this is provable by the fact that we can predict (probabilistically) what path p-photon will take one it emerges from the polarizer...(if we had information about s-photon, of course)...
 
  • #29
San K said:
you can call that random if you wish however:

you can predict that with X probablitiy p-photons will be this
you can predict that with Y probablitiy p-photons will be that
you can predict that with Z probablitiy p-photons will be this-that
etc

Yes, so, if I put the eraser on pluto how does the s-photon know which p-photon to match to?
 
  • #30
San K said:
ok fine...i swapped the idlers and signals by mistake...but still they are going to different detectors...not sure what you are asking...

are you asking why they are matching (results/patterns)?...well they are matching because they are entangled...

Fair enough, I accept entanglement. But that suggests to me that when the p-photon meets the final BS, the chances of it passing through or reflecting are not 50/50, but are influenced somehow by where the s-photon landed.
 
  • #31
Joncon said:
Fair enough, I accept entanglement. But that suggests to me that when the p-photon meets the final BS, the chances of it passing through or reflecting are not 50/50, but are influenced somehow by where the s-photon landed.

yes, when s-photon landed, the entanglement was broken since s-photon position has been fixed/locked...and the p-photon also becomes probabilistically determinable...that’s how I think...some on the forum agree, some disagree and some have another explanation, etc
 
  • #32
San K said:
yes, when s-photon landed, the entanglement was broken since s-photon position has been fixed/locked...and the p-photon also becomes probabilistically determinable...that’s how I think...some on the forum agree, some disagree and some have another explanation, etc

Well what happens if I just (by mechanical means) put the eraser in place (by microseconds) after EACH s-photon is detected, or if I remove the eraser (by microseconds) before the entangled p-photon can reach it EACH time.

Or I put the eraser so remotely that all s-photons are measured before a single p-photon can reach the eraser, and then I put the eraser in place just before they reach it?

Will the (timing offset if necessary) coincidence counts show interference?

You see, not so simple is it?
 
  • #33
unusualname said:
Well what happens if I just (by mechanical means) put the eraser in place (by microseconds) after EACH s-photon is detected, or if I remove the eraser (by microseconds) before the entangled p-photon can reach it EACH time.

Or I put the eraser so remotely that all s-photons are measured before a single p-photon can reach the eraser, and then I put the eraser in place just before they reach it?

Will the (timing offset if necessary) coincidence counts show interference?

You see, not so simple is it?

the eraser is simply a filter/sieve, so is the co-incidence counter and all of the above can be explained by this/that
 
  • #34
San K said:
the eraser is simply a filter/sieve, so is the co-incidence counter and all of the above can be explained by this/that

NO IT CAN'T!


Until photons are MEASURED/DETECTECTED they have UNKNOWN quantum properties (like polarisation). This is a well establsished experimental fact, it is highly nonintuitive and unsettling but it is the way the world is.

Sorry San K, but this experiment is kinda old school compared to what's been shown with multi-entangled states recently. Reality just ain't really real the way you think. You can allow a non-local interpretation to retain some idea of reality, which is what I thought you were suggesting.
 
  • #35
unusualname said:
NO IT CAN'T!

Sure it can. It is of course necessarily a non-local filter, as you would need to get the information about which state photon B has to go to once A is detected across instantly, but definitely sufficient to explain all results. For the first experiments on DCQE (before 2004) the results were, however, not tested for violations of Bell or GHZ inequalities and could therefore indeed have been explained completely classically in terms of a common history in the SPDC process. Or in other words: hidden variables have not been ruled out back in the early days of DCQE. These tests started in 2004. By the way I still do not get your resistance against the subsampling picture. You do realize that also the subsampling explanation requires a non-classical description and non-locality, do you?

unusualname said:
Until photons are MEASURED/DETECTECTED they have UNKNOWN quantum properties (like polarisation). This is a well establsished experimental fact, it is highly nonintuitive and unsettling but it is the way the world is.

I do not care much about interpretations, but a lot of people caring about them on these forums will tend to disagree with that statement. It might be that nature is non-realistic and there are no properties before measurement. However, at current it is not experimentally possible to distinguish between non-realistic and non-local descriptions of qm as both can lead to the same results. It is a well established experimental fact that Bell and other inequalities are violated for entangled photons. Non-realism is one possible and appealing conclusion, but not a fact that can be tested experimentally at current.
 

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