Mach-Zehnder interferometer, but with an entangled beam?

In summary, the conversation discusses the use of the Mach-Zehnder interferometer to highlight the corpuscular and undulatory aspect of light, and the possibility of using it with entangled photons. The question is raised about the presence of an obstacle in one of the paths and its effect on the interference pattern on the screens. It is suggested that with the BBO crystal in the path, both screens would show a diffuse result, regardless of whether or not someone uses "Bob's photons" to determine the path traveled. The person asking the question has not yet conducted the experiment and is seeking more information before proceeding.
  • #1
Marilyn67
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TL;DR Summary
Presentation of two identical experiments : One with "pure" photons and one with entangled photons.
Hello,

The Mach-Zehnder interferometer makes it possible to highlight the corpuscular and undulatory aspect of light, (in particular using single photons).

By using the "continuous" beam from a coherent source, one is able to visualize directly on two screens E1 and E2, an interference pattern or a diffuse pattern by means of two lenses L1 and L2, depending on whether the light borrows the two paths through the mirrors M1 and M2 or an obstacle is placed on one of the two paths.

I did this fairly simple experiment a few weeks ago as said in my previous thread, and I have some photos and a short video if you're interested :wink:

In the following diagram, light takes two paths :

MZI - 532 Nm.JPG
The source is a 20 mW green laser diode at 532 Nm, and thanks to the BE beam expander, the fine tuning of the interferometer is facilitated, and the appearance of the patterns is relatively easy to obtain.

This tricky step was much more difficult without a beam expander to properly "match" too small "spots" on the BS1 and BS2 beamsplitters.

We observe very clearly with the naked eye in E1 and E2, pretty interference fringes like these (several millimeters of inter-fringes) :

Interference.JPG


By placing an obstacle on one of the two paths, these fringes disappear and are replaced by a diffuse pattern like this :

Fog.JPG


So far, no mystery.

Now, let's modify the source as shown in the following diagram :
MZI - One beam 810 NM entangled.JPG

This time, a beam is exclusively composed of entangled photons (all members of the same subset) at 810 Nm (near infrared) and its power is about a million times weaker, or even a little less.

The downstream device has not changed, except that the two screens E1 and E2 have been replaced by two CCD screens (cameras), to make it possible to "see" patterns invisible to the naked eye (near infrared spectrum and low power detectable with peltier-cooled cameras).

This device is of course in total darkness (box), and sheltered from any parasitic emission.

The upstream device is made up of a violet "pump" laser diode of 100 mW at 405 Nm, the beam of which passes through a BBO crystal (type I), after being polarized, stripped of its sub-harmonics (blue), then a birefringent crystal like a properly tilted quartz plate, for the phase matching needed for the SPDC.

The main beam is completely stopped and absorbed after passing through the BBO crystal, and the two entangled beams at 810 Nm each move away by 3° from the axis of the main beam.

So one of the two beams is sent into my Mach-Zehnder interferometer.

I am "Alice" and I perform exactly the same experiments as before, with the previous green diode.

The second beam is sent "elsewhere" (say to "Bob"), and I don't care.

I know, and we all know, that whatever I do with my interferometer can in no way be used to communicate "FTL" information to Bob, we agree (it is'nt my question).

My question is the following (I couldn't find it anywhere explicitly) :


By carrying out the same experiments as before (with the green diode), on my side only, (Alice) will the experimental results be (theoretically) the same here ? (always on my side, Alice), that is to say :

- One of my paths is closed and I directly observe my two diffuse patterns...(?)

- The light follows my two paths and I directly observe my two interference patterns...(?)

In other words, does the presence of an obstacle on one of my two paths always give (in theory) a difference in pattern on my two screens, indicating to me that the obstacle is present or not at my place ? :oldconfused:

What answer does quantum mechanics provide to this question ? :headbang:

In advance, thank you for your informed answers.

Cordially,
Marilyn
 
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  • #2
With the BBO crystal in the path, I would expect both screens to show a diffuse result with no interference fringes. That's because somebody could use "Bob's photons" to determine which path each photon detected at Alice's screens travelled. Whether or not anyone does that makes no difference. Alice's screens will show no interference pattern regardless of whether or not she blocks one of the paths.

I am moderately confident of that answer, but certainly open to correction.

Have you done the experiment? If so, what did you see?
 
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  • #3
Hello @andrewkirk

andrewkirk said:
I would expect both screens to show a diffuse result with no interference fringes.

I am moderately confident of that answer, but certainly open to correction.

Have you done the experiment? If so, what did you see?

Thank you for your reply.

So you're not sure, and neither am I...:oops:

No, I haven't done this experiment, because I need to know more before investing in building an entangled photon source and refining the protocol I have in mind because this schematic is an option for now.

If other experienced speakers could give their informed opinion on the subject... :oldconfused:

This is an important question for the sequel.

I'm not sure that the systematic presence of fog in Alice is a necessary condition to prevent Bob from reading Alice's information :olduhh:
That's my problem :headbang:

All I've read in the literature is that Bob will never see interference, OK.

But we are not talking about Alice, who is free to do what she wants...right ?
She must be able to observe the wave behavior of her own photons, right?

Cordially,
Marilyn
 
  • #4
What are you hoping to achieve by doing the experiment? Test QM predictions?

To be honest, I've seen a lot of posts about these - and similar - experimental setups on this forum as of late. Not too sure what the craze is, though.
 
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  • #5
Hello @StevieTNZ ,

StevieTNZ said:
What are you hoping to achieve by doing the experiment? Test QM predictions?

To be honest, I've seen a lot of posts about these - and similar - experimental setups on this forum as of late. Not too sure what the craze is, though.

I have to be honest too.
I get the feeling you think I don't trust QM predictions.

It's the exact opposite !

I would have a hard time, with my modest level, to question them !

I have the greatest admiration for all the physicists who have contributed for nearly a century to developing this fundamental pillar of physics and thanks to whom I have very modest knowledge in this field today.

I am trying to deepen them, because I do not have the knowledge of many very cultured speakers on this forum.

I dedicate a lot of my person to deepening this knowledge, both theoretically and practically.
(we learn a lot in practice).

I'm just interested in QM predictions because I don't know all of them and that's exactly why I asked this question.

Is it a bad thing ?
I thought she was more interesting than knitting.

I am French and if I come on www.Physicsforums.com it's because in France, the level on the forums is extremely low, I am sorry to have to admit it to you.

I also come to exchange knowledge :
The little that I could learn and share with you, and all those that you bring me and for which I am grateful to you.

Excuse me, but I'm really disappointed by your answer which discredits my remarks, and which is not fair.

I too have seen a lot of discussion threads devoted to pure speculation and I have no interest in them.
I saw one again very recently.
I will not quote the link...:bow:
These discussions don't interest me..:sleep:

Getting back on topic, I'm sure the QM predictions are very clear as far as my question is concerned.

I just don't have the answer and apparently you don't either.

@andrewkirk gave me a hint and had the honesty to admit that he'd also like confirmation of what he thinks.
Nobody has the infused science, and certainly not me.

I hope I'm not mistaken in thinking that prominent members of this forum have the answer and can confirm (or deny) @andrewkirk's opinion.

I continue to think that this question is interesting and that many participants are interested in the answer.

Cordially,
Marilyn
 
  • #6
andrewkirk said:
With the BBO crystal in the path, I would expect both screens to show a diffuse result with no interference fringes. That's because somebody could use "Bob's photons" to determine which path each photon detected at Alice's screens travelled. Whether or not anyone does that makes no difference. Alice's screens will show no interference pattern regardless of whether or not she blocks one of the paths.
Yes, this is correct. The OP configuration is actually fairly close to the scenario in problem 9.6 of Ballentine, and the answer is the same.
 
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  • #7
Marilyn67 said:
The light follows my two paths and I directly observe my two interference patterns...(?)
That's what I'd expect, looking at your diagrams. If you look at only one of the entangled photons, it behaves like an unentangled photon would. Normal Mach-Zehnder effect.
But apparently the issue is not decided yet. Good luck!
 
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  • #8
WernerQH said:
If you look at only one of the entangled photons, it behaves like an unentangled photon would.
No, it doesn't. See posts #2 and #6.
 
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  • #9
Not too sure what you mean by deepening quantum mechanical predictions.

All I asked was what you hoped to achieve by doing the experiment and remarks about posts similar to yours, but appear to have received a long winded response that makes me even more suspicious about your intents (whether right or wrong). Had the reply been shorter, maybe I wouldn't lean towards that viewpoint.
 
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  • #10
Thank you @PeterDonis for confirming this :

PeterDonis said:
Yes, this is correct. The OP configuration is actually fairly close to the scenario in problem 9.6 of Ballentine, and the answer is the same.

@andrewkirk had therefore found the correct answer 👍
There are many popular interpretations, but I had never heard of the "ensemble interpretation".

Thank you both again :wink:

Cordially,
Marilyn
 
  • #11
Hello @WernerQH,

WernerQH said:
That's what I'd expect, looking at your diagrams. If you look at only one of the entangled photons, it behaves like an unentangled photon would. Normal Mach-Zehnder effect.

It was my first intuition to me also not so long ago.

It was on thinking about it, knowing that the literature specified that Bob would never see interference, no matter what Alice did, that I began to have my doubts about what Alice might see :

I said to myself : why would Alice have a privileged role compared to Bob ?

Intuition being a bad adviser in quantum mechanics, I wanted to know the opinion of specialists to be completely sure.

Cordially,
Marilyn
 
  • #12
@StevieTNZ

StevieTNZ said:
Not too sure what you mean by deepening quantum mechanical predictions.

All I asked was what you hoped to achieve by doing the experiment and remarks about posts similar to yours, but appear to have received a long winded response that makes me even more suspicious about your intents (whether right or wrong). Had the reply been shorter, maybe I wouldn't lean towards that viewpoint.

Short answer :

I suggest you re-read my "lengthy response", post #5 :wink:

I wonder what makes you react this way.
No one saw you arguing on the question asked...

Thanks for your help 😞
 
  • #13
Marilyn67 said:
It was on thinking about it, knowing that the literature specified that Bob would never see interference, no matter what Alice did, that I began to have my doubts about what Alice might see :

I said to myself : why would Alice have a privileged role compared to Bob ?
Alice does not have a privileged role. To see an interference pattern, you need of course many photons, not just one. An obstacle on one of the paths will destroy the interference pattern. You don't need quantum theory for that because classical optics is sufficient to describe a Mach-Zehnder interferometer. I'm not a specialist in quantum optics, but what I have understood from the discussions here is that entanglement can be used to "sort" photons into different groups that either show an interference pattern or don't. But this requires to combine the data of both Alice and Bob.

Marilyn67 said:
In other words, does the presence of an obstacle on one of my two paths always give (in theory) a difference in pattern on my two screens, indicating to me that the obstacle is present or not at my place ?
I think the answer is yes.
 
  • #14
Hello @WernerQH

WernerQH said:
You don't need quantum theory for that because classical optics is sufficient to describe a Mach-Zehnder interferometer. I'm not a specialist in quantum optics, but what I have understood from the discussions here is that entanglement can be used to "sort" photons into different groups that either show an interference pattern or don't. But this requires to combine the data of both Alice and Bob.I think the answer is yes.

The beam sent into the interferometer is an entangled beam, and the entanglement comes under quantum mechanics.
Therefore, classical physics can no longer deal with this problem.

Yes, it's always necessary to establish correlations a posteriori to sample the patterns which, individually, can differentiate a fog from an interference pattern.

I don't have the level of the participants who together provided a clear answer to my question, and their answer is NO.

@PeterDonis indirectly alluded to the "ensemble interpretation " championed by Leslie A. Ballentine.
I trust him.
I read some information about this interpretation that I did not know.

Detailed source from @PeterDonis (what problem 9.6 ?) would be welcome.

Cordially,
Marilyn
 
  • #15
Marilyn67 said:
The beam sent into the interferometer is an entangled beam, and the entanglement comes under quantum mechanics.
Sorry, I can't understand in what sense the beam sent to Alice is entangled. With itself? I thought it was correlated/entangled with the beam sent to Bob. And if you ignore the other beam, I can't see how the results on Alice's side could deviate from what you would derive from classical optics. I'm confused ... o_O Perhaps I should read up on what is labelled "bbo" in your diagram.
 
  • #16
Hello @WernerQH,

WernerQH said:
Sorry, I can't understand in what sense the beam sent to Alice is entangled. With itself? I thought it was correlated/entangled with the beam sent to Bob. And if you ignore the other beam, I can't see how the results on Alice's side could deviate from what you would derive from classical optics. I'm confused ... o_O Perhaps I should read up on what is labelled "bbo" in your diagram.

The beam sent to Alice is entangled with the second beam (whether it's sent to Bob or in nature, it's the same...).

From what I understand, in this situation (with both paths open), there are always "participations" in building individual interference patterns.

Contrary to the first situation, all these interference patterns are randomly phase-shifted and superimposed on each other so as to give a fog pattern in the end.

These patterns are no longer in phase and they are therefore indistinguishable from the fog pattern corresponding to a closed path.

Entanglement "clouds the cards" and the only way to order them is to reconstruct previous models to establish correlations a posteriori (coincidences point by point).

The BBO crystal (nonlinear crystal) simply makes it possible to entangle about one millionth of the number of incident photons.

Most of the incident beam passes through the crystal in the center and is not entangled.

As this beam would be detrimental to the weak flux of entangled photons (noise), it's stopped and absorbed so as not to disturb reception on CCD screens.

Cordially,
Marilyn
 
  • #17
Marilyn67 said:
Detailed source from @PeterDonis (what problem 9.6 ?)
It's problem 9.6 in the textbook by Ballentine, as I said.
 
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  • #18
WernerQH said:
I can't understand in what sense the beam sent to Alice is entangled.
The photon source is producing entangled pairs of photons--for example by parametric down conversion. One photon of each pair goes to Alice and the other goes to Bob. This is not at all problematic; plenty of existing experiments utilize the same kind of setup.
 
  • #19
PeterDonis said:
The photon source is producing entangled pairs of photons--for example by parametric down conversion. One photon of each pair goes to Alice and the other goes to Bob.
That's what I understood. But if you use only one photon of each pair, how do the results then differ from what you get with a "normal" source. In which sense are the photons of a single beam correlated?
 
  • #20
WernerQH said:
if you use only one photon of each pair, how do the results then differ from what you get with a "normal" source.
With a "normal" (i.e., non-entangled) source, Alice will see interference at her detector. With an entangled source, she won't. That is what @andrewkirk said in post #2 and I confirmed (and gave a reference for) in post #6.
 
  • #21
WernerQH said:
In which sense are the photons of a single beam correlated?
They aren't. In the entangled case, the Alice photons are correlated with the Bob photons.
 
  • #22
PeterDonis said:
The OP configuration is actually fairly close to the scenario in problem 9.6 of Ballentine, and the answer is the same.
How close is "fairly close"? There is no mention of a polarizing beam splitter in Marilyn's setup, and she talks about a simple "obstacle" in one of the paths of the interferometer, not a spin flipper (half-wave plate). It's a completely different experiment.
 
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  • #23
WernerQH said:
How close is "fairly close"?
Close enough for the same logic that is given in the solution of problem 9.6 in Ballentine to apply to the scenario given in the OP. That logic is quite general and is not limited to the specific case of entanglement and measurements described in the specific problem in Ballentine.
 
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  • #24
WernerQH said:
There is no mention of a polarizing beam splitter in Marilyn's setup
What kind of beam splitter is in a Mach-Zehnder interferometer? (This question is actually for the OP @Marilyn67 as well.)
 
  • #25
Hello @PeterDonis ,
Hello @WernerQH ,

Your exchanges are really very instructive (the questions and the answers).

PeterDonis said:
Yes, this is correct. The OP configuration is actually fairly close to the scenario in problem 9.6 of Ballentine, and the answer is the same.

Does the scenario you are talking about correspond to the figure in the following document ? (page 271 of the pdf = page 256 of the paper) :

https://faculty.washington.edu/seattle/physics441/online/Ballentine.pdf

WernerQH said:
How close is "fairly close"? There is no mention of a polarizing beam splitter in Marilyn's setup, and she talks about a simple "obstacle" in one of the paths of the interferometer, not a spin flipper (half-wave plate). It's a completely different experiment.

PeterDonis said:
What kind of beam splitter is in a Mach-Zehnder interferometer? (This question is actually for the OP @Marilyn67 as well.)

I confirm that my beamsplitters are not polarizing.

My two separating blades are standard and correspond exactly to this reference :

Edit : French or English

https://www.edmundoptics.fr/p/20-x-27mm-50r50t-plate-beamsplitter/2245/

https://www.edmundoptics.com/p/20-x-27mm-50r50t-plate-beamsplitter/2245/

You can recognize them here, on the right.
As you can see, my "montages" are very "professional" :wink:

I borrowed them from my 12 year old son :biggrin: as they are cheap and very accurate for interferometer adjustment.

It's the result that counts...!

Mirrors and beam splitters.JPG


@PeterDonis :

The question asked by @WernerQH remains interesting :

Would your answer to post #6 be different with polarizing beamsplitters ?

Would the patterns that Alice sees be fundamentally different and would the predictions of quantum mechanics be fundamentally different ?

Cordially,
Marilyn
 
Last edited:
  • #26
Marilyn67 said:
Does the scenario you are talking about correspond to the figure in the following document ? (page 271 of the pdf = page 256 of the paper)
Yes.

Marilyn67 said:
I confirm that my beamsplitters are not polarizing.
Then that explains why you don't have to have a spin flipper (half-wave plate) in one path of the interferometer, as the setup in Ballentine does: the polarizations aren't being split. In Ballentine, they are. Your scenario determines whether there is or is not interference (in the absence of entanglement) by the absence or presence of a blocker in one arm of the interferometer; Ballentine's does it by whether both ##z## polarizations are present in the input beam to the interferometer, or only one. Those details don't affect the overall logic that is relevant to your question in this thread.

Marilyn67 said:
Would your answer to post #6 be different with polarizing beamsplitters ?
No. With polarizing beam splitters, since you do not have a spin flipper (or rather the corresponding device for photons that swaps polarizations) in one arm of the interferometer, you would never measure interference even if both arms were open. So the whole setup would be pointless.
 
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  • #27
Thank you very much @PeterDonis for your really very detailed explanations (I really appreciate all these important clarifications).

I think I have understood the meaning of your comparison and to be sure of having understood everything, I will still study in depth the configuration of L. E. Ballentine.

It's very instructive !

Cordially
Marilyn
 
  • #28
@Marilyn67 Bonjour!

My apologies for having allowed your threads to slip away from me. I was away, and all too many things slipped away.

I now see that you are well beyond the pay grade of a simple spectroscopist. Thus, I will sit by paying close attention -- kouign-amann and bowl of cafe in hand. Fear not, I promise not to get crumbs on the optical table.

Bonne chance!
 
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  • #29
Hello @Hyperfine,

Thank you for your sincere response :smile:

Don't worry about my optical table, it didn't cost me much.

I used a simple very large mirror as an optical table for its remarkable qualities of surface flatness and for its reflections which allowed me to facilitate the adjustments.

I don't have much merit, anyone can do it :wink:

Sincerely
Marilyn
 
  • #30
Hello,

I'm not sure I understood the whole transposition of L.E. Ballentine's problem 9.6, to that of the Mach-Zehnder interferometer but it seems to me that the main thing to remember, (the same thing) is this :

The probability density on the screen for the whole ensemble will be : [|ψ1(x)|² + |ψ2(x)|²]/2
which has no interference.

I discovered 2 rather interesting things :

1/ It seems that a beam of entangled particles carries a message ! This message is "I am entangled".
In other words, trying to create interference by all means without success, with a beam of which we don't know the source would be the signature of an entangled beam, right ?

2/ It seems that the beam has lost all its wave properties to retain only its particle properties.
This is reminiscent of the double slit experiment where the photons behave like balls when performing measurements in S1 and S2:

Impossible to create interference with balls.
There it's the same thing.

I wonder how a physicist from the beginning of the 20th century (before the birth of quantum mechanics, of course) would have reacted if given a sealed box with such a beam coming out ?

He would have discovered a really very special light, capable of producing photosynthesis, photoelectric effects, captured by the retina, releasing energy by being absorbed, reflected, diffracted in the same way in media of index n, and yet, impossible to produce interference with this beam...!

I wonder what his conclusions would have been on that laser beam !

What might his conclusions have been in your opinion ?

Cordially,
Marilyn
 
  • #31
Marilyn67 said:
1/ It seems that a beam of entangled particles carries a message ! This message is "I am entangled".
In other words, trying to create interference by all means without success, with a beam of which we don't know the source would be the signature of an entangled beam, right ?
Although the PDC source is a laser, the entangled beam that comes from the crystal lacks one attribute that is typical of laser light: coherence. Coherence is needed for the interference to appear. You *can* make the light be coherent, but then it will lose some of its entangled properties. To the extent you get more of one attribute, you get less of the other.
 
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  • #32
Hello @DrChinese ,

Thank you for your quick response.

Something escapes me:

In 1803, Thomas Young performed his experiment of the two slits with a beam of sunlight and found interference.
A reproduction of the experiment can be seen here as well as the reading of Thomas Young's manuscripts :



I thought sunlight wasn't coherent light, like from an incandescent bulb ?

Cordially,
Marilyn
 
  • #34
Thank you @DrClaude, this discussion is indeed very interesting ! :smile:
 

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