# Dumb'n Down EPR-Bell Entanglement

1. Mar 30, 2006

### RandallB

LnGrrR
Since the old thread is well off topic from the OP
I’ll continue comments on Entanglement in a new thread And try to “Dumb it down" here

I assume your have trouble following the details on probabilities and why calculating probabilities for an event of grater than 100% or less than Zero is such a problem for a local realist.
So let me try to “Dumb it down” a bit at the risk of maybe going too dumb,
you tell me.

Take two flashlights designed to emit polarized light, filtered or what ever. However, when you want you can turn down the brightness until only one photon at a time is sent.
Each pointed at a detector with a polar filter lined up with its flashlight.
#1 is aligned at 00 for vertical. #2 flashlight and its filtered detector is turned to 900 for horizontal we know as long as the test filters are aligned this way all the light will be detected. If we turn the test 900 none of the light will be detected. So, we will only do one test to keep this very simple.

Turn the test filter on #1 to 450 and on # 2 to 1350. (Notice the tests are still 900 apart from each other).
Know just like when you play with polarized sunglasses you know just 50% of the light gets through for both (A paradox in it own right). Even before we get to entanglement, how do you explain this??

The only thing that explains this is the uncertainty principle. Doesn’t matter which one you use; HUP or BM uncertainty, both are unable to say which part of the original light gets though or not. Uncertainty is defined into both theories to account for this, making them both equivalently non-local IMO. How does a realist explain polarized sunglasses, they cannot for them it still a paradox.

Back to dumb’n this down, let’s look at separating that light out in detail by sending exactly one photon per second. We only one test per second, we know that if we don’t get detection, that the one photon had to be blocked by the filter.
Now 100 seconds, 100 photons – how many get though? 50 of course, for both flashlights. But if we keep track by which second 1 though 100; in exactly which positions do we predict photons will be detected? QM or BM may use different styles of uncertainty to spread the detections out randomly, but that “non-local” uncertainty is all they have to distribute those positions no predits which time 'slots' will be filled.

With both #1 and #2 showing 50 hits spread out randomly in 100 positions, how often do they land in the same numbered positions between #1 and #2 – all agree here it's a pure coin flip 25 out of 50 or half the time.

NOW comes the fun part we hook a little wire between the two flashlights, and label it “entanglement”(And from reading DrC's stuff, you should see that this is a tricky wire to hook up)

Here is the neat part about having an uncertainty principle in your theory, you get to define just how that uncertainty works – and both QM and BM say that when #1 gets a hit so will #2! That is always coordinated so they detect together, but never knowing what order the detections will occur. As long as they hit together all 50 of them, both QM & BM are satisfied.

But how does the realist explain it?
While you’re at it, explain just how those polarized sunglasses work too. If you can do the sunglasses as a realist, you should be able explain both entanglement and the double slit without an uncertainty principle and much more.

Finding how a realist can solve this with a variable defined as part of the photon, (from its beginning, separate from its twin with no interconnection) is what John Bell’s Theorem was trying to do. But experiments have shown that for a variable to do that it must produce probabilities that are nonsense (negative or more than 100%) as the price for being real and local. QM and BM have no such restriction.
Seeing how this works will come from the notes DrC has at :
http://drchinese.com/Bells_Theorem.htm

I can think of nothing better than those notes, except give yourself some time with them.

Last edited: Mar 30, 2006
2. Mar 30, 2006

### LnGrrrR

RandallB,

Ok, I've been reading it a bit, and some questions arise...

Isn't there a way to determine the validity of locality (if even a thought experiment?)

For instance, in the delayed choice quantum eraser shown here ( http://images.google.com/imgres?img...choice+quantum+eraser&svnum=10&hl=en&lr=&sa=N ) sorry for the long link! Anyways, may I propose a thought experiment which places the coincidence counter a light year away from the actual experiment being done.

Now, the actual experiment takes place on Jan 1st, 2005. The scientist sets it up, and doesn't bother to determine 'which way' the particle goes, so he sees an interference pattern. Meanwhile, the man at the coincidence counter patiently waits a light year, and starts to tally them up on Jan 1st, 2006. As soon as he starts 'tallying' them up, will the scientist on the other side see his interference screen start to take on the 'particle clumping' type qualities IMMEDIATELY (ie. faster than the light year it should take) instead of an interference pattern?

Would that rule out the... theory? of locality entirely? Or would the probabilies of 100+ and negative mean that wouldn't 'make sense' either?

Is it that non-locality isn't a problem, but FTL communication is?

I'm sure there's something I'm not quite getting about all this, and yet I'm not sure where it is.

Lastly, thanks for the help. I've read up lots on it, but my brain seems to want to work on this puzzle without looking for the edge pieces first. If I can lay out the frame, I'm relatively sure I can fill in the rest.

3. Mar 31, 2006

### vanesch

Staff Emeritus
There are a gazillion threads on that topic. For instance, here:

The point is that a LOCAL observation does not depend on what you do with the "other" particle, so *locally* you won't see anything happening.

In https://www.physicsforums.com/showpost.php?p=849418&postcount=12" post, I give (see attachment) a formal proof of this property of quantum theory.

So, no matter what you do with the other particle, how you measure it, and whatever you decide to do with a coincidence counter or not, if you only look LOCALLY on one side of it, it won't make a difference.
So no interference pattern will appear or disappear.

Last edited by a moderator: Apr 22, 2017
4. Mar 31, 2006

### LnGrrrR

Vanesch,

Ok, this is the part I'm having trouble with. You say a local observation does not depend on what we do with the other particle, correct? How so?

Let me try to take an easier tact....a light years long double slit experiment.

Without a measurement device, an interference screen is shown, right? WITH a measurement device, the particle 'bar' effect is shown.

So what would happen if we had, say, a light year's difference between the slit and the photoreceptive plate? We start up the experiment and let it run for a year, so the scientist on the other end can detail whether he sees an interference on his screen or the bar pattern. Then the scientist on the side of the photon gun places a measurement device on the slits. Will the scientist on the end of the experiment suddenly see his photoreceptor plate go to a particle pattern instead of a wave pattern (non-locality)? Or will the results take a year to get there? (locality)

5. Mar 31, 2006

### vanesch

Staff Emeritus
No, it doesn't depend upon the existence or not of a measurement device. Look at the threads I gave as a reference, because it is discussed there (and elsewhere too).

The situation is this:
Situation 1:
Alice gets her photons, and does, say, a 2-slit interference experiment on *her* photons. She observes an interference pattern. This means, that the STATE of the other photon DOES NOT CONTAIN ANY INDICATION of which slit Alice's photons went through. You can measure what you want on the other photons, you'll NEVER find out which slit Alice's photons went true. For instance, this could be because the quantum states of the photons at Bob's are all identical.
This is what you have, for instance, when the two photons are NOT entangled, but are in a product state.

Situation 2:
Alice does NOT get an interference pattern for her photons. This can happen because she gets a MIXTURE of 2 different photonstates presented, and this washes out the interference pattern.

THIS is the interesting situation, and splits into two subcases:
2A) Bob could do a measurement and tell us, from his outcome, through which slit Alice's photon went (for instance, because of the polarisation direction, or whatever).

2B) Bob could do an interference experiment on HIS side, to take apart which one of the 2 different photon states Alice got. And it is in THIS CASE that something funny happens. If he SENDS HIS LIST TO ALICE, so that she can know which of her photons belong to which of the 2 different states, well then she can CORRELATE her outcomes with this result, and for instance, only consider the outcomes of, say, the first kind. And, MIRACLE, in this SUBSET, an interference pattern occurs. While in the complementary subset, the complementary interference pattern occurs (which explains why, in the mixture, she got the sum of both and hence no interference pattern).

And now, the trick is, that the measurement Bob has to perform for 2A and 2B (finding out which slit, or finding out which of the states) are INCOMPATIBLE measurements ; meaning: if he does one, he has irreversibly made the second one impossible.

And THIS is the essential point: the fact of MEASURING 2A, and hence finding out which slit for Alice, MAKES IT IMPOSSIBLE FOR BOB TO DO 2B. And as such, to find the list against which Alice could subselect her sample to make an interference pattern appear.

cheers,
Patrick.

6. Mar 31, 2006

### LnGrrrR

Vanesch,

(Note: This example doesn't have alot to do with entanglement....I'm dropping that out of the question for now, so I can determine this bit which seems a bit easier. I'm hoping THIS example (once I receive the answer) will help me to understand the Bell experiment as well.)

First off, let me say thanks for the help. I know I might seem a bit of a dullard, (and maybe I am) but I am doing my best to figure this out, honest! *I have read many different links, but it seems no matter how many different things I read, I still have trouble putting it together. I know I need more study in probabilities and GR and SR, and have been studying these as well.*

When you say Alice and Bob, I assume you mean that Alice is setting up the experiment, and Bob is a light year away where the 'photoreceptor plate' is. When I say a 'year' later in this thought process, I mean a 'light year' away, not our normal years.

I'm going to take this slow, that way you can point out the exact point in my thinking where I'm going wrong (and hopefully I'll be able to understand!)

I have read that, if you take a single wavelength beam of light and shine it through the two slits, you will receive the interference pattern. Correct, yes?

Even when this occurs one photon at a time (which is regarded as a photon 'interfering' with itself, since there is only one photon shooting at a time and yet the interference pattern still occurs)?

Now, some sort of way to measure which slit each photon goes through is applied.

A particle pattern should appear, correct? Instead of the previous interference pattern witnessed?

Given this, again I bring up the theoretical experiment. Alice, at End A, sets up the equipment a light year away from Bob, who is at the photoreceptor plate. We will also assume for the sake of this experiment that email is now carried on massless (but information containing) particles.

Alice starts up the two slit experiment on Jan 1st, 2050 and starts up the single wavelength photon shooter. Now, according to the previous experiment, without a way to determine 'which-path' information, it should produce an interference pattern on the plate (no matter how far theoretically). Even if said plate is a light year away.

On Jan 3rd, 2050, she sets up a measuring device to determine which-path information on the photons emitted from her photon-gun. She then correlates this information in a day *determining where each photon goes*.

On Jan 5th, 2050, she sends the information to Bob regarding which-path information.

Bob is sitting on his side, and waiting patiently. A light year after Jan 1st, 2050, he sees the photons hitting his plate. What does he see?

Now it is a light year after Jan 3rd, and the photons that went through the measurement device are just getting to him. What does he see?

Two more days past, and Bob receives the 'fast as light' email from Alice sharing the 'which-path' information. What does he see?

I hope I've made this clear enough so that someone can point out the exact point in my thinking that I'm not getting clear.

Thanks again!

7. Mar 31, 2006

### Staff: Mentor

Yes, with the stipulation that the inteference pattern builds up one dot at a time, as photons arrive randomly on the detector screen. The pattern gradually becomes apparent as more dots accumulate. Imagine Georges Seurat painting one of his masterpieces, one dot at a time.

In both cases, you get a "particle pattern," that is, an accumulation of dots on the detector screen, randomly appearing at various locations. The shape of the final pattern differs in the two cases. In the first case, you get the two-slit interference pattern.

In the second case, you get two single-slit interference (diffraction) patterns superimposed on each other. The photons that went through one slit form one single-slit pattern, and the photons that went through the other slit form another single-slit pattern. If the two slits are the same width, and close together, the two patterns have the same shape, but are offset very slightly from each other. On the detector screen, the offset is usually much smaller than the width of the maxima in each pattern, so the two patterns practically coincide and we see what amounts to one single-slit pattern.

Note that in the linked diagrams, the width and spacing of the slits is tremendously exaggerated in comparison with the interference patterns. Also, the width of the central maximum of the single-slit pattern is normally rather larger than the spacing of the maxima in the double-slit pattern. The ratio of widths depends on the ratio of the slit width to the slit spacing.

Last edited: Mar 31, 2006
8. Mar 31, 2006

### LnGrrrR

JTBell,

Yes, I see what you're saying. But interference 'disappears' when it can be determined which path/slit the photon took to get to the photoreceptor plate, correct?

9. Mar 31, 2006

### Staff: Mentor

The "two-slit" aspect of the interference disappears, yes.

10. Mar 31, 2006

### LnGrrrR

JTbell,

So, in theory, if the photoreceptor plate was placed a light year away...when exactly would the interference pattern disappear?

11. Mar 31, 2006

### RandallB

Distance has nothing to do with the opportunity to test a particle. Each particle can only be "tested" once even if it just detected as passing by. That one test must be where the pattern is being formed, if in any way that test for the pattern could be a second test on the particle, that particle cannot be part of building a pattern.

Including if the “first opportunity” to be tested was at the slit the particle did not even go though. By that I mean only checking for particles going by at one slit does not mean half of them we know would be going though the other untested slit can still form a pattern, they do not. Notice the point is that they “do not”. We do not know why they “cannot” except to follow the rules as best we can define in uncertainty.

Even if we don’t understand uncertainty, no one truly does. You should be able to see that as well in entanglement.

12. Mar 31, 2006

### vanesch

Staff Emeritus
This is where things go wrong. You cannot require to have photons in "identical enough" states so that they produce an interference pattern and have somewhere a quantum state, of another (entangled) particle which indicates which slit the original photon took.

*IF* such a possibility exists, then the photons arriving at the two slits are "different enough" to blur out the interference pattern.

So the very fact of setting up a 2-slit experiment which shows interference is enough to be SURE that nowhere else in the universe exists a system with a quantum state that can tell you through which of the slits each of the photon went.

It was the essence of the argument in the thread:

I could explain it using the quantum formalism (I tried to do that in that thread) but I don't know how much that will be helpful.

The case you describe is in fact my "situation 1". It is "uninteresting" on the level of the quantum erasure thing, because you HAVE THE GUARANTEE that nowhere in the universe, the POTENTIAL INFORMATION of which-slit information is available, through entanglement or whatever.

So this cannot happen if Bob was going to see an interference pattern in the overall dataset...

cheers,
Patrick.

13. Mar 31, 2006

### LnGrrrR

Vanesch,

Ok...wait. When we do the two-slit experiment in a 'normal' fashion, we see an interference screen if we do not receive any 'which-path' information, correct?

And as soon as there is some way to determine the 'which-path' information, then those photons 'collapse', correct?

But I'm not asking whether or not there can be a simultaneous interference pattern AND knowledge of the slits. I want to know when the waveform will 'collapse'...when Bob looks at it on the 1st, the 3rd, or the 5th...or some other time I'm not aware of.

From what you seem to be saying, as soon as Alice has some way of determining 'which-path' information, than Bob will see the 'interference' on his side 'disappear', and instead show a pattern as if the the photon were behaving like particles. Is this correct?

Edit: Additionally, if I'm hearing you right....as soon as Alice measures which slit each photon goes through, Bob's side must IMMEDIATELY collapse...and not have to worry about locality or light speed or anything like this, because there can not be any point in the universe when there is measurement simultaneous with interference...right?

Last edited: Mar 31, 2006
14. Apr 1, 2006

### vanesch

Staff Emeritus
Well, that depends of what light you send on the two slits. Imagine you send the light of two different lightbulbs, the light of lightbulb1 on slit1, and the light of lightbulb2 on slit2. You won't have an interference pattern, right ?

It is only when the incident mode of light (let's talk classically) "covers" both slits that we may hope to have "interference". Well, the point is, when this is so, there's no experiment you could do on whatever other system (say, another photon with which the photons in this beam are entangled) that could find out "which slit" information.

Don't think of collapse - it is an interpretation-dependent concept.

Well, I'm an adept of the MWI view where collapse never happens - one of the reasons I prefer this view is exactly that the above considerations are misleading. But you're not obliged to follow this view - then you'll have to resolve for yourself what collapse means and when it happens (good luck :-)

If you insist on it: the safest way to apply collapse is when a measurement is "printed out on paper somewhere" ; that collapse then applies immediately and in a non-local way. But you cannot locally "see something collapse" ; you can only have probabilities of measurement outcomes.

15. Apr 1, 2006

### RandallB

That’s not true, two independent & separate light sources can interfere with each other. Just like making waves in a pool from opposite ends can create a pattern on the sides. IF there are synchronized! I.e. The same Hz. And how the pattern presents will also depend of phase difference between them. The point is your dealing with too many photons to address quantum issues.
For quantum level issues we need to deal with one, just one, photon at a time in the double slit problem.

Entanglement, is unique in that you get to work with two photons (near twins to each other when created).

LnGrrrR
I’m not sure but you seem to be thinking of two photons, (entanglement), while working on a single photon (double slit) problem.
That will give you a headache – be sure your clear what your thinking though with on a problem one (2-slit) or two (entanglement) photons.
Granted there are ways to combine the two issues (when they do I consider those as fundamentally entanglement problems).

But if you’re not clear on each type problem first, you will need some aspirin for that headache.

My advice – focus on understanding the two photon entanglement issue.

16. Apr 1, 2006

### vanesch

Staff Emeritus
Two *independent* light sources are not "synchronized" of course, and IF they are "synchronized" then the "photon" is not originating from ONE source. I gave the example of two lightbulbs in order NOT to have to address this issue, but now that you bring it up, here goes the explanation:
the "phase" of a classical EM wave is quantum-mechanically coded into a multiple-photon superposition, which is called a coherent state, and which is a superposition of vacuum, 1-photon, 2-photon, 3-photon etc... states Two *synchronized* classical beams are quantum-mechanically hence described as the tensor product of the individual coherent states, which means that the one-photon state is the quantum-mechanical superposition of the two corresponding one-photon states of the beams ; in other words, there's not "a photon coming from one source" OR "a photon coming from the other source" but a photon which is a superposition of the two sources:

Coherent state of beam 1 (mode and amplitude/phase symbolized by v):
$$|v> = c_0 \sum_{n=0}^\infty \frac{v^n}{\sqrt{n!}} |v_n>$$

c_0 is a normalization factor, and we sum over 0, 1, 2, ... n,... photon states in the EM mode corresponding to beam 1, |v_n>

Νοte that the v_0 state is the vacuum state, the same for everybody.

Coherent state of beam 2 (symbolized by w):
$$|w> = d_0 \sum_{m=0}^\infty \frac{w^m}{\sqrt{m!}} |w_m>$$

The case of the two beams which are "locked in"

$$|v>|w> = c_0 d_0 \sum_{n=0}^\infty \sum_{m=0}^\infty (stuff) |v_n> |w_m>$$

In the last expression, there is a term |v_0>|w_1> and a term |v_1>|w_0> ; noting that |vacuum> = |v_0> = |w_0> (and the fact that photons are bosons), this reduces to a term |v_1> + |w_1>

So the 1-photon state in this (also coherent!) state corresponds to |v_1> + |w_1> ; in other words, this photon comes FROM THE TWO SOURCES.

But this explanation will only bring confusion to the OP, I'm affraid.

17. Apr 1, 2006

### LnGrrrR

Damn Quantum Mechanics to hell and back! YEarrghhh! :)

Ok Vanesch, I think you might have hit on one of the things that's really hanging me up. You said we can't see the waveform 'collapse' and that it is merely an interpretation, correct?

I was under the impression that you can 'see' the collapse on the photoreceptor plate, on whether or not there is an interference pattern or a regular 'particle' pattern (as if light was only being shot through one slit.)

In the example I gave above, you said, "When it's printed out on paper" roughly. Let me change the experiment around slightly. Alice is merely setting up the equipment, but not actually 'taking anything down on paper' so to speak, and merely forwarding the results to Bob with ls communication.

Once Alice sets up the equipment, that waveform won't 'collapse' until Bob receives the data from the measuring equipment and determines the probabilities...correct?

Which would explain why superluminal communication is impossible...

18. Apr 1, 2006

Staff Emeritus
I don't think vanesch or anyone else would claim we have ever "seen" the wave function do anything.

There is no empirical warrant for regarding the whole quantum apparatus, Hilbert spaces, Banach algebras and all, as anything more than a purely conceptual machine, with dials on which we can (conceptually) set up a description of our experiment, and windows in which we can read out the set of possible outcomes with their probabilities.

19. Apr 1, 2006

### LnGrrrR

But can't we 'see' the results of whether a waveform has collapsed or not? (Interference or no interference?)

20. Apr 1, 2006

### RandallB

How can A "photon" originate from more than “ONE source”?
You stopped using “A photon” as soon as you used two light bulbs.
And two light sources ARE "synchronized" if they are running at exactly the same frequency, how could they not. The only issue is “phase” or where do their waves meet to establish a line of interference. That interference line will remain stationary as long as the phase is constant.
So, what’s with all the hieroglyphics on phase, just to showing off how math can be confusing?
There is nothing hard about visualizing phase. In the waves of a pool for example, as you move the source of one wave generator forward or back (same Hz), the phase of how and where their waves meet to draw that stationary interference line moves as well. No need to make it complex and introduce confusion.
The point is you have to deal with one photon at a time, not two –

And LnGrrrR take careful note: you must deal with one photon at a time (in two slit), each one completing its path (with wave collapse or guide wave or whatever) each on its own. Then you must measure literally hundreds of them to decide what kind of pattern you get. Where do you have any chance to “SEE” a “wave collapse” or whatever?
You don’t -- you only have your speculations on how the individual photons maneuvered as they do. No individual landing is unexpected, the sum creating an interference pattern is.

These things are called Paradoxes for a reason.

Different speculations – good ones – generate different theories like QM and BM that “solve” or at least resolve these paradoxes “within whatever boundaries of the theory”.

For Einstein and “Local Realist”, like myself, these are unsolved paradoxes – we don’t get to use “non-local” solutions like uncertainty that somehow coordinates. We expect to be able to visualize a “real” solution.
I still recommend DrC’s notes.
Read them more than a couple times, until you “get” the difference between “local” & “non-local”.