Quantum Interpretations history

[vanesch]

I don't care if we can tell _which slit_ the photons went through. The fact that they only hit one detector or the other is what's remarkable to me.

Indeed. But that is already the case in any optical interference. Classically, you have an image. Quantum-mechanically, you have single impact upon single impact. The simple double slit experiment already does this.

 

[vanesch]

That is always the case, but what Afshar introduced - which, to my knowledge, no one else had before - was the fact that the photons _physically avoid_ the grid like a wave would, but still only hit one detector or the other, like a particle would. As I see it, that's the "duality" being exemplified in toto. True, "which-way" is gone, but it's still "one or the other" and not "both."

Or, you can see it differently: the photons don't avoid the wires ! The photon coming from slit 1 scatters on the wires, and the photons coming from slit 2 also scatter on the wires, but it happens that they scatter with opposite phases. So at the detector, these two scattering amplitudes cancel perfectly. That's actually what happens when you apply the evolution operator to:

|slit 1 > + |slit 2> ---> |image1> + |scatter > + |image2> - |scatter>

 

[peter0302]

Indeed. But that is already the case in any optical interference.

"Any optical interference" does not test single photons. Let's try this experiment with single electrons then.

The photon coming from slit 1 scatters on the wires,

Not buying it. The wires are opaque. They absorb photons as shown in the other part of the experiment where one slit is closed. They are not lenses (which would always pass ~100% of the photons but alter their trajectory). They are more akin to the double slits themselves - filters - that happen to filter NO photons when placed in the interference minima.

Unless you're suggesting that the scatter actually does reduce the image quality?

 

[vanesch]

"Any optical interference" does not test single photons. Let's try this experiment with single electrons then.


Not buying it. The wires are opaque. They absorb photons as shown in the other part of the experiment where one slit is closed. They are not lenses (which would always pass ~100% of the photons but alter their trajectory). They are more akin to the double slits themselves - filters - that happen to filter NO photons when placed in the interference minima.

Unless you're suggesting that the scatter actually does reduce the image quality?

I thought they scattered, because they are metallic wires, but in fact, it doesn't really matter what kind of channels they scatter into. "absorption" is indeed nothing else but another channel (excitation of an electron cloud or whatever). Read in the state "scatter", instead then "excited electron cloud" or something. The opposite slit will put that electron cloud then in the same state, but with a 180 degree phase shift. The story remains the same.

 

[peter0302]

No the story does not remain the same. Let's assume (you may be right about the metal wires - I'm now unsure) that the experiment uses thin strips of glass mirrors at the interference minima that actually reflect back any photons. Do we agree then that no photon that hits the wire should in any way be seen by the detector? Yet in that case the result will still be the same as in Afshar's experiment - but now you can't argue there was scattering of any photons by the grid.

Or another possibility - the metal wires, due to the photoelectric effect, would generate current when a photon strikes, right? Let's measure the current to see if there actually was any excitation of the electrons in the metal. Bet you a steak dinner there wouldn't be any.

 

[vanesch]

No the story does not remain the same. Let's assume (you may be right about the metal wires - I'm now unsure) that the experiment uses thin strips of glass mirrors at the interference minima that actually reflect back any photons. Do we agree then that no photon that hits the wire should in any way be seen by the detector? Yet in that case the result will still be the same as in Afshar's experiment - but now you can't argue there was scattering of any photons by the grid.

Or another possibility - the metal wires, due to the photoelectric effect, would generate current when a photon strikes, right? Let's measure the current to see if there actually was any excitation of the electrons in the metal. Bet you a steak dinner there wouldn't be any.

Of course there won't be any, because that current itself is the result of an interference, so in the end there won't be any!
What I mean is:

unitary evolution of slit 1: |slit 1> ---> |detector 1> + |otherstuff>

that "otherstuff" can be scattering (photons in different directions, emission of an electron, backscattering... whatever: it is another quantum state). It can be a sum of different states (some spatial scattering, some electron emission, some atom excitation, ... whatever: the different channels of interaction with the wire)

unitary evolution of slit 2: |slit 2> ---> |detector 2> - |otherstuff>

It is exactly 180 degrees shifted in phase with the first one, exactly because of the position of the wires (the "empty" interference lines are namely those where both states are 180 degrees out of phase). So this means that the quantum state of, say, the scattered electron will also be 180 degrees out of phase with the quantum state of the scattered electron from the first slit.

Hence, if you make the sum (superposition principle of time evolution):

|slit 1> + |slit 2> evolves into |detector 1> + |otherstuff> + |detector 2> - |otherstuff>

So you get destructive interference in all the "other channels": the emitted electron quantum state from slit 1 interferes with the emitted electron quantum state from slit 2, the excited atom state from slit 1 interferes with the exited atom state from slit 2 etc... and nothing remains (simply due to the 180 degree phase shift of the incoming states).

 

[peter0302]

Where is "otherstuff" coming from in the evoluion of Slit 1 if the photon that went through slit 1 doesn't interact with the grid?

You're _assuming_ what you're trying to prove - that the phtoons interact with the grid. Part of Afshar's experiment was showing that when one slit was closed, the grid blocked photons, but when both slits were open, there was no blockage. Neither you nor colorspace have provided any evidence or argument why that part of the experiment is wrong and, in fact, there is interaction between the photons and the grid.

And you also didn't address my point about using mirror strips instead of metal wires. Or put photon detectors at the interference minima instead of wires. They'll never (or rarely) go off.

 

[vanesch]

Where is "otherstuff" coming from in the evoluion of Slit 1 if the photon that went through slit 1 doesn't interact with the grid?

You're _assuming_ what you're trying to prove - that the phtoons interact with the grid. Part of Afshar's experiment was showing that when one slit was closed, the grid blocked photons, but when both slits were open, there was no blockage. Neither you nor colorspace have provided any evidence or argument why that part of the experiment is wrong and, in fact, there is interaction between the photons and the grid.

And you also didn't address my point about using mirror strips instead of metal wires. Or put photon detectors at the interference minima instead of wires. They'll never (or rarely) go off.

I think you're entirely missing my point. I'm not claiming that photons are interacting or not interacting with the wires ; I'm showing you that you can consider it both ways, because of the linearity of the unitary evolution operator.

We know that if we have a photon in state |slit 1> (slit 2 is closed), that we have a certain amplitude of arriving in detector 1, and "other stuff". That other stuff can be: "absorption" (that is: excitation of another system, such as the electron cloud in the wire), or "scattering" (blurring of the image, going elsewhere, or hitting detector 2 or ...), or "ionisation" (that is, the emission of an electron), or whatever.
But all these other channels just give rise, also, to an AMPLITUDE for this to happen. For instance, the emitted electron is a wavefunction of an outgoing electron and so on.

So we have that the "physics of what happens behind the screen" can be written as:

U( |slit 1> ) = |detector 1> + |otherstuff>

In a similar way, if we have a photon in state |slit 2> (slit 1 is closed) we have a similar story. But, and that's the important part, the wires being placed at exactly those points where the wavefunction from slit 1 is in antiphase with the wavefunction from slit 2 (that's what it means to be in the troughs of the interference pattern), we know that the interaction of the |slit2> photon with the grid will be identical but in opposite phase with the interaction of a |slit1> state.

As such, we have that:

U ( |slit 2> ) = |detector 2> - |otherstuff>

For instance, the electron wavefunction of an emitted electron by the slit-2 photon on the wire will be in exact anti-phase with the wavefunction of an emitted electron by a slit-1 electron. As such the emitted electron by the slit1 state will interfere destructively with the emitted electron of the slit 2 state, and in fine, no electron will be emitted.

So, using the superposition principle, we have:

U (|slit1> + |slit2> = |detector1> + |otherstuff> + |detector2> - |otherstuff> =
|detector1> + |detector2>

This is the view that the two components |slit1> and |slit2> each interacted with the slit, but them being in opposite phases there, gave rise to anti-phases in the resulting channels, which hence cancel. (so, in the end, 0 amplitude for an emitted electron, or an excited atom or some scattered photons)

The other view is that we FIRST look at what |slit1> + |slit2> looks like, and then decide that this state has no photon amplitude at the wires, and will hence not scatter.

These two views are equivalent, given the linearity of the time evolution operator U.

 

[peter0302]

I'm not even sure your equation is right though. This only works if the wire grid is capable of directing any photon it absorbs to one or the other detector. If we make the grid out of black rock, or mirror, this will not happen - the vast majority of photons hitting the grid will be absorbed and not re-emitted, or reflected back the directoin they came - either way, no possibility of being detected, regardless of the phase. Thus, the equation should be:

U( |slit 1> ) = |detector 1> + |blocked by grid> + |otherstuff>
U( |slit 2> ) = |detector 2> + |blocked by grid> - |otherstuff>

U( |slit 1> + |slit 2> ) = |detector 1> + |detector 2> + 2*|blocked by grid>

Yet that is not what we see.

 

[colorSpace]

Part of Afshar's experiment was showing that when one slit was closed, the grid blocked photons, but when both slits were open, there was no blockage. Neither you nor colorspace have provided any evidence or argument why that part of the experiment is wrong and, in fact, there is interaction between the photons and the grid.

Not any argument? I've said multiple times that diffraction applies to photons which *pass* the grid, not to any blocked ones. It alters the direction of photons that pass the grid next to it. This argument doesn't depend on whether there are any blocked photons or not. The way the direction is affected depends on the interference at that point, and can be very different from how a single-slit pattern is affected by the grid, since the grid is placed at a location very specific to the interference pattern. On Afshar's website, I couldn't find any indication that he has considered interference-specific diffraction.

Furthermore, Bohm writes in "The Undivided Universe", that even a "negative" measurement causes "collapse"-like effects elsewhere. Such as, in a different experiment, if a counter is *not* triggered, it will still cause a collapse-like effect. This is because the wavefunction is affected even if the particle isn't [directly, so to speak]. It would seem to me this is easier to understand in an interpretation where the wavefunction is non-local, although I wouldn't really know how a local interpretation might handle this, I've read it just in passing.

 

[peter0302]

I've said multiple times that diffraction applies to photons which *pass* the grid, not to any blocked ones. It alters the direction of photons that pass the grid next to it.

That's not my point. You're arguing that the grid's presence destroys which-way (even though there's no evidence of this). My point is that the photons avoid the space occupied by the grid.

 

[vanesch]

I'm not even sure your equation is right though. This only works if the wire grid is capable of directing any photon it absorbs to one or the other detector. If we make the grid out of black rock, or mirror, this will not happen - the vast majority of photons hitting the grid will be absorbed and not re-emitted, or reflected back the directoin they came - either way, no possibility of being detected, regardless of the phase. Thus, the equation should be:

U( |slit 1> ) = |detector 1> + |blocked by grid> + |otherstuff>
U( |slit 2> ) = |detector 2> + |blocked by grid> - |otherstuff>

U( |slit 1> + |slit 2> ) = |detector 1> + |detector 2> + 2*|blocked by grid>

Yet that is not what we see.

No, you still don't understand what I said. While we can keep your first line:
U( |slit 1> ) = |detector 1> + |blocked by grid> + |otherstuff>

we have to write the second line as follows:

U( |slit 2> ) = |detector 2> - |blocked by grid> - |otherstuff>

because whatever state is described by "blocked" (an excited atom, a backscattered photon, an emitted photo-electron), it will be in anti-phase with what came from the first slit. I took the example of the emitted photo-electron: the wavefunction of that emitted electron for slit 2 will be in anti-phase with the wavefunction of the photo-electron for slit 1. There will hence be destructive interference, this time not by the photon, but by the photo-electron. As such, there will in the end not be a state anymore with an emitted photo-electron, because the state emitted by slit 1 was canceled by the state emitted by slit 2.

Same can be said about an absorbing atom (which must hence get into an excited state). The wavefunction of the excited atom from slit 1 will be in anti-phase with the wavefunction of the excited atom from slit 2. Hence, the total wavefunction of an excited atom will vanish.

 

[colorSpace]

That's not my point. You're arguing that the grid's presence destroys which-way (even though there's no evidence of this). My point is that the photons avoid the space occupied by the grid.

Well, I'm not questioning that that they do so (mostly, since that area won't be completely 'black' even when there is interference).

 

[colorSpace]

Furthermore, Bohm writes in "The Undivided Universe", that even a "negative" measurement causes "collapse"-like effects elsewhere. Such as, in a different experiment, if a counter is *not* triggered, it will still cause a collapse-like effect. This is because the wavefunction is affected even if the particle isn't [directly, so to speak]. It would seem to me this is easier to understand in an interpretation where the wavefunction is non-local, although I wouldn't really know how a local interpretation might handle this, I've read it just in passing.

To extend my point above: the grid, even if no hit by the photons, still makes a ("negative") position measurement, so to speak reducing the uncertainty of the position. If this reasoning is correct, this will have an (increasing) effect on the uncertainty of the momentum, and due to the very specific position of the grid, it is a very specific measurement with a very specific effect.

 

[peter0302]

because whatever state is described by "blocked" (an excited atom, a backscattered photon, an emitted photo-electron), it will be in anti-phase with what came from the first slit.

The "anti-phase" of a blocked photon is still a blocked photon. Either way it's not being detected.

The reason your equation is wrong is obvious as soon as you re-write it in terms of detector1 and detector2:

U( |slit 1> ) = |detector 1> + |blocked by grid> + |otherstuff>
...
U( |slit 2> ) = |detector 2> - |blocked by grid> - |otherstuff>

|detector 2> = U( |slit 2> ) + blocked by grid> + |otherstuff>

In other words, you have detector2 detecting all the photons that went through slit2 _AND_ all the photons that were blocked by the grid that were not detected by detector1. That makes no sense - we can look at the detection count of detector1 and detector2 and see that they're equal. The only way your equation works is if |blocked by grid> == 0 (and if |otherstuff> equals zero also).

 

[vanesch]

The "anti-phase" of a blocked photon is still a blocked photon. Either way it's not being detected.

Yes, but the SUPERPOSITION of a blocked photon state and an anti-phase blocked photon state is a photon that isn't blocked, because the channel has amplitude 0.

This is btw exactly as if you would say: a photon amplitude here, and a photon anti-phase amplitude here, is still a photon amplitude here... while we have destructive interference.

Again, I'm not claiming that the grid is scattering/absorbing or not in this case. I am simply illustrating that you can view it both ways!

You can consider first the sum, and then the U, or you can consider first the U, and then the sum. Both are equivalent, given that U is linear.

In the first case, you FIRST consider "interference" of the photon states, and you find that the wires are exposed to 0 total photon amplitude. Hence they don't do much. OR, you can consider that we first apply U individually to each term in the overall photon state |slit 1> + |slit 2>, and then we see that the term from slit 1 does things (scatter, absorb...), while the second term does exactly the same thing, but in anti-phase. So the scattering, absorption and so on cancels after making the superposition of the results.

It is just two ways of looking at exactly the same process.

The reason your equation is wrong is obvious as soon as you re-write it in terms of detector1 and detector2:


|detector 2> = U( |slit 2> ) + blocked by grid> + |otherstuff>

In other words, you have detector2 detecting all the photons that went through slit2 _AND_ all the photons that were blocked by the grid that were not detected by detector1. That makes no sense - we can look at the detection count of detector1 and detector2 and see that they're equal. The only way your equation works is if |blocked by grid> == 0 (and if |otherstuff> equals zero also).

Huh ?

The elements you have there are VECTORS IN HILBERT SPACE. As such, you cannot assume that they are just scalar additional quantities. The equation above is correct, because U(|slit 2> ) contains a component in the negative direction (in hilbert space) of |blocked by grid> and |otherstuff>. As such, the total length (hilbert norm) of |detector 2> can be SHORTER than the hilbert norms of the 3 terms on the right.

The kets don't represent necessarily detectable states!

|blocked by grid> is a quantum state, and it can hence have a phase. For instance, in Fock space it can be represented by the vacuum state (zero photon state). But of course, a photon cannot "just disappear in a puff of logic" and hence a photo-electron or an excited state of an electron cloud must be created.

Overall, we can then write:

|unexcited atom in wire> |photon at slit 1> evolves into:
a | photon_at_detector2>|unexcited atom> + b |no_photon> |excited atom> + c |photon_scattered_in_space>|unexcited atom>

Now, this is the result of an interaction at the wires: the "non-interacting" component of the beam will give rise to the first term with amplitude a, while the interaction with the atoms will give rise to the second and the third terms.

If we do the calculation again for a photon at slit 2 we find:
|unexcited atom in wire> |photon at slit 2> evolves into:
a | photon_at_detector2>|unexcited atom> - b |no_photon> |excited atom> - c |photon_scattered_in_space>|unexcited atom>

The minus sign comes simply from the fact that at the point where the wire is, the incoming wave (the incoming photon) is 180 degrees out of phase with what we had for the first case. So this phase will be transmitted also to the outgoing states.

Now, if we make the sum of both, we have:

|unexcited atom in wire> |photon at slit 1> + |unexcited atom in wire> |photon at slit 2> = |unexcited atom in wire> (|photon at slit 1> + |photon at slit 2>)
= a | photon_at_detector2>|unexcited atom> + b |no_photon> |excited atom> + c |photon_scattered_in_space>|unexcited atom>
+ a | photon_at_detector2>|unexcited atom> - b |no_photon> |excited atom> - c |photon_scattered_in_space>|unexcited atom>

= a (|photon_at_detector1> + |photon_at_detector2>) |unexcited atom>

 

[gptejms]

The "anti-phase" of a blocked photon is still a blocked photon. Either way it's not being detected.

The question of a photon or no photon comes up only when the wavefunctions have interfered-in this case there is destructive interference resulting in no photo-electron(hence blocked photon).I think it's wrong to talk of a photon, before you have detected(or not detected) it.

The whole argument is retroactive-detect a photon at the detector and wonder from which slit it came.Only interactions(in this case with detector) are quantum--what happens in between is better left to wave theory/picture.

 

[peter0302]

Vanesh, if I'm understanding you right (and I have to admit you're really starting to confuse me), the gist of your argument is that this other view you're throwing out there still has the presence of the grid affecting the photons but in equal ways that happen to cancel each other out. And I keep coming back to the fact that you have no evidence whatsoever that the grid affects the photons. The detection rate is the _same_.

This second interpretation simply introduces unnecessary terms for no other reason than to preserve CI. But it's even worse - because when you look at one slit or the other in isolation, we see the grid having bizarre effects on the photons which we never see otherwise. The grid is not magical. It is a filter that is designed to _block_ or _not block_ photons. This idea that "blocking" one set of photons and "blocking 90 degrees out of phase" for another set somehow adds up to "no blocking" is a mathematical trick that, when you think about its meaning for the real world, makes no sense. That would be fine if the grid's effect on the photons was random, like a half-silvered mirror - but that's not the case. The grid is simple - "block" if you hit me, "don't block" if you don't hit me. Where do you get the justification, based on any coherent physical theory or empirical evidence, that "block some photons" + "block some other photons" adds up to "block NO photons"??

[Edit]
I re-re-read your last post because I really am trying to understand this but I'm just not convinced. Taking your equations again:

|unexcited atom in wire> |photon at slit 1> evolves into:
a | photon_at_detector2>|unexcited atom> + b |no_photon> |excited atom> + c |photon_scattered_in_space>|unexcited atom>

...

|unexcited atom in wire> |photon at slit 2> evolves into:
a | photon_at_detector1>|unexcited atom> - b |no_photon> |excited atom> - c |photon_scattered_in_space>|unexcited atom>

Am I correct in assuming that a^2+b^2+c^2 = 1 (i.e. all possible states are accounted for)? Ok, good, so let's give non-zero values to "b" and "c" (we'll assume your "other view" is right and that there is some interaction between the grid and the photons' states). That means "a" is less than 1. Then in your final equation:

|slit1>+|slit2> = a (|photon_at_detector1> + |photon_at_detector2>) |unexcited atom>

"a" is still less than 1, and therefore if there was any magnitude to the |blocked> and |scatterd> states, then |detector1> and |detector2> do not account for all possible states of the photons that went through the slits, and while we cannot know what happened to them because that information is lost, we would expect that the two detectors won't show us all the photons that went through the slits.

Yet we _know_ this is wrong because the image quality was not significantly degraded. So tell me where my math is wrong because I'm just taking your equations to the logical conclusion.

 

[vanesch]

But it's even worse - because when you look at one slit or the other in isolation, we see the grid having bizarre effects on the photons which we never see otherwise. The grid is not magical. It is a filter that is designed to _block_ or _not block_ photons. This idea that "blocking" one set of photons and "blocking 90 degrees out of phase" for another set somehow adds up to "no blocking" is a mathematical trick that, when you think about its meaning for the real world, makes no sense. That would be fine if the grid's effect on the photons was random, like a half-silvered mirror - but that's not the case. The grid is simple - "block" if you hit me, "don't block" if you don't hit me. Where do you get the justification, based on any coherent physical theory or empirical evidence, that "block some photons" + "block some other photons" adds up to "block NO photons"??

No, no: block a photon + block the same photon with 180 degrees phase shift is don't block that photon.

Now, it is funny, but there exists actually an experimental technique that is exactly based upon that.

It is a well-known technique, called EXAFS. I even build such an instrument once. You find them around synchrotrons. EXAFS is nothing else but an extremely sensitive X-ray absorption spectrum as a function of incoming photon energy.

What happens in EXAFS ? A certain atom in a sample (usually a metal atom or something) undergoes a photo-electric effect. As a function of photon energy, you see the famous K-edge: the sudden rise in absorption when the energy of the photons reach the K-shell ionisation potential. In other words, the photo-electric effect: the photons get absorbed, and an electron is emitted.
Well, if you analyse (and that's the experimental challenge in EXAFS) finely the absorption spectrum just beyond the K-edge, you will see a wiggle superimposed on the smooth falling curve after the edge. That means that for certain energies slightly higher than the K-edge, sometimes you get a bit more absorption, and sometimes a bit less absorption than would a "gas" of the atoms in question. How come ?

Well, the atoms you are aiming at are within a material, and the photo-electron that gets emitted gets emitted in a certain quantum state with an energy which is the difference between the actual photon energy and the K-edge. You can think of it as a wave emitted by the photo-electric effect. And it can partially be scattered by the surrounding atoms of different kinds, in such a way, that you can get the emitted photo-electron state BACK to your atom, but with a phase shift that depends on the distance to the reflecting atom and the energy of the electron state.
In other words, you can get constructive or destructive interference for the emitted photo-electron. And lo and behold, if you get destructive interference, you get LESS photon absorption, and if you get constructive interference you get MORE photon absorption than you would, when the atoms would be alone. This explains the wiggles on the absorption spectrum. So you have a kind of photo-electron interferrometer.
And this works very well: by analysing the wiggles (actually, essentially a Fourier analysis after a normalisation), you can get very accurate inter-atomic distances to the nearest neighbours, the next-to-nearest neighbours etc...

So this experimental technique is exactly based upon constructive and destructive interference of "absorption states of photons".

"a" is still less than 1, and therefore if there was any magnitude to the |blocked> and |scatterd> states, then |detector1> and |detector2> do not account for all possible states of the photons that went through the slits, and while we cannot know what happened to them because that information is lost, we would expect that the two detectors won't show us all the photons that went through the slits.

This is correct: with the grid in place, you will loose a very tiny bit of photons, but you won't notice it, because it is a second-order effect:

a^2 = 1 - b^2 --> a = sqrt(1 - b^2) ~ 1 - 1/2 b^2

BTW, in classical optics too: if the grid has any "coverage" it will cover the holes in the interference pattern where the intensity is quadratic (good approximation of 1-cos(x) if 1-cos(x) represents the interference pattern intensity).

Yet we _know_ this is wrong because the image quality was not significantly degraded. So tell me where my math is wrong because I'm just taking your equations to the logical conclusion.

Indeed, the image will NOT degrade, and the intensity will only suffer a second-order decrease (exactly as classical optics predicts).

 

[peter0302]

Ok now we're getting somewhere! :) We agree that the magnitudes of the two |detect> states is almost 1 when both slits are open, and is equal to 1 when the grid is not there. And I appreciate and now agree with your explanation that any interaction caused by the grid is qualitatively canceled out due to the phase difference and therefore not seen in the image, but is still quantatatively seen in the final photon count at each detector, which I apparently erroneously thought you were arguing was not the case.

But we should also agree that the |detect> states are _significantly_ smaller when one slit is closed. And so we are now left with my original point - which is - the photons tend to avoid the grid wires due to the fact that they had a choice of which slit to go through! The choice actually altered their path as though they were waves. Yet we still detect only a single photon at a time in each detector (never both at once). Therefore, wave-particle duality is shown in the same experiment. Despite your statement that classical optics predicts this, classical optics does NOT predict that only one detector or the other will go off at any given time.

So the punch line - let's try Afshar's experiment with a single photon source, and let's compare average detection rates over a long period of time, rather than fuzzy concepts like image quality. Then arguments that "this is just classical optics" go out the window, and taking into account the second order effects of the grid, we'll have proof that the grid didn't significantly affect the photon count - i.e. - the photons tended to avoid the grid when both slits were open.

Tell me how that result is not remarkable and does not support a hidden variable or Bohmian theory?

 

[colorSpace]

Tell me how that result is not remarkable and does not support a hidden variable or Bohmian theory?

The discussion has become a little hard to follow. What, at this point, is your argument that this is difficult for the Copenhagen Interpretation, and/or MWI?

 

[peter0302]

Simple. We see the wave (in the form of physical space avoided by the photons without being detected) and particle (in the form of discreet detections) in the same experiment and in the same photons.

 

[vanesch]

But we should also agree that the |detect> states are _significantly_ smaller when one slit is closed.

On BOTH detectors ? Because of course (and I didn't include that in my calc :-( ) the grid is a diffraction grating for the light from one hole, which will scatter onto the other detector (it is exactly at the Bragg condition!).

EDIT: OOPS! I made a serious mistake... see next post...

 

[vanesch]

You indicated an interesting point (which, I admit, bothered me also when I was typing my post yesterday), and indeed, I made a big mistake.

I wrote:

(1)

U(|slit 1>|unexcited atom>) =
a | photon_at_detector2>|unexcited atom> + b |no_photon> |excited atom> + c |photon_scattered_in_space>|unexcited atom>

and:

(2)

U(|slit 2> |unexcited atom>) =
a | photon_at_detector2>|unexcited atom> - b |no_photon> |excited atom> - c |photon_scattered_in_space>|unexcited atom>

from which followed: (3)
U(|slit1>+|slit2>) = a (|photon_at_detector1> + |photon_at_detector2>) |unexcited atom>

WHICH IS OF COURSE IMPOSSIBLE.

The reason is that U is a UNITARY operator - which preserves hilbert norm, and if all kets are of unit length (are normalized), and if (1) is respected, then a^2 + b^2 + c^2 = 1 (which agrees with (2) )

but then we have a serious problem with (3), because on the left side the norm is sqrt(2) while on the right side, the norm is a^2 sqrt(2)

which is impossible for a unitary operator!

And what I forgot (and for a particle physicist, shame on me!) is that unitarity puts severe constraints on what can happen in interactions (scattering especially!).

So what we see, is what you pointed out correctly: IF in (1) nothing goes into detector 2, then it follows that a = 1 and b = c = 0. In other words, you cannot have absorption or wild scattering WITHOUT also coherent diffraction!

So (1) should be:

U(|slit 1>|unexcited atom>) =
a | photon_at_detector2>|unexcited atom> + b |no_photon> |excited atom> + c |photon_scattered_in_space>|unexcited atom> + a' |photon_scattered_at_detector1> |unexcited atom>

and (2) should be:

U(|slit 2> |unexcited atom>) =
a | photon_at_detector2>|unexcited atom> - b |no_photon> |excited atom> - c |photon_scattered_in_space>|unexcited atom> + a'|photon_scattered_at_detector2>|unexcited atom>

which leads then to the new (3):

U(|slit1>+|slit2>) = (a+a') (|photon_at_detector1> + |photon_at_detector2>) |unexcited atom>

and unitarity requires that |a+a'|^2 = 1.

So we now have that the intensities of:

only slit 1 open:
intensity on detector 2: |a|^2
intensity on detector 1: |a'|^2

and |a|^2 + |a'|^2 < 1

only slit 2 open:
intensity on detector 1: |a|^2
intensity on detector 2: |a'|^2

However,
both slits open:

intensity on detector 1: 1/2|a+a'|^2 = 1/2
intensity on detector 2: 1/2|a+a'|^2 = 1/2

Sorry for the confusion.

So this explains the eventual diminishing of the total intensity on both detectors with 1 slit open, and not with both.

EDIT: what is interesting here, is that we see an upper limit for the total amount of "lost" light. In the worst case, a = a' = 1/2. This is the lowest value that one can have for a and a' and still satisfy the condition that |a + a'|^2 = 1.

In that case, the intensity on detector 1 (with 1 slit open) is 1/4 and the intensity on detector 2 is 1/4, while half of the intensity is absorbed or scattered in space.

You cannot absorb more than half of the slit intensity as long as the phase relationship is to hold, which is normal: you cannot be in "destructive interference" for more than half of the interference pattern!

EDIT 2:
At first you might object why we flip signs of b and c between (1) and (2), but why we keep the same sign for a'. The reason is that while the states with coefficients b and c are the SAME in the two expressions, the states with the coefficient a' are symmetrical with the same symmetry as the one between "slit1" and "slit2". So we get TWO TIMES a phase flip: the amplitude to go from slit 1 to the grid to detector 1 is the same (phase included) in the symmetrical picture: the amplitude to go from slit 2 to the grid to detector 2. One could say that this is a reflection of the conservation of parity of EM interactions

 

[peter0302]

Ok wow. :) First off thanks for the lengthy correction; I definitely appreciate your taking so much time to explain this.

Second, are we still agreed that "b" and "c" are negligible when both slits are open and large when one slit is closed?

[Edit]
What's bothering me about this equation is that you could have
a=.5
a'=.5
b=.5
c=.5

Then |a+a'|^2 = 1 and
a^2+b^2+c^2+a'^2 = 1 also

So you could still have a quarter of the image being blocked by the grid, and another quarter being scattered into space, which means the total intensity of the two detectors _should be_ 1/2, but yet by your equation the intensity is undiminished.

That's why I'm still uncomfortable with the way you're adding the U|slit1> and U|slit2> together. I just don't see how photons being blocked (and therefore not being detected) in both states could constructively interfere to "unblock" those same photons. I guess this is just a matter of interpretation but even if that works mathematically it seems like adding unnecessary terms and therefore should be the disfavored viewpoint.

Bottom line: is the grid blocking photons when both slits are open or not? Are you going to say "Yes, they're being blocked, and those same photons are still being detected."?

What if we put a photon detector at every point where there's a grid wire? Would those detectors go off ever?

[Edit2]

At first you might object why we flip signs of b and c between (1) and (2), but why we keep the same sign for a'.

I actually understand that part. :) I see the reason for flipping the signs for b and c and not "a" or "a'".

I think the part that's bothering me is simply that I'm fixated with deriving physical meaning from your equations. Your |slit1> and |slit2> equations are obviously right when one slit is closed. Your |slit1>+|slit2> equation is obviously right when both slits are open. So on paper, you get the right result, but when it comes to actually interpreting "what happened" you simply cannot derive any physical interpretation from the derivation - it's just a mathematical trick. Do you agree?

 

[vanesch]

Ok wow. :) First off thanks for the lengthy correction; I definitely appreciate your taking so much time to explain this.

Second, are we still agreed that "b" and "c" are negligible when both slits are open and large when one slit is closed?

[Edit]
What's bothering me about this equation is that you could have
a=.5
a'=.5
b=.5
c=.5

Then |a+a'|^2 = 1 and
a^2+b^2+c^2+a'^2 = 1 also

So you could still have a quarter of the image being blocked by the grid, and another quarter being scattered into space, which means the total intensity of the two detectors _should be_ 1/2, but yet by your equation the intensity is undiminished.

I guess you're talking when 1 slit is open.
The intensity is given by the amplitudes SQUARED

In this case, detector 1 has an intensity of a^2 = 1/4 and detector 2 has an intensity of a'^2 = 1/4. So yes, the total intensity seen by the two detectors together is then only half of the incoming intensity.

When the TWO slits are open, however, we get destructive interference in the channel "scattered into space" and in the channel "blocked by the grid", which means that in the result, nothing is scattered into space, and nothing is blocked by the grid, because the END AMPLITUDE of these channels is 0.

That's why I'm still uncomfortable with the way you're adding the U|slit1> and U|slit2> together. I just don't see how photons being blocked (and therefore not being detected) in both states could constructively interfere to "unblock" those same photons.

That's because you are thinking in statistical ensembles, not in amplitudes. The example I gave of EXAFS does exactly that (although I guess you can also interpret it in any different way, as you can interpret any quantum phenomenon in any of the interpretations that go with it). You seem to make a difference between "blocked" and "scattered" on one hand, and "making an interference pattern" on the other hand. It seems as if you consider these events as "irreversible and classical": so what "is absorbed" cannot be "unabsorbed" by another term. It is also probably (I guess that has to do with your hang for a Bohmian view - which is ALSO a possible way of seeing things) related to the fact that you want to give trajectories to photons.

But when you just consider unitary evolution of a quantum state, then "absorption" is nothing else but a quantum interaction between a photon and another system (say, an atom) which just ends up in an end state (here: one photon less, and the atom goes into an excited state or an ionised state). Now, if you add together this resulting state with the resulting state when exactly the same photon state comes in, up to a phase flip, then the result will be 0. In other words, the amplitude to have an excited atom (or an emitted electron) will be 0.

But you could have made the sum of the incoming amplitudes of course BEFORE applying the interaction, and then you would have found of course that the incoming amplitude was already 0.

That's the same as saying "the total photon state (the sum of both incoming sides) didn't interact". Or, you can say, each TERM interacted, but the results were in anti-phase, and hence there was no net result in the interaction channel.

This simply comes about because U is linear (and unitary).

I guess this is just a matter of interpretation but even if that works mathematically it seems like adding unnecessary terms and therefore should be the disfavored viewpoint.

No, I like it for several reasons. The "mystery" of the Afshar experiment is somehow: how come that photons that come through a slit and of which we can see the image of the slit somehow, should hit the wires, interact with it, be absorbed, scattered whatever, but if they come through the two slits, they seem magically to AVOID the wires.

My answer is: well, if it BOTHERS YOU to think that they AVOID suddenly the wires, and nevertheless "continue onto their slit images as before", then look upon it as this:
They DO interact with the wires as before, but the interaction channels now suffer destructive interference. So what is "absorbed" by one, is "unabsorbed" by the other, and what is scattered by one is "unscattered" by the other.

Bottom line: is the grid blocking photons when both slits are open or not? Are you going to say "Yes, they're being blocked, and those same photons are still being detected."?

They are "blocked" and "unblocked" together, so that the "blocking channel" suffers destructive interference and doesn't happen anymore, if you FIRST apply U and THEN the sum ; while they never got blocked in the first place if you FIRST apply the sum over slits, and THEN apply U.

What if we put a photon detector at every point where there's a grid wire? Would those detectors go off ever?

No, they wouldn't, because (and I guess here my MWI preference shows :-) the detectors would also show up "clicking" and "clicking with 180 degree phase shift" so that they wouldn't click. But you don't have to go so far if you don't like MWI. You can just say that any photon detector is somehow based upon the photo-electric effect, and if you can keep "quantum mechanics running" upto the emission of the photo-electron, then you will find a photo-electron going out and a photo-electron going out with anti-phase for the two slit contributions, so that in the end the amplitude for an outgoing photo-electron is 0 (and the detector doesn't click). This is the same kind of photo-electron quantum mechanical interference as you can find in EXAFS: destructive interference of the outgoing photo-electron wave gives rise to a diminished absorption.

I think the part that's bothering me is simply that I'm fixated with deriving physical meaning from your equations. Your |slit1> and |slit2> equations are obviously right when one slit is closed. Your |slit1>+|slit2> equation is obviously right when both slits are open. So on paper, you get the right result, but when it comes to actually interpreting "what happened" you simply cannot derive any physical interpretation from the derivation - it's just a mathematical trick. Do you agree?

This is personally how I look "physically" upon things in quantum mechanics. I don't see "bullets with uncertainties" running around in the experiment, but I see "vectors in hilbertspace wiggle about" (again, hence, my preference for MWI).

The only thing I wanted to illustrate is that in quantum-mechanical interference with annihilation, you can do the annihilation where-ever you want (as long as you consider unitary evolution). As such, the question of "did the photon MISS the wires" is in fact arbitrary. It "misses" the wires if you want to, or it interacts with it if you want to!
It just comes down to what you apply first: the plus or the U.

Another thing which is interesting in this analysis is that you see (and that's what I forgot) that it is ESSENTIAL that the photons from slit 1 are scattered onto the OTHER detector and vice versa if absorption and so on has to take place. As such, the claim that we "know the which-way information" afterwards is wrong too.
But that could already have been seen by the fact that the grid is a diffraction grating which scatters photons from slit 1 also onto detector 1 and vice versa.

Mind you, I'm not claiming that the photons DO scatter from the wires and then interfere destructively in the scatter and absorption channels. I'm just saying that this is *a possible way of looking at things*.

This is to show that you cannot draw CONCLUSIONS based upon the experiment to EXCLUDE certain views, unless those views were already erroneous from the start (such as naive statistical hidden variable views which mix up superpositions and mixtures).

If you analyse an experiment such as Afshar's within each interpretation, and you don't make errors (like I did!) then you come to a coherent view in each case.

EDIT:

A classical analogue (which is actually pretty close to Afshar's experiment!) of this arbitrariness of interference is this:

Look at a classical light wave coming through two slits and making an interference pattern on a white diffusing paper. A camera with a lens is looking at the paper. Now, imagine that we calculate this as follows:

We have the EM wave coming from the first slit, and we calculate the propagating E and B fields through space, and we calculate the diffusion of the EM field by the paper, as a further propagating EM wave in space. We call the total solution of the EM field for this case E1(r,t). We can calculate the image that E1 causes through the lens upon our camera CCD element.
And next, we calculate the EM wave coming through the second slit, and do the same thing. We find the total solution for this case E2(r,t).

Now, if we open both slits, we have, in space: Etot = E1 + E2.

Now, did the interference happen at the CCD camera where we had to add E1 and E2, or did the interference already happen on the diffusing paper, and we shouldn't have calculated the diffusing fields E1 and E2 independently, but first have made the sum of E1 and E2 on the paper, and then have calculated the diffusion of the NEW field that was the result there ?

The answer is that you get twice the same thing of course: the diffusion by the paper of the interference pattern will give rise to an EM field in space which is exactly equal to E1 + E2.

 

[colorSpace]

I guess you're talking when 1 slit is open.
The intensity is given by the amplitudes SQUARED

In this case, detector 1 has an intensity of a^2 = 1/4 and detector 2 has an intensity of a'^2 = 1/4. So yes, the total intensity seen by the two detectors together is then only half of the incoming intensity.

If both detectors show the same intensity when only 1 slit is open, then Afshar couldn't claim that they give which-way-information in the first place.

 

[vanesch]

If both detectors show the same intensity when only 1 slit is open, then Afshar couldn't claim that they give which-way-information in the first place.

Yes, but the example given by peter0302 was the extreme limit authorised by unitarity. It is the one where the wires, seen as a diffraction grid, have a the diffraction peak is as intense as the direct peak. This is difficult to realize in practice.

I guess that in the usual experiment, the a is much closer to 1 and the a' is relatively small. Moreover, a and a' are complex numbers in general.
You could have something like:

a = 0.95 (a^2 = 0.9025)
a' = 0.05 (a'^2 = 0.0025)
b = 0.308 (b^2 = 0.095)

(we drop c, consider it included in b).

we have a^2 + a'^2 + b^2 = 1 (check it!)

This means that with 1 slit, we have:

detection rate at "good detector" lowered by about 10% (a^2 = 90%)
detection rate at "bad detector" only 0.25 % (a'2 = 0.0025), so one seems to "miss" this entirely
absorption/scattering in space (blurring?): 9.5% (the rest).

But with 2 slits, we have:

-detection rate at both detectors: (a+a')^2 = 100%
-no absorption/scattering.

Doesn't this look awfully like Afshar's results ?

 

[colorSpace]

This means that with 1 slit, we have:

detection rate at "good detector" lowered by about 10% (a^2 = 90%)
detection rate at "bad detector" only 0.25 % (a'2 = 0.0025), so one seems to "miss" this entirely
absorption/scattering in space (blurring?): 9.5% (the rest).

But with 2 slits, we have:

-detection rate at both detectors: (a+a')^2 = 100%
-no absorption/scattering.

Doesn't this look awfully like Afshar's results ?

That looks much better. Just a side note: The images I've seen indicate that there is at least some distortion even with both slits open.

 

[colorSpace]

BTW, very basic question: Wave functions are apparently derived from our "measurements" and knowledge of the experimental setup, which in Bohmian Mechanics would correspond to particle positions. So aren't the wave functions, at least in a practical sense, deducible from the particle positions? :)