Quantum Interpretations history

[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? :)

 

[peter0302]

Vanesch, thanks to your explanations, this view seems to make perfect sense in MWI. :)

I just still can't see how it's consistent with any sensible single-world interpretation. You have a choice between assuming the photons are magically avoiding the grid, or assuming the grid is magically re-directing / re-emitting photons that it otherwise should be blocking. (Or just giving up a la CI). Of these, the former is more palletable, but I don't really love either of them.

You might be making an MWI believer out of me...

Anyway thanks again. Much appreciated.

 

[vanesch]

Vanesch, thanks to your explanations, this view seems to make perfect sense in MWI. :)

I just still can't see how it's consistent with any sensible single-world interpretation. You have a choice between assuming the photons are magically avoiding the grid, or assuming the grid is magically re-directing / re-emitting photons that it otherwise should be blocking. (Or just giving up a la CI). Of these, the former is more palletable, but I don't really love either of them.

You might be making an MWI believer out of me...

I'm not really an MWI *believer* you know. I think we don't have all the pieces of the puzzle yet to raise it to the ultimate goal of "the meaning of life, the universe and everything". But I like the MWI view on quantum mechanics: I find that it gives the "explanation" that is most in sinc with the formalism, and is most of the time, once understood, rather crystal-clear (no ambiguity left in the interpretation of a setup). The only problem is of course that it is quite crazy! However, its crazyness can be dealt with if you delve into some philosophy. That is to say, one can have a logically sound reasoning which helps you overcome a *logical* objection to the crazy world view of MWI. That doesn't avoid that intuitively you say: "bollocks!".

However, I have to say that in many quantum-mechanical situations where one delves into "metaphysical" considerations about certain aspects, sticking to an MWI view on it "for the sake of argument" helps me to get a quite clear view on it. Especially EPR experiments and delayed choice quantum eraser experiments become "totally logical" in this view.

The intuitively most acceptable view to me is BM. However, I find it formally a monstruousity, because of the explicit impossibility of formulating its machinery in a lorentz-invariant way (which is NOT the case for the pure wavefunction part). That said, when you look closer at BM, its intuitiveness suffers a bit. I would say that if we never had relativity, and we were stuck with NR QM, then BM is really very nice. But I hate its clash with the spirit of relativity. Also, many situations are intuitively difficult to analyse in BM - at least for me. I have first to switch to an MWI kind of picture, understand it from that PoV, and THEN I'm sometimes able to understand the BM picture. But getting an intuition for BM as such is not easy!

Anyway thanks again. Much appreciated.

thanks

 

[colorSpace]

The wave functions are of course common among all interpretations. The difference arises when there is a change from superposition and/or interference to a specific outcome relative to an observer or measurement device, meaning: at which point will the parts of the wave function stop to interfere. It seems the discussion of this experiment has avoided this crucial question by looking only at the wave function.

But it doesn't take MWI to look only at the wave function, since, again, it is common to all interpretations. Or?

 

[colorSpace]

[Continued]

Also, one shouldn't overlook the potential of interpretations such as BM to show possibilities of explaining specific outcomes in a more precise way than "randomly", at least for specific situations. Perhaps an analysis of particle dynamics will show that it is possible to create situations in which a specific outcome is predictable. CI and MWI have in common that they exclude such a possibility in regards to a specific observer, since from a subjective point of view randomness is built-in, and seemingly unavoidable. So they will discourage research into looking for models which might be more precise (at least in some situations) than the wave function.

 

[vanesch]

But it doesn't take MWI to look only at the wave function, since, again, it is common to all interpretations. Or?

Eh, MWI *does* only look at the wavefunction. It declares (coarse grained) quantum states of a body as "subjectively experienced" (like classical brain configurations are to be "experienced") and the probability to be experienced by you is given by the amplitude squared of the product state in which this body state appears.
The difference with other interpretations is that it doesn't ADD anything to the wavefunction (such as BM, or TI), and that it doesn't fiddle with it or declares it non applicable (CI) to certain systems. Wavefunction all the way!

So if, in the wavefunction, two terms devellop (though normal unitary evolution), one which contains a body state which is close to a classical body state with a brain that sees a red light flashing, and another term with a body state which is close to a classical body state with a brain that sees a green light flashing, then in MWI you say that these two terms are two branches, one with an observer seeing a red light, and the other with an observer seeing a green light flashing, and that the probability for YOU to be/experience that first or that second observer state is given by the amplitude squared of each of the two terms.

Now, you can think it to need a kind of miracle to develop such classically-looking terms, but in fact, decoherence shows us that this happens in fact quite fast and that these terms are then stable (that is, they become classically tracable throughout the further evolution under U).

 

[colorSpace]

Eh, MWI *does* only look at the wavefunction.

What I said is that you don't need MWI in order to look at the wave function.

 

[vanesch]

What I said is that you don't need MWI in order to look at the wave function.

Sure ! The only thing you need to avoid is to do OTHER things than to look at the wavefunction, such as blah blah about through which slit the photon came, and whether or not it "behaves as a particle or a wave" and so on!

The analysis of the Afshar experiment I presented here is - as I tried to point out - independent of interpretational issues, but is probably easier to follow in an MWI mindset where you are not *perturbed* by other, often misled considerations, such as "did it hit the wires or not".

The nice thing about this experiment is that it helps you to understand better the intimate workings of each interpretation, which is in any case always based on the wavefunction. It might chase erroneous views on certain interpretations. For MWI, this is very easy, as once we have the wavefunction, things are over! That's why I like that view, btw.

 

[colorSpace]

Sure ! The only thing you need to avoid is to do OTHER things than to look at the wavefunction, such as blah blah about through which slit the photon came, and whether or not it "behaves as a particle or a wave" and so on!

My understanding is that "behaving as a wave" means there is interference seen by a specific observer; "behaving as a particle" means there isn't (at least not seen by a specific observer). That difference must also exist in MWI.

 

[vanesch]

My understanding is that "behaving as a wave" means there is interference seen by a specific observer; "behaving as a particle" means there isn't (at least not seen by a specific observer). That difference must also exist in MWI.

But these statements "behaving as a wave" or "behaving as a particle" are relics of a time when one didn't understand much about quantum dynamics, and one wanted to attach it to classical notions. The *actual* difference is: when does a superposition become equivalent to a statistical ensemble ? That is:

when can we consider something like u |a> + v|b> as:
|u|^2 chance to have |a> and |v|^2 chance to have |b> ?

Answer: when we do an irreversible measurement. Period.

Now, if the measurement is in the position representation, then this means:
when does a "superposition of position states" (a wave) can be seen as a statistical mixture of different positions (possible "particle" positions) ?
Answer: when we do a position measurement !

Now, and this is where much confusion rules, especially in "which way" experiments and so on, SOMETIMES we get the right results out, even if we PREMATURELY consider a superposition of position states (a wave) as "already a statistical mixture". So, though this is IN PRINCIPLE wrong, we might get out nevertheless the right result this way. If this happens, we say that we don't see any interference. The point is, we would have obtained the right result also of course if we would have followed the *correct* procedure: namely: keep superpositions as such, until we do an irreversible measurement.

Why do we sometimes switch "too early" to the statistical mixture ? First of all, of course, because it is closer to our classical intuition. But second, because especially the CI introduces an ambiguity, by this "switching to classical when a measurement is performed" clause. Indeed, this should read: an IRREVERSIBLE measurement (for instance: with a result printed on paper and so on). So some people extend that concept and call everything that interacts with the system under study "a measurement". Like the grid which performs "a measurement" on the "positions" of the photons - although no result is printed out! As I said, sometimes one can get away with this: the results come out the same. But sometimes, not. In any case, it is an error of principle to do the superposition -> ensemble transition BEFORE an irreversible measurement is done.

MWI has the advantage of clearly not introducing a different concept of interaction for a "measurement" or for an "interaction", and so one is not tempted to make too early a transition to a statistical ensemble. But even in CI, if interpreted correctly, one shouldn't have done that!

 

[colorSpace]

Answer: when we do an irreversible measurement. Period.

MWI has the advantage of clearly not introducing a different concept of interaction for a "measurement" or for an "interaction", and so one is not tempted to make too early a transition to a statistical ensemble. But even in CI, if interpreted correctly, one shouldn't have done that!

If "irreversible measurement" means 'a measurement which leaves a trace of information' (roughly) then I guess this puts MWI and "modern" CI back on the same page in this specific regard.

 

[vanesch]

If "irreversible measurement" means 'a measurement which leaves a trace of information' (roughly) then I guess this puts MWI and "modern" CI back on the same page in this specific regard.

Yes! That's why it is kind of silly to try to distinguish between them experimentally!
Caveat: at least, that is how *I* understand CI, because you never find two identical accounts of what is exactly CI...

 

[Hans de Vries]

With all the different quantum interpretations out there, which is your favourite " ?

You asked me for my favorite interpretation in a PM, Now, since I use
my own interpretation I'll post the elementary ideas here.


Just to give it a name: Sea-of-particle interpretation


Q: What is a particle?
A: A single surplus/absent particle in the see-of-particles


Q: Why can a particle turn up everywhere in the wave-function?
A: First this:

00_000000000000
000000000000_00

How long does it take for the 'gap' to go from the left to the right?
In principle not more as it takes the ten '0' take to go one step to
the left. There is in principle no SR limitation for the 'gap' to be either
at the left or the right.


Q: What is the wave function?
A: A single absent/surplus particle will distort the whole 'grid'
Instead of a single gap of say '40' there will be many small gaps
like for instance:

1-2-4-6-7-7-6-4-2-1

This is the wave function. The bigger the gap, the higher the chance
of detection. No single gap corresponds to the single deficit particle.
No single 'sea-of-particles' particle corresponds to the one surplus
particle.


Q: What is the detection of a particle?
A: The removal of the single surplus / deficit particle from the sea-of-
gates. Fixed and localized for instance to an atom.


Q: What happens with the wave function?
A: Without the single surplus/absent particle there is nothing anymore
to 'distort' the grid of the sea of particles. The wider wave-function will
die out. The wave-function will become fixed /localized to an atom for
instance.


Q: Shouldn't the wave-function disappear instantaneously? A remaining
gap could be detected as a particle which would violate unitarity.

A: This describes particle-pair creation, it leaves an extra surplus particle.
This happens all the time, temporary, because of energy conservation.
Single particle unitarity does not exist.


Q: What is the physical interaction?
A: The physical interaction is generally with the whole wave-function.

example: In Afshar's experiment there are more than 10^23 interactions
(~Avogadro's number) between the single photon's wave-function and
the dielectric molecules of the lens. These are real physical interactions.
Atoms are displaced back and forward. They detect the wave-function
but they do not -detect- the photon in the sense that they remove a
single surplus/absent particle from the see-of-particles.



For so far, In the hope to give you something which is more acceptable... :^)

Regards, Hans

 

[akhmeteli]

You asked me for my favorite interpretation in a PM, Now, since I use
my own interpretation I'll post the elementary ideas here.


Just to give it a name: Sea-of-particle interpretation


Q: What is a particle?
A: A single surplus/absent particle in the see-of-particles

Interesting. Actually, I discussed a similar interpretation in http://arxiv.org/abs/quant-ph/0509044 .
Could you give references to your (or other people's) publications on this approach?

 

[vanesch]

You asked me for my favorite interpretation in a PM, Now, since I use
my own interpretation I'll post the elementary ideas here.


Just to give it a name: Sea-of-particle interpretation

It is difficult to imagine how this is in 1-1 correspondence with the quantum formalism. After all, what you describe is almost litterally the dynamics of crystal defects in materials (Frenkel defects and all that). For instance, how does one get simply, say, the helium spectrum out of such a picture ?

 

[Hans de Vries]

For instance, how does one get simply, say, the helium spectrum out of such a picture ?

That shouldn't be so difficult.

The atomic spectra arise when the Laplacian is combined with a central potential
and the Laplacian occurs almost everywhere where you have particle assembles
since it's part of the classical wave equation.


Regards, Hans