# Thought experiment concerning worldlines

1. Jul 18, 2009

### SWwright

I have a thought experiment that may be of interest to some of you. If this has been discussed before, please direct me to the discussion. Thanks!

The experiment involves a beam generator consisting of a laser, a parametric downconverter cut for operation at the 2nd harmonic, and mirrors to direct the output in opposite directions. The laser is ultraviolet, the downconverter output is green. If you are not familiar with downconverters, the basic idea is that one photon goes in and two come out in slightly different directions but with the same total energy. At the 2nd harmonic they each have half the frequency of the incoming photon.

Now, put the generator in a spaceship, with the beams directed fore and aft. Accelerate the ship to 1/3 c and run it past an observer. The observer observes a blue beam directed toward him and a red beam directed back along the ship's path. Lorentz transformation applied to the perceived wavelength of the downconverted light: the forward beam is blue-shifted, the backward beam is red-shifted, from the observer's perspective.

Now let us have a second generator, stationary with respect to the observer, but it uses an asymmetric downconverter at the 3rd harmonic, so the output is naturally blue and red with the blue photon twice the frequency of the red photon. The sum of their frequencies is still equal to the frequency of the incoming ultraviolet photon (conservation of energy). Is there any way for the observer to distinguish which beams came from the stationary generator and which from the moving generator, if the observer is not able to see the generators themselves?

The reason this is of interest, is another relativistic effect combined with a quantum effect. Parametric downconverters produce pairs of photons that are polarization entangled. Because they are entangled, if you measure one photon, its partner also collapses into a definite state (Einstein's "spooky action at a distance"). Now draw a worldline between the two collapse events. If you are at rest with respect to the source of the photons, then that worldline is "flat" and the events are perceived as simultaneous. If you are in motion with respect to the source, then a Poincare rotation applies to your perception of that worldline: if the source is approaching you, then the near collapse (of the "blue" photon's state) is rotated slightly into the future, and the far collapse (the "red" photon's state) is rotated slightly into the past, as far as your perception is concerned. You observe the "red" photon collapsing before the "blue" one does.

Finally, and with apologies for the lengthy setup, here's my point (and my real question). If one cannot distinguish whether the redshift/blueshift of a pair of beams of light is due to motion, or due to the nature of the downconverter, then should there not be a Poincare rotation of the worldline between the collapse events, in either case? And is there an experiment that could verify (or falsify) this question?

2. Jul 20, 2009

### Ich

Not photons are collapsing, the wavefunction does. Nobody ever observed a wavefunction collapsing, it is not an observable. There's no such thing as "collapse events".

3. Aug 18, 2009

### SWwright

Ich,
I beg to differ. What does one call it when a photon of undetermined polarization, encounters a polarizing filter? The photon's polarization wavefunction collapses to a specific real polarization (vertical or horizontal, to oversimplify) and the photon is either absorbed by the filter, or passes through. What does one call it when another photon is entangled with that photon, and its polarization "simultaneously" changes from an undetermined superposition of states, to a definite state? I call these "collapse events". They are real events in spacetime, and one can draw a world line from one event to the other.

4. Aug 18, 2009

### Ich

You can draw lines at will. Still, there are no collapse events. Study QM to find out what collapse means and how you can measure it. Do you think you can measure whether the polarization is "defined" or not?

5. Aug 18, 2009

### SWwright

As you surmise, I am not a physicist, I am but an interested amateur. So I will ask you, what is the correct nomenclature for that which happens when one photon of an entangled pair encounters something that causes it to change state in some way, and its entangled partner, elsewhere (and assumed to be traversing empty space, so that there is no thing with which it may interact) likewise changes state (according to the rules of quantum mechanics)?

I don't think that Clauser, Gisin, and others who have tested Bell's Theorem, were just waving their hands in the air. Something measurable happened. If I don't have the right name, tell me what the right name is.

6. Aug 18, 2009

### Ich

It's not about names. We're talking about the collapse of a wavefunction. It's just that a wavefunction is nothing measurable, let alone its collapse.
The point is: it does not change state in any measurable way. Its polarization is from the beginning antiparallel to the other, and it stays so all the time, measured or not. It's common sense wihch insists that the measurement of the first photon somehow instantly changes the second. Physics has no such claim.
So I ask you (rethorical question of course, to invite you to find out on your own): How would you, via a local measurement of photon 2, find out that photon 1 has been measured "right now"?

7. Aug 18, 2009

### AllanGoff

Actually, this is exactly what physics claims. The implication of Bell's inequality is that collapse not only happens but that no common cause can account for it. The challenge in spacelike separated entanglements is to explain the correlations. Consider a series of experiments where a succession of EPR photons (polarization entangled) are directed from a central source to equidistant detectors. Because they are spacelike separated, timelike causality cannot explain the correlations exhibited between the EPR pairs. Bell's inequality shows that common cause, some earlier 3rd cause in the past light cones of both photons, cannot explain the correlations either. These are the only two types of causality in classical physics. Mermin has presented the clearest popular level description of Bell's inequality by the simple expedient of restricting the measurement bases to being multiples of 60 degrees.

In order to account for the correlations one of the measurements is treated as occurring first (it doesn't matter which) which puts the other photon in one of two determined but unknown polarization states. The statistics of a measurement on either state is well known and straight forward. Records are kept on both sides for a long run, logging the measurement basis and the polarization results. When the two lists are later compared the only explanation is that the wave function collapsed because of a measurement. Which measurement is not specified in the current formulation of quantum mechanics. The collapse does not have to happen "instantaneously", insistence on that is a common sense trap, but it must happen in a spacelike way. What this means is that any speed between +c, up to infinity, and back down to -c (backwards in time) are physically consistent.

This so bothered Suarez & Scarani that in 1999 they proposed the speculative idea that since the measurement mechanism is unknown, perhaps the particulars in any given situation implied a preferred but temporary frame of reference, and in this frame of reference the first measurement was the cause. They hypothesized that if the two measurements could be put into relative motion so that each in their own frame of reference occurred before the other, then it became ambiguous which measurement was the cause and they predicted that the correlations would disappear. This was a brilliant if unorthodox idea, and its very craziness is an indicator of the conundrum that nonlocality presents. In 2002, Nicholas Gisin tested this conjecture over a distance of 10.6 km using the Swiss fiber optic telecon network in Geneva. He got the null result, the correlations persisted. He later concluded that some new 3rd type of causality is required.

We already know of a potential 3rd type of causality, spacelike causality, but this is rejected a priori because of the twin problems of temporal paradox & indeterminacy, the difficulty in making it Lorentz invariant, and the violation it does to our common sense notions of causality. It seems to me that theory and experiment are compelling us to consider spacelike causality, despite the conceptual challenges. This means that collapse of the wave function will have to be made consistent with relativity, something that was achieved for the evolution of the wave function with Schroedinger's equation, but that still eludes us 80 years later. Suarez & Scarani may have been on the right track. Perhaps the key to the measurement problem lies in part within relativity rather than residing exclusively in quantum physics.

(Note, the original author used the term worldline for intervals that were clearly spacelike. With this change the original query may be more insightful than it first appears.)

8. Aug 19, 2009

### Austin0

Is equidistance a neccessary condition or is it like simultaneity ,not really relevant????

9. Aug 19, 2009

### Ich

No. Physics claims that if you measure the polarization of the photons, the results will be correlated. And that's it.
Anything else is interpretation. With an appropriate "shut up and calculate" - interpretation as philosophical background, there is no problem.
There is a description, according to which the - spatially extended - wavefunction of the photons collapses instantly. The wavefunction represents our knowledge about the quantum system, and its collapse represents our change in knowledge. You can't measure the collapse.
Exactly: it doesn't matter which. It's a broad hint that it won't help to think in terms of cause and effect if the order of events is completely irrelevant. Musing at which speed a nonexistent cause travels to reach a nonexistent effect is bad philosophy, not physics.
If you drop the unwarranted assumption that measurement here changes the photon there, and instead simply buy it that the results are correlated, everything is ok. You don't have to explain the correlation in some classical terms, that'll always go wrong in QM.

So my question still holds: "How would you, via a local measurement of photon 2, find out that photon 1 has been measured "right now"?"
If you can't, it makes no sense to specify "collapse events" and their ordering.

10. Aug 19, 2009

### Austin0

How do you view the measuring??? If measuring inherently means changing what is being measured then the ordering becomes less relevant. It doesn't matter if the events are simultaneous for the assumption of some causal relationship between the events even if it cant be given a definite direction.
SO if measuring polarization actually shifts some intermediate orientations during passage as well as absorbing some intermediates then the resulting correlation would not neccessarily have been present from the beginning .
If there was some totally passive way to measure without change then what you are saying would seem to apply.
It is simply a case of tangled photons are correlated . Done

IMO

11. Aug 19, 2009

### AllanGoff

This has been a fun thread but it occurs to me that our discussion is handicapped by our notation, plain English text. Does the physics forum support equations? We could then give clear examples in Dirac notation of the various superposition, entangled & collapsed states. Does anyone know how to insert, say a Microsoft equation? If any internet forum should support this, it should be this one.

Thanks.

12. Aug 19, 2009

### DrGreg

Yes. Click on an equation like this

$$E=mc^2$$

to find out how to do it. Or use the $\Sigma$ button in the "Go Advanced" Editor.

13. Aug 19, 2009

### SWwright

They use $\LaTeX$ for equations here.

There is an discussion dating to when this feature was introduced, https://www.physicsforums.com/showthread.php?t=8997". The first post contains links to some PDF's that list examples of common stuff. Mouse-over any equation and the LaTex source will display in a popup briefly, or quote the message and you can examine or copy the source.

Edit: Or do what DrGreg suggests.

Last edited by a moderator: Apr 24, 2017
14. Aug 20, 2009

### AllanGoff

Thanks.

$$\Psi^{P60}_1=\frac{1}{\sqrt2}\{|h\uparrow>+|v\downarrow>\}$$

$$B=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]$$

OK, that should get us started.

The current topic of this thread is whether or not the wave function collapses. Let's start with the simpler question of whether it changes when a measurement occurs.

We'll need to agree on a number of concepts. Measurements are made in some basis and it may be necessary to consider multiple bases in order to fully appreciate a particular phenomenon. Notation is not well standardized in QM, we will try to stick to one, but one no more complicated than necessary. In general the state of a system will have to indicate the basis in which it is being represented, the eigenvectors for that basis and which particles have which values. In anticipation of later developments we'll also insist that the normalization constant be explicit. Since the operators, eigenvectors, & states have to be represented mathematically in some basis we'll call it the lab basis. Any expression not in the lab basis must explicitly identify which basis it is being expressed in.

Because of the statistical nature of quantum systems experiments have to consist of many trials. A system is prepared in a known state over and over again and measurements made. A run consists of a enough trials to produce sound statistics. The resulting statistics, not any one particular outcome, is the result of the run.

Experiment Set #1. Individual photons are prepared in a known polarization state and sent to a polarizing beam splitter with two detectors. Let the beam splitter be in the H/V basis (horizontal & vertical polarization).

Experiment #1-1. The photons are all horizontally polarized.
$$\Psi^{L}_h=\frac{1}{\sqrt1}\{|H>\}$$
Outcome #1-1. All photons exit the |H> port. Furthermore, if run into a second beam splitter, they continue to take the |H> port.

Experiment #1-2. The photons are all vertically polarized.
$$\Psi^{L}_v=\frac{1}{\sqrt1}\{|V>\}$$
Outcome #1-2. All photons exit the |V> port. Furthermore, if run into a second beam splitter, they continue to take the |V> port.

Experiment #1-3. The photons are randomly prepared in either polarization. That is, their polarization's are determined but unknown.
$$\Psi^{L}_m=\left\{\begin{array}{cc}\frac{1}{2}\{|H>\}\\\frac{1}{2}\{|V>\}\end{array}\right$$
Outcome #1-3. Half the photons exit the |H> port, the other half exit the |V> port. Furthermore, if run into a second beam splitter they don't change ports. The dogma here is that each photon is simply shuttled to one port or the other depending on its polarization.

Now this may all seem a bit tedious, but tedium is a sound strategy for avoiding the introduction of hidden assumptions.

Experiment #1-4. The photons are prepared in a diagonal polarization, a superposition of horizontal & vertical polarizations. That is, their polarization (with respect to the H/V basis) is indeterminate.
$$\Psi^{L}_d=\frac{1}{\sqrt2}\{|H>+|V>\}$$
Outcome #1-4. Half the photons exit the |H> port, the other half exit the |V> port. Furthermore, if run into a second beam splitter they don't change ports. The dogma here is that each photon exits both ports, in a superposition of states until finally detected at one and only one detector.

Now the really interesting question. Is it possible to distinguish between photons prepared in the #1-3 state (mixture, determined but unknown) versus those prepared in the #1-4 state (superposition, indeterminate)? Yes. Let the polarizing beam splitter be the entry port to an interferometer. Experiment #1-3 will not produce interference, #1-4 will. It is this result that supports the previous dogmas.

Experiment #1-5. The photons are prepared in a slant polarization, a superposition of horizontal & vertical polarizations (but with a different phase angle). That is, their polarization (with respect to the H/V basis) is indeterminate.
$$\Psi^{L}_s=\frac{1}{\sqrt2}\{|H>-|V>\}$$
Outcome #1-5. Half the photons exit the |H> port, the other half exit the |V> port. Furthermore, if run into a second beam splitter they don't change ports.

As before, this state can be distinguished from the mixture of states in Experiment #1-3 through the use of an interferometer.

Experiment #1-6. The photons are randomly prepared as either |D> or |S>. That is, their polarization's are determined but unknown (in the diagonal/slant basis).
$$\Psi^{L}_M=\left\{\begin{array}{cc}\frac{1}{2}\{|D>\}\\\frac{1}{2}\{|S>\}\end{array}\right$$
Outcome #1-6. Half the photons exit the |H> port, the other half exit the |V> port. Furthermore, if run into a second beam splitter they don't change ports.

Can this state be distinguished from the state in experiment #1-3? No! The reason is that the interference patterns for |D> & |S> photons have their peaks & troughs swapped so the sum is just a diffraction pattern. This is the gist of the phenomenon of quantum erasure.

Now the point of all this is that when a particle is measured in a basis in which its state is indeterminate, the measurement process changes its state. This change in state is discrete; a common term is that the wave function 'jumps' from one state to another. The temporal aspect of such jumps is a bit ill-defined in the current formulation of QM. However, it should be clear that this is a physical change in the particle and not merely a change in our knowledge. A typical example of this in freshman physics is the crossed pair of polarizing filters that are dark, until a third polarization filter, at 45 degrees to both is slipped in between them. The third filter changes the polarization from a known state to an indeterminate state and it is this indeterminism that allows a few (1/8th) to get through.

This post has gotten a tad long, so I'll continue with it for the case of EPR style entangled photon pairs in a day or two, and then consider Mermin's presentation of Bell's inequality. That will finally give us the tools to investiate whether it is reasonable to regard collapse as occurring over spacelike intervals.

15. Aug 21, 2009

### AllanGoff

When we last left our heroes...

Now lets consider two photons, 1 & 2, both in an equally weighted superposition of polarization. Their states are
$$\Psi_1=\frac{1}{\sqrt2}\{|H>_1+|V>_1\}$$
and
$$\Psi_2=\frac{1}{\sqrt2}\{|H>_2+|V>_2\}$$

When the two particles are independent, when a measurement on either one does not have an effect on the outcome of a measurement on the other, they are separable and the system state is just the product of the states of the two particles
$$\Psi_1=\frac{1}{\sqrt4}\{|H>_1|H>_2+|H>_1|V>_2+|V>_1|H>_2+|V>_1|V>_2\}$$
The state subscripts refer to the individual particles.

EPR states represent the highest possible degree of entanglement between a pair of particles, and so they have received most of the attention. However, a full continuum of partially entangled states exist between independent and EPR style states. Any state that is not independent is non separable. About halfway are states that play a central role in abstract quantum systems, such as http://www.paradigmpuzzles.com" [Broken], they are called semi-entanglements, and have not been much studied in the lab. An example is
$$\Psi_1=\frac{1}{\sqrt3}\{|H>_1|H>_2+|H>_1|V>_2+|V>_1|H>_2\}$$
If the outcome of a measurement on this system finds photon 1 in a |V> state, then photon 2 will be in an |H> state. However, when photon 1 is measured to be in the |H> state, then the state of photon 2 depends on the type of measurement made. The two extremes are a classical state of either |H> or |V> or a superposition of |H> & |V>. I don't know if points in between are possible but suspect they are.

The following fully entangled non separable state is the classical EPR state.
$$\Psi_1=\frac{1}{\sqrt2}\{|H>_1|V>_2-|V>_1|H>_2\}$$
Because of the minus sign connecting the two possible outcomes this state actually has the same mathematical form in every basis.

Before continuing, something about the semi-entangled state has caught my eye. If photon 2 is measured to be in state |V>, then photon 1 must be in state |H>, and when photon 2 is found in the |H> state, photon 1 is either a mixture or full superposition or somewhere in between. The system is in fact symmetric with respect to an H/V measurement on either particle but it is not clear to me that this can be done consistently.

I'll be back after some calculations.

Last edited by a moderator: May 4, 2017
16. Aug 21, 2009

### Ich

Hi,

thanks for your effort, but before you continue I'd like to clarify:
No. The question is whether you can identify the "moment when the wavefunction collapses" by a local measurement at photon 2's position. You can't, and that's why I call that whole concept an interpretation rather than "actual physics" - whatever that is.
Like the Lorentz Ether Theory: That's a wonderful theory which gives totally accurate predictions. Still, there's no point discussing our speed relative to the ether. That is a basic ingredient of the theory, but you can't measure it and might as well say that there is no ether, and use a different description instead.

I don't exactly see what you're getting at; there seems to be some confusion at the end. But let me comment on an intermediate step:
While it should be clear, it isn't.
What you're describing all the time here are Stern-Gerlach type "measurements", where you split the ensemble into subsets. "collapse of the wavefunction" has nothing to do with it at this stage, because you can always reunite those subsets and restore the initial state.
It's different with your freshman example: the filters get translucent because, with the third filter, you choose a specific subset of photons that enters the last filter.
It doesn't matter whether you treat the problem deterministically to the end and let the wavefunction collapse then, or whether you let it collapse before the photons enter the last filter. The result is the same, there is no such thing as a collapse changing anything on the outcome. What you attribute to the "collapse" is simply the decision to let only a specific subset enter the last filter - and that you implicitly drop the other photons in an irreversible way. As long as they are around, you can still undo all the "measuring" and "collapsing".

17. Aug 21, 2009

### SWwright

OK, wait a minute. I have done this three-filters experiment (albeit with lenses from some cheap Polaroid sunglasses, which seem to have a limit on what they can filter; even with two crossed filters, I could see blue light from my flashlight -- I guess the shorter wavelengths pass through without being affected).

So when I did this, some of the light got through the first lens, and when the second lens was rotated 90 degrees from the first, no light got through (except the blue wavelengths that didn't get filtered, DO NOT try this with the sun, I'm sure the ultraviolet gets through the same way and you'll fry your eye. Use a flashlight like I did.)

So then I put a third lens in between the one at zero degrees and the one at 90 degrees, this third being rotated to 45 degrees, and now some light got through all three lenses! More than just the blue light. What AllanGoff said sounds about right, one-eighth of the total light from the flashlight.

So, I am not clear on what you are talking about when you say "selecting a subset." My 0-degrees lens selected some subset. OK. My 90-degrees lens selected from among what was left, and there wasn't anything left (except the blue light that wasn't affected anyway 'cause Polaroid lenses are cheap). Anyway: How can the 45-degrees lens select some SUBSET of what came through the first lens, so that that subset also gets through the 90-degrees lens? When ALL of what gets through the first lens, including that subset, gets blocked by the second lens when there's only the two?

Last edited: Aug 21, 2009
18. Aug 21, 2009

### Ich

I don't disagree with Allan about the outcome of the experiment. I just pointed out that it's completely irrelevant to which event out attribute the "collapse" of the wavefunction.
Your 90° lens selected from about half of the original photons those can be measured with 90° polarization. Exactly 0, call it destructive interference at this orientation.

The third lens removes some of the 0°-photons. Those were needed to make the destructive interference complete, without them some light gets through again.

With the usual caveats (works also with only one photon at a time and so on).

19. Aug 22, 2009

### Austin0

I thought that in this particular situation it was not a matter of destructive interference
but a question of obsorbtion or non-absorbtion.?????
A single photon can alter its path through interference but can it self destruct , ????

Interfence enters into it when considering the interaction of the photons with the structure of the material. Ie: the paths of some photons may be changed [through interference] along with their polarization orientation,, effecting the outcome as to absorbtion .

SO the first lens removes 50% of the photons but were they all 0$$^{o}$$ ???
And when they emerged were they all at the same orientaion or were they a spread within the range between 315$$^{o}$$ and 45$$^{o}$$
This range would still be totally absorbed by an orthogonal filter.
But suppose that an intermediate lens would eliminate a percentage of the first lens output
but what did pass would be spread through some part of the range between 0$$^{o}$$ and 90$$^{o}$$ in which case the final lens would pass part of this [between 45$$^{o}$$ and 90$$^{o}$$] and absorb part.[between 0$$^{o}$$ and 45$$^{o}$$]
I dont know enough about the actual physics of polarization filters and wave plates to put any quantitative guesses to the outcomes and this whole picture may be completely out of the park but this is how I came to a comfortable acceptance of the phenomenon and it didn't seem completely unreasonable.

Last edited: Aug 22, 2009
20. Aug 22, 2009

### SWwright

Ich,

I am confused by how you are using the word "interference". Based on my understanding of the word, if the lack of light passing through two Polaroid filters (one oriented at 0 degrees, the other at 90 degrees) was a result of interference, then I should have been able to move a bit closer or farther from the flashlight/filters assembly and see light again, as my eye moved out of the "minimum" area where no light is perceived. It should be a pure accident that I happened to be at the right distance for my eye to be in the minimum. What happens is, at any distance, no light passes through the two crossed Polaroid filters, some light passes through when a third filter oriented at 45 degrees is inserted between the 0-degree filter and the 90-degree filter (as always, ignoring the blue wavelengths that are too short to be affected by cheap Polaroid filters).

Am I completely misunderstanding what you said?

Last edited: Aug 22, 2009
21. Aug 22, 2009

### Ich

Scroll down a bit to http://en.wikipedia.org/wiki/Interference_(physics)#General_quantum_interference" in Wikipedia.

Mathematically, it's like vector addition. For example, if you split a vector (x,y)

$$\left(\begin{array} &1 \\0\end{array}\right) = \left(\begin{array}0.5\\-.5\end{array}\right) + \left(\begin{array}0.5\\+.5\end{array}\right)$$
(that's what the Stern-Gerlach setup does), you suddenly have a y-component where there was none before.
Your filters work like this, but additionally throw away one of the decomposed vectors. The 90° filter would extinguish all x-components, leaving nothing from the original vector. But if you insert the diagonal filter, it splits the vector just like in the equation, and kills the second one. The 90° filter will now let something through, namely the new y-component.
If you'd keep the second vector, send both through the 90° filter, and add them afterwards, they would canel again. That means that you can still undo the so-called measurement and collapse until you deliberately (or accidentally) throw away a part of the original information.

Last edited by a moderator: Apr 24, 2017
22. Aug 22, 2009

### SWwright

We are way far afield from my original question, but that's OK. Once I understand more of the nomenclature I mis-used, I will ask again correctly. In the mean time...

OK, you really are saying that the results I saw with those Polaroid lenses and a flashlight, were the result of interference, not a quantum-mechanical effect.

I have been trying to understand what happened in terms of polarization states that were undetermined, and conjugate bases. In those terms, the photons that got through the first (0-degree) filter had become determined in the 0/90 basis (and were completely undetermined in the 45/235 basis). Then when they encountered the second (90-degree) filter, they were all absorbed because their polarization was 90 degrees out of phase with the orientation of that filter. Finally, when I inserted the 45-degree filter in between, the photons that got through were determined in the 45/235 basis, and undetermined again in the 0/90 basis. And that is why some of them could get through the 90-degree filter.

Now, is the quantum mechanical description of those photons, mathematically equivalent to your interference description?

23. Aug 24, 2009

### Ich

No, I'm talking about the quantum mechanical effect that, after superposition, probabilities don't add. At least that's what I thing "Interference" means in that context, and wikipedia seems to use it similarily.

All right, you have to understand - in the context of this thread - that the filter does two things:
1. It projects the quantum state onto a new basis. In this example (my equation), the vector (1,0) is represented as a combination of the new basis vectors (.5,.5) + (.5,-.5). You could imagine that as a physical splitting of the beam, where the two new states follow different paths, like in the Stern-Gerlach experiment. That doesn't matter, still all the information is there. And still, if you recombine both new states again, either in front of or behind the last filter, the result is the same: no transmission.
2. It shuts off one of the new states, by letting it run into its lattice or whatever. So you send only one of the new states through the filter, which contains a y-component, and there is some transmission. So that's kind of the "measuring process", you drop some information, and there is transmission where none was before. Some would say the Wavefunction has collapsed at the diagonal filter, and you would like to somehow measure/prove/find out that it "really has collpased".

My point: If the diagonal filter, instead of removing and "destroying" one of the new states (you can't really destroy a state, just swamp it with a multitude of undefined other states), kept it in a secret drawer, you could retrieve it much later, combine it with the other new state, and undo that "collapse". No transmission again. Like a quantum eraser.
Thus you can't measure the collapse.