# Why don't the Slits collapse the wave function?

1. Sep 11, 2006

### kliide

I found a layman's explanation of the wave characteristics of subatomic particles in the form of a "Dr.Quantum" video from "What the Bleep do we know?". Aside from the parapsychological junk in the last 2/3rds of the movie, the explanations of quantum properties seemed mostly accurate and concise. The way they described subatomic behavior was easy to digest from a layman's perspective.

One thing I don't understand is how the screen with the slits do not collapse the wave function until there is a measurement. Wouldn't bouncing off the sides of the slit constitute some sort of interaction?

What is it about the "observer" or measurement device that would interact differently with the photon than how the photon interacts with the slit?

2. Sep 11, 2006

### ZapperZ

Staff Emeritus
The issue here isn't the slit. The issue of importance here is the PATH, and that there are two paths that it can go through. So the superposition is the path it took through the slit. I can easily do the same type of experiment using a superconductor having a split path - you get the same type of "interference" pattern. This is what you get from Superconducting Quantum Interference Devices (SQUIDs).

Zz.

3. Sep 12, 2006

### kliide

Thanks for the quick reply. After some mulling over your response, I think I asked the wrong question, but it's been sorted out now and I appreciate the help.

Last edited: Sep 12, 2006
4. Sep 12, 2006

### J77

I hope I get this right, I to am an amateur but quantum enthusiast...

The wave is like a corkscrew, representing the momentum.

When this goes through the slits you have two corkscrews upon which changes of the phase give the differing interference patterns.

If you measure the position, you lose the corkscrew - the Fourier transform of which is the position: a Dirac delta function.

Therefore, you lose the interference pattern.

Someone please correct me if wrong - I'm only recalling what I've read recently in Penrose's books.

5. Sep 12, 2006

### jpr0

I think the question he is asking is why interaction with the slit doesn't constitute a measurement, whereas interaction with some kind of "detection" device does. What's the difference between interacting with the slit and the thing we use to measure the electron?
The double slits can be anything - they can be two impurities in parallel from which the electrons scatter, in which case the above question looks more reasonable.

I think the answer is because the slit (or impurities) has no internal degrees of freedom, so as the electron passes through the slit arrangement there is no randomization of its phase, and at the detector the parts of the wave function arriving from both paths have a definite phase relationship, which produces interference.

If you give the impurity (or slit) some internal degree of freedom, so that it can change its state upon interaction with the electron, then you will (may...) randomize the phase of the electron, and there will be no definite phase relationship between the two paths.

This is explained in a nice way in the book by Datta, "Electron transport in mesoscopic systems" (the experiment he talks about is an electron in an Ahronov-Bohm ring). The point is that if the impurity, or slit, has a internal degrees of freedom, then it's possible, by measuring the state of the impurity, to tell which path the electron took. Hence any interference would be destroyed.

This might be incorrect in some way, but it's what I understand from what I've read.

6. Sep 12, 2006

### ZapperZ

Staff Emeritus
The "slit" in the idealized case is simply an illustration of separate paths for the photon, electron, neutron, buckyball, etc. As I've said, that isn't the real issue. Now, if the slit happens to be a metallic device that can somehow detect things like electric field of an electron or a photon, THEN it is now a detector that can tell you if a photon, electron, or whatever, passed through it. This is now a different set up. In that case, yes, the slit will cause an interaction that is now part of the system to be considered.

Zz.

7. Sep 12, 2006

### jpr0

Hi,
I agree that the slits are there only to spatially seperate trajectories, and highlight the effect of interference, which is fine. But the question he was asking was

'What is it about the "observer" or measurement device that would interact differently with the photon than how the photon interacts with the slit?'

The answer here is that one destroys phase coherence and the other does not.

8. Sep 12, 2006

Staff Emeritus

I like this answer, but could you or someone else flesh it out a bit showing how the different actions on the wave function look?

9. Sep 12, 2006

### masudr

It's just like how you need phase coherence for there to be any useful interference. The reason for this is that to get Fraunhofer diffraction, we require the phase difference between two bits of wave to be a linear function of position, otherwise the Fourier integral doesn't give a nice answer.

10. Sep 15, 2006

### blackwizard

Bloody hell, its hard 2 figure out wat ye guys r talkin bout sometimes! Internal degrees of freedom?

Im just gonna ask if my view is right or not. Which is double slit does collapse the wavefunction. When the wavefunction hits the double slit plate it can collaspe to a point on the plate (particle hits the plate itself) or to the area covered by the two slits (particle passes through the slits). So in the second possibility only the part of the wavefunction that hit the plate collapsed and the remaining part, that emerges the other side is the shape of the slits. Then the 2 parts of the wavefunction (slit 1 & 2) spread out in2 each other to cause the interferance.

11. Sep 15, 2006

### ZapperZ

Staff Emeritus

Now was that simple enough?

Zz.

12. Sep 15, 2006

### blackwizard

13. Sep 15, 2006

### ZapperZ

Staff Emeritus
Tell me how to possibly do that without using "big words".

You freely used the terms "wavefunction" and "collapse" as IF they were "simple" words, where in reality, there's a LOT of physics that comes with such terminology. Those are not simply words that can be used weely neely. Everything in physics has CLEAR and unambiguous definition with underlying mathematical description.

I have no ability to explain that without using "big words", or without refering to physics papers that have been cited many times on here.

Zz.

14. Sep 16, 2006

### Chronos

OK, I'm tempted beyond restraint to toss a coin into the fountain. The wavefunction, in quantum speak, is a mathematical construct that explains 'how' a photon, electron or other quantum entity 'chooses' the path it takes to the target. The concept is perfectly logical when you accept it is a probability, not a deterministic prediction. Quantum entities, by definition, are fuzzy to begin with [the uncertainty principle thing], so it is not [at least to me] very surprising their paths also appear to be fuzzy. In the very unintuitive realm of quantum spacetime, photons, et. al., are free to be in more than one place at the same time - re: shooting electrons through a double slit one at a time.

15. Sep 18, 2006

### blackwizard

Ah, well that does make things clearer, thanks. If i picked you up right that is. I was tryin 2 understand wavefunctions as physical entities. Probly 2 used 2 bein able to visualise physics.

No idea, just hopin. Only started 2 learn bout Theoretical Physics a month ago

16. Sep 18, 2006

### Careful

Hi,

The question is not silly at all There is nothing in the theoretical construct of quantum mechanics which tells you when and how measurement occurs; at least not in those interpretations where the wave function is imagined to collapse such as in the copenhagen interpretation. As a rule of thumb you might use the reduction rule when a macroscopic apparatus is put in front of the setup (but in this case you are partially right, even the plate with the two slits will reflect a large part of the wavefunction and fringe effects at the slits will occur). But then again, you have macroscopic samples which might occur in superposition and claims exist that this superposition of macroscopic states can be observed (in other words, observation is not going to disturb the system too much)! Measurement is a tricky business and one should think carefully about it.

Careful

17. Sep 18, 2006

Staff Emeritus
I note that in http://realityconditions.blogspot.com/2006/09/on-price-and-penrose-on-time-asymmetry_18.html (which I have also posted about in the Philosophy of Science and Math Forum), it is claimed that all the cases where the wave function is supposed to collapse occur in statistical contexts where entropic time asymmetry is to be expected. The author cites decoherence in justification, and notes that there is no asymmetry when two electrons interact by exchanging a (virtual) photon.

What does anybody think of this idea? Would it have any bearing on the OP question?

18. Sep 19, 2006

### Chronos

I think this is a very good idea. A virtual photon represents the average of all possible superposition states that can be shared between the interacting electrons. Collapsing the wavefunction of one of the interacting electrons automatically collapses the other, the way I interpret matters.

19. Sep 19, 2006

### Careful

Hi, my first reaction would be a hmmmm, entropy of what ?'' For example, lets do a double slit experiment with electrons. The electron wave function will slow down in front of the plate with the slits resulting in electromagnetic radiation being created at uncertain times. Assuming the plate to be a perfect reflector, the entropy'' of the localized'' state particle + EM field goes down since radiation escapes (the expectation values of the Von Neumann entropy operator : entropy particle + EM field + universe should'' remain the same) --> measurement ?! I did not calculate this, but you might want to check if this radiation can also effect that part of the electron wave function which does not slow down and goes through the slit (it seems so to me). This also shows that only that part of it which is involved in strong'' interactions can be measured. However, it seems to me that the decoherence argument is far from sufficient either since (a) it does not help you with closed systems (b) there is a debate about the ontology of the density matrix (c) with a density matrix you can only compute statistical expectation values and you do not arrive at a single event interpretation in any case (hence it is FAPP). In case of our reflecting plate, the way out of course is that reflection is not constraining the wave package in any way. Anyway, measurement only occurs when the spatial wave function can be written as a sum over disjoint localized wave packages which (a) remain localized AND (b) more or less disjoint for a sufficient amount of time (the position basis is physically preferred). That is, the de Broglie current = more or less sum over currents. This explains why the reflected, nor the other part are measured when passing the double slit plate. This is different from the decoherence interpretation, I do not need to trace out the environment, I need to study the time evolution of the different spatial parts of the particle wave though! So, one should study in detail how the interaction with the final plate is going to decohere the different spatial parts of the interfering wave prior to measurement, I guess de Broglie has said something similar already in the 1920 ties. I remember disagreeing with Penrose too about this *particular* example of his when reading the book; I have more sympathy for the thoughts of Price (which I recommended already a few times here before on this site) though. Anyway, I can only repeat myself when saying make a decent theory of single events !!''.

Careful

Last edited: Sep 19, 2006
20. Sep 19, 2006

Staff Emeritus
Reading your argument Careful I am a little puzzled, you do not seem at all to have coupled with the author's point that the slits and the screen are both macroscopic objects, and you just can't assume in the real world that "the plate is a perfect reflector". The formalism you cite was all developed with just such toy models and the author claims that that, ignoring the unavoidable coupling to the statistical environment in real interactions is the reason that the formalism has the awkward time asymmetry appearing.

21. Sep 19, 2006

### Careful

Excuse me ? :grumpy: I treated this point with due care and explained how one can avoid these presumed difficulties, it seems you did not get that at all. And of course you can assume the plate with the slits to be a perfect conductor, that will do just fine for all practical purposes.

** The formalism you cite was all developed with just such toy models and the author claims that that, ignoring the unavoidable coupling to the statistical environment in real interactions is the reason that the formalism has the awkward time asymmetry appearing **

I explicitely commented upon the decoherence interpretation which is a working tool and is for example useless in the context of closed quantum systems. I even specifically referred to the latter again in the context of a de Broglie like interpretation which is actually superior to decoherence and does not need a conscious choice'' to one of the worlds at all - and no time asymmetry is involved here either. Moreover, the effective dynamics in the decoherence interpretation is not unitary, so no wonder you can make of an AND an (approximate) OR. Actually, although the entropy in the decoherence interpretation increases during the loss of coherence process, it decreases again due to Poincare recurrence times. Moreover, selection of one macroscopic possibility (that is measurement) is decreasing entropy (at least from this point of view, in the Copenhagen interpretation entropy remains constant after renormalization).

For the rest, I am commenting upon the question posed in this tread from my own point of view - there is no reason why I should agree (which you call to couple'') with what is written by someone else.

Careful

Last edited: Sep 19, 2006
22. Sep 19, 2006

### Hans de Vries

I have been playing with the following argument why there should be
interaction between the photon and the (two) slits even if the photon
doesn't "collapse" there.

Imagine the experiment floating in zero-g: The screen + arm holding the
plate with the two slits. The photon hitting the screen will give the setup
an arbitrary angular momentum depending on where it hits the screen...

Now this is not allowed. (Say we are shooting the photon along the
center of gravity line). Angular momentum should be conserved. The
only other place where momentum transfer could take place is, yes,
at the two slits. This is where the photon was "bend", but then, it
doesn't know it is bend because it hasn't collapsed yet at the screen.

The two interactions (at screen and slits) together would conserve
angular momentum, but the interaction at the slits is at odds with
many interpretations. The idea however is ofcourse to examine how
well the interpretations hold up...

Regards, Hans

Last edited: Sep 19, 2006
23. Sep 20, 2006

### DrChinese

When the setup is to detect a photon passing 2 slits: in the path integral view, you consider all possible ways that the photon could arrive there. Those possibilities give rise to interference, and a pattern results.

That does not mean that every photon emitted from the source made such a journey. And those "other photons" would have the property of having had their wave function collapsed elsewhere, and would not contribute to the interference pattern observed. So in a way, the slits could also act as the "observer" if a source photon was absorbed there and not re-emitted. This is simply a relatively less likely outcome.

In sum: The interference pattern represents the subset of photons that went through the slits. In effect, the detection mechanism is designed to be a fertile target for the source photons and the rest of the apparatus is not. That is the only distinction.

24. Sep 22, 2006

### nrqed

But that begs an obvious question. Any physical slit is composed of atoms. So why can we say that if there is a metallic device (i.e. a macroscopic measuring device) then we now have a detector and the interference patterm will disappear, and yet say that if there is no metallic device, there is no collapse of the wavefunction, even though the atoms in the slits should in principle be "detecting" whethere an electron passed or not. So what's the difference? Of course, in the second case the information is unavailable to me (at the macroscopic scale). But in the end, any actual measuring device is made of atoms, so at what point can we say that there is a collapse? When the atoms are arranged in just a specific way as to make a recording possible? What is the exact rule that dictates when a collapse will occur and when it won't?

Maybe I am being too dense but that is really not clear to me.

Cheers

25. Sep 22, 2006

### ueit

I've had this idea some time ago and I've even debated it on iidb forums. This is the first time I've seen it presented by someone else and this gives me hope that I'm up to something.
I've found yet another problem with this experiment. Two assumptions are made:

1) the wall does not interact with the particles
2) the wall is perfectly opaque.

It is easy to see that these assumptions are logically incompatible.

I think that the whole "collapse" thing is misleading. The wave function is a mathematical abstraction and that's why it can be non-local. We need an explanation in terms of what is real, a local-realistic model. I think it is pretty obvious that the particles interact at the slits and the real question is how this interaction produces the interference patterns. "When the wave collapses" is a non-question on par with "when the edges of a cube produce the volume inside the cube?".