# 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.

Please dont use big words!

11. Sep 15, 2006

### ZapperZ

Staff Emeritus
Your view is wrong.

Now was that simple enough?

Zz.

12. Sep 15, 2006

### blackwizard

So your not gonna let me know wats wrong about it?

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