Simple example of the collapse of the wavefunction?

[Ulysees]

I'm not sure the double-slit experiment is one such example, maybe I have not understood it yet.

This experiment shows the wave nature of light due to the wavefunctions of photons. But how does it show a particle of zero size that is not just a burst of waves?

 

[andrewm]

First, do not be quick to assume photons have zero width, because otherwise the uncertainly principle would have \Delta x = 0, which is forbidden.

The collapse of the wavefunction can be imagined by detecting the photon as it passes through one slits. With the detector off, the photon is in a superposition of paths 1 and 2 and the experimenter observes an interference pattern. With the detector on, the photon travels either path 1 or path 2,and no interference pattern is observed.

 

[Ulysees]

The collapse of the wavefunction can be imagined by detecting the photon as it passes through one slits. With the detector off, the photon is in a superposition of paths 1 and 2 and the experimenter observes an interference pattern. With the detector on, the photon travels either path 1 or path 2,and no interference pattern is observed.

Ok, so here's what happens when bursts of waves come to the entrance of a naval port. Let's say the entrance is very narrow. Each burst comes in, diffuses, and ends up pretty much all over the place inside of the port. No interference pattern observed. This does not prove sea waves have a particle nature. It is just short bursts that produce no interference.

So where do I find an undisputable example of the collapse of the wavefunction?

 

[monish]

I'm quite sure the double-slit experiment with light is NOT a good example of the collapse of the wavefunction. At least it's not a simple example. The famous experiment where you observe the particle as it passes through the slits is actually a THOUGHT experiment proposed by Feynmann, and it uses electrons, not light. I don't think it's ever been done with electrons, and I don't think it really works with light because there is no way of detecting single photons as they pass through slits.

I'd really like to see what people suggest as a simple example of the "collapse of the wave function."

 

[monish]

Ok, so here's what happens when bursts of waves come to the entrance of a naval port. Let's say the entrance is very narrow. Each burst comes in, diffuses, and ends up pretty much all over the place inside of the port. No interference pattern observed. This does not prove sea waves have a particle nature. It is just short bursts that produce no interference.

So where do I find an undisputable example of the collapse of the wavefunction?

No, it's not because the burst is short that the wave ends up all over the place. It's because you have it passing through a single slit. A burst of waves through two slits will indeed show an interference pattern.

I agree with you that I'd like to see a good example of the collapse of the wavefunction.

 

[Ulysees]

No, it's not because the burst is short that the wave ends up all over the place.

You missed the point here, that it ends up all over the place is just additional information. The point is that the lack of interference does not prove particle-ness.

Can't believe no one noticed this.

 

[Ulysees]

First, do not be quick to assume photons have zero width, because otherwise the uncertainly principle would have \Delta x = 0, which is forbidden.

Sure when they're in their normal condition, the probabilistic wavy one where the uncertainty principle is defined. But when they collapse?

Do they collapse to something that has a size?

 

[andrewm]

I would just like to correct my \Delta x = 0 assertion:

In order to measure the position of a particle, one needs a probe (say a photon, if we use Compton scattering) of sufficient momentum. As the position measurement becomes more ideal (to the limit where we know exactly the particle's position and \Delta x = 0) one needs a photon of infinite momentum.

So the ideal position measurement is not forbidden, but it is strictly theoretical, as an infinite photon momentum implies an infinite photon energy. For all experimental measurements, the wavefunction does not collapse to a perfect position eigenfunction and so the resulting wavefunction has nonzero width.

A good discussion of this subject is given in Shankar, 2e, Section 4.2 "Collapse of the State Vector".

 

[andrewm]

I'm quite sure the double-slit experiment with light is NOT a good example of the collapse of the wavefunction. At least it's not a simple example. The famous experiment where you observe the particle as it passes through the slits is actually a THOUGHT experiment proposed by Feynmann, and it uses electrons, not light. I don't think it's ever been done with electrons, and I don't think it really works with light because there is no way of detecting single photons as they pass through slits.

I'd really like to see what people suggest as a simple example of the "collapse of the wave function."

I am not involved in quantum optics, but I am skeptical that there is impossibility in detecting single photons as they pass through slits.

See "Heralded Generation of Ultrafast Single Photons in Pure Quantum States" by Mosley in PRL.

 

[monish]

I am not involved in quantum optics, but I am skeptical that there is impossibility in detecting single photons as they pass through slits.

See "Heralded Generation of Ultrafast Single Photons in Pure Quantum States" by Mosley in PRL.

I can't comment on your reference except to note that it is a very modern one whereas the collapse of the wave function was something that bothered people in 1927. So I think if we are looking for a simple example of the phenomenon we should be going back to the source rather than arguing after the fact.

 

[Ulysees]

For all experimental measurements, the wavefunction does not collapse to a perfect position eigenfunction and so the resulting wavefunction has nonzero width.

So collapse is when the wavefunction becomes smaller in extent? (Due to a device that detects the entity).

I found this in wikipedia, not sure if it says anything useful about the collapse of the wavefunction:

Electrons hit a screen that detects where they hit. But first they pass through a double slit. So collapse is when they hit the screen, right? I've read somewhere a statement by R Feynman I think that no one really understands the collapse of the wavefunction. But here things look so simple, too simple to be true. Dots on the screen is where the wavefunction of each electron collapses. Is it just that?

 

[andrewm]

I think so, yes.

From my memory of Feynman's books, he may have been trying to express the fact that most physicists postulate that the wavefunction collapses after measurement. If we assume that it is true, we can derive many of the experimental predictions such as the uncertainty relation.

As for the actual clockwork that occurs when a wavefunction collapses, your guess is as good as anyone else's. I think that's what he was saying.

 

[monish]

So it has nothing to do with the double slits or the interference pattern? You could shine a weak light on a piece of photographic film and watch the dots appear one at a time, and that would be the collapse of the wave function?

 

[andrewm]

Yes, when the photographic film "measures" the position of the photon it collapses the particle's wavefunction.

There is a nice example of the "dots appearing one at a time" in the book Principles of Quantum Physics by French and Taylor. I do not have the book with me, but the figure is 2-2 (or 12-2, maybe?) showing a photograph that is "built up" over time by using a weak light source (10^5 photons or so).

 

[pmb_phy]

First, do not be quick to assume photons have zero width, because otherwise the uncertainly principle would have \Delta x = 0, which is forbidden.

I disagree. It's not forbidden. It just that you'd have \Delta p_x = infinity.

Pete

 

[andrewm]

I disagree. It's not forbidden. It just that you'd have \Delta p_x = infinity.

Pete

Yes, you're right. (I realized this in post #8)

 

[monish]

Yes, when the photographic film "measures" the position of the photon it collapses the particle's wavefunction.

There is a nice example of the "dots appearing one at a time" in the book Principles of Quantum Physics by French and Taylor. I do not have the book with me, but the figure is 2-2 (or 12-2, maybe?) showing a photograph that is "built up" over time by using a weak light source (10^5 photons or so).

The appearance of dots doesn't mean the wave function has collapsed. The molecules on the photographic film (silver nitrate??) are presumably happier after they change color, and it is perfectly reasonable to think that they only need a bit of a quiver from the light wave to induce them to change states. To show the collapse of the wave function you would need one more step: you'd need to show that once a dot has appeared somewhere, that no other dots appear elsewhere at the same time...in other words, that the same wave can't stimulate the molecular transition in two places at once.

What is the evidence to show that it works this way?

 

[Ulysees]

So exactly what is going on at the dot? Surely the electron is not just observed, it is absorbed, it stops existing. So collapse requires that the electron goes out of existence?

And reading Monish's post, I'd ask the same question he asks.

 

[muppet]

The appearance of dots doesn't mean the wave function has collapsed. The molecules on the photographic film (silver nitrate??) are presumably happier after they change color, and it is perfectly reasonable to think that they only need a bit of a quiver from the light wave to induce them to change states. To show the collapse of the wave function you would need one more step: you'd need to show that once a dot has appeared somewhere, that no other dots appear elsewhere at the same time...in other words, that the same wave can't stimulate the molecular transition in two places at once.

What is the evidence to show that it works this way?

Well, you can use a source with a sufficiently low intensity that only one particle is emitted at a time, and observe that only one dot is produced. In the images of the two slit experiments in which the diffraction pattern has been built up from a large number of individual dots, this is what has been done!

So exactly what is going on at the dot? Surely the electron is not just observed, it is absorbed, it stops existing. So collapse requires that the electron goes out of existence?

To be absorbed is not necessarily to stop existing. A photon ceases to exist when it impinges upon a photographic plate, and its energy is used to promote electrons in a silver halide crystal into a different energy state. But what I suspect happens with an electron is that it ceases to propagate as a free particle, but occupies an empty state in a conduction band or similar. I don't know much about the specifics of detectors, so you'd have to ask Zapper if you're really fascinated by the details.

 

[Ulysees]

what I suspect happens with an electron is that it ceases to propagate as a free particle, but occupies an empty state in a conduction band or similar.

So it's not collapse then is it? Looks like the wide wavefunction becomes much narrower as it is constrained to orbit an atom. But if it becomes a free electron in a metal, does this count as collapse of the wavefunction?

 

[Ulysees]

Isn't collapse something much more fundamental than just a change of shape?

The shape changes anyway in many ways according to Schrödinger's equation, so what's the big deal with collapse that makes it hard to understand, what's counter-intuitive about a probability wave that changes shape? Collapse must be something different from just changing shape.

Does anyone understand this statement in wikipedia?

"When an external agency measures the observable associated with the eigenbasis then the state of the wave function changes from |psi> to just one of the |i>'s with Born probability |ψi|^2. This is called collapse because all the other terms in the expansion of the wave function have vanished or collapsed into nothing."

http://en.wikipedia.org/wiki/Wavefunction_collapse

 

[monish]

Well, you can use a source with a sufficiently low intensity that only one particle is emitted at a time, and observe that only one dot is produced. In the images of the two slit experiments in which the diffraction pattern has been built up from a large number of individual dots, this is what has been done!


To be absorbed is not necessarily to stop existing. A photon ceases to exist when it impinges upon a photographic plate, and its energy is used to promote electrons in a silver halide crystal into a different energy state. But what I suspect happens with an electron is that it ceases to propagate as a free particle, but occupies an empty state in a conduction band or similar. I don't know much about the specifics of detectors, so you'd have to ask Zapper if you're really fascinated by the details.

The specifics of the detector become pretty important when you're trying to understand these things. You suggest that the energy of the light promotes electrons to a higher energy state. I think it's actually the other way around. Under your hypothesis, the light energy would have to be sufficiently concentrated to activate the chemical transition; therefore you need the "collapse" (concentration) of the e-m wave to explain the dots. Under my explanation the molecules are in a metastable state, ready to transition (with diminishing probability) at the slightest stimulation. So dots will appear here and there even if the field is very weak, because they don't need to absorb the full quantum of energy.

So if the photographic plate works the way I think it does, then your idea of making the light source very weak doesn't show the collapse of the wave function because my alternative explanation gives the same qualitative results.

 

[muppet]

I think you may misunderstand what the term "collapse of the wavefunction" means.
Prior to an act of measurement, a particle is described by a linear combination of eigenfunctions of any particular operator. The superposition of these functions is what leads to interference effects. But when a measurement of a property is made and the particle is known to be in a definite place or posess some definite momentum, it can no longer be described as a linear combination of eigenfunctions of the operator corresponding to that observable. It doesn't mean that it's no longer described by any kind of eigenfunction which may or may not posess wavelike properties. I admit to not knowing how the detection process works specifically. But if you see a localised dot then you know that the electron has had an interaction there, so it's not in a superposition of being spread over the face of the detector, or going through both slits at once.

 

[muppet]

The specifics of the detector become pretty important when you're trying to understand these things. You suggest that the energy of the light promotes electrons to a higher energy state. I think it's actually the other way around. Under your hypothesis, the light energy would have to be sufficiently concentrated to activate the chemical transition; therefore you need the "collapse" (concentration) of the e-m wave to explain the dots. Under my explanation the molecules are in a metastable state, ready to transition (with diminishing probability) at the slightest stimulation. So dots will appear here and there even if the field is very weak, because they don't need to absorb the full quantum of energy.

So if the photographic plate works the way I think it does, then your idea of making the light source very weak doesn't show the collapse of the wave function because my alternative explanation gives the same qualitative results.

http://en.wikipedia.org/wiki/Silver_halide is whence I got such explanation as I have offered (silver halides being the key ingredient in photographic plates).
Can I check at this point: how much do you know about 1) quantum mechanics 2) the photoelectric effect?

 

[monish]

http://en.wikipedia.org/wiki/Silver_halide is whence I got such explanation as I have offered (silver halides being the key ingredient in photographic plates).

I looked in your reference and it doesn't explicitly say whether the Gibbs Free Energy of the transition is positive or negative. In the latter case the crystals are metastable and only need a slight disturbance to make the transition. So you can't conclude that the energy of the light needed to be concentrated in order to make a dot.

Can I check at this point: how much do you know about 1) quantum mechanics 2) the photoelectric effect?

What's your point?

 

[muppet]

My point is that I'm skeptical about the idea of "not needing to absorb the full quantum of energy"; the electrons don't have a choice about absorbing the full quantum of energy, overkill or otherwise; surplus is re-radiated by another photon. Also, if you have a plate of atoms in a metastable state and you have a non-localised source of sufficient energy to disturb this happy arrangement, then you don't get a localised dot- you'd get a collapse over the entirewavefront.

 

[monish]

My point is that I'm skeptical about the idea of "not needing to absorb the full quantum of energy"; the electrons don't have a choice about absorbing the full quantum of energy, overkill or otherwise; surplus is re-radiated by another photon. Also, if you have a plate of atoms in a metastable state and you have a non-localised source of sufficient energy to disturb this happy arrangement, then you don't get a localised dot- you'd get a collapse over the entirewavefront.

If I'm reading between the lines, you're interpreting the camera as working based on a type of photo-electric effect, where the film absorbs energy. I used to think it worked this way too, but one day it occurred to me that maybe it didn't. And the clue to me was what you said about re-radiating excess energy: I don't think that would actually work. If the crystal was actually absorbing a specific transition, then I would expect it to be really frequency-selective. Which it doesn't seem to be. (I have no idea how color film works).

If it comes down to a question of how a camera works, then we ought to be able to settle it one way or another. You're saying the metastable system wouldn't work because one hit would destabilize the entire arrangement; I don't know if that's necessarily true.

The funny thing about our opposite theories of film is that in both cases there is a metastable state: in your theory the EXPOSED film must be metastable because it absorbed energy to make the transition, and it will want to go back. So how does old film go bad? If you take a picture and leave it in the camera for ten years, does it gradually white over or does it just return to black, ready to take another picture?

 

[fanchini]

In the recent years, with advanced experiments, it is possible to create single photons (in avarage) with a very small light intensity. It is possible, thus, to observe dots appearing one at time over a photographic film, and in fact, it represents the wave function collapse. Before the dot emergence, the probability to find the photon in a especific position is given by the wave function and, in this case, no deterministic photon position is capable.

 

[monish]

The original poster asked for a "simple example" of the collapse of the wave function. I have to object to any attempt to settle this question by invoking advanced experiments that have been developed only in recent years. The "collapse of the wave function" was something that bothered the hell out of people in 1927, and I think it makes a lot of sense to look for the kind of examples that existed back then.

Also, there is still the open question of which state of silver chloride is metastable: the exposed or unexposed molecule? I hope someone can answer this. It is relevant to the question as to whether the mere appearance of dots on a weakly illuminated photographic plate might constitute evidence of wave function collapse.

 

[Ulysees]

Why are the dots so much bigger than an atom, if they indicate collapse of the wavefunction of a photon?