# Collapse of a wave function

1. Oct 26, 2012

### hover

Here is a question I would like answered. The wave function I am going to use for this example is the ground state of the infinite well but I assume the outcome to this problem will apply to any wave function.

Ground State:
ψ = √(2/a)*sin(∏nx/a) , where 0 < x < a

Lets say that I know a particle has this particular wave function. I wish to try to measure where the particle is. Lets say I stick in a device that can detect the particle between a/4 and 3a/4. Lets say that for the first experiment, I detect the particle. What will the resulting wave function look like or what will it collapse to? Lets say during the second experiment, I don't detect the particle. What will the resulting wave function look like or what will it collapse to?

What I am trying to get at is what wave functions look like in both cases. I'm very curious because I have never been told what the wave function is suppose to look like after it collapses. I hope I am making sense.

Thanks

2. Oct 27, 2012

### andrien

Collapse of a wave function means that there is no wave function defining it after you have made your measurements.

3. Oct 27, 2012

### Alesak

According to my textbook after obtaining some eigenvalue in measurement the state vector should collapse to corresponding eigenstate. So upon measuring position it should be dirac delta function.... wierd, I know.

4. Oct 27, 2012

### Ilmrak

I think the OP question is a bit more difficult to answer.

If I understood well what he's asking, he wants to know what the wavefunction would be after a "partial" measurement of position, i.e his experimental apparatus can only say whether the particle is in some finite space region or not.

I think that the only answer one could give without knowing the exact interaction between the experimental device and the particle measured is that the final wave function, if you found the particle in your detector, is some linear combination of those position eighenstates $|x \rangle$ with $\frac{a}{4} < x < \frac{3}{4}a$.

Obviously if the interval of positions that you are capable to measure is "very small" the wavefunction is "very similar" to a Dirac's delta, but this is actually true only in the limit of an ideal measurement of a single position eigenstate.

Ilm

5. Oct 27, 2012

### RedSonja

If the wavefunction is interpreted as a (time-evolving) probability distribution for the position of the particle, then immidiately after the measurement the wavefunction is restricted to the state corresponding to the measured eigenvalue. The shape of this wavefunction will depend strongly on how the measurement is performed. After a while the state will again be an eigenstate of the well.

6. Oct 27, 2012

### Alesak

You are right, I missed this subtlety.

There is also a nice article on wiki on wavefunction collapse. Some interpretations apparently doesn't require a collapse.

7. Oct 27, 2012

### Fredrik

Staff Emeritus
The first experiment proves that the particle certainly isn't outside the region from a/4 to 3a/4, so if the measuring device hasn't absorbed the particle, it has left it in a state such that the probability that an immediate position measurement will find the particle outside that region is 0. If it's a state that can be represented by a wavefunction at all, it will be 0 outside that region. Without more information about what exactly the measurement does, we can't know the values of the wavefunction inside that region.

Suppose e.g. that the device has a 50% chance of leaving the system in a state represented by a wavefunction $\psi_1$ and a 50% chance of leaving it in a state represented by $\psi_2$. This is the type of situation where the new state can't be represented by a wavefunction at all. You would have to use a density matrix.

8. Oct 27, 2012

### Staff: Mentor

The collapsed wavefunction will be a superposition of different eigenstates (actually, every physical wavefunction can be expressed as one), and they evolve independently in time afterwards.
Unless you add some external interaction, it will stay in a superposition of different eigenstates.

Which is very fortunate. They don't need (non-local,) non-unitary processes with an ad-hoc introduction of a process of unclear physical meaning.

9. Oct 27, 2012

### hover

There are some interesting responses here. Some of you say that the wavefunction will just turn into a Dirac Delta function. Others disagree and say that it is more complex than that. I wanted to see what people would come up with before I reveal what I thought would happen.

The way I pictured the situation was that if the particle was detected within a/4 < x < 3a/4, then the wavefunction would just move its nodes originally at x = 0 and x = a to x = a/4 and x = 3a/4. In other words, the wavefunction would shrink to fit within the new bounds.

In the second experiment where the particle isn't detected, I imagined that you would have two smaller versions of the original wavefunction. One in the bound of 0 < x < a/4 and one in the bound of 3a/4 < x < a. You could think of it as splitting the wavefunction.

I hope what I said makes sense. I certainly don't know if it is right so that is why I asked my original question. Since my answer didn't appear, I'm guessing I'm not right. Come to think of it, I don't know if it would be that simple. I mean if you know more about the position of the particle, don't you know less about its momentum and hence wavelength? If that is the case then I'm not sure my answer is correct since my answer has sine waves which you can calculate the exact wavelength to.

10. Oct 28, 2012

### Fredrik

Staff Emeritus
You only get a delta function (or more realistically, an actual function with one tall and narrow peak) if the particle is detected by a very small detector within that region. If a single detector fills up the entire region from a/4 to 3a/4, and doesn't create any records of information about how the particle was found in a smaller subregion, then you're definitely not getting a delta function.

The answer you suggested yourself is something I thought about, but I couldn't figure out what the properties of the detector would have to be to ensure that the new state would be a wavefunction with its peaks in the region at the same places as the old one. Your own counterargument about momentum sounds valid, and there's also the fact that a wavefunction must be continuous (because it's a solution of the Schrödinger equation and therefore differentiable), so ψ(x)→0 as x→a/4+ and as x→3a/4-. (The + and the - mean "from the right" and "from the left" respectively).

Edit: About the momentum/energy argument. What we have done is to make a measurement on a particle in a box that effectively shrinks the box (temporarily). It's still possible (in principle) to have the particle end up in a state such that the restriction of the wavefunction to the smaller region is an eigenstate of the Hamiltonian for the smaller box. Then we would have a "well-defined wavelength" within the region. So your argument doesn't seem to rule out that. However, this would be a shorter wavelength, so the peaks will typically be in different places. (The ground state is an exception because it has only one peak right at the center, but if you move the boundaries of the smaller region, not even the ground state will have its peak at the same place)

I'm not sure what this means. For example, is it likely that the particle will end up with a wavefunction that in the three relevant regions is 0, equal to the ground state for the small box, and 0, just because we started with the ground state for the larger box? All I can say is that it will depend on how the particle interacts with the detector. I think the fact that it does interact with the detector suggests that its state will change significantly.

Last edited: Oct 28, 2012
11. Oct 28, 2012

### hover

The fact that I don't have a clear picture of what the final wavefunction will look like makes me determined to figure this out. Fredrik, you say that the resulting wavefunction will depend on the nature of how the particle will interact with the detector itself. Lets create a detector. How about we use good old photons to detect the particle? A simple way to detect if the particle was inbetween the bounds is to fire photons from x = a/4 to the right and if there was a detector at both x = a/4 and x = 3a/4. If the particle is inbetween the bounds then we should expect a photon fired from x = a/4 to return with less energy. If we detect the photon with the same energy at x = 3a/4 then the particle isn't between the bounds.

If I remember correctly, I believe that if you want to measure the position of a particle accurately using photons then the photons must have a small wavelength. Of course, a smaller wavelength means that you put more energy into the particle. Putting more energy into the particle will cause the particle to change into a different eigenstate(at the very least, it won't be the ground state).

In the first experiment, since you are increasing the eigenstate, the "particle" will actually look more like a classical particle since you would be increasing the number of waves between a/4 < x < 3a/4. I'm not exactly sure what the wave function will look like between the other regions 0 < x < a/4 and 3a/4 < x < a. A part of me wants to say that the wavefunction in those regions is zero but I'm not quite sure.

In the second experiment, if we fire a photon at x = a/4 and detect the same photon at x=3a/4, then I feel that we have something that looks like a ground state at 0 < x < a/4 and 3a/4 < x < a with the wave function at zero between a/4 < x < 3a/4.

Does what I said make logical sense??

The more I think about this, the more I almost want to treat this problem as a double delta potential when the detectors are present but who knows if that is right...