QM: Measurement & Wave Function Change

[Silviu]

<Moderator's note: two essentially very similar threads merged.>

Hello! I am a bit confused about the idea of measurement in QM. As far as I understand, if you measure the position of a particle, the wavefunction of that particle changes into a delta function, and thus the particle gets localized. Now, let's say we have a particle in a box in a state with 3 main peaks ($\psi_3$). If we look at the main peak let's say (so the center of the box) and see the particle, the wave function collapsed at that point. But what happens if we don't see it there? Obviously, the wavefunction changes, as we know for sure that the probability of the particle being at that point is 0 now, but how is the wave function changed? Does it turn into a delta function at a random point, different from the one where we measured, or what?

 

[Silviu]

As far as I understood, in real world the wave function of a particle can have non-zero probability at any point in space. So if 2 observers at 2 ends of the universe observe this particle and one of them sees it (so the wave function collapses at his location) what happens to the other observer? Does the information within the wavefunction travels at an infinit speed (i.e. the value of the wavefuntion at the second observer gets instantaneously 0?)?

 

[Jilang]

Now just how are you planning to enter that box?

 

[Silviu]

Now just how are you planning to enter that box?

That was just an example. My question was what happens to the wave function if you don't see the particle at a point where the probability to find it was different from zero, before measurement.

 

[Jilang]

The wavefunction will evolve from that starting point and quickly settle down again.

 

[Silviu]

The wavefunction will evolve from that starting point and quickly settle down again.

But what is that starting point? Like assuming that before the measurement, the probability of finding the particle at position $x_0$ was 1/4. After the measurement, as the observer didn't see the particle, the probability goes to 0 (right?). However the wavefunction still needs to be continuous, so yeah, it gets equal to 0 at $x_0$, but how does it behave at all the other points, what shape does it take?

 

[Orodruin]

Measurements and decoherence are generally complicated issues. Using some simplification, the state is projected onto the subspace of states compatible with your measurement.

 

[Silviu]

Measurements and decoherence are generally complicated issues. Using some simplification, the state is projected onto the subspace of states compatible with your measurement.

What about the speed at which the wave function collapses?

 

[Jilang]

But what is that starting point? Like assuming that before the measurement, the probability of finding the particle at position $x_0$ was 1/4. After the measurement, as the observer didn't see the particle, the probability goes to 0 (right?). However the wavefunction still needs to be continuous, so yeah, it gets equal to 0 at $x_0$, but how does it behave at all the other points, what shape does it take?

I would think it is the same as before with a little sliver missing.

 

[Silviu]

I would think it is the same as before with a little sliver missing.

Well I think so, too. But how do u express it mathematically? Like when u see it, it becomes a delta function. But when you don't see it, how do you describe it mathematically. So in the example I gave with 3 peaks in a particle in a box (assuming you can get inside the box :D), if I measure at the center of the box and I don't see anything and right after that I measure somewhere random in the first 1/3 of the box. Initially I had a chance of 1/3 to find the particle there, but now, after I measure the center without seeing anything, what are the chances to find the particle in the first 1/3 of the box (I assume it can't be also 1/3, otherwise it would mean that my first measurement had no influence on the wave function).

 

[Jilang]

I would go with the initial wavefunction less the delta function.

 

[Silviu]

I would go with the initial wavefunction less the delta function.

What do you mean by "less the delta function"? Something like $\psi(x) - \delta(x_0)$?

 

[Jilang]

Yes.

 

[Silviu]

Yes.

But this is wrong. First, if you do the integral over space, you don't get 1 as the integral of $\psi$ is 1 (as it is the same function as before, except for 1 point), but the integral of delta function is also 1, so you get 0 in the end (let me know if there is something wrong about this reason). Secondly if you take the limit of $\psi(x)$ as $x \to x_0$, one can see that the function is not continuous at $x_0$, but it should be even after the measurement. So I am pretty sure this is not the mathematical form of the wavefunction after the measurement

 

[Jilang]

Agreed, If the amplitude reduces in a certain region it would need to increase elsewhere to be normalised. Why do you think the wavefunction is then discontinuous?

 

[Silviu]

Agreed, If the amplitude reduces in a certain region it would need to increase elsewhere to be normalised.

Yes! This is what I am asking. Where and why is that part of the wavefunction (which is now 0 for sure) distributed?

 

[Jilang]

Yes! This is what I am asking. Where and why is that part of the wavefunction (which is now 0 for sure) distributed?

All over, it all increases a bit.

 

[Silviu]

All over, it all increases a bit.

No! This is not possible, because of the continuity. At the points around the point where we measured, the value of the wavefunction must decrease, otherwise we have a discontinuity at $x_0$, so it can't increases all over the place. So there are points where it goes up and points where it goes down, but again, how do you express it mathematically?

 

[Jilang]

You will appreciate that the wavefunction would evolve as expected (by the equations) only if the system is not tampered with?

 

[Silviu]

You will appreciate that the wavefunction would evolve as expected (by the equations) only if the system is not tampered with?

I am not sure I understand your question.

 

[Jilang]

I wondered if you were worried about the discontinuity in the time evolution.

 

[Silviu]

I wondered if you were worried about the discontinuity in the time evolution.

I am talking about the space discontinuity. Assuming the box has length 1, after our measurement we have $\psi(0.5)=0$. So if immediately after this I measure let's say $\psi(0.501)$, what value do I get? Of course it must be very close to 0, but how can I calculate it?

 

[Jilang]

Uncertainty in the initial measurement will screw up this situation.

 

[Silviu]

Uncertainty in the initial measurement will screw up this situation.

What uncertainty?

 

[PeterDonis]

What about the speed at which the wave function collapses?

There is no such thing. Wave function collapse is an adjustment of our model to take into account new information--the result of the measurement. It is not a change in the system being measured. (If there is such a change as a result of the measurement, and you really wanted to model it instead of just approximating it with a wave function collapse, you would have to switch to a much more complicated model.)

 

[PeterDonis]

what happens if we don't see it there?

Then you have the two other peaks left. Since you started with a combination of three delta functions, it doesn't make sense to talk about normalization, since delta functions are not normalizable. So the best you can do is to say that the measurement eliminates one of the three delta functions, leaving the other two.

 

[Silviu]

Then you have the two other peaks left. Since you started with a combination of three delta functions, it doesn't make sense to talk about normalization, since delta functions are not normalizable. So the best you can do is to say that the measurement eliminates one of the three delta functions, leaving the other two.

Thank you for your reply. This is true if we had 3 delta functions. But I was asking for a more general case. In my example I was talking about a particle in a box with 3 regions (so a sin function). If we don't see the particle at the top of one of the regions, we can't assume that the whole region goes to 0. So I was wondering how does the wavefunction changes after the measurement.

 

[PeterDonis]

If we don't see the particle at the top of one of the regions, we can't assume that the whole region goes to 0.

So what can you assume? How much is excluded by an actual position measurement? (Hint: the answer is not "just one point"; such a measurement is impossible.)

 

[Silviu]

So what can you assume? How much is excluded by an actual position measurement? (Hint: the answer is not "just one point"; such a measurement is impossible.)

Well yeah, it is not just one point it is a region dx. So $\psi$ in that region becomes 0. So what happens to the rest?

 

[PeterDonis]

So what happens to the rest?

You just renormalize to a total measure that now excludes the region $dx$ that was ruled out by the measurement.

 

[Silviu]

You just renormalize to a total measure that now excludes the region $dx$ that was ruled out by the measurement.

But the shape of the wavefunction will not be the same as before, minus the region dx. So I can't just multiply everything by a constant.

 

[Orodruin]

But the shape of the wavefunction will not be the same as before, minus the region dx. So I can't just multiply everything by a constant.

Why would you think so? Did you read post #7?

 

[Silviu]

Why would you think so?

Because the wavefunctions still needs to be continuous. If we just remove the region dx then the right limit as $x \to x_0-dx$ will be the value before the measurement while the left limit will be 0 (as all the region between $x_0-dx$ and $x_0+dx$ is 0). So if we just remove that region and multiply everything else by a normalization constant, we will have 2 discontinuities at $x_0 \pm dx$, right?

 

[Orodruin]

Because the wavefunctions still needs to be continuous

Says who? I suggest you start working out the case of a finite dimensional Hilbert space instead of looking at the case of a wavefunction in position representation. It is conceptually simpler.

 

[romsofia]

You say "obviously the wavefunction changes" but what's so obvious about it? I'm not sure I follow your logic... how can this part be the obvious part, but then you don't understand how it changes??

Instead of using a cop out by saying obviously x occurs, I'd ask you to show us WHY you think it should change.