# I Measurement in QM

1. Jun 23, 2017

### 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?

Last edited by a moderator: Jun 23, 2017
2. Jun 23, 2017

### 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?)?

Last edited by a moderator: Jun 23, 2017
3. Jun 23, 2017

### Jilang

Now just how are you planning to enter that box?

4. Jun 23, 2017

### Silviu

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.

5. Jun 23, 2017

### Jilang

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

6. Jun 23, 2017

### Silviu

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?

7. Jun 23, 2017

### Orodruin

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

8. Jun 23, 2017

### Silviu

What about the speed at which the wave function collapses?

9. Jun 24, 2017

### Jilang

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

10. Jun 24, 2017

### Silviu

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).

11. Jun 24, 2017

### Jilang

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

12. Jun 24, 2017

### Silviu

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

13. Jun 24, 2017

### Jilang

Yes.

14. Jun 24, 2017

### Silviu

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

15. Jun 24, 2017

### 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?

16. Jun 24, 2017

### Silviu

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

17. Jun 24, 2017

### Jilang

All over, it all increases a bit.

18. Jun 24, 2017

### Silviu

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?

19. Jun 24, 2017

### Jilang

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

20. Jun 24, 2017

### Silviu

I am not sure I understand your question.