Undergrad A question about the Collapse of a Wavefunction

[PORFIRIO I]

I’m new in QM. I have a simple question: when one says that the wavefunction collapses, is it the same as saying that the variance of an observable is 0? Thanks.

 

[Isaac0427]

Pretty much, but with a caveat. Technically, it's not just saying that the variance is zero, but the state of the particle has changed to the state where the variance is zero (there's still a caveat). Although I think this does tread a little bit into interpretations, as some interpretations disagree with the whole concept of collapse.

In the case of spin or angular momentum, your statement about the variance is correct. If you measure, say, the z-component of the spin of an electron to be up, then the electron, no matter what state it was in before, changes to be in the state of being spin up in the z-component (100% probability). This would also work if you measure the energy of a particle in a bound state (which could be in a superposition of multiple energy states).

Here's the caveat: a particle can never have a precisely defined position. The variance can never be zero, and therefore one can never precisely measure the position of a particle. The wavefunction would collapse to a spike around the point it is measured at, but it wouldn't be exactly a Dirac delta, which is the only "wavefunction" with a precisely defined position. Additionally, this collapsed state would be short-lived, due to Schrodinger evolution.

To see this in action, look at this simulation. To see it best, switch to a constant potential, pause time, and click "make quantum measurement" (and notice how the spike still isn't a perfect delta function). Then progress time to see what happens.

 

[PORFIRIO I]

Pretty much, but with a caveat. Technically, it's not just saying that the variance is zero, but the state of the particle has changed to the state where the variance is zero (there's still a caveat). Although I think this does tread a little bit into interpretations, as some interpretations disagree with the whole concept of collapse.

In the case of spin or angular momentum, your statement about the variance is correct. If you measure, say, the z-component of the spin of an electron to be up, then the electron, no matter what state it was in before, changes to be in the state of being spin up in the z-component (100% probability). This would also work if you measure the energy of a particle in a bound state (which could be in a superposition of multiple energy states).

Here's the caveat: a particle can never have a precisely defined position. The variance can never be zero, and therefore one can never precisely measure the position of a particle. The wavefunction would collapse to a spike around the point it is measured at, but it wouldn't be exactly a Dirac delta, which is the only "wavefunction" with a precisely defined position. Additionally, this collapsed state would be short-lived, due to Schrodinger evolution.

To see this in action, look at this simulation. To see it best, switch to a constant potential, pause time, and click "make quantum measurement" (and notice how the spike still isn't a perfect delta function). Then progress time to see what happens.

Thanks, that was very helpful!