- #1

- 676

- 46

If I am correct so far ; what happens when the wavefunction is observed ? Surely it collapses to one of its eigenstates with k and hence momentum now precisely known which violates the H.U.P. as Δx is fixed ?

Thanks

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter dyn
- Start date

- #1

- 676

- 46

If I am correct so far ; what happens when the wavefunction is observed ? Surely it collapses to one of its eigenstates with k and hence momentum now precisely known which violates the H.U.P. as Δx is fixed ?

Thanks

- #2

- 13

- 1

Now due to H.U.P. if Δx is fixed as the infinite well size we can't know the exact momentum.

It sounds like you are assuming the uncertainty in the position of the particle Δx is set to the size of the well, but this is something that depends on the state of the particle. But yes, by the uncertainty principle, Δp has a lower bound.

I presume this is because the wavefunction exists as a superposition of all the wavefunctions with k quantized so the momentum is not precisely known ?

When you solve the Schrodinger equation and find ψ

- #3

- 300

- 22

http://www.wolframalpha.com/input/?i=Plot+%28integrate+sin%28Pi+x%29+e^%28-i+k+x%29+from+0+to+1%29*%28integrate+sin%28Pi+x%29+e^%28i+k+x%29+from+0+to+1%29

The problem is that the bound state is not an eigenstate of the momentum operator. The values of momentum you can measure are the eigenvalues of the eigenstates. So you take your state, find out how much of each momentum eigenstate is in your state and that gives you the probability of finding it in that momentum state, that means the probability of measuring it to have that momentum.

- #4

- 17,994

- 8,959

http://cds.cern.ch/record/493028/files/0103153.pdf

- #5

- 676

- 46

Thanks for all your replies. I don't totally understand all of it but obviously the problem was more complex than I first thought. If the particle is confined to the well , how can Δx be anything other than the size of the well ?

- #6

- 13

- 1

If the particle is confined to the well , how can Δx be anything other than the size of the well ?

You can say that the uncertainty Δx can certainly not be

[tex]

\Delta x \equiv \sqrt{\langle \psi | \hat{x}^2 |\psi\rangle -\langle \psi |\hat{x} |\psi \rangle^2}

[/tex]

In particular, if you calculate this for the states of the square well (with length L and number n) they look like:

[tex]

\Delta x = \sqrt{\frac{L^2}{12}\left(1-\frac{6}{n^2\pi^2}\right)}

[/tex]

Share: