Uncertainty Principle and the Infinite Well

[dyn]

For the infinite square well in one-dimension the wavefunctions have the form Acos(kx) where k is the wavenumber which is proportional to momentum. Now due to H.U.P. if Δx is fixed as the infinite well size we can't know the exact momentum. 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 ?
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

 

[jk86]

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_n(x)=Asin(k_nx), you are finding energy eigenstates not momentum eigenstates (check it with an application of the momentum operator). It may seem like these states are momentum eigenstates considering that a momentum eigenstate goes as ~e^ikx and you can write cos(kx)=(e^ikx+e^-ikx)/2, but that is a combination of two states going in opposite directions, even though they have the same magnitude. Furthermore that doesn't even really work because it doesn't respect the same boundary conditions as the infinite well states---a state with definite momentum is necessarily infinite in spatial extent. The reality is you have a continuous probability density of momenta given by |φ(p)|², where φ(p) is the momentum space wavefunction.

 

[chingel]

Yeah you just need to find the probability density of various momenta using the standard formula. For a particle with a position wavefunction of sin(Pi x) between 1 and 0, and 0 otherwise, the momentum probability density comes out to be (in some units, not normalized):

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.

 

[vanhees71]

There is no momentum observable for a particle in an infinite-square well. That's a well-known fact, often overlooked in introductory quantum-theory lectures. Please see, e.g.,

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

 

[dyn]

Hi ,
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 ?

 

[jk86]

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 larger than the size of the well, but that doesn't imply that Δx=L. Remember that Δx is the standard deviation of the position given the wavefunction of the particle ψ:
$$\Delta x \equiv \sqrt{\langle \psi | \hat{x}^2 |\psi\rangle -\langle \psi |\hat{x} |\psi \rangle^2}$$
In particular, if you calculate this for the states of the square well (with length L and number n) they look like:
$$\Delta x = \sqrt{\frac{L^2}{12}\left(1-\frac{6}{n^2\pi^2}\right)}$$