Momentum eigenstates particle in box

[clacker]

Quantum mechanics says measurement of observable always produces result that is one of eigenvalues of that observable. Subsequent measurement yields same value. For a particle in a box with infinite potential barriers if measure momentum doesn't that put system in eigenstate of momentum insuring subsequent same value of momentum. Doesn't this then violate uncertainty since know particle's postion with certainty of width of box. I don't understand how this doesn't violate uncertainty, in any event you could always measure momentum and with the particle in box it seems you can always violate uncertainty.

 

[saaskis]

For a particle in a box with infinite potential barriers if measure momentum doesn't that put system in eigenstate of momentum insuring subsequent same value of momentum..

The energy eigenstates are, for example, sine functions. Operating on them with the momentum operator produces cosines, so these are not momentum eigenstates. Also the expectation value of momentum squared yields a non-zero number, and therefore the uncertainty in p is not zero. I guess.

 

[diazona]

Since momentum eigenstates are of the form \psi = C e^{i p x/\hbar}, it would seem that there are no momentum eigenstates that satisfy the boundary conditions - you can only have linear combinations of states of different momenta. Although that does bring up the question of what the state of a particle in a box actually is after you measure its momentum... maybe it's not actually the case that (theoretical infinitely fast) repeated measurements are guaranteed to give the same result? I think I should know this but it's escaping me at the moment :/