Expectation value of momentum in discrete states

[Heirot]

Is there any way of proving <p> = 0 for a discrete (bound) state given it's wave function? I've seen proofs using the hermitian properties of p but I'm interested in proving that the integral of Psi*(x) Psi'(x) is identically zero regardless of Psi(x) as long as it's a solution of Schroedinger's equation.

Thanks

 

[Heirot]

Can the wave function of a bound state always be chosen to be real?

 

[Dr Transport]

From symmetry arguments alone it is easy to prove.

If [tex] \Psi(x) [/tex] is even(odd) symmetry and [tex] p [/tex] is always odd, then [tex] <\Psi | p | \Psi> [/tex] is even*odd*even or odd*odd*odd either case is of odd symmetry which will always be zero over all space.

 

[Heirot]

Yes, that's true, but in general case, parity need not be a good quantum number.

 

[DrDu]

You could also argue that p/m is the expectation value of the commutator [r/i,H], at least when a magnetic field is absent. The expectation value of this operator vanishes for a bound state. When a magnetic field is present, you will have to consider also the hidden momentum of the field.

 

[Heirot]

Let's see...

By Ehrenfest theorem we have <p>/m = d/dt <r> = 1/ih <[x,H]> = 1/ih <n|xH-Hx|n> = 1/ih E_n (<n|x|n> - <n|x|n>).
OK, so for bound states we have <n|x|n> = finite so <p> = 0. What about for stationary states with E>0? Why doesn't the argument apply there? Is <n|x|n> infinite or not defined well?

 

[DrDu]

In orthodox Hilbert space formalism, there are no eigenstates corresponding to the continuous spectrum. In rigged Hilbert space formalism, they are defined, but, nevertheless, you are not allowed to form expectation values of them.

 

[Petr Mugver]

What do we mean by "bound state"? I guess a sinultaneous eigenket of a complete set of commuting operators, belonging to the discrete spectrum. If parity is in this set, then all is easy. Otherwise, take total angular momentum, this must be conserved for a physical system. Perform a pi-rotation around any axis. A bound state transforms by an inconsequential phase factor that goes out of the expectation value <psi|p|psi>, while p changes sign...