# Smoothness of potential energy and powers of momentum

1. Oct 13, 2014

### sweet springs

Hi. In case potential energy V(x) in one dimensional stationary Shrodinger equation does not have smoothness of C-infinity, I assume that some power n of momentum for an energy eigenstate, <p^n>, diverge. Finite square well potential gives infinite <p^n> for n=6,8,10,.. for example. <p^4> also diverges for infinite square well. Could you advise me if the assumption is right? Thanks in advance.

2. Oct 13, 2014

### dextercioby

There's no apriori requirement that the arbitrary power of momentum exists, much less the energy eigenvectors being all in the domain of pn. Speaking of p in case of the infinite square well (with 'physical boundary conditions') is a little illegal, since p = -ihbar d/dx is not an observable (it's not selfadjoint).

3. Oct 13, 2014

### sweet springs

Thanks. You taught me that I should not be upset in case <p^n.> diverges.
Whether we can define momentum operator properly or not is influenced by Hamiltonian or potential energy V(x) of the system, right? Some documents, e.g. http://en.wikipedia.org/wiki/Particle_in_a_box, discuss momentum in infinite square well. Some difficulty seems to lie here.
Thank you again.

Last edited: Oct 13, 2014
4. Oct 14, 2014

### sweet springs

I assumed that square well potential like C-infinity smooth functions, e.g.
$$V(x)=\frac{V_0}{2}(tanh\frac{x-a}{\delta}-tanh\frac{x+a}{\delta}),\delta>0$$
give finite <p^n>. It diverges as delta goes to zero.
That's the reason why I was keen on smoothness of V(x) that could originate smoothness of n-derivative of wave functions.