Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Smoothness of potential energy and powers of momentum

  1. Oct 13, 2014 #1
    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. jcsd
  3. Oct 13, 2014 #2


    User Avatar
    Science Advisor
    Homework Helper

    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).
  4. Oct 13, 2014 #3
    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
  5. Oct 14, 2014 #4
    I assumed that square well potential like C-infinity smooth functions, e.g.
    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook