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Expectation value of momentum in discrete states

  1. Sep 16, 2010 #1
    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.

  2. jcsd
  3. Sep 16, 2010 #2
    Can the wave function of a bound state always be chosen to be real?
  4. Sep 16, 2010 #3

    Dr Transport

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    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.
  5. Sep 17, 2010 #4
    Yes, that's true, but in general case, parity need not be a good quantum number.
  6. Sep 17, 2010 #5


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    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.
  7. Sep 17, 2010 #6
    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?
  8. Sep 17, 2010 #7


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    In orthodox Hilbert space formalism, there are no eigenstates corresponding to the continous spectrum. In rigged Hilbert space formalism, they are defined, but, nevertheless, you are not allowed to form expectation values of them.
  9. Sep 18, 2010 #8
    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...
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