1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Expected value inequality

  1. Oct 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Why is [itex]\langle p^2\rangle >0[/itex] where [itex]p=-i\hbar{d\over dx}[/itex], (noting the ***strict*** inequality) for all normalized wavefunctions? I would have argued that because we can't have [itex]\psi=[/itex]constant, but then I thought that we can normalize such a wavefunction by using periodic boundary conditions... So I don't how to argue that the inequality should be strict... Is it that otherwise it would be trivial?

    2. Relevant equations
    [itex]p=-i\hbar{d\over dx}[/itex]

    3. The attempt at a solution
    clearly, [itex]\langle \psi|p^2|\psi\rangle = \langle p\psi|p\psi\rangle \geq 0[/itex] since p is Hermitian. But why the strict inequality??
  2. jcsd
  3. Oct 26, 2011 #2
    When would that inner product with zero? and what would happen then? ;)
  4. Oct 26, 2011 #3


    User Avatar
    Science Advisor
    Homework Helper

    Well, if there exists [itex] \psi \in D(p) [/itex], so that [itex] \langle \psi,p^2 \psi\rangle [/itex] = 0, then [itex] ||p\psi|| [/itex] has 0 norm. Which vector has 0 norm ? Is it normalizable ?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Expected value inequality
  1. Expectation values (Replies: 2)

  2. Expectation Value (Replies: 1)

  3. Expectation values (Replies: 1)

  4. Expectation values (Replies: 4)