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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

    dextercioby

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    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 ?
     
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