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Expected value inequality

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  • #1
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Homework Statement


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?


Homework Equations


[itex]p=-i\hbar{d\over dx}[/itex]


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??
 

Answers and Replies

  • #2
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When would that inner product with zero? and what would happen then? ;)
 
  • #3
dextercioby
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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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