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Why use a fourier transform to change PSI (x) to PSI (p) for expectation value of momentum?

  1. Oct 30, 2014 #1
    we have a wavefunction [tex]\psi (x)[/tex] the question asks for [tex]\psi (p)[/tex] and says to use this to calculate the expectation value of momentum. The problem is the expectation value of momentum is integrated over dx so after transforming how do you get the integral to be over dp?

    thanks for any help with this, I don't see the point in doing the transform if the integral is over dx, is there a different integral for the expectation value of momentum over dp?

    ps the [tex]\psi (p)[/tex] should have a tilda over it
     
  2. jcsd
  3. Oct 30, 2014 #2

    WannabeNewton

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    The expectation value is [tex]\langle \psi|P|\psi \rangle \\= \langle \psi |P|\int dp |p\rangle \langle p|\psi \rangle \\= \langle \psi| \int dp \psi(p) P|p \rangle \\= \langle \psi| \int dp \psi(p)p |p\rangle \\= \int dp \psi(p) p \langle \psi| p \rangle \\= \int dp p |\psi(p)|^2 [/tex] where in the second line I've inserted a complete set of momentum eigenstates.
     
  4. Oct 30, 2014 #3
    That helps greatly, thanks!
     
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