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Operator dependent functions in QM

  1. Oct 13, 2009 #1
    1. The problem statement, all variables and given/known data

    Show that [tex]\widehat{A}\Psi(x) = \Psi(x + b)[/tex], where [tex]b[/tex] is a constant.



    2. Relevant equations

    Given: [tex]\widehat{A} = exp(b[d/dx])[/tex]



    3. The attempt at a solution

    I know I'm supposed to write out the function as a power series expansion, though I'm not sure what I am exactly to do after this. What is (d/dx) of? What information do the above expressions convey?
     
  2. jcsd
  3. Oct 13, 2009 #2

    Dick

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    Write out a taylor expansion of psi(x+b) around x. Now compare that with the power series expansion of A applied to psi(x). The derivative operators will operate on psi(x). Don't you see a similarity in the two sides?
     
  4. Oct 13, 2009 #3
    Pardon my ignorance, but what would I be expanding since I am not explicitly given a function for [tex]\Psi(x)[/tex]?
     
  5. Oct 13, 2009 #4

    Dick

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    Write them out without assuming you know psi(x). The first derivative of psi(x) is psi'(x), the second is psi''(x) etc etc.
     
  6. Oct 14, 2009 #5
    Just to make sure my work is correct, would I get {[(b^n)/n!]*(d/dx)^n}*[tex]\Psi(x)[/tex]?
     
  7. Oct 14, 2009 #6

    Dick

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    Well, yeah. Isn't that the Taylor series of psi(x+b) expanded around x?
     
  8. Oct 14, 2009 #7
    Okay, so I did it right. Though, I am curious about [tex]\Psi(x + b)[/tex]. If [tex]\Psi(x)[/tex] is an eigenstate of [tex]\widehat{A}[/tex], then the former is equal to some factor times [tex]\Psi(x)[/tex] right?
     
  9. Oct 14, 2009 #8

    Dick

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    That's the definition of an eigenstate alright, A(psi(x))=k*psi(x)=psi(x+b).
     
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