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The Displacement Operator

  1. Mar 15, 2013 #1
    Consider the displacement operator Tψ(x)=ψ(x+a). Is T Hermitian?
     
  2. jcsd
  3. Mar 15, 2013 #2
    To get you started, what's the definition of a Hermitian operator?
     
  4. Mar 15, 2013 #3
    here is the definition: <f│Ag>=<Af│g> always if A is Hermition.
    Don't know how to start.
     
  5. Mar 16, 2013 #4
    Suppose the wave function of the state |g> is ##\psi_g(x)## and the wave function of state |f> is ##\psi_f(x)##. Then what does the above definition of an operator A being Hermitian say if you translate the inner product into an integral of wave functions, using

    ##\langle a | b \rangle = \int dx \psi_a(x)^* \psi_b(x)##

    and plug in A = T?
     
  6. Mar 16, 2013 #5
    Is this what you are suggesting?

    <f│Tg>=∫dxΨ(x)*ψ(x+a)

    If so I don't know how to proceed.

    It does not look like this would give <Tf│g> but don't know why.
     
  7. Mar 16, 2013 #6
    Yup.

    If you think they aren't equal, perhaps you can find an explicit counterexample?
     
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