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My bra-ket calcs seem to be going wrong - help!

  1. Feb 17, 2012 #1

    andrewkirk

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    What am I doing wrong here?

    Let [itex]\psi[/itex] be a ket whose representation in the X basis is given by
    [itex]\psi(x)\ =\ \langle x|\psi\rangle\ =\ e^{-x^{2}/2}[/itex]

    Then
    [itex]\psi(-x)\ =\ \langle -x|\psi\rangle\ =\ e^{-x^{2}/2}\ = \psi(x)[/itex] (1)

    But we also have:
    [itex]\psi(-x)\ =\ \langle -x|\psi\rangle[/itex] (2)
    [itex]\ =\ \langle (-1)\times x)|\psi\rangle[/itex] (3), by the linearity of the inner product
    [itex]\ =\ (-1)^*\times\langle x|\psi\rangle[/itex] (4)
    [itex]\ =\ -\langle x|\psi\rangle[/itex] (5)
    [itex]\ = -\psi(x)[/itex] (6)

    and this contradicts (1).

    I must have gone wrong here somewhere. I think it might be in (2) or (3). But I can't see the problem.

    Thank you very much for any help.
     
  2. jcsd
  3. Feb 17, 2012 #2
    Going from (3) to (4) is wrong. The ket |x> is the eigenket of the position operator with eigenvalue x. The notation generally |-x> means the eigenket of the position operator with eigenvalue -x. It is NOT "the operator -1 acting on the eigenket |x>." This confusion is understandable since we sometimes write "the ket obtained by acting on the ket |ψ> with the operator A" as |Aψ>; however that is not the meaning intended in this case.

    So you need to be clear on the distinction between -|x> and |-x>. For instance, while both are eigenstates of the position operator, the first has eigenvalue x, while the second has eigenvalue -x. The first one is a multiple of the ket |x>, while the second one is completely orthogonal to |x>.
     
  4. Feb 17, 2012 #3

    andrewkirk

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    Thank you Duck. It all makes sense now.

    I always thought using numbers to label kets was a bit dicey, and now I see why.
    From now on I'll remind myself that
    [itex]|x\rangle[/itex] is really [itex]|\delta_x\rangle[/itex]
     
  5. Feb 17, 2012 #4
    I agree with what The Duck said. I will take this a bit further for your curiosity's sake.

    Define the Parity operator P by

    P|x> = |-x>.

    The eigenvalues of P can be either 1 or -1 (try and prove this or tell me if you can't. Hint for proof: consider P^2 and it's eigenvalues).

    These are the ONLY eigenvalues P can take.

    In the -1 (odd) case, we have P|x> = -|x>, and since P|x> = |-x>, we get |-x> = -|x>.

    This is the case you have described here.

    In the other case, (eigenvalue of P is +1), you would get

    -|-x> = |x>.

    Edit: I'm not sure if what I've described is exactly the same as what you have in the OP because you are doing the inner product whereas I am acting an operator on |x>. Food for thought...
     
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