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Operator, eigenstate, small calculation

  1. Aug 12, 2016 #1
    Hello :-) I have a small question for you :-)

    1. The problem statement, all variables and given/known data


    The Operator [tex] e^{A} [/tex] is definded bei the Taylor expanion [tex] e^{A} = \sum\nolimits_{n=0}^\infty \frac{A^n}{n!} . [/tex]
    Prove that if [tex] |a \rangle[/tex] is an eigenstate of A, that is if [tex] A|a\rangle = a|a\rangle[/tex], then [tex] |a\rangle[/tex] is an eigenstate of [tex] e^{A}[/tex] with the eigenvalue of [tex] e^{a}.[/tex]

    3. The attempt at a solution

    I show you a very bad attempt:
    [tex] A|a \rangle = a|a \rangle\quad\quad\quad| : |a\rangle |e^{...} [/tex]
    [tex] e^A =e^a \quad\quad\quad| * |a\rangle[/tex]
    [tex] e^A|a\rangle = e^a|a\rangle [/tex]

    I would be glad about an info how to do it right... thank you :)
     
    Last edited: Aug 12, 2016
  2. jcsd
  3. Aug 12, 2016 #2

    DrClaude

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    Staff: Mentor

    I don't even understand what that means.

    Start with ##e^A|a\rangle## and use the Taylor expansion of the exponential.
     
  4. Aug 12, 2016 #3

    Krylov

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    This is a correct definition when ##A## is a bounded operator. However, operators occurring in QM (is this a QM exercise?) are typically not bounded, but they are usually self-adjoint, so you can use a suitable functional calculus.

    In view of the above remark, let us assume in addition that ##A## is bounded. Then just apply the series definition, so the first two steps become:
    $$
    e^A|a\rangle = \left(\sum_{n=0}^{\infty}{\frac{A^n}{n!}}\right)|a\rangle = \sum_{n=0}^{\infty}{\frac{A^n|a\rangle}{n!}}
    $$
    You continue.
     
    Last edited: Aug 12, 2016
  5. Aug 12, 2016 #4
    ok, thank you :-)

    yes it is QM. ok thanks


    I think you mean [tex] e^A |a\rangle = (\sum\nolimits_{n=0}^\infty \frac{A^n}{n!} )|a\rangle ?[/tex]
    DrClaude used it and you also used it in the second term.

    So I will try now:

    [tex] e^A |a\rangle = (\sum\nolimits_{n=0}^\infty \frac{A^n}{n!}) |a\rangle = \sum\nolimits_{n=0}^\infty \frac{A^n|a\rangle }{n!} = \sum\nolimits_{n=0}^\infty \frac{a^n|a\rangle }{n!} = (\sum\nolimits_{n=0}^\infty \frac{a^n}{n!}) |a\rangle = e^a |a\rangle [/tex]

    It think, that looks fine? :-)
     
  6. Aug 12, 2016 #5

    Krylov

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    Thank you, fixed. I think you would have known how to do this exercise.
    Yes, much better! :smile:
     
  7. Aug 12, 2016 #6
    thank you both :-)
     
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