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Unitary and linear operator in quantum mechanics

  1. Dec 6, 2015 #1
    Given a transformation ##U## such that ##|\psi'>=U|\psi>##, the invariance ##<\psi'|\psi'>=<\psi|\psi>## of the scalar product under the transformation ##U## means that ##U## is either linear and unitary, or antilinear and antiunitary.

    How do I prove this?

    ##<\psi'|\psi'>##
    ##= <U\psi|U\psi>##

    For ##U## unitary and linear, we have
    ##<U\psi|U\psi>##
    ##<U(\alpha\psi_{a}+\beta\psi_{\beta})|U(\alpha\psi_{a}+\beta\psi_{\beta})>##
    ##<(\alpha U\psi_{a}+\beta U\psi_{\beta})|(\alpha U\psi_{a}+\beta U\psi_{\beta})>##
    ##=(<\psi_{a}|(U^{\dagger})\alpha^{*}+<\psi_{b}|(U^{\dagger})\beta^{*})(\alpha U|\psi_{a}>+\beta U|\psi_{\beta}>)##
    ##=|\alpha|^{2}(<\psi_{a}|(U^{\dagger})(U)|\psi_{a}>+|\beta|^{2}(<\psi_{b}|(U^{\dagger})(U)|\beta>##
    ##=|\alpha|^{2}+|\beta|^{2}##

    Is this how the proof should go for ##U## linear and unitary?
     
    Last edited: Dec 6, 2015
  2. jcsd
  3. Dec 6, 2015 #2

    dextercioby

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    Yes. That's right.
     
  4. Dec 6, 2015 #3

    A. Neumaier

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    But you don't need to go into a basis: ##\langle U\phi|U\psi\rangle =\langle \phi|U^*U\psi\rangle=\langle \phi|\psi\rangle## is enough.
     
  5. Dec 7, 2015 #4
    I wanted to show the property of linearity.
     
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