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Proving Cauchy-Schartz inequality in Brak-ket notation

  1. Feb 13, 2008 #1
    [SOLVED] Proving Cauchy-Schartz inequality in Brak-ket notation

    [itex](\langle \alpha | + \lambda^\ast\langle \beta |)(|\alpha\rangle+ \lambda|\beta\rangle) = \langle \alpha |\alpha \rangle + |\lambda|^2\langle \beta | \beta \rangle + \lambda \langle \alpha | \beta \rangle + \lambda^\ast \langle \beta | \alpha \rangle \geq 0[/itex].

    Put [itex]\lambda = - \langle \beta| \alpha\rangle/\langle \beta|\beta \rangle[/itex]. Then

    [itex] \langle \alpha | \alpha \rangle \langle\beta|\beta\rangle + |\langle\beta|\alpha\rangle|^2 - 2\langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0[/itex].

    Not quite what I wanted.

    Any help would be appreciated.
     
  2. jcsd
  3. Feb 13, 2008 #2

    kdv

    User Avatar

    the last line implies

    [itex] \langle \alpha | \alpha \rangle \langle\beta|\beta\rangle - \langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0[/itex]
    so
    [itex] \langle \alpha | \alpha \rangle \langle\beta|\beta\rangle \geq \langle \alpha |\beta\rangle \langle \beta | \alpha \rangle[/itex]
    Isn't this what you wanted to show?
     
  4. Feb 13, 2008 #3
    Thanks for replying.

    How do you get [itex] \langle \alpha | \alpha \rangle \langle\beta|\beta\rangle - \langle \alpha |\beta\rangle \langle \beta | \alpha \rangle\geq 0[/itex]?
     
  5. Feb 13, 2008 #4
    OMG I'm so dumb. No need to answer that. Thanks.
     
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