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Variance of an eigenvector

  1. Aug 28, 2009 #1
    1. The problem statement, all variables and given/known data
    On page 5 of http://arcsecond.wordpress.com/2009...uality-and-heisenbergs-uncertainty-principle/ the author states (w/o proof) that if [tex]\psi[/tex] is an eigenvector (say with eigenvalue [tex]\lambda[/tex]) of an Hermitian operator A (I don't think the Hermitian-ness matters here), then its variance is 0; that is, [tex]\langle \psi| A^2\psi\rangle = \langle \psi| A\psi\rangle^2[/tex]. However, I've not been able to show this.


    2. Relevant equations



    3. The attempt at a solution
    I keep getting
    [tex]\langle \psi|A^2\psi\rangle = \lambda\langle \psi|A\psi\rangle = \lambda^2\langle\psi |\psi\rangle[/tex]​
    and
    [tex]\langle \psi|A\psi\rangle^2 = \left(\lambda\langle \psi|A\psi\rangle\right)^2 = \lambda^2\langle\psi |\psi\rangle^2.[/tex]​
    Where am I going wrong?
     
  2. jcsd
  3. Aug 28, 2009 #2

    gabbagabbahey

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    Eigenvectors are typically normalized, so [itex]\langle\psi\vert\psi\rangle=1[/itex]
     
  4. Aug 28, 2009 #3
    D'oh! Of course! Thanks!
     
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