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Quantum and Matrix Mechanics

  1. Oct 8, 2004 #1
    A is an operator and Uinv is the inverse of operator U

    How am I to show that A and Uinv(A)U have the same eigenvalues? Must U be unitary for this to be true?

    And if the eigenvectors of A are (Psi n), what are the eigenvectors of Uinv(A)U?

    Help anyone? Not sure what to do here.
     
  2. jcsd
  3. Oct 8, 2004 #2
    Unitarity of U is not required. To prove your assertion, try to start with the equation [tex]Av=\lambda v[/tex]. How may you manipulate this equation to obtain [tex]U^{-1}AU[/tex]?
     
  4. Oct 8, 2004 #3
    So can I just do something like following:?

    We know [tex]Av=\lambda 2 v[/tex]

    We then let [tex]Uv=\lambda 1 v[/tex]

    So then [tex]Av \lambda1=\lambda 2 \lambda1 v[/tex]

    Then since we know
    [tex]U^{-1}U=1[/tex], we know

    [tex]U^{-1}v=v/(\lambda 1 )[/tex]

    Thus[tex]U^{-1}AUv=\lambda 2 v[/tex]
     
  5. Oct 8, 2004 #4

    Dr Transport

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    Not quite, but close. Multiply the original equation on the left by [tex] U [/tex] then insert [tex] 1 = U^{-1}U [/tex] and work from there.....
     
  6. Oct 9, 2004 #5
    Can you explain to me why what I did is wrong? And when you say to multiply thee original equation by U, and then use the fact that 1=U multiplied by its inverse, it seems that the order of terms in the equation is altered from the expected result [tex]U^{-1}AU[/tex]

    And I know when it comes to matrix multiplication AB does not equal BA, so I am a tad confused as to what you want me to do. I apologize for my slowness.
     
  7. Oct 9, 2004 #6
    To get you started,

    [tex]Av=\lambda v[/tex]
    [tex]U^{-1}Av=\lambda U^{-1}v[/tex]

    Now how may you manipulate the last equation to get [tex]U^{-1}AU[/tex]?
     
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