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Unitary Transformations proof

  1. Mar 7, 2007 #1
    1. The problem statement, all variables and given/known data

    Show that the eigenvectors of a unitary transformation belonging to distinct eigenvalues are orthogonal.

    2. Relevant equations

    I know that U+=U^-1 (U dagger = U inverse)


    3. The attempt at a solution

    I tried using a similar method to the proof which shows that the eigenvectors of hermitian transformations belonging to distinct eigenvalues are orthogonal.

    So assume our eigenvectors are a and b. I assumed U(a)=xa and U(b)=yb

    x<a|b>=<Ua|b>=<a|U^-1b>= ???

    Help anyone. I know this probably isn't too rough.
     
  2. jcsd
  3. Mar 7, 2007 #2

    Dick

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    If U(a)=xa and you act on both sides with U^(-1), what does that say about eigenvectors of U^(-1)???
     
  4. Mar 7, 2007 #3
    They are the reciprocals. U^(-1)a=1/x
     
  5. Mar 7, 2007 #4
    So

    x<a|b>=<Ua|b>=<a|U^-1b>=<a|(1/y)b>=1/y<a|b>

    So (x - 1/y)<a|b>=0

    Now how do I know x - 1/y cannot equal 0?
     
  6. Mar 8, 2007 #5

    Dick

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    Be a little careful. You are probably dealing with a complex inner product. If it's real then this is fine. As U is orthogonal, what do you know about the absolute value of x and y?
     
    Last edited: Mar 8, 2007
  7. Mar 8, 2007 #6
    The absolute values of x and y must be real.
     
  8. Mar 9, 2007 #7

    dextercioby

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    HINT: The spectrum of a unitary operator in a complex Hilbert space is the unit circle...
     
  9. Mar 9, 2007 #8

    Dick

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    <a,b>=<Ua,Ub>. Apply that to an eigenvector. As dextercioby says...
     
  10. Mar 9, 2007 #9
    Ok, so I get that the norm of the eigenvalues must equal 1.

    <a|b>=<Ua|Ub>=x*y<a|b>

    x*y=1?
     
  11. Mar 9, 2007 #10

    Dick

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    As I've said, be a little careful. You are correct in the case if U is real. But if U is complex, the condition is [tex]x^* y=1[/tex]. So if x=y, then [tex]x^* x=1[/tex] and the eigenvalues are unit complex numbers. How does this help you with the original problem?
     
  12. Mar 9, 2007 #11
    So x*y does not equal 1 unless y=x.
     
  13. Mar 9, 2007 #12

    Dick

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    If you mean x and y being real numbers with norm 1, then yes.
     
    Last edited: Mar 9, 2007
  14. Mar 10, 2007 #13
    I'm still confused man. I want to show that x*y<a,b>-<a,b>=0

    I want to show then that x*y does not equal 1. Where do the norms fit in?
     
  15. Mar 10, 2007 #14

    Dick

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    You've gotten this far and have assumed x and y are DIFFERENT eigenvalues of U. If U is real this is super easy, since x and y are both in the set {+1,-1} and DIFFERENT. What about U complex? Then you have to mend your ways and remember [tex]<c x,y>=c^*<x,y>[/tex].
     
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