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Eigenvalues of an unitary operator

  1. Mar 11, 2007 #1
    1. The problem statement, all variables and given/known data
    A unitary operator U has the property
    U(U+)=(U+)U=I [where U+ is U dagger and I is the identity operator]

    Prove that the eigenvalues of a unitary operator are of the form e^i(a) with a being real.

    NB: I haven't been taught dirac notation yet. Is there a way i can do this without it?



    2. Relevant equations
    U(U+)=(U+)U=I [where U+ is U dagger and I is the identity operator]



    3. The attempt at a solution
    Assume eigenvalues exist
    U(a)=x(a) => (U+)U(a)=(U+)x(a) => (a)=(U+)x(a)??
     
  2. jcsd
  3. Mar 11, 2007 #2

    mjsd

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    note that you don't need to understand Dirac notation, all you need to know is some basic linear algebra in finite dimensional space. hint: "of the form [tex]e^{i\theta}[/tex]" means that magnitude of complex e-vals are 1
     
  4. Mar 12, 2007 #3

    dextercioby

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    HINT: U unitary means U isometry. Assume the spectral equation

    [tex] U\psi =a\psi [/tex] (1)

    has solutions in a Hilbert space [itex] \mathcal{H} [/itex].

    Then use (1), the assumption regarding the space of solutions and the isometry condition to get the desired result.
     
  5. Mar 13, 2007 #4
    Sorry i've never heard of isometry or the name spectral equation. I just know it as the eigenvalue equation.
     
  6. Mar 13, 2007 #5

    Dick

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    Isometry means <x,y>=<Ux,Uy>. Why is this true for U unitary? Once you believe it's true set y=x and x to be an eigenvector of U. What do you conclude?
     
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