Eigenvalues of an unitary operator

  • Thread starter jnazor
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  • #1
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Homework Statement


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?



Homework Equations


U(U+)=(U+)U=I [where U+ is U dagger and I is the identity operator]



The Attempt at a Solution


Assume eigenvalues exist
U(a)=x(a) => (U+)U(a)=(U+)x(a) => (a)=(U+)x(a)??
 

Answers and Replies

  • #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
 
  • #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.
 
  • #4
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Sorry i've never heard of isometry or the name spectral equation. I just know it as the eigenvalue equation.
 
  • #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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