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Let $U$ be a unitary operator on a Hilbert Space, and let $\phi$ be an eigenvector of $U$ with eigenvalue $\mu$.

Show that $|\mu|=1$ ?

I know that if $U$ is unitary then $UU^{+}=UU^{-1}=I$

but I'm not really sure how to use it in this case?

sarahusher
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2 Answers2

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Hint:

  • Take the defining identity for a unitary operator $U$ $$(U v, U w) = (v,w)$$ with $(\cdot, \cdot)$ the (complex)-scalar product and $v,w$ arbitrary elements of your vector (or function)-space.

  • Set $v= w = e_\mu$ where $e_\mu$ is the eigenvector to the eigenvalue $\mu$.

Fabian
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  • So using the notation as $\phi$ is the eigenvector corresponding to $\mu$ I'll get $<\phi,\mu \phi> = \mu <\phi,\phi>$

    Is that right?

     – sarahusher May 20 '14 at 10:22
  • Not exactly, you have $\langle \mu \phi, \mu \phi \rangle = \langle \phi, \phi \rangle$. Now you should use the sesquilinearity of the inner product. – Fabian May 20 '14 at 11:54
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We start with

$U \left| \phi \right\rangle = \mu \left| \phi \right\rangle$.

This implies that

$\left\langle \phi \right| U^\dagger = \mu^\star \left\langle \phi \right|$.

Here, we multiply the second expression times the first:

$\left \langle \phi \right| U^\dagger U \left| \phi \right\rangle = \mu^\star \mu \left\langle \phi | \phi\right\rangle$,

but since $U^\dagger U = I$ and $\left\langle\phi|\phi\right\rangle = 1$, we have that

$1=\left|\mu\right|^2$.

QED.

jordix
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