# Unitary operator and eigenvalue

• I
• Peter_Newman

#### Peter_Newman

Hello,

I recently saw ##U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}## and am wondering how to come up with this or how to show this.

My first thought is based on the definition of unitary operators (##UU^\dagger = I##), I would show it something like this:

##(U|v\rangle)^\dagger = \langle v| U^\dagger, \quad (e^{ia}|v\rangle)^\dagger = \langle v|e^{-ia}##

Now ##\langle v|v\rangle = \langle v|I|v\rangle = \langle v| U^\dagger U|v\rangle = \langle v|e^{-ia}e^{ia}|v\rangle = \langle v|1|v\rangle = \langle v|v\rangle##

Is it possible to write it this way? I am not sure if I have really shown ##U|v\rangle= e^{ia}|v\rangle## with this.

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There has to be some more constraints on the problem to show what you want show. It isn't generally true. What else should we know about the problem?

Hi @Haborix, the futher context comes from this:
The eigenvalues of a unitary matrix ##U \in \mathbb{C}^{n \times n}## also all have complex magnitude one, so are of the form ##\lambda = e^{it}##. The question is how one then come up with ##U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}##

I see. Well, let ##\ket{v}## be a normalized eigenvector of ##U## with eigenvalue ##\lambda##, then try computing the inner product of ##U\ket{v}## with itself. See what kind of condition that gives you on ##\lambda##.

Ok, if I understand you right, you mean this ##\langle v | U^\dagger U | v \rangle## and ##\langle v|\lambda^\dagger\lambda |v\rangle## (last because you say ##|v\rangle## with eigenvalue ##\lambda##, so we can write ##\lambda |v\rangle##) right ?

Ok, if I understand you right, you mean this ##\langle v | U^\dagger U | v \rangle## and ##\langle v|\lambda^\dagger\lambda |v\rangle## (last because you say ##|v\rangle## with eigenvalue ##\lambda##, so we can write ##\lambda |v\rangle##) right ?
That's part of the answer. You need to do a bit more. Note that showing ##|\lambda| = 1## is equivalent to showing that ##\lambda = e^{ia}##.

OK, we have ##\langle v | v \rangle= \langle v | U^\dagger U | v \rangle= \langle v | \lambda^* \lambda | v \rangle=|\lambda|^2 \langle v | v \rangle## When I exclude the case ##\lambda \neq 0## then ist must be the case that ##|\lambda|^2 = 1##. Form this I would argue, and follow first ##\vert \lambda\vert^2=1\implies \vert \lambda\vert=1## and second that the eigenvalues have norm 1, and since we know this famous equation ##e^{ia}##, which is always one for any ##a## (lies on unit circle). We can write ##|\lambda| = e^{ia}##. But how do we come than to ##U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}##?

OK, we have ##\langle v | v \rangle= \langle v | U^\dagger U | v \rangle= \langle v | \lambda^* \lambda | v \rangle=|\lambda|^2 \langle v | v \rangle## When I exclude the case ##\lambda \neq 0## then ist must be the case that ##|\lambda|^2 = 1##. Form this I would argue, and follow first ##\vert \lambda\vert^2=1\implies \vert \lambda\vert=1## and second that the eigenvalues have norm 1, and since we know this famous equation ##e^{ia}##, which is always one for any ##a## (lies on unit circle). We can write ##|\lambda| = e^{ia}##. But how do we come than to ##U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}##?
That's essentially the proof that the eigenvalues of a unitary operator must have modulus ##1##. If ##|\lambda| = 1 ##, then ##\lambda = e^{ia}## for some ##a \in \mathbb R##.

Yes ok, but how do you derive this connection ##U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}##, this is for me not clear. Is that then apply the definition (eigenvalue problem) ## U|v\rangle = \lambda|v\rangle ##

Yes ok, but how do you derive this connection ##U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}##, this is for me not clear.
##v## is an eignevector there.

Does this turn out to be applying the definition of the eigenvalue problem? Generally ##Ax = \lambda x##, now ##A = U## and the eigenvalues of ##U## are, as argued before then ##\lambda = e^{ia}##?

Does this turn out to be applying the definition of the eigenvalue problem? Generally ##Ax = \lambda x##, now ##A = U## and the eigenvalues of ##U## are, as argued before then ##\lambda = e^{ia}##?
Yes. One problem with your work is that you don't tend to define things or say what that are. For example:

OK, we have ##\langle v | v \rangle= \langle v | U^\dagger U | v \rangle= \langle v | \lambda^* \lambda | v \rangle=|\lambda|^2 \langle v | v \rangle## When I exclude the case ##\lambda \neq 0## then ist must be the case that ##|\lambda|^2 = 1##. Form this I would argue, and follow first ##\vert \lambda\vert^2=1\implies \vert \lambda\vert=1## and second that the eigenvalues have norm 1, and since we know this famous equation ##e^{ia}##, which is always one for any ##a## (lies on unit circle). We can write ##|\lambda| = e^{ia}##. But how do we come than to ##U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}##?
A more complete proof would be something like:

Let ##|v\rangle## be an eigenvector of a unitary operator, ##U##, with eigenvalue ##\lambda##. As ##U## preserves inner products we have:
$$\langle v | v \rangle= \langle v | U^\dagger U | v \rangle= \langle v | \lambda^* \lambda | v \rangle=|\lambda|^2 \langle v | v \rangle$$And, as ##|v \rangle \ne 0##, we have ##|\lambda| = 1##, hence ##\lambda = e^{ia}## for some ##a \in \mathbb R##. And, we may write: $$U|v \rangle = e^{ia}|v \rangle$$

Peter_Newman
@PeroK, thank you for your answer. I was relatively close to the solution :)