I How to Show U|v⟩ = e^(ia)|v⟩ for Unitary Operators?

[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.

 

[Haborix]

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?

 

[Peter_Newman]

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 $U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}$

 

[Haborix]

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 $\lambda$.

 

[Peter_Newman]

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 ?

 

[PeroK]

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 $\lambda = e^{ia}$.

 

[Peter_Newman]

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 $U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}$?

 

[PeroK]

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 $a \in \mathbb R$.

 

[Peter_Newman]

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) $U|v\rangle = \lambda|v\rangle$

 

[PeroK]

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.

 

[Peter_Newman]

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 $\lambda = e^{ia}$?

 

[PeroK]

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 :)