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

In summary, the conversation discusses the connection between the eigenvalues of a unitary matrix and the complex number ##e^{ia}##, and how to derive the equation ##U|v\rangle= e^{ia}|v\rangle, \, a \in \mathbb{R}##. The conclusion is reached by applying the definition of the eigenvalue problem where ##A = U## and the eigenvalues of ##U## are shown to be of the form ##\lambda = e^{ia}##.
  • #1
Peter_Newman
155
11
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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  • #2
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?
 
  • #3
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}##
 
  • #4
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##.
 
  • #5
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 ?
 
  • #6
Peter_Newman said:
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}##.
 
  • #7
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}##?
 
  • #8
Peter_Newman said:
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##.
 
  • #9
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 ##
 
  • #10
Peter_Newman said:
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.
 
  • #11
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}##?
 
  • #12
Peter_Newman said:
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:

Peter_Newman said:
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: $$U|v \rangle = e^{ia}|v \rangle$$
 
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  • #13
@PeroK, thank you for your answer. I was relatively close to the solution :)
 

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

1. What is a unitary operator?

A unitary operator is a linear transformation that preserves the inner product of a vector space. This means that the length of a vector and the angle between two vectors are not changed by the unitary operator.

2. How is a unitary operator different from a Hermitian operator?

A unitary operator is a type of linear operator that preserves the inner product, while a Hermitian operator is a type of matrix that is equal to its own conjugate transpose. In other words, a unitary operator is a type of transformation, while a Hermitian operator is a type of matrix.

3. What is an eigenvalue?

An eigenvalue is a scalar value that represents the scaling factor of an eigenvector when it is multiplied by a linear transformation. In other words, it is a value that remains unchanged when the eigenvector is transformed by the linear operator.

4. How are unitary operators related to eigenvalues?

Unitary operators have the property that their eigenvalues have an absolute value of 1. This means that when an eigenvector is transformed by a unitary operator, its length remains unchanged. Additionally, the eigenvectors of a unitary operator form an orthonormal basis.

5. What are some applications of unitary operators and eigenvalues?

Unitary operators and eigenvalues have many applications in quantum mechanics, signal processing, and linear algebra. They are used to solve differential equations, analyze quantum systems, and compress data in signal processing. They are also important in the study of symmetry and group theory.

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