Unitary operator eigenvectors

In summary, the eigenvalues of a unitary operator are complex numbers of modulus 1. Unitary operators produce orthogonal eigenvectors.
  • #1
82
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Homework Statement


I know that Unitary operators act similar to hermitean operators.
I want to prove that the eigenvalues of unitary operators are complex numbers of modulus 1, and that Unitary operators produce orthogonal eigenvectors.

Homework Equations


UU = I
U-1=U
λ = eiΦ{/SUP] (eigenvalue form)


The Attempt at a Solution


For Eigenvalues being modulus 1, I wasn't sure if i started far enough back, but i have:
  • I read in Sakurai that Eigenvalues of unitary operators have form λ = eiΦ{/SUP]

U |a> = λ |a>
<a| U = λ*<a|
<a| U U |a> = λ*λ<a|a>
<a|I|a> = |λ|2<a|a>
  • But then I get to the point where I am already trying to assume Orthonormality and not showing it is complex.
1 = |λ|2 where <a|a> = 1
  • The only other way i can assume to to show this, is by asserting that the form IS: λ = eiΦ{/SUP]. Which then i can expand with Euler's Formula and use the basic Mod sq formatting. This gives me 1. At this point I don't know where to start when trying to show orthogonality...
    [*]We used this argument in lecture to show orthogonality of a hermitian matrix

<a'|A|a>- <a'|A|a> = (λ*-λ)<a'|a>
And since A is self adjoint The LHS equals zero..
Any tips on how i should be viewing this?​
 
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  • #2
Lawrencel2 said:

Homework Statement


I know that Unitary operators act similar to hermitean operators.
I want to prove that the eigenvalues of unitary operators are complex numbers of modulus 1, and that Unitary operators produce orthogonal eigenvectors.

Homework Equations


UU = I
U-1=U
λ = e (eigenvalue form)​

The Attempt at a Solution


For Eigenvalues being modulus 1, I wasn't sure if i started far enough back, but i have:
  • I read in Sakurai that Eigenvalues of unitary operators have form λ = e
U |a> = λ |a>
<a| U = λ*<a|
<a| U U |a> = λ*λ<a|a>
<a|I|a> = |λ|2<a|a>
  • But then I get to the point where I am already trying to assume Orthonormality and not showing it is complex.
No: so far you have used ##\ U^\dagger = U^{-1}## to conclude ##\lambda^*\lambda = 1##
1 = |λ|2 where <a|a> = 1
  • The only other way i can assume to to show this, is by asserting that the form IS: λ = e. Which then i can expand with Euler's Formula and use the basic Mod sq formatting. This gives me 1. At this point I don't know where to start when trying to show orthogonality...
  • We used this argument in lecture to show orthogonality of a hermitian matrix
<a'|A|a>- <a'|A|a> = (λ*-λ)<a'|a>
And since A is self adjoint The LHS equals zero..
Any tips on how i should be viewing this?​
(with a magnifying glass, all these ##[##SUP##]## ! :) )

Corny, sorry.

Your lefthand side is zero, your righthand side is ##(\lambda '^* - \lambda) <a'|a>## (not ##(\lambda ^* - \lambda)...## ) .
So one of the two factors must be zero.
From there, you can proceed to demonstrating orthogonality​
 
  • #3
Sorry, I have no clue how all those superscripts got in there. I tried to remove them.

As per your note, I see now that I have used that to show the modulus sq is one, but I think i have failed to show that it is complex? Only other way I can imagine showing that, is assuming it has the e^(i λ φ), and using Euler's to expand it.

I was trying to start with the same argument they did with showing hermitian operators were Orthogonal (where I used operator A)..

I tried to follow the same steps and i arrive at the same basic eq:
<a'|U|a> = λ<a'|a> ; <a'|U|a> = λ' * <a'|a>
Subtracting the two
<a'|U|a> - <a'|U|a> =( λ' *-λ)<a'|a>
Now, because i know they ARE orthogonal, I should be able to make an argument about the LHS equaling zero, but I can't say why it is..
 
  • #4
First, eigenvalues are in general complex - you only need to prove that the modulus is one, which you have. For the orthogonality of the eigen-basis, just due the exact same thing you did to show that the modulus of the eigenvalues is one - except use two different eigenvectors. You get a relation of the form x = ( # different than 1)*x, and there is only one value of x for which this holds true.

edit: of course, this is assuming non-degeneracy
 
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  • #5
DelcrossA said:
First, eigenvalues are in general complex - you only need to prove that the modulus is one, which you have. For the orthogonality of the eigen-basis, just due the exact same thing you did to show that the modulus of the eigenvalues is one - except use two different eigenvectors. You get a relation of the form x = ( # different than 1)*x, and there is only one value of x for which this holds true.

edit: of course, this is assuming non-degeneracy
Only thing is, where you call it "# different than 1" isn't. λ'* λ works out to be one due to its Euler's type form
 
  • #6
Lawrencel2 said:
Only thing is, where you call it "# different than 1" isn't. λ'* λ works out to be one due to its Euler's type form

How? The two Eigenvalues are different and so the phases are different.
 
  • #7
DelcrossA said:
How? The two Eigenvalues are different and so the phases are different.
right, so when you multiply the the Cos()+i Sin() to the other, the sum-difference formula just takes it in as a new angle.. Or just adding the exponents while its in the form of an exponential.
 
  • #8
ooooo But you make a point that I glanced over. let's see if it helps. What i said was correct, but what you said was also correct.. so thank you.. ill see if it works
 
  • #9
So you are saying that exp(iθ) is always equal to 1 regardless of θ? That isn't true
 
  • #10
DelcrossA said:
So you are saying that exp(iθ) is always equal to 1 regardless of θ? That isn't true
no, that modulus of exp(i(theta) is always 1. But I worded some things wrong and convinced myself of the error i had written. Thank you for that enlightenment. Idk how i convinced myself of what I had on my paper. I blame it on the caffeine.
 
  • #11
No problem - mix ups like that happen to me all the time.
 

1. What is a unitary operator in terms of eigenvectors?

A unitary operator is a linear transformation on a vector space that preserves the inner product between vectors. In terms of eigenvectors, this means that the unitary operator will preserve the length and angle of the eigenvectors.

2. How are unitary operators related to eigenvalues?

Unitary operators are closely related to eigenvalues because unitary operators preserve the lengths and angles of eigenvectors, which are associated with specific eigenvalues. This means that the eigenvalues of a unitary operator will remain the same after the transformation.

3. Can unitary operators have complex eigenvalues?

Yes, unitary operators can have complex eigenvalues. This is because unitary operators are defined over complex vector spaces, and complex eigenvalues correspond to complex eigenvectors. However, the unitary property ensures that the complex eigenvalues will still remain preserved after the transformation.

4. How are unitary operators used in quantum mechanics?

In quantum mechanics, unitary operators are used to represent time evolution of quantum systems. The unitary property ensures that the probabilities of different outcomes of a quantum measurement are preserved over time, making unitary operators crucial for understanding the dynamics of quantum systems.

5. What are some examples of unitary operators?

Some common examples of unitary operators include rotation matrices, which preserve the length and angle of vectors, and Fourier transforms, which preserve the inner product between vectors. Other examples include reflection and translation operators in 2D and 3D spaces.

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