Unitary operator eigenvectors

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?​

BvU
Homework Helper

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​

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..

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

Last edited:
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

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.

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.

ooooo But you make a point that I glanced over. lets see if it helps. What i said was correct, but what you said was also correct.. so thank you.. ill see if it works

So you are saying that exp(iθ) is always equal to 1 regardless of θ? That isn't true

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.

No problem - mix ups like that happen to me all the time.