Are Unitary Operator Eigenvalues Always Modulus 1 and Eigenvectors Orthogonal?

[Lawrencel2]

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

U^†U = I
U^-1=U^†
λ = e^{iΦ{/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 = e^iΦ{/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 = e^{iΦ{/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 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

U^†U = I
U^-1=U^†
λ = e^iΦ (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^iΦ

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^iΦ. 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​

 

[Lawrencel2]

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..

 

[DelcrossA]

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

 

[Lawrencel2]

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

 

[DelcrossA]

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.

 

[Lawrencel2]

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.

 

[Lawrencel2]

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

 

[DelcrossA]

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

 

[Lawrencel2]

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.

 

[DelcrossA]

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