# Homework Help: Unitary Transformations proof

1. Mar 7, 2007

### Ed Quanta

1. The problem statement, all variables and given/known data

Show that the eigenvectors of a unitary transformation belonging to distinct eigenvalues are orthogonal.

2. Relevant equations

I know that U+=U^-1 (U dagger = U inverse)

3. The attempt at a solution

I tried using a similar method to the proof which shows that the eigenvectors of hermitian transformations belonging to distinct eigenvalues are orthogonal.

So assume our eigenvectors are a and b. I assumed U(a)=xa and U(b)=yb

x<a|b>=<Ua|b>=<a|U^-1b>= ???

Help anyone. I know this probably isn't too rough.

2. Mar 7, 2007

### Dick

If U(a)=xa and you act on both sides with U^(-1), what does that say about eigenvectors of U^(-1)???

3. Mar 7, 2007

### Ed Quanta

They are the reciprocals. U^(-1)a=1/x

4. Mar 7, 2007

### Ed Quanta

So

x<a|b>=<Ua|b>=<a|U^-1b>=<a|(1/y)b>=1/y<a|b>

So (x - 1/y)<a|b>=0

Now how do I know x - 1/y cannot equal 0?

5. Mar 8, 2007

### Dick

Be a little careful. You are probably dealing with a complex inner product. If it's real then this is fine. As U is orthogonal, what do you know about the absolute value of x and y?

Last edited: Mar 8, 2007
6. Mar 8, 2007

### Ed Quanta

The absolute values of x and y must be real.

7. Mar 9, 2007

### dextercioby

HINT: The spectrum of a unitary operator in a complex Hilbert space is the unit circle...

8. Mar 9, 2007

### Dick

<a,b>=<Ua,Ub>. Apply that to an eigenvector. As dextercioby says...

9. Mar 9, 2007

### Ed Quanta

Ok, so I get that the norm of the eigenvalues must equal 1.

<a|b>=<Ua|Ub>=x*y<a|b>

x*y=1?

10. Mar 9, 2007

### Dick

As I've said, be a little careful. You are correct in the case if U is real. But if U is complex, the condition is $$x^* y=1$$. So if x=y, then $$x^* x=1$$ and the eigenvalues are unit complex numbers. How does this help you with the original problem?

11. Mar 9, 2007

### Ed Quanta

So x*y does not equal 1 unless y=x.

12. Mar 9, 2007

### Dick

If you mean x and y being real numbers with norm 1, then yes.

Last edited: Mar 9, 2007
13. Mar 10, 2007

### Ed Quanta

I'm still confused man. I want to show that x*y<a,b>-<a,b>=0

I want to show then that x*y does not equal 1. Where do the norms fit in?

14. Mar 10, 2007

### Dick

You've gotten this far and have assumed x and y are DIFFERENT eigenvalues of U. If U is real this is super easy, since x and y are both in the set {+1,-1} and DIFFERENT. What about U complex? Then you have to mend your ways and remember $$<c x,y>=c^*<x,y>$$.