• Support PF! Buy your school textbooks, materials and every day products Here!

Unitary Transformations proof

  • Thread starter Ed Quanta
  • Start date
  • #1
297
0

Homework Statement



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

Homework Equations



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


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.
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
26,258
618
If U(a)=xa and you act on both sides with U^(-1), what does that say about eigenvectors of U^(-1)???
 
  • #3
297
0
They are the reciprocals. U^(-1)a=1/x
 
  • #4
297
0
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
Dick
Science Advisor
Homework Helper
26,258
618
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:
  • #6
297
0
The absolute values of x and y must be real.
 
  • #7
dextercioby
Science Advisor
Homework Helper
Insights Author
12,981
540
HINT: The spectrum of a unitary operator in a complex Hilbert space is the unit circle...
 
  • #8
Dick
Science Advisor
Homework Helper
26,258
618
<a,b>=<Ua,Ub>. Apply that to an eigenvector. As dextercioby says...
 
  • #9
297
0
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
Dick
Science Advisor
Homework Helper
26,258
618
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 [tex]x^* y=1[/tex]. So if x=y, then [tex]x^* x=1[/tex] and the eigenvalues are unit complex numbers. How does this help you with the original problem?
 
  • #11
297
0
So x*y does not equal 1 unless y=x.
 
  • #12
Dick
Science Advisor
Homework Helper
26,258
618
So x*y does not equal 1 unless y=x.
If you mean x and y being real numbers with norm 1, then yes.
 
Last edited:
  • #13
297
0
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
Dick
Science Advisor
Homework Helper
26,258
618
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?
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 [tex]<c x,y>=c^*<x,y>[/tex].
 

Related Threads on Unitary Transformations proof

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
0
Views
1K
Replies
5
Views
4K
  • Last Post
Replies
5
Views
16K
  • Last Post
Replies
1
Views
899
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
10
Views
4K
  • Last Post
Replies
4
Views
11K
  • Last Post
Replies
5
Views
4K
Top