Are Eigenvectors of Unitary Transformations Orthogonal?

Click For Summary
SUMMARY

The eigenvectors of a unitary transformation corresponding to distinct eigenvalues are orthogonal. This conclusion is derived from the properties of unitary operators, specifically that their eigenvalues lie on the unit circle in the complex plane. The discussion highlights the relationship between the eigenvalues and their norms, confirming that if the eigenvalues are distinct, their inner product must equal zero. The proof utilizes the fact that for unitary transformations, the eigenvalues satisfy the condition \(x^* y = 1\), reinforcing the orthogonality of the eigenvectors.

PREREQUISITES
  • Understanding of unitary transformations and their properties
  • Familiarity with eigenvalues and eigenvectors in linear algebra
  • Knowledge of complex inner products and Hilbert spaces
  • Basic principles of Hermitian operators and their orthogonality conditions
NEXT STEPS
  • Study the properties of unitary operators in quantum mechanics
  • Learn about the spectral theorem for unitary matrices
  • Explore the implications of eigenvalue norms in complex vector spaces
  • Investigate the relationship between Hermitian and unitary transformations
USEFUL FOR

Students and professionals in mathematics, physics, and engineering, particularly those focusing on quantum mechanics, linear algebra, and operator theory.

Ed Quanta
Messages
296
Reaction score
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.
 
Physics news on Phys.org
If U(a)=xa and you act on both sides with U^(-1), what does that say about eigenvectors of U^(-1)?
 
They are the reciprocals. U^(-1)a=1/x
 
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?
 
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:
The absolute values of x and y must be real.
 
HINT: The spectrum of a unitary operator in a complex Hilbert space is the unit circle...
 
<a,b>=<Ua,Ub>. Apply that to an eigenvector. As dextercioby says...
 
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
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
So x*y does not equal 1 unless y=x.
 
  • #12
Ed Quanta said:
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
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
Ed Quanta said:
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 &lt;c x,y&gt;=c^*&lt;x,y&gt;.
 

Similar threads

  • · Replies 15 ·
Replies
15
Views
3K
Replies
10
Views
8K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 22 ·
Replies
22
Views
6K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K