Are Eigenvectors of Unitary Transformations Orthogonal?

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Homework Help Overview

The discussion centers around the properties of eigenvectors associated with unitary transformations, specifically focusing on whether eigenvectors corresponding to distinct eigenvalues are orthogonal. The subject area involves linear algebra and quantum mechanics concepts related to unitary operators.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between eigenvectors and eigenvalues of unitary transformations, drawing parallels to hermitian transformations. Questions arise regarding the implications of acting on eigenvectors with the inverse of the transformation and the nature of the eigenvalues.

Discussion Status

The discussion is active, with participants providing hints and considerations about the properties of eigenvalues in complex spaces. Some guidance has been offered regarding the norms of eigenvalues and the implications of their absolute values, but no consensus has been reached on the final argument.

Contextual Notes

Participants note the complexity of dealing with unitary operators in a complex Hilbert space, raising questions about the nature of eigenvalues and the conditions under which they can be considered distinct. There is an emphasis on the need to be cautious with assumptions regarding the inner product and the nature of the eigenvalues.

Ed Quanta
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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.
 
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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?
 
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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.
 
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  • #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;.
 

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