Operator that interchanges variables

1. Sep 3, 2009

ELESSAR TELKONT

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

I suppose that this is not directly a quantum mechanical problem, but this have been assigned as homework for the Quantum Mechanics course.

Let be an operator L and eigenvalue equation $$Lf=\lambda f$$. This operator, applied to a function $$f(x,y)$$, interchanges the variables i.e. $$Lf(x,y)=f(y,x)$$. What's the general property of the eigenfunctions of this problem? Get the possible eigenvalues.

2. Relevant equations

3. The attempt at a solution

Well. I think that if $$Lf(x,y)=f(y,x)$$ then if $$f$$ is an eigenfunction, obviously, $$\lambda f(x,y)=f(y,x)$$. One possible kind of $$f$$ that fills conditions is one that is symmetric, that is $$f(x,y)=f(y,x)$$ then $$\lambda$$ for this kind of eigenfunctions will be $$\lambda=1$$. Others are the antisymmetric ones, those for is true $$f(y,x)=-f(x,y)$$ and then the eigenvalue for this kind is $$\lambda=-1$$

But I'm sure that there are more conditions that generate other kinds of eigenfunctions, not only symmetric nor antisymmetric. My question is: there are more or the antisymmetric and symmetric ones are the only ones, and if there are more how I get them and their eigenvalues?

2. Sep 3, 2009

gabbagabbahey

If $f(y,x)=\lambda f(x,y)$ is to be true for all $(x,y)$, then surely it must be true for $x=y$....what does that tell you?

3. Sep 3, 2009

ELESSAR TELKONT

it tells me $$\lambda=1$$?

4. Sep 3, 2009

Avodyne

Here's another hint: what happens if you act on f with L twice? What does that tell you about $\lambda^2$?

5. Sep 3, 2009

Avodyne

Or that f(x,x)=0.

6. Sep 3, 2009

ELESSAR TELKONT

Or that operator $$L$$ is an involution, that's, it's its own inverse.

7. Sep 3, 2009

ELESSAR TELKONT

and then $$\lambda=\pm 1$$ necessarily and takes all values only if $$f\equiv 0$$. In fact zero function is symmetric and antisymmetric function at the same time.

8. Sep 3, 2009

gabbagabbahey

f(x,y)=0 is a trivial solution, and not really an eigenfunction...so it should be discarded.

That leaves you with $\lambda=1$ and symmetric eigenfunctions, or $\lambda=-1$ and antisymmetric eigenfunctions.