Operator that interchanges variables

[ELESSAR TELKONT]

Homework Statement

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.

Homework Equations

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?

 

[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?

 

[ELESSAR TELKONT]

it tells me \lambda=1?

 

[Avodyne]

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

 

[Avodyne]

it tells me \lambda=1?

Or that f(x,x)=0.

 

[ELESSAR TELKONT]

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

 

[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.

 

[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.