Operator that interchanges variables

  • Thread starter Thread starter ELESSAR TELKONT
  • Start date Start date
  • Tags Tags
    Operator Variables
Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
7 replies · 2K views
ELESSAR TELKONT
Messages
39
Reaction score
0

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 [tex]Lf=\lambda f[/tex]. This operator, applied to a function [tex]f(x,y)[/tex], interchanges the variables i.e. [tex]Lf(x,y)=f(y,x)[/tex]. 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 [tex]Lf(x,y)=f(y,x)[/tex] then if [tex]f[/tex] is an eigenfunction, obviously, [tex]\lambda f(x,y)=f(y,x)[/tex]. One possible kind of [tex]f[/tex] that fills conditions is one that is symmetric, that is [tex]f(x,y)=f(y,x)[/tex] then [tex]\lambda[/tex] for this kind of eigenfunctions will be [tex]\lambda=1[/tex]. Others are the antisymmetric ones, those for is true [tex]f(y,x)=-f(x,y)[/tex] and then the eigenvalue for this kind is [tex]\lambda=-1[/tex]

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?
 
Physics news on Phys.org
If [itex]f(y,x)=\lambda f(x,y)[/itex] is to be true for all [itex](x,y)[/itex], then surely it must be true for [itex]x=y[/itex]...what does that tell you?:wink:
 
it tells me [tex]\lambda=1[/tex]?
 
Here's another hint: what happens if you act on f with L twice? What does that tell you about [itex]\lambda^2[/itex]?
 
ELESSAR TELKONT said:
it tells me [tex]\lambda=1[/tex]?
Or that f(x,x)=0.
 
Or that operator [tex]L[/tex] is an involution, that's, it's its own inverse.
 
and then [tex]\lambda=\pm 1[/tex] necessarily and takes all values only if [tex]f\equiv 0[/tex]. In fact zero function is symmetric and antisymmetric function at the same time.
 
f(x,y)=0 is a trivial solution, and not really an eigenfunction...so it should be discarded.

That leaves you with [itex]\lambda=1[/itex] and symmetric eigenfunctions, or [itex]\lambda=-1[/itex] and antisymmetric eigenfunctions.