Hermiticity of permutation operator

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The discussion centers on the properties of the permutation operator defined on the Hilbert space of two-variable complex functions, specifically its linearity, hermiticity, and eigenfunctions. The operator is confirmed to be linear and its square equals the identity operator, indicating that its eigenvalues are ±1, which are real. The participants clarify that while real eigenvalues suggest hermiticity, they do not guarantee it without further verification. To demonstrate hermiticity, the integral condition involving the functions is established, confirming that the operator is indeed hermitian. The conversation concludes with suggestions on deriving the eigenfunctions through symmetry and direct matrix representation.
issacnewton
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Hi

here's a problem I am having.

Consider the hilbert space of two-variable complex functions \psi (x,y).

A permutation operator is defined by its action on \psi (x,y) as follows.

\hat{\pi} \psi (x,y) = \psi (y,x)

a) Verify that operator is linear and hermitian.

b) Show that

\hat{\pi}^2 = \hat{I}

find the eigenvalues and show that the eigenfunctions of \hat{\pi} are given by

\psi_{+} (x,y)= \frac{1}{2}\left[ \psi (x,y) +\psi (y,x) \right]

and

\psi_{-} (x,y)= \frac{1}{2}\left[ \psi (x,y) -\psi (y,x) \right]

I could show that the operator is linear and also that its square is unity operator I . I did
find out the eigenvalues too. I am having trouble showing that its hermitian and the
part b about its eigenfunctions.

Any help ?
 
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After you already show that \hat{\pi} is linear. Take a look at \hat{\pi}^2. What is it? From this what can you infer about the possible eigenvalues for \hat{pi}? What properties for the eigenvalues for an operator implies Hermiticity?
 
Hi , eigenvalues are \pm 1 they are real. I know that hermitian operators have real eigenvalues but does it mean that if operator has real eigenvalues then its a hermitian
operator ( this is converse statement , so not necessarily true) ?
 
IssacNewton said:
Hi , eigenvalues are \pm 1 they are real. I know that hermitian operators have real eigenvalues but does it mean that if operator has real eigenvalues then its a hermitian
operator ( this is converse statement , so not necessarily true) ?

If all the eigenvalues of an operators are real.

More explicitly, you are trying to show that:
\int_{-\infty}^{\infty} \psi(r)^{*} \phi(-r) dr = \int_{-\infty}^{\infty} \psi(-r)^{*} \phi(r) dr
which is easy by simple change of variable.
 
Last edited:
So since all eigenvalues of \hat{\pi} are real , its hermitian. fine that's solved.
what about the second part about the eigenfunctions ? can you help ?
 
Just apply the operator to the functions and show that you get an eigenvalue times the function.
 
vela, yes that's one approach. but how do I derive these eigenfunctions in the first place ?
 
IssacNewton said:
vela, yes that's one approach. but how do I derive these eigenfunctions in the first place ?

There are really no general approach for arbitrary operator. A good guess is a start.
 
I suppose you could make a hand-waving argument that the Hilbert space can be written as a direct sum of subspaces, where each subspace is the span of Φ(x,y) and Φ(y,x) for a particular Φ(x,y). Then find the matrix representing \hat{\pi} in one of these subspaces and diagonalize it. Blah, blah, blah...

It's more useful, though, to recognize this pattern of combining elements to achieve a certain symmetry.
 
  • #10
thanks fellas. makes some sense now...
 

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