Homework Help: Hermiticity of permutation operator

1. Aug 11, 2011

issacnewton

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

Any help ?

2. Aug 11, 2011

mathfeel

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?

3. Aug 11, 2011

issacnewton

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) ?

4. Aug 11, 2011

mathfeel

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: Aug 11, 2011
5. Aug 11, 2011

issacnewton

So since all eigenvalues of $\hat{\pi}$ are real , its hermitian. fine thats solved.
what about the second part about the eigenfunctions ? can you help ?

6. Aug 11, 2011

vela

Staff Emeritus
Just apply the operator to the functions and show that you get an eigenvalue times the function.

7. Aug 12, 2011

issacnewton

vela, yes that's one approach. but how do I derive these eigenfunctions in the first place ?

8. Aug 12, 2011

mathfeel

There are really no general approach for arbitrary operator. A good guess is a start.

9. Aug 12, 2011

vela

Staff Emeritus
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. Aug 12, 2011

issacnewton

thanks fellas. makes some sense now...........