# 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...........