# Hermiticity and expectation value

1. Oct 7, 2006

### holden

is there a better way to check for hermicity than doing expecation values? for example, what if you had xp (operators) - px (operators), or pxp (operators again); how can you tell if these combos are hermetian or not, without going through the clumsy integration (that doesn't give a solid result, as far as i can tell)? thanks.

2. Oct 7, 2006

### Euclid

$$(AB+CD)^\dagger = B^\dagger A^\dagger + D^\dagger C^\dagger$$

3. Oct 7, 2006

### holden

heh, thanks. i suck.

4. Oct 7, 2006

### holden

although, actually, i still have a question.. where does that come from, is it just an accepted property? i can't seem to derive it or find a place where it has been derived.

Last edited: Oct 7, 2006
5. Oct 8, 2006

### Euclid

$$\left< (AB)x,y \right> = \left< A(Bx),y \right> = \left< Bx, A^\dagger y \right> = \left< x, (B^\dagger A^\dagger) y \right>$$

By definition, this shows that $$(AB)^\dagger=B^\dagger A^\dagger$$.

Last edited: Oct 8, 2006
6. Oct 9, 2006

### dextercioby

What Euclid wrote is valid only for bounded operators. The "x" and "p_x" operators are unbounded, so care is needed in order not to write some garbage.

Daniel.

7. Oct 17, 2006

### kcirick

However, in the context of what holden wrote, what Euclid stated is valid enough, because holden wants to check for hermicity and hermitian operators are symmetric and bounded.

On the other note, I want to see a somewhat rigorous proof that
$$\left(A+B\right)^{\dagger} = A^{\dagger}+B^{\dagger}$$

All the sources say this property, but they don't bother proving it.