Hermiticity and expectation value

[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.

 

[Euclid]

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

 

[holden]

heh, thanks. i suck.

 

[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.

 

[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.

 

[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.

 

[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.