Hermitian Operators and Inner Products

[csnsc14320]

Homework Statement

Consider the vector space of square-integrable functions \psi(x,y,z) of (real space) position {x,y,z} where \psi vanishes at infinity in all directions. Define the inner product for this space to be

<\phi|\psi> = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \phi^* \psi dx dy dz

Show that the operator

\hat{L_z} = -i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right)

is a self-adjoint (Hermitian) operator on this space.

Homework Equations

The Attempt at a Solution

So I know that for an operator to be Hermitian that \hat{H} = \hat{H}^\dagger

so just starting from what I know...

<\hat{L} \phi|\psi> = <\phi|\hat{L}^\dagger|\psi>

<\hat{L} \phi|\psi> = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \left(-i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right) \phi \right)^* \psi dx dy dz = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right) \phi^* \psi dx dy dz

and

<\phi|\hat{L}^\dagger|\psi> = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \phi^* \left(i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right)\psi \right) dx dy dz = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right) \phi^* \psi dx dy dz

But I guess these are my two main questions:

For a non-matrix operator

\hat{L_z} = -i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right)

does

\hat{L_z}^\dagger = i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right)

??

and, can I move around my partials in the second equation to match that of my first or do I have to carry them out?

Thanks, and I'm sorry if what I said is complete junk seeing as how I'm not THAT strong in linear algebra... yet.

 

[CompuChip]

But I guess these are my two main questions:

For a non-matrix operator

\hat{L_z} = -i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right)

does

\hat{L_z}^\dagger = i \hbar \left(x \frac{\delta}{\delta y} - y \frac{\delta}{\delta x}\right)

Actually, L_z^\dagger = L_z -- that is what you are trying to show (it's precisely what being Hermitian means).

So the idea is that you start from
\langle L_z \phi \mid \psi \rangle
where L works on the first function, and use partial integration to show that it is equal to
\langle \phi \mid L_z \psi \rangle
where L works on the second function.

(Then, of course, since you know that it is also equal to \langle \phi \mid L_z^\dagger \psi \rangle you can conclude that L is equal to its Hermitian conjugate and you are done).