# Hermitian Operators and Inner Products

1. Oct 18, 2009

### csnsc14320

1. The problem statement, all variables and given/known data

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.
2. Relevant equations

3. 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.

2. Oct 19, 2009

### CompuChip

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).