# Simple angular momentum problem. QM

1. Nov 15, 2008

### armis

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

Show that Lx is a Hermitian operator

2. Relevant equations

Well, since Lx is an observable it must be represented by an Hermitian operator.

$$L = -i\cdot\hbar\cdotr\times\nabla$$

If an operator is Hermitian, then it's equal to it's Hermitian conjugate

3. The attempt at a solution

I do realise what I have to do, however there are holes in my math.

So in order to show that the Lx operator is Hermitian I could show that:

$$< L_{x} f | f > = < f | L_{x} f >$$ is that correct?

If I assume that's correct and since $$< L_{x} f | f > = (< f | L_{x} f >)^*$$ I have a complex conjugate missing somewhere. Thus there is an error

2. Nov 15, 2008

### CompuChip

You don't have a conjugate missing, the identity on the last line is true.
The point is that, by definition of Hermitian conjugate, if you want,
$$\langle f \mid L_x f \rangle = \langle L_x^\dagger f \mid f \rangle$$
so if you show that it's also $\langle L_x f \mid f \rangle$ then effectively you are showing that $L_x^\dagger = L_x$.

3. Nov 15, 2008

### armis

Thanks. That I understand but I probably don't understand something else or I don't really understand either, anyway to be more specific here is what I mean

$$\langle f \mid L_{x} f \rangle$$ which is

<f|-ih(y*d/dz - z*d/dy)f>. (I am sorry, Latex doesn't show up for this one. I guess I typed it wrong)

now if $$L_{x}$$ is Hermitian then as far as I can understand the above expression must be equal to

$$\langle L_{x} f \mid f \rangle$$ which is $$\langle f \mid L_{x}f \rangle^*$$

and here is my problem, I get that $$\langle L_{x} f \mid f \rangle$$ is equal to $$\langle f \mid L_{x} f \rangle$$ conjugated and because of the i in the $$L_{x}$$ expression the two don't really match, do they? What am I missing?

Just to be absolutely clear I don't like the fact that $$\langle L_{x} f \mid f \rangle$$ is <-ih(y*d/dz - z*d/dy)f|f> doesn't seem to match with <f|-ih(y*d/dz - z*d/dy)f>

4. Nov 16, 2008

### malawi_glenn

you can give a proof of this since the L_x operator is = y*p_z - z*p_y

now you proove that x_ip_j is hermitian. i.e show that

$$\int ( x_i\hat{p}_j \phi)^*\psi d^3r = \int \phi ^*x_i\hat{p}_j\psi d^3r$$