Simple angular momentum problem. QM

[armis]

Homework Statement

Show that Lx is a Hermitian operator

Homework 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

The Attempt at a Solution

I do realize 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

 

[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.

 

[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 <br /> \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>

 

[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