Homework Help: Show Lx is Hermitian

1. Nov 19, 2007

physgirl

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

I have to show that in 3-d, Lx (angular momentum) is Hermitian.

2. Relevant equations

In order to be Hermitian: Integral (f Lx g) = Integral (g Lx* f)
Where Lx=(hbar)/i (y d/dz - z d/dy)
and f and g are both well behaved functions: f(x,y,z) and g(x,y,z)

3. The attempt at a solution

I know to do this I have to do integration by parts. I got to the point where I had to figure out, using integration by parts,: Integral [f(x,y,z) y (dg(x,y,z)/dz) dx]

And I cannot figure this out :(

I set:
u=f(x,y,z) y
dv=(dg(x,y,z)/dz) dx

So then I get: du=[df(x,y,z)/dx]y + f(x,y,z)
But what is v then?? Unless I'm completely off-track already, in which case, help would be great!

2. Nov 19, 2007

Avodyne

You need to multiply by dx dy dz, and integrate over all three, not just dx. This should make the integration by parts much easier.

3. Nov 19, 2007

physgirl

As in...

u=f(x,y,z)y
dv=(dg(x,y,z)/dz) dxdydz

So that...

du = (df/dx)y + (df/dx)y + f + (df/dz)y
v=...?

Still not sure :(

4. Nov 19, 2007

Avodyne

Focus on the z integral (because the derivative is with respect to z). So du = (df/dz)y.

5. Nov 19, 2007

physgirl

oh, and then v is just g(x,y,z)...

6. Nov 19, 2007

Avodyne

Yep!

7. Nov 19, 2007

physgirl

great, thanks!!

8. Nov 21, 2007

dextercioby

Since L_{x} (or rather its closure in the strong topology of L^{2}(R^3)) generates a uniparametric subgroup of the group of unitary operators which represent a rotation (around an arbitrary axis) in a Hilbert space, then, by Stone's theorem, L_{x} is e.s.a. and its closure is s.a. But all e.s.a. operators are hermitian/symmetric. QED

9. Nov 21, 2007

Avodyne

Yeah, that's what I *meant* to say ...