# 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 ...