Show Lx is Hermitian: Homework Equation Integration

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To demonstrate that Lx is Hermitian in 3D, the integral condition Integral(f Lx g) = Integral(g Lx* f) must be satisfied, where Lx is defined as (hbar)/i (y d/dz - z d/dy). The discussion emphasizes the need for integration by parts to evaluate the integrals involving functions f and g. A participant struggles with determining the correct expressions for u and dv in the integration process but eventually clarifies that v should be g(x,y,z). The conclusion drawn is that Lx generates a uniparametric subgroup of unitary operators, confirming its Hermitian nature through the application of Stone's theorem. The conversation ultimately affirms that Lx is indeed Hermitian as all e.s.a. operators are symmetric.
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Homework Statement



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

Homework 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)

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!
 
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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.
 
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 :(
 
Focus on the z integral (because the derivative is with respect to z). So du = (df/dz)y.
 
oh, and then v is just g(x,y,z)...
 
Yep!
 
great, thanks!
 
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
 
Yeah, that's what I *meant* to say ...
 
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