Square of x component of Orbital Angular momentum?

[d3nat]

Homework Statement

Solve for ##L^2_x##
##L_x = \frac{\hbar}{i} (-sin(\phi)\frac{d}{d\theta} - cos(\phi)cot(\theta)\frac{d}{d\phi}##** the d's should be partial derivatives, but I'm not sure how to do that symbol. Sorry!

Homework Equations

Solve for ##L^2_x##
##L_x = \frac{\hbar}{i} (-sin(\phi)\frac{d}{d\theta} - cos(\phi)cot(\theta)\frac{d}{d\phi}##** the d's should be partial derivatives, but I'm not sure how to do that symbol. Sorry!

The Attempt at a Solution

I'm trying to square everything
But can I square partial derivatives? Is that the same thing as a second derivative?

so ##-\hbar^2 \Big( sin^2(\phi)\frac{d^2}{d^2\theta} + cos^2(\phi)cot^2(\theta)\frac{d^2}{d^2\phi} +sin(\phi)\frac{d}{d\theta}(cos(\phi)cot(\theta)\frac{d}{d\phi}) +cos(\phi)cot(\theta)\frac{d}{d\phi}(sin(\phi)\frac{d}{d\theta}) \Big)##

Can I take a partial derivative of a partial derivative?
I'm so stumped. Can't seem to find any explanations online. Help please?

[EDIT:]

Is this the simplest form?

##-\hbar^2 \Big( sin^2(\phi)\frac{d^2}{d^2\theta} + cos^2(\phi)cot^2(\theta)\frac{d^2}{d^2\phi} -sin^2(\phi)cos(\phi)cot(\theta)\frac{d}{d\phi}+cos^2(phi)cot(\theta)\frac{d}{d\theta}##

I still don't think this is right.

 

[fzero]

This can be tricky when you come across it for the first time. The problem is that the derivatives also act on the functions appearing in the operator, so it can be useful to put in an arbitrary placeholder function to help remember how to apply the product rule for derivatives. For example, suppose we have

$$ \hat{T} = \sin x \frac{d}{dx}.$$

Then to compute ##\hat{T}^2##, we act on a function ##F(x)##:

$$ \hat{T}^2 F(x) = \sin x \frac{d}{dx}\left( \sin x \frac{dF(x)}{dx} \right) = \sin x \left(\frac{d}{dx} \sin x\right) \frac{dF(x)}{dx} + \sin^2 x \frac{d^2F(x)}{dx^2} .$$

Completing the algebra and removing the placeholder function, we'll find that

$$ \hat{T}^2 = \sin^2 x \frac{d^2}{dx^2} + \sin x \cos x \frac{d}{dx} .$$

Your example is a bit more complicated, but hopefully this gives you an idea of how to proceed.