the Momentum operator in spherical coordinate

\nabla^2 = \frac {1}{r^2}\frac {\partial } { \partial r} \left ( r^2 \frac {\partial} {\partial r} \right ) - \frac {1}{r^2} L^2

where L is the Reduced angular momentum operator. the minus sign is very important for giving a correct sign. the original angular momentum operator J is related by:

J=\hbar^2 L

by compare the Laplacian in spherical coordinate, the L is

L^2 = - \frac {1}{sin(\theta)} \frac {\partial}{\partial \theta} \left( sin(\theta) \frac {\partial}{\partial \theta} \right ) - \frac {1}{sin(\theta)} \frac{\partial^2} {\partial \phi ^2}

But this complicated form is rather useless, expect you are mathematic madman.

we can start from classical mechanic

\vec{L} = \vec {r} \times \vec{p}

L_x = y \frac {\partial} {\partial z} - z \frac {\partial}{\partial y }

L_y = z \frac {\partial} {\partial x} - x \frac {\partial}{\partial z }

L_z = x \frac {\partial} {\partial y} - y \frac {\partial}{\partial x }

with the change of coordinate

\begin {pmatrix} x \\ y \\ z \end{pmatrix} = \begin {pmatrix} r sin(\theta) cos(\phi) \\ r sin(\theta) sin(\phi) \\ r cos(\theta) \end{pmatrix}

and the Jacobian Matrix M_J , which is used for related the derivatives.

since

\frac {\partial}{\partial x} = \frac {\partial r}{\partial x} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial x} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial x} \frac {\partial} {\partial \phi}

\frac {\partial}{\partial y} = \frac {\partial r}{\partial y} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial y} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial y} \frac {\partial} {\partial \phi}

\frac {\partial}{\partial z} = \frac {\partial r}{\partial z} \frac {\partial} {\partial r} +\frac {\partial \theta}{\partial z} \frac {\partial} {\partial \theta}+\frac {\partial \phi}{\partial z} \frac {\partial} {\partial \phi}

which can be simplify

\nabla_{(x,y,z)} = M_J^T \nabla_{(r, \theta, \phi )}

M_J = \frac {\partial ( r, \theta, \phi) }{\partial (x,y,z)}

M_J^{\mu\nu} = \frac {\partial \mu}{\partial \nu}

then, we have

L_x = i sin(\phi) \frac {\partial }{\partial \theta} +i cot(\theta) cos(\phi) \frac { \partial }{\partial \phi}

L_y =-i cos(\phi) \frac {\partial }{\partial \theta} + i cot(\theta) sin(\phi) \frac { \partial }{\partial \phi}

L_z = - i \frac {\partial }{\partial \phi}

However, even we have the functional form, it is still not good.  we need the ladder operator

L_+ = L_x + i L_y = Exp(i \phi) \left( \frac {\partial }{\partial \theta} + i cot(\theta) \frac { \partial }{\partial \phi} \right)

L_- = L_x - i L_y = Exp(-i \phi) \left( \frac {\partial }{\partial \theta} - i cot(\theta) \frac { \partial }{\partial \phi} \right)

notice that

L_+^\dagger = L_-

so, just replacing i \rightarrow -i .

when we looking for the Maximum state of the spherical Harmonic Y_{max}(\theta, \phi)

L_+ Y_{max}(\theta,\phi) = 0 *)

use the separable variable assumption.

Y_{max}(\theta, \phi) = \Theta \Phi

L_+ \Theta \Phi = 0 = - Exp(i \phi) \left( \frac {d\Theta}{d \theta} \Phi + i cot(\theta) \frac { d\Phi}{d\phi} \right) \Theta

\frac {tan(\theta)}{\Theta} \frac { d \Theta} {d \Theta } = - \frac {i}{\Phi} \frac {d \Phi} {d \phi} = m

the solution is

Y_{max}(\theta,\phi) = sin^m(\theta) Exp(i m \phi )

L^2 Y_{max}(\theta, \phi) = m(m+1) Y_{max}(\theta,\phi)

an application on Hydrogen wave function is here.

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