🍎 Classically $\vec L=\vec r\times \vec p$ so we can write a vector-values quantum operator
$\widehat{\underline{L}}=\widehat{\underline{r}}\times\widehat{\underline{p}}$
$$ \widehat{\underline{L}}=\left|\begin{array}{ccc}\underline{i}&\underline{j}&\underline{k}\\\widehat{x}&\widehat{y}&\widehat{z}\\\widehat{p}_x&\widehat{p}_y&\widehat{p}_z\end{array}\right| $$
From this we can show that
$$ \begin{aligned} \widehat{L}_z = \widehat{x}\widehat{p}_y-\widehat{y}\widehat{p}_x \\ \phantom{so}\ \widehat{L}_x = \widehat{y}\widehat{p}_z-\widehat{z}\widehat{p}_y \\ \phantom{so}\ \widehat{L}_y = \widehat{z}\widehat{p}_x-\widehat{x}\widehat{p}_z \end{aligned} $$
We are able to construct eigenstates of the orbital angular momentum vector only if its components commute (otherwise we wouldn’t be able to measure all of them at once)
$$ \begin{aligned}\left[\widehat{L}_z,\widehat{L}_x\right]\psi=& \left(\widehat{L}_z\,\widehat{L}_x-\widehat{L}_x\,\widehat{L}_z\right)\psi \\=& -\hbar^2\left(x\frac{\partial}{\partial y}-y\frac{\partial}{\partial x}\right)\left(y\frac{\partial\psi}{\partial z}-z\frac{\partial\psi}{\partial y}\right)\\&+\hbar^2\left(y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}\right)\left(x\frac{\partial\psi}{\partial y}-y\frac{\partial\psi}{\partial x}\right) \\=& \hbar^2\left(z\frac{\partial\psi}{\partial x}-x\frac{\partial\psi}{\partial z}\right) = i\hbar\widehat{L}_y\psi\end{aligned} $$
We can find the others by cyclic symmetry:
$$ \begin{aligned}\left[\widehat{L}_x,\widehat{L}_y\right]&=i\hbar\widehat{L}_z \\\left[\widehat{L}_y,\widehat{L}_z\right]&=i\hbar\widehat{L}_x \\\left[\widehat{L}_z,\widehat{L}_x\right]&=i\hbar\widehat{L}_y\end{aligned} $$
💎 Conclusion: Since the values are non zero we cannot know two components of $\vec L$ simultaneously
🗒️ Note: by convention we will usually define the value that we measure as $L_z$
The magnitude-squared of the orbital angular momentum $\widehat{L}^2=\widehat{L}_x^2+\widehat{L}_y^2+\widehat{L}_z^2$ does commute with the individual components
$$ \begin{aligned} \left[\widehat{L}^2,\widehat{L}_z\right] &= \left[\widehat{L}_x^2,\widehat{L}_z\right] + \left[\widehat{L}_y^2,\widehat{L}_z\right] + \left[\widehat{L}_z^2,\widehat{L}_z\right] \\&= \widehat{L}_x\left[\widehat{L}_x,\widehat{L}_z\right] + \left[\widehat{L}_x,\widehat{L}_z\right]\widehat{L}_x + \widehat{L}_y\left[\widehat{L}_y,\widehat{L}_z\right] + \left[\widehat{L}_y,\widehat{L}_z\right]\widehat{L}_y \\&= -i\hbar\left(\widehat{L}_x\widehat{L}_y+\widehat{L}_y\widehat{L}_x\right) +i\hbar\left(\widehat{L}_y\widehat{L}_x+\widehat{L}_x\widehat{L}_y\right) \\&= 0\end{aligned} $$
💎 Conclusion: we can know $|\vec L|$ and $L_z$ simultaneously but not $\vec L$
lets convert to polar coordinates using since most potentials are $V(x,y,z) = V(r)$
$$ \begin{aligned} z &= r\cos\theta \\ y &= r\sin\theta\,\sin\phi \\ x &= r\sin\theta\,\cos\phi\end{aligned} $$
In spherical polar coordinates we have
$$ \widehat{L}_z=-ih\frac{\partial}{\partial\phi} $$
For $\widehat{L}_y$ we have
$$ \begin{aligned} \widehat{L}_y &= \widehat{z}\widehat{p}_x-\widehat{x}\widehat{p}_z=\Bigl(r\cos\theta\Bigr)\Bigl(-i\hbar\Bigr)\left(\frac{\partial r}{\partial x}\,\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial x}\,\frac{\partial}{\partial\theta}+\frac{\partial\phi}{\partial x}\,\frac{\partial}{\partial\phi}\right)\\&\phantom{\widehat{z}\widehat{p}_x-}-\Bigl(r\sin\theta\cos\phi\Bigr)\Bigl(-i\hbar\Bigr)\left(\frac{\partial r}{\partial z}\,\frac{\partial}{\partial r}+\frac{\partial\theta}{\partial z}\,\frac{\partial}{\partial\theta}+\frac{\partial\phi}{\partial z}\,\frac{\partial}{\partial\phi}\right) \\&= -i\hbar\left(\phantom{-{}}\cos\phi\frac{\partial}{\partial\theta}-\cot\theta\sin\phi\frac{\partial}{\partial\phi}\right)\end{aligned} $$
Similarly for $\widehat L_x$ we have
$$ \widehat{L}_x= -i\hbar\left(-\sin\phi\frac{\partial}{\partial\theta}-\cot\theta\cos\phi\frac{\partial}{\partial\phi}\right) $$
We can find the value for $\widehat L^2$
$$ \widehat{L}^2=-\hbar^2\left(\frac{\partial^2}{\partial\theta^2}+\cot\theta\frac{\partial}{\partial\theta}+\frac1{\sin^2\theta}\,\frac{\partial^2}{\partial\phi^2}\right) $$
🗒️ Note: Its easier to see that $\widehat{L}^2$ commutes with $\widehat{L}_z$ because derivatives always commute with each other. $\widehat{L}_z$ only takes the derivative with respect to $\phi$ and none of the coefficients in $\widehat L^2$ are $\phi$-dependent, and although $\widehat L^2$ takes derivatives with respect to both $\theta$ and $\phi$, the coefficient in $\widehat L_z$ does not depend on either
$$ \left[\widehat{L}^2,\widehat{L}_z\right] = 0 $$
The TISE for a central potential is
$$ \left(\frac{-\hbar^2}{2m}\nabla^2+V(r)\right)\psi=E\psi $$
🧠 Remember: we can express $\nabla^2$ in spherical polar coordinates
$$ \nabla^2=\frac{\partial^2}{\partial r^2}+\frac2r\,\frac{\partial}{\partial r}+\frac1{r^2}\left[\frac{\partial^2}{\partial\theta^2}+\cot\theta\frac{\partial}{\partial\theta}+\frac1{\sin^2\theta}\,\frac{\partial^2}{\partial\phi^2}\right] $$
Using this we can write the TISE as
$$ \frac{-\hbar^2}{2m}\left(\frac{\partial^2}{\partial r^2}+\frac2r\,\frac{\partial}{\partial r}\right)\psi+\frac{\widehat{L}^2}{2mr^2}\psi+V(r)\psi=E\psi $$
We notice the solutions are seperable
$$ \psi(r,\theta,\phi)=R(r)\,Y(\theta,\phi) $$
Moreover if $Y$ is chosen to be the eigenfunction of $\widehat{L}^2$ then we have
$$ \widehat{L}^2\psi=L^2\psi $$
The two separable TISE’s become
$$ \begin{aligned} \left(\frac{-\hbar^2}{2m}\left(\frac{\partial^2}{\partial r^2}+\frac2r\,\frac{\partial}{\partial r}\right)+\frac{L^2}{2mr^2}+V(r)\right) R(r)&=ER(r) \\\widehat{L}^2Y(\theta,\phi)&=L^2Y(\theta,\phi) \end{aligned} $$
💎 Conclusion: Therefore, the eigenfunctions and eigenvalues of orbital angular momentum assume crucial importance in the solution of the Schrödinger equation for any central potential.
if a mass $m$ is at some point described by $r$ and $\phi$ its velocity vector can be decomposed into components in the $\widehat r$ and $\widehat \theta$ directions $v_r$ and $v_\phi$ respectively
Angular momentum is conserved in a central potential so
$$ L=mrv_\phi $$
We replace trade $v_\phi$ for $L$ in the expression for kinetic energy
$$ T = {\textstyle{\frac12}}m\left(v_r^2+v_\phi^2\right) = {\textstyle{\frac12}}m\dot{r}^2+\frac{L^2}{2mr^2} $$
The problem is effectively one-dimensional in the $\vec r$ direction but with $V(r)$ replaces by
$$ V_{ef\!f}(r) = V(r)+\frac{L^2}{2mr^2} $$
which is the effective potential