A central force is one that comes from a potential energy function with spherical symmetry, that is, a function that only depends on the distance to the origin of the particle: $V=V(r)$ Thus: $\left(\frac{ \partial V}{\partial\theta}\right)_{r,\varphi}=0 ;$ and $\left(\frac{\partial V}{\partial\varphi}\right)_{r,\ theta}=0 ;$
Let us now consider the quantum mechanics of a simple particle subjected to a central force:
\begin{equation} \hat{H}=\hat{T}+\hat{V}=-\frac{\hbar^2}{2m}\nabla^2+V(r) \end{equation}
Let's express $\nabla^2$ in spherical polar coordinates:
\begin{equation} \nabla^2=\frac{\partial^2 }{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial^2}{\partial \theta^2}+\frac{1}{r^2}\cot\theta\frac{\partial}{\partial\theta}+\frac{1}{r^2 sin^2\theta}\frac {\partial^2}{\partial\varphi^2} \end{equation}
Remembering the expression of the operator $\hat{l}^2$:
\begin{equation}\label{ec3} \hat{l}^ 2=-\hbar^2\left(\frac{\partial^2}{\partial\theta^2}+\cot\theta\frac{\partial}{\partial\theta}+\frac{1}{sin^2\theta}\frac{\partial^2}{\partial\varphi^2}\right) \end{equation}
Substituting equation (3 ) in (2) we get:
\begin{equation}\label{ec4} \nabla^2=\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac {\partial}{\partial r}-\frac{1}{\hbar^2 r^2}\hat{l}^2 \end{equation}
Taking equation (4) to (1) the Hamiltonian for a one-particle system becomes:
\begin{equation}\label{ec5} \hat{H}=-\frac{\hbar^2}{2m}\left(\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}\right)+\frac{1}{2mr^2}\hat{l}^2+V(r ) \end{equation}
In quantum mechanics, we ask if the existence of states with definite values for energy and angular momentum is possible. For the set of eigenfunctions of $\hat{H}$ to be eigenfunctions of $\hat{l}^2$, the commutator $[\hat{H},\hat{l}^2]$ must be canceled :
\begin{equation} [\hat{H},\hat{l}^2]=[\hat{T}+\hat{V},\hat{l}^2]=[\hat{T} ,\hat{l}^2]+\underbrace{[\hat{V},\hat{l}^2]}_{0} \end{equation}
The commutator $[\hat{V},\hat {l}^2]$ is zero since the operators $\hat{l}(\theta,\varphi)$ and $V(r)$ do not have common variables.
\begin{equation} [\hat{T},\hat{l}^2]=\left[-\frac{\hbar^2}{2m} \left(\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}\right)+\frac{1}{2mr^2}\hat{l}^2,\hat{l }^2\right] \end{equation}
Applying commutator properties:
\begin{equation} [\hat{T},\hat{l}^2]=-\frac{\hbar^2}{2m}\left[\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r},\hat{l}^2\right]+\frac{1}{2m}\left[\frac{1}{r^2}\hat{l}^2,\hat{l}^2\right]=0 \end{equation}
Similarly, it can be prove that: $[\hat{H},\hat{l}_z]=0$ and $[\hat{l}^2,\hat{l}_z]$.
This means that we can have a set of simultaneous eigenfunctions of $\hat{H},\hat{l}^2,\hat{l}_z$ for the central forces problem.
If we denote by $\psi$ the common eigenfunctions
\begin{eqnarray} \hat{H}\Psi & = & E\Psi\\ \hat{l}^2 \Psi & = & \hbar^2l(l+ 1)\Psi\;\;l=0,1,2...\\ \hat{l}_z \Psi & = & m\hbar\Psi\;\;m=-l,...,0 ....+l \end{eqnarray}
The Schrödinger equation can be written:
\begin{equation} \left[-\frac{\hbar^2}{2m}\left(\frac{\partial^2} {\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}\right)+\frac{1}{2mr^2}\hat{l}^2+V (r)\right]\Psi=E\Psi \end{equation}
The wave function can be written as the product of a radial part $R(r)$ and a spherical harmonic $Y_{l,m}(\theta ,\varphi)$. This separation of variables is possible because the potential has spherical symmetry (only dependent on r).
\begin{equation} \Psi = R(r)Y_{l,m}(\theta,\varphi) \end{equation} Substituting into the Schrödinger equation \begin{equation} \left[-\frac{\hbar ^2}{2m}\left(\frac{\partial^2}{\partial r^2}+\frac{2}{r}\frac{\partial}{\partial r}\right)+\frac {1}{2mr^2}\hat{l}^2+V(r)\right]R(r)Y_{l,m}(\theta,\varphi)=ER(r)Y_{l,m }(\theta,\varphi) \end{equation}
\begin{equation} -\frac{\hbar^2}{2m}\left(\frac{\partial^2 R(r)}{\partial r^ 2}+\frac{2}{r}\frac{\partial R(r)}{\partial r}\right)Y_{l,m}(\theta,\varphi)+\frac{l(l+ 1)\hbar^2}{2mr^2}R(r)Y_{l,m}(\theta,\varphi)+\\+V(r)R(r)Y_{l,m}(\theta ,\varphi)=ER(r)Y_{l,m}(\theta,\varphi) \end{equation}
Dividing by $Y_{lm}(\theta,\varphi)$
\begin{equation} -\frac {\hbar^2}{2m}\left(\frac{\partial^2 R(r)}{\partial r^2}+\frac{2}{r}\frac{\partial R(r)} {\partial r}\right)+\frac{l(l+1)\hbar^2}{2mr^2}R(r)+V(r)R(r)=ER(r) \end{equation}
In this last equation, called the radial Schrödinger equation, it is observed that for any problem of a particle, with a function potential energy of spherical symmetry V(r), the eigenfunction $\Psi$ is the product of a radial factor and a spherical harmonic.