Proving an identity for the quantum angular momentum operator I am trying to prove the following identity for the quantum angular momentum operator:
$$\vec L^2=-\hbar^2r^2(\hat r \times \boldsymbol{\nabla})\bullet (\hat r \times \boldsymbol{\nabla})=-\hbar^2r^2\left[\nabla^2-\dfrac{1}{r^2} \dfrac{\partial}{\partial r}(r^2\dfrac{\partial}{\partial r})\right]$$
So basically I have to prove:
$$(\hat r \times \boldsymbol{\nabla})\bullet (\hat r \times \boldsymbol{\nabla})=\nabla^2-\dfrac{1}{r^2} \dfrac{\partial}{\partial r}(r^2\dfrac{\partial}{\partial r})$$
I was trying to use vector identities like $A \times (B \times C)=(A \cdot C)B-(A\cdot B)C$ , $A\cdot(B\times C)=C\cdot(A\times B)$ or $A\times B=-B\times A$, but I am unsure if they hold when $\boldsymbol{\nabla}$ is one or more of the vectors. Any help please?
 A: We'll work in Cartesian coordinates, summing repeated indices. If $f$ is a scalar function,$$\partial_k(\tfrac{x_l}{r})=\tfrac{r^2\delta_{kl}-x_kx_l}{r^3}\implies\partial_k(\tfrac{x_l}{r}\partial_mf)=\tfrac{(r^2\delta_{kl}-x_kx_l)\partial_mf+r^2x_l\partial_k\partial_mf}{r^3}.$$Hence$$\begin{align}[(\hat{r}\times\nabla)f]_i&=\epsilon_{ilm}\tfrac{x_l}{r}\partial_mf,\\(\hat{r}\times\nabla)\cdot(\hat{r}\times\nabla)f&=\epsilon_{ijk}\epsilon_{ilm}\tfrac{x_j}{r}\partial_k(\tfrac{x_l}{r}\partial_mf)\\&=(\delta_{jl}\delta_{km}-\delta_{jm}\delta_{kl})\tfrac{x_j(r^2\delta_{kl}-x_kx_l)\partial_mf+r^2x_jx_l\partial_k\partial_mf}{r^4}\\&=\nabla^2f-\tfrac{2(x\cdot\nabla)f+x_jx_k\partial_j\partial_kf}{r^2}.\end{align}$$Since $\partial_r=\tfrac{x_i}{r}\partial_i$,$$\tfrac{1}{r^2}\partial_r(r^2\partial_rf)=\tfrac{x_i}{r^3}\partial_i(x_jr\partial_jf)=\tfrac{x_i}{r^3}(r\partial_i+\tfrac{x_ix_j}{r}\partial_j+x_jr\partial_i\partial_j)=\tfrac{2(x\cdot\nabla)f+x_ix_j\partial_i\partial_jf}{r^2},$$so $(\hat{r}\times\nabla)\cdot(\hat{r}\times\nabla)f=\nabla^2f-\tfrac{1}{r^2}\partial_r(r^2\partial_rf)$ as claimed.
