show that $-(\vec{r} \cdot \nabla)^2\psi = -r^2\frac{\partial^2\psi}{\partial r^2} -2r\frac{\partial\psi}{\partial r}$

Question

So, I'm trying to proof the relation $(\vec{r}\times\nabla)\cdot (\vec{r}\times\nabla)\psi = r^2\nabla^2\psi -r^2\frac{\partial^2\psi}{\partial r^2} -2r\frac{\partial\psi}{\partial r}$, where $\psi$ is a scalar (wave) function. That's is the problem 1.9.9 from "George B Arfken, Hans J Weber - Mathematical Methods For Physicists- Sixth edition". I know that it would be easier to proof using spherical polar coordinates, but until that chapter of the book he hasn't mention anything about spherical polar coordinates, so because of that, I would like to try that using only cartesian coordinates.

So, I was thinking about use the relation $(\vec{A}\times\vec{B})\cdot (\vec{A}\times\vec{B}) = (AB)^2 - (\vec{A}\cdot\vec{B})^2$, so the first term come easily out:

$\begin{equation}(\vec{r}\times\nabla)\cdot (\vec{r}\times\nabla)\psi = r^2\nabla^2\psi - (\vec{r}\cdot\nabla)^2\psi \end{equation}$

But now I'm having some trouble tho show that:

$\begin{equation}-(\vec{r} \cdot \nabla)^2\psi = -r^2\frac{\partial^2\psi}{\partial r^2} -2r\frac{\partial\psi}{\partial r}\end{equation}$

I think that it should be possible to show using only cartesian coordinates, but I'm not seeing this throw... Could someone please help me?

Answer 1

Let $r_1,r_2,r_3$ denote the Cartesian components of $\vec{r}$. All that is needed about spherical coordinates is

$$r = |\vec{r}|= \sqrt{r_1^2+ r_2^2+r_3^2}, \quad \frac{\partial r}{\partial r_j}= \frac{2r_j}{2 \sqrt{r_1^2+ r_2^2+r_3^2}}= \frac{r_j}{r}, \quad \frac{\partial f}{\partial r_j} = \frac{\partial f }{\partial r} \frac{\partial r}{\partial r_j} = \frac{\partial f }{\partial r}\frac{r_j}{r}$$

Using the Levi-Civita symbol, $\varepsilon_{ijk}$ and its property $\varepsilon_{ijk}\varepsilon_{imn} = \delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} $ we have with the Einstein summation convention

$$(\vec{r}\times\nabla)\cdot (\vec{r}\times\nabla)\psi = \varepsilon_{ijk}r_j\frac{\partial}{\partial r_k}\varepsilon_{imn}r_m\frac{\partial \psi}{\partial r_n} = (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} )r_j\frac{\partial}{\partial r_k}\left(r_m\frac{\partial \psi}{\partial r_n} \right) \\ = (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} )r_jr_m\frac{\partial}{\partial r_k}\left(\frac{\partial \psi}{\partial r_n} \right) + (\delta_{jm}\delta_{kn} - \delta_{jn}\delta_{km} )r_j\frac{\partial r_m}{\partial r_k}\frac{\partial \psi}{\partial r_n}\\ = \underbrace{r_j^2\frac{\partial\psi}{\partial r_k^2}}_A- \underbrace{r_jr_k\frac{\partial}{\partial r_k}\left(\frac{\partial \psi}{\partial r_j} \right)}_B+ \underbrace{r_j\frac{\partial r_j}{\partial r_k}\frac{\partial \psi}{\partial r_k}}_C- \underbrace{r_j\frac{\partial r_k}{\partial r_k}\frac{\partial \psi}{\partial r_j}}_D $$

As a reminder, the Einstein summation convention gives $r_j^2 = r_1^2 + r_2^2 + r_3^2 = r^2$. Simplifying each of the four terms on the RHS, we get

$$A= r_j^2\frac{\partial\psi}{\partial r_k^2} = r^2 \nabla^2\psi$$

$$B = r_jr_k\frac{\partial}{\partial r_k}\left(\frac{\partial \psi}{\partial r_j} \right) = r_jr_k\frac{\partial}{\partial r_k}\left(\frac{\partial \psi}{\partial r}\frac{r_j}{r} \right)\\= r_jr_k\frac{\partial}{\partial r_k}\left(\frac{\partial \psi}{\partial r} \right)\frac{r_j}{r} + r_jr_k\frac{\partial \psi}{\partial r}\frac{\partial r_j}{\partial r_k}\frac{1}{r}+ r_j^2r_k\frac{\partial \psi}{\partial r}\frac{-1}{r^2}\frac{\partial r}{\partial r_k}\\ = \frac{r_j^2r_k^2}{r^2}\frac{\partial^2 \psi}{\partial r^2} + \frac{r_j^2}{r}\frac{\partial \psi}{\partial r}- \frac{r_j^2r_k^2}{r^3} \frac{\partial \psi}{\partial r}= r^2 \frac{\partial^2 \psi}{\partial r^2}$$

$$C= r_j\frac{\partial r_j}{\partial r_k}\frac{\partial \psi}{\partial r_k}= r_j\frac{\partial \psi}{\partial r}\frac{r_j}{r} = r \frac{\partial \psi}{\partial r}$$

$$D = r_j\frac{\partial r_k}{\partial r_k}\frac{\partial \psi}{\partial r_j}=3 r_j \frac{\partial \psi}{\partial r} \frac{r_j}{r}= 3r \frac{\partial \psi}{\partial r}$$

Thus,

$$(\vec{r}\times\nabla)\cdot (\vec{r}\times\nabla)\psi = A- B+ C- D = r^2 \nabla^2\psi - r^2 \frac{\partial^2 \psi}{\partial r^2}+ r \frac{\partial \psi}{\partial r}- 3r \frac{\partial \psi}{\partial r} \\ =r^2 \nabla^2\psi - r^2 \frac{\partial^2 \psi}{\partial r^2}- 2r \frac{\partial \psi}{\partial r}$$