How can you use the Levi-Civita symbol (index notation) to derive an expression for $(\hat{r}\times\nabla)\cdot(\hat{r} \times\nabla)$?

I was studying the expression for the momentum operator in quantum mechanics, and found a text that described this operator in spherical coordinates. However, during the derivation, the text states that:

$$(\hat{r}\times\nabla)\boldsymbol{\cdot}(\hat{r} \times\nabla)=\nabla^2-\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) \tag{1}$$

Previously I have used the Binet-Cauchy identity which states;

$$(\textbf{a} \times \textbf{b})\boldsymbol{\cdot}(\textbf{c} \times \textbf{d}) = (\textbf{a} \boldsymbol{\cdot} \textbf{b})(\textbf{c} \boldsymbol{\cdot} \textbf{d}) - (\textbf{a} \boldsymbol{\cdot} \textbf{d}) (\textbf{b} \boldsymbol{\cdot} \textbf{c})$$

Applying this rule to (1), we get:

$$(\hat{r}\times\nabla)\boldsymbol{\cdot}(\hat{r} \times\nabla)=\nabla^2-(\hat{r}\boldsymbol{\cdot}\nabla)(\nabla \boldsymbol{\cdot}\hat{r})$$

However, I cannot go further with this expression, and I struggle to get (1) using Binet-Cauchy. I therefore choose to use the Levi-Civita symbol (index notation) to derive the correct answer, but I want help to derive (1) using this formalism.

Eq. (1) is used during the derivation of the squared angular momentum operator. The angular momentum operator is given as;

$$\hat{L}=\hat{R} \times \hat{P}=(-i \hbar r)\hat{r} \times \boldsymbol{\nabla}$$

And the square of the angular momentum operator is thus [using (1)];

$$\hat{L}^2=(- \hbar^2 r^2)(\hat{r} \times \boldsymbol{\nabla})(\hat{r} \times \boldsymbol{\nabla})=(-\hbar^2r^2)\left[\nabla^2-\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)\right]$$

Now since we know the laplace operator in spherical coordinates, $$\nabla^2=\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right)+\frac{1}{r^2 \text{sin}\theta}\frac{\partial}{\partial \theta}\left(\text{sin}\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{r^2\text{sin}^2\theta}\frac{\partial^2}{\partial\phi^2}$$, we can easily see that the $$r$$-component disappears in the expression of the squared angular momentum operator, leaving;

$$\hat{L}^2=(-\hbar^2r^2)\left[\frac{1}{r^2 \text{sin}\theta}\frac{\partial}{\partial \theta}\left(\text{sin}\theta\frac{\partial}{\partial\theta}\right)+\frac{1}{r^2\text{sin}^2\theta}\frac{\partial^2}{\partial\phi^2}\right]$$

But this all rests on eq. (1) to be true, so how can we use the index-notation to derive (1)?

• It might be relevant that right side is the purely angular part of the Laplacian. Commented Jul 14, 2023 at 21:47
• Have you tried showing this by using $\nabla$ in spherical coordinates? Commented Jul 14, 2023 at 21:51
• This question is about math, not physics. Commented Jul 14, 2023 at 22:02
• The BCI doesn't always work when differential operators are involved, as its proof makes commutativity assumptions.
– J.G.
Commented Jul 14, 2023 at 22:52

Recall that $$\frac{\partial}{\partial r}= \hat r\cdot \nabla,\\ \left [ \frac{\partial}{\partial r}, \hat r\right ]=0.$$
You then simply slug through $$(\hat{r}\times\nabla)\boldsymbol{\cdot}(\hat{r} \times\nabla)\\ = \epsilon^{ijk}\epsilon_{imn}\frac{r_j}{r}\partial_k \left (\frac{r^m}{r}\partial^n\right)\\ =(\delta^j_m \delta^k_n - \delta^j_n \delta^k_m)\left (\delta^m_k\frac{r_j}{r^2}\partial^n- \frac{r_jr_kr^m}{r^4}\partial^n+\frac{r_jr^m}{r^2}\partial_k\partial^n\right )\\ = \partial^n \partial_n - \left (\frac{r^jr^n}{r^2}\partial _j \partial_n+\frac{2}{r} \frac{\partial}{\partial r} \right)\\ =\nabla^2-\left (\frac{r^n}{r}\frac{\partial}{\partial r} \partial_n+\frac{2}{r} \frac{\partial}{\partial r} \right)= \nabla^2-\left ( \frac{\partial^2}{\partial r^2}+\frac{2}{r} \frac{\partial}{\partial r} \right)\\ =\nabla^2-\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) .$$
• Indeed, for a flat space as here, upper and lower indices are equivalent. The second identity provided, provable from WP, allows you to commute $\hat r$ with $\partial_r$, so as to fuse it with the rightmost gradient: $(\hat r \partial _r)\cdot \nabla=\partial_r (\hat r \cdot \nabla)$. Commented Jul 16, 2023 at 10:08
• Ah I see! So we have $\frac{r^n}{r}\partial_r\partial_n=\frac{r}{r}[\hat{r}\partial_r]\cdot\nabla=\partial_r[\hat{r}\cdot\nabla]=\partial_r^2$ because $[r^n\partial_r]\partial_n=[\textbf{r}\partial_r]\cdot\nabla$ Commented Jul 16, 2023 at 10:18