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)?
Best Answer
Recall that $$ \frac{\partial}{\partial r}= \hat r\cdot \nabla,\\ \left [ \frac{\partial}{\partial r}, \hat r\right ]=0. $$
You then simply slug through $$(\hat{r}\times\nabla)\boldsymbol{\cdot}(\hat{r} \times\nabla)\\ = \epsilon^{ijk}\epsilon_{imn}\frac{r_j}{r}\partial_k \left (\frac{r^m}{r}\partial^n\right)\\ =(\delta^j_m \delta^k_n - \delta^j_n \delta^k_m)\left (\delta^m_k\frac{r_j}{r^2}\partial^n- \frac{r_jr_kr^m}{r^4}\partial^n+\frac{r_jr^m}{r^2}\partial_k\partial^n\right )\\ = \partial^n \partial_n - \left (\frac{r^jr^n}{r^2}\partial _j \partial_n+\frac{2}{r} \frac{\partial}{\partial r} \right)\\ =\nabla^2-\left (\frac{r^n}{r}\frac{\partial}{\partial r} \partial_n+\frac{2}{r} \frac{\partial}{\partial r} \right)= \nabla^2-\left ( \frac{\partial^2}{\partial r^2}+\frac{2}{r} \frac{\partial}{\partial r} \right)\\ =\nabla^2-\frac{1}{r^2}\frac{\partial}{\partial r}\left(r^2\frac{\partial}{\partial r}\right) . $$