When doing some simple quantum mechanics problem involving commutators, I forgot the result of this expression
$$\left[\vec{r} ,\hat{p}\right]$$
Thus I then brute force it using the definition of commutator and vector calculus
$$=-i\hbar[\vec{r}, \nabla]=-i\hbar(\vec{r}\nabla-\nabla(\vec{r}\hspace{4mm})=\nabla\vec{r}($$
I then compute the gradient of $\vec{r}$ using various methods
Method 1: Brute force (J is Jacobian matrix)
$$\nabla\vec{r}=J(\vec{r})=\begin{pmatrix} \frac{\partial x}{\partial x} && \frac{\partial x}{\partial y} && \frac{\partial x}{\partial z} \\ \frac{\partial y}{\partial x} && \frac{\partial y}{\partial y} && \frac{\partial y}{\partial z} \\ \frac{\partial z}{\partial x} && \frac{\partial z}{\partial y} && \frac{\partial z}{\partial z}\end{pmatrix}=\begin{pmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && 1 \end{pmatrix}=I_3$$
Method 2: Index notation
$$\nabla\vec{r}=\frac{\partial x_i}{\partial x_j}=\delta_{ij}$$
Method 3: Vector or Tensor Calculus (???)
We know in spherical coordinates
$$\nabla=\frac{\partial }{\partial r}\hat{r}+\frac{1}{rsin\theta}\frac{\partial }{\partial \phi}\hat{\phi}+\frac{1}{r}\frac{\partial }{\partial \theta}\hat{\theta}$$
and the following results (organised into a block of numbers to look neat)
$$\left[\begin{matrix}\partial_r \hat{r} && \partial_r \hat{\phi} && \partial_r \hat{\theta} \\ \partial_{\phi} \hat{r} && \partial_{\phi} \hat{\phi} && \partial_{\phi} \hat{\theta} \\ \partial_{\theta} \hat{r} && \partial_{\theta} \hat{\phi} && \partial_{\theta} \hat{\theta}\end{matrix}\right]=\left[\begin{matrix}\vec{0} && \vec{0} && \vec{0} \\ sin\theta\hspace{1mm} \hat{\phi} && -(sin\theta\hspace{1mm}\hat{r}+cos\theta\hspace{1mm}\hat{\theta}) && cos\theta\hspace{1mm} \hat{\phi} \\ \hat{\theta} && \vec{0} && -\hat{r}\end{matrix}\right]$$
Thus
$$\nabla\vec{r}=\frac{\partial \vec{r}}{\partial r}\otimes\hat{r}+\frac{1}{rsin\theta}\frac{\partial \vec{r}}{\partial \phi}\otimes\hat{\phi}+\frac{1}{r}\frac{\partial \vec{r}}{\partial \theta}\otimes\hat{\theta}$$ $$=\hat{r}\otimes\hat{r}+\frac{1}{rsin\theta}rsin\theta\hspace{1mm}\hat{\phi}\otimes\hat{\phi}+\frac{1}{r}r\hat{\theta}\otimes\hat{\theta}$$ $$=\hat{r}\otimes\hat{r}+\hat{\phi}\otimes\hat{\phi}+\hat{\theta}\otimes\hat{\theta}=\delta_{ij}$$
So I am pretty convinced mathematically that the result is the identity tensor (Kronecker delta)
However I still have confusion when the result is interpreted geometrically
We know that a tensor is a multilinear map and for our case since it is a matrix, it is just a linear map
So our result of the gradient of the radial vector is suggesting that the infinitesimal change at any point in the vector field is the identity map, which means the vectors don't change in the neighbourhood.
But when looking at the plot of the function itself, and choosing an arbitrary point and we move infintesimally in the x,y,z directions respectively (or r, $\phi$, $\theta$ diections) our vectors are clearly changing
Q: Why geometrically the gradient of $\vec{r}$ is the identity map?
