In Quantum Mechanics, the translation operator $\hat{T}$ can be written as
$$\hat{T}(\boldsymbol{x}) = 1 – \dfrac{ix\cdot \hat{p}}{\hbar} – \dfrac{i(x\cdot \hat{p})^2}{2\hbar^2} – \dfrac{i(x\cdot \hat{p})^3}{6\hbar^3} + \ldots$$
with
$$\hat{p} = -i\hbar \nabla $$
This question isn't really about the translation operator itself. I just wanted to mention it as an example. Something that is really bothering me in that Taylor expansion are the expressions
$$\hat{p}^n = (-i\hbar)^n \nabla^n$$
More specifically I wanted to ask what
$$\nabla^n $$
means? From what I know, $\nabla f = \begin{pmatrix}\partial_x f \\\partial_y f \\ \partial_z f\end{pmatrix}$ for a scalar function $f$. This makes sense. But then what is $\nabla^2 f$ supposed to be? From the kinetic energy operator I know that $\nabla^2 = \Delta$ should be the Laplacian-Operator. But this isn't how to product of two operator is defined. By the definition I should apply the nabla operator to $\nabla f = \begin{pmatrix}\partial_x f \\\partial_y f \\ \partial_z f\end{pmatrix}$ again:
$$\nabla \nabla f = \nabla \begin{pmatrix}\partial_x f \\\partial_y f \\ \partial_z f\end{pmatrix} = \begin{pmatrix}\partial_x \begin{pmatrix}\partial_x f \\\partial_y f \\ \partial_z f\end{pmatrix} \\\partial_y \begin{pmatrix}\partial_x f \\\partial_y f \\ \partial_z f\end{pmatrix} \\ \partial_z \begin{pmatrix}\partial_x f \\\partial_y f \\ \partial_z f\end{pmatrix} \end{pmatrix}$$
which could probably be interpreted as the Jacobian matrix.
Best Answer
Well, how do you define $p^n$ for classical $p$, or $\hat{p}^n$ for $\hat{p}$ with operator-valued components? It's that question you really need to answer; $\vec{\nabla}$ isn't the thorny part here. Indeed, we could instead write $\hat{p}_j=-i\hbar\partial_j$, so all the exponentiate-a-vector work is done elsewhere.
For a vector $\vec{v}$ we define $\vec{v}^2:=\vec{v}\cdot \vec{v}=\sum_iv_i^2$, so $\vec{v}^0:=1,\,\vec{v}^1:=\vec{v},\,\vec{v}^{n+2}:=(\vec{v}\cdot\vec{v})\vec{v}^n$ defines all non-negative integer powers of $\vec{v}$ recursively, with $\vec{v}^{2n}=(\vec{v}\cdot\vec{v})^n,\,\vec{v}^{2n+1}=\vec{v}^{2n}\vec{v}$. There's no problem using that with anything here, especially since $[\hat{p}_j,\,\hat{p}_k]=0$. So $\nabla^2=\sum_j\partial_j^2$.