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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.

ook
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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 $\nabla^2=\sum_j\partial_j^2$.

J.G.
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  • So, $\hat{p}^n$ is a scalar quantity for even powers and a vector quantity for odd powers? – ook May 15 '20 at 15:26
  • @ook Yes. This also comes up when you exponentiate imaginary quaternions with a power series. (That's probably not of interest to you, but "Yes" was too short for a comment.) – J.G. May 15 '20 at 15:29
  • But shouldn't all the terms in the expansion have the same dimension? Otherwise I wouldn't be able to add the terms together. – ook May 15 '20 at 15:35
  • @ook $x\cdot p/\hbar$ is a dimensionless scalar. – J.G. May 15 '20 at 15:36
  • What I mean is, the expansion would then consist of Scalar + Vector terms that can't be added together. – ook May 15 '20 at 15:38
  • @ook As I've said, that's not true of $\hat{T}$. Which expansion are you talking about? If you're picking up on my comment about quaternions, see here. – J.G. May 15 '20 at 15:42
  • No, I am still talking about $\hat{T}$. Well, I don't see why this isn't true for $\hat{T}$. The quantities $x$ and $\hbar$ are scalars, $\hat{p}$ is a vector, so the second term is a vector and it can't be added to the first Term, which is the scalar $1$. Similarly I get a scalar for the third term and again a vector for the fourth term. – ook May 15 '20 at 15:48
  • @ook No, in this context $x$ means the vector containing $x$, $y$ and $z$, and this is often called $r$. So $x\cdot p$ is a dot product, which is why I said it was a scalar. – J.G. May 15 '20 at 16:27