# How is the $\nabla^n$ operator defined?

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.

• The powers in the taylor expansion are multiplicative powers, not repeated applications May 15, 2020 at 15:15

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

• So, $\hat{p}^n$ is a scalar quantity for even powers and a vector quantity for odd powers?
– ook
May 15, 2020 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, 2020 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, 2020 at 15:35
• @ook $x\cdot p/\hbar$ is a dimensionless scalar.
– J.G.
May 15, 2020 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, 2020 at 15:38