Identity in quantum operator tutorial

Question

I'm reading this tutorial by Ben Simons entitled Operator methods in quantum mechanics in connection with his course in advanced QM, and I'm a bit puzzled by an identity in page 25, a bit above relation (3.3):

With the momentum operator $\textbf{p}=-i\nabla$ and the vector $\textbf{a}$ we have

$e^{-i\textbf{a}\cdot\textbf{p}}=e^{\ \textbf{a}\cdot\nabla}=\displaystyle\sum_{n=0}^{\infty}\frac{1}{n!}a_{i_1}\cdots a_{i_n}\nabla_{i_1}\cdots\nabla_{i_n}\ ,$

where repeated indices are summed over. What confuses me here is that (in 3D) $\textbf{a}$ and $\nabla$ have 3 components while this expression seems to refer to component $i_n$ where $n\rightarrow\infty$, and I don't recognize the usual expansion of the exponential.

What am I missing here?

Answer 1

In this case the index can vary on the number of spatial dimensions (three if you are in 3D). The $i_n$ notation refers to the fact that you are "repeating" the scalar product "$\mathbf{a}\cdot\nabla$" n times: $$ (\mathbf{a}\cdot\nabla)^n = (a_i\nabla_i)^n = (a_{i_1}\nabla_{i_1})(a_{i_2}\nabla_{i_2})\dots (a_{i_n}\nabla_{i_n}) = a_{i_1}a_{i_2}\dots a_{i_n}\nabla_{i_1}\nabla_{i_2}\dots \nabla_{i_n} $$

Answer 2

Notice that \begin{align} e^{\mathbf a\cdot\nabla} = \sum_{n=0}^\infty \frac{1}{n!} (\mathbf a\cdot\nabla)^n \end{align} and \begin{align} (\mathbf a \cdot\nabla)^n &= \underbrace{(\mathbf a\cdot\nabla)\cdots(\mathbf a\cdot\nabla)}_{\text{$n$ factors}} = (a_{i_1}\nabla_{i_1})\cdots (a_{i_n}\nabla_{i_n}) = a_{i_1}\cdots a_{i_n}\nabla_{i_1}\cdots \nabla_{i_n} \end{align} where we have made sure to use a different dummy index for each $\mathbf a\cdot \nabla$, hence the extra index on each index.