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While working on a mathematical physical problem, i came across seemingly contradictory results.

Notations

Let's consider $\mathbf{x}_1$ to be the origin of a spherical coordinate system and $\mathbf{x}_2$ a point located at the $z$ axis a distance $R$ from the origin, i.e. $R = \left| \mathbf{x}_2 - \mathbf{x}_1 \right|$. We introduce the unit vector $\mathbf{d} = (\mathbf{x}_1 - \mathbf{x}_2)/R = -\mathbf{\hat{e}}_z$, where $\mathbf{\hat{e}}_z$ is the unit vector along the $z$ axis.

We denote by $s$ the distance from a given point $\mathbf{x}$ from $\mathbf{x}_2$ such that $s = |\mathbf{s}|$ with $\mathbf{s} = \mathbf{x} - \mathbf{x}_2$.

Problem statement

On the one hand, it can readily be checked that $$ \left( \mathbf{e} \cdot \mathbf{\nabla}_2 \right) \left( \frac{1}{s} \right) = \frac{\mathbf{e} \cdot \mathbf{s}}{s^3} \, , \tag{1} $$

where $\mathbf{\nabla}_2$ denotes the gradient taken along $\mathbf{e}$ with respect to $\mathbf{x}_2$.

We note that $\mathbf{e} \perp \mathbf{d}$.

On the other hand, by expressing $1/s$ in terms of harmonics based at $\mathbf{x}_1$ such that $$ \frac{1}{s} = \sum_{n=0}^\infty \frac{R^n}{r^{n+1}} \, P_n(\cos\theta) $$ where $r = \left| \mathbf{x}-\mathbf{x}_1 \right|$, $\theta$ is the polar angle in spherical coordinates (physics convention), and $P_n$ denotes Legendre polynomial of degree $n$.

Then, $$ \left( \mathbf{e} \cdot \mathbf{\nabla}_2 \right) \left( \frac{1}{s} \right) = \sum_{n=0}^\infty -n R^{n-1} (\mathbf{e} \cdot \mathbf{d}) = 0 \, , $$ since we have imposed that $\mathbf{e} \perp \mathbf{d}$. However, by direct calculation in Eq. $(1)$, the gradient does now necessarily vanish.

i have spend a couple hours today and most of the afternoon looking at this without understanding why.

In fact, i need the spherical harmonic representation for further calculations.

Any help or hint is highly appreciated.

Thank you,

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The answer is straightforward. Here you have forgotten to take the derivative with respect to the spherical harmonic. We have $$ \frac{1}{s} = \sum_{n=0}^\infty R^n \frac{\left( \mathbf{d} \cdot \mathbf{\nabla} \right)^n}{n!} \frac{1}{r} \, . $$

By noting that $$ \left( \mathbf{e} \cdot \mathbf{\nabla}_2 \right) \mathbf{d} = -(1/R) \mathbf{e} \, , $$ then \begin{align} \left( \mathbf{e} \cdot \mathbf{\nabla}_2 \right) \frac{1}{s} &= \sum_{n=0}^\infty R^n \left( - \frac{\mathbf{e} \cdot \mathbf{\nabla}}{R} \right) n \, \frac{\left( \mathbf{d} \cdot \mathbf{\nabla} \right)^{n-1}}{n!} \frac{1}{r} \, , \\ &= - \left( \mathbf{e} \cdot \mathbf{\nabla} \right) \sum_{n=0}^\infty R^n \frac{\left( \mathbf{d} \cdot \mathbf{\nabla} \right)^n}{n!} \frac{1}{r} \end{align} leading to the desired result your equation (1).

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