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The field of the electric dipole is $\displaystyle\vec{E}=\frac{3(\vec{p}\cdot\hat{r})\hat{r}-\vec{p}}{4\pi\epsilon_{0}r^3}$, show that it can be written as linear combination of $\displaystyle\frac{Y_{2m}(\theta,\phi)}{r^3}$, i.e, $$\vec{E}=\frac{3(\vec{p}\cdot\hat{r})\hat{r}-\vec{p}}{4\pi\epsilon_{0}r^3}=\sum_{m=-2}^{2}\vec{C}_m\frac{Y_{2m}(\theta,\phi)}{r^3}$$

I start from the potential of the dipole $\displaystyle V=\frac{\vec{p}\cdot\hat{r}}{4\pi\epsilon_0r^2},$ since it satisfies Laplace Equation, it can be expressed as $$V=\frac{\vec{p}\cdot\hat{r}}{4\pi\epsilon_0r^2}=\sum_{l=0}^{\infty}\sum_{m=-l}^{l}(A_{lm}r^l+B_{lm}\frac{1}{r^{l+1}})Y_{lm}(\theta,\phi)$$

Then I tried to take the gradient of $V$:$$\vec{E}=-\vec{\nabla}V=-\sum_{l=0}^{\infty}\sum_{m=-l}^{l}\left[\left(\frac{\partial f}{\partial r}\hat{r}Y_{lm}\right)+\left(\frac{1}{r}\frac{\partial Y_{lm}}{\partial \theta}\hat{\theta}\right)+\left(\frac{1}{r\sin\theta}\frac{\partial Y_{lm}}{\partial \phi}\hat{\phi}\right)\right]\quad,$$ where $f=A_{lm}r^l+B_{lm}\frac{1}{r^{l+1}}$.

Updates

By inspection deduce that $$V=\sum_{m=-1}^{1}B_{1m}\frac{1}{r^2}Y_{1m}(\theta,\phi)$$

Taking the gradient and a minus sign gives: $$\sum_{m=-1}^{1}B_{1m}\left(\frac{2Y_{1m}}{r^3}\hat{r}-\frac{1}{r^3}\frac{\partial Y_{1m}}{\partial \theta}\hat{\theta}-\frac{1}{r^3\sin\theta}\frac{\partial Y_{1m}}{\partial \phi}\hat{\phi}\right)$$

Why this can be written as a linear combination of $\displaystyle\frac{Y_{2m}(\theta,\phi)}{r^3}$. Is there anything wrong about my way? Or is there other way to do it? Thanks.

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You're taking the gradient too early. Start by finding $A_{lm}$ and $B_{lm}$ for $V$, and then take the gradient. You should be able to see by inspection that most values of $l$ don't work (hint: look at $r$ dependence).

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  • $\begingroup$ Thanks for your answer. I found that only $B_{1m}$ is non-zero. Why the subscript $l$ of the spherical harmonics is $2$ after taking the gradient? I see that the $\hat{r}$ and $\hat{\phi}$ component is the linear combination of $\frac{Y_{1m}}{r^3}$. But the $\hat{\theta}$ components involves $\frac{\partial Y_{1m}}{\partial \theta}$. $\endgroup$
    – Dumbbbb
    May 19, 2020 at 1:22
  • $\begingroup$ What happens if you take that derivative? You should get something like $Y_{2m}$ out. $\endgroup$ May 19, 2020 at 16:48
  • $\begingroup$ Sorry, but I don't understand why it becomes linear combinations of $\frac{Y_{2m}}{r^3}$? Since by inspection from the expression for $V$, the index $l$ is $1$. Why it becomes $2$ after taking the gradient? Thanks. $\endgroup$
    – Dumbbbb
    May 23, 2020 at 3:14

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