Why is the potential field inside a quadrupole field given by $V=k(\alpha x^{2} +\beta y^{2} +\gamma z^{2})$ for constants $k,\alpha,\beta,\gamma$?

Question

I am seeking an explanation regarding the origin of the potential inside a quadrupole field. The common understanding is that the potential can be described by the equation: $$V=k(\alpha x^{2} +\beta y^{2} +\gamma z^{2})$$ where $k$, $\alpha$, $\beta$, and $\gamma$ are constants. This potential offers confinement in two dimensions, and three dimensions if the field is allowed to vary (as is the case in a quadrupole ion trap). I'm curious about the derivation of this potential form. Can it be derived similarly to how the far outside field of an electric dipole is given by: $$ V(\textbf{r})=\frac{\textbf{p}\cdot\hat{\textbf{r}}}{4\pi\epsilon_{0}|\textbf{r}|^{2}}$$

Answer 1

The usual approach is to assume that the potential is quadratic about the origin. This is justified by a Taylor expansion about the origin, valid for small displacements.

The constant part can be removed up to a change of the potential’s origin. Since by definition there is no electric field at the origin, this kills the linear terms as well. Thus at leading order, you are left only with the quadratic terms. Finally, using the spectral theorem, you can find a cartesian coordinate system where the quadratic part is diagonal: $$ V=\frac{V_{xx}}{2}x^2+ \frac{V_{yy}}{2} y^2+ \frac{V_{zz}}{2} z^2 $$ This already gives you your potential.

It's called quadrupole due to the angular dependence. In terms of spherical harmonics, $l=0$ is the monopole term, $l=1$ is the dipole term and $l=2$ is the quadrupole term. Generally, the $l$ term is the $2^l$-pole term. The radial dependence is given by $r^l$ or $r^{-(l+1)}$ according to the Laplace equation. The former is useful about the origin, while the latter is what you were referring to, and is useful at large distances. With the usual multipole expansion, the general field is written as an infinite series, but you usually keep only the leading order term for simplicity.

Note that the potential is usually created by external charges like in the Paul trap. Thus, according to Gauss law, there is no charge at the origin so: $$ \nabla^2V=0 $$ giving: $$ V_{xx}+V_{yy}+V_{zz}=0 $$ which means that the origin is a saddle point if the eigenvalues are non zero.

Hope this helps.