The alpha-independent Ewald potential and renormalized energy

Write $Q={\sum }_i\mathrm{\mid }{q}_i\mathrm{\mid }$. For $\alpha >0$ define the periodized Gaussian source ${\gamma }_{\alpha }(x)={\sum }_i{q}_i{\sum }_{\lambda \in \mathrm{\Lambda }}{g}_{\alpha }(x-x_i+\lambda )\mathrm{.}$ By [F2] applied to each translate and then summed over the finite source set, ${\gamma }_{\alpha }(x)=\frac{1}{V}{\sum }_{k\in {\mathbb{Z}}^3}{e}^{-\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2\mathrm{/}(4{\alpha }^2)}{\sigma }_k{e}^{i{\kappa }_k\cdot x}\mathrm{.}$ Neutrality gives ${\sigma }_0=0$, so the $k=0$ term vanishes. Both series are absolutely convergent; [F4] supplies uniform Gaussian tails for the lattice sums and their derivatives.

Define $S_{\alpha }(x)={\sum }_i{q}_i{\sum }_{\lambda \in \mathrm{\Lambda }}{\varphi }_s(\mathrm{\mid }x-x_i+\lambda \mathrm{\mid })$ and $L_{\alpha }(x)=\frac{4\pi }{V}{\sum }_{k\mathrm{\ne }0}\frac{e^{-\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2\mathrm{/}(4{\alpha }^2)}}{\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2}{\sigma }_k{e}^{i{\kappa }_k\cdot x}$. The finitely many nonzero modes with $\mathrm{\mid }{\kappa }_k\mathrm{\mid }<1$ cause no difficulty, and [F4] bounds the remaining Gaussian shells; with $\mathrm{\mid }{\sigma }_k\mathrm{\mid }\le Q$, the reciprocal series is absolutely and uniformly convergent. Thus termwise integration over the finite-volume cell is valid, and [F10] gives $\frac{1}{V}{\int }_F{L}_{\alpha }(x)dx=\frac{4\pi }{V}{\sum }_{k\mathrm{\ne }0}\frac{e^{-\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2\mathrm{/}(4{\alpha }^2)}}{\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2}{\sigma }_k\left(\frac{1}{V}{\int }_F{e}^{i{\kappa }_k\cdot x}dx\right)=0$. For the real part, [F4], [F5], and Tonelli [F8] give ${\int }_F{\sum }_{\lambda \in \mathrm{\Lambda }}{\varphi }_s(\mathrm{\mid }x-x_i+\lambda \mathrm{\mid })dx={\int }_{{\mathbb{R}}^3}\frac{\mathrm{erfc}(\alpha \mathrm{\mid }y\mathrm{\mid })}{\mathrm{\mid }y\mathrm{\mid }}dy=4\pi {\int }_0^{\mathrm{\infty }}r\mathrm{erfc}(\alpha r)dr=\frac{\pi }{{\alpha }^2}\mathrm{.}$ For the last equality, set $s=\alpha r$ and use the tail identity in [F1]. Tonelli [F8] gives ${\int }_0^{\mathrm{\infty }}s\mathrm{erfc}(s)ds=\frac{2}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{e}^{-t^2}\left({\int }_0^{t}sds\right)dt=\frac{1}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}{t}^2{e}^{-t^2}dt\mathrm{.}$ Integration by parts [F7] on $[0,R]$ gives ${\int }_0^R{t}^2{e}^{-t^2}dt=-R{e}^{-R^2}\mathrm{/}2+(1\mathrm{/}2){\int }_0^R{e}^{-t^2}dt$; the boundary term tends to zero since $e^{R^2}\ge R^4\mathrm{/}2$ for $R>0$. Taking $R\to \mathrm{\infty }$ and using [F6] yields ${\int }_0^{\mathrm{\infty }}{t}^2{e}^{-t^2}dt=\sqrt{\pi }\mathrm{/}4$, so ${\int }_0^{\mathrm{\infty }}s\mathrm{erfc}(s)ds=1\mathrm{/}4$. Multiplying by $4\pi {\alpha }^{-2}$ gives the displayed value. Hence ${\int }_F{S}_{\alpha }=(\pi \mathrm{/}{\alpha }^2){\sum }_i{q}_i=0$, and $S_{\alpha }+L_{\alpha }$ has mean zero.

On compact sets avoiding all source images, the real sum and each differentiated tail converge uniformly by the $\mathrm{erfc}$ bounds of [F1] and the shell controller [F4]. The reciprocal series and every derivative converge uniformly because each derivative adds only a fixed power of $\mathrm{\mid }{\kappa }_k\mathrm{\mid }$ to a Gaussian tail. Thus both are continuous off-source, periodic, and their Fourier operations below are legitimate.

The distributional identity in [F1], summed over the locally finite source images, gives $-\mathrm{\Delta }{S}_{\alpha }=4\pi \rho -4\pi {\gamma }_{\alpha }$. Absolute tail control from [F4] lets the distribution pairing pass through the image sum. Termwise differentiation of the uniformly convergent reciprocal series for $L_{\alpha }$ gives $-\mathrm{\Delta }{L}_{\alpha }=\frac{4\pi }{V}{\sum }_{k\mathrm{\ne }0}{e}^{-\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2\mathrm{/}(4{\alpha }^2)}{\sigma }_k{e}^{i{\kappa }_k\cdot x}=4\pi {\gamma }_{\alpha },$ where the last equality is step 1.1 and ${\sigma }_0=0$. Therefore $-\mathrm{\Delta }(S_{\alpha }+L_{\alpha })=4\pi \rho$.

The sum $U_{\alpha }=S_{\alpha }+L_{\alpha }$ is a mean-zero periodic distribution solution by steps 3.1 and 1.2. Uniqueness [F3] makes it equal to the same Fourier distribution $u$ for every $\alpha >0$. Its absolutely convergent representative is continuous off-source by step 2.1. If two such representatives differed at a point, one real or imaginary component of their difference would keep a strict sign on a small ball; choose a nonnegative smooth bump supported there using [F9]. Their distribution pairings with that bump would differ, contradicting equality to $u$. Thus the displayed potential is independent of $\alpha$.

Near $x_i$, the unique term with its own zero image is $q_i{\varphi }_s(\mathrm{\mid }x-x_i\mathrm{\mid })$. By [F1], subtracting $q_i\mathrm{/}\mathrm{\mid }x-x_i\mathrm{\mid }$ leaves the finite limit $-2\alpha q_i\mathrm{/}\sqrt{\pi }$ from that term; every other real term is continuous near $x_i$ because positions are distinct modulo $\mathrm{\Lambda }$, and the reciprocal term is smooth. This proves existence and alpha-independence of the regular part by step 4.1. Multiplying those regular parts by $q_i\mathrm{/}2$ gives the real-space sum with $(j,\lambda )\mathrm{\ne }(i,0)$ and the self contribution $-\alpha {\sum }_i{q}_i^2\mathrm{/}\sqrt{\pi }$. The reciprocal contribution is $\frac{1}{2}{\sum }_i{q}_i{L}_{\alpha }(x_i)=\frac{1}{2}{\sum }_{k\mathrm{\ne }0}\frac{4\pi }{V\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2}{e}^{-\mathrm{\mid }{\kappa }_k{\mathrm{\mid }}^2\mathrm{/}(4{\alpha }^2)}\left({\sum }_i{q}_i{e}^{i{\kappa }_k\cdot x_i}\right)\left({\sum }_j{q}_j{e}^{-i{\kappa }_k\cdot x_j}\right)=\frac{1}{2}{\sum }_{k\mathrm{\ne }0}{c}_k(\alpha )\mathrm{\mid }{\sigma }_k{\mathrm{\mid }}^2\mathrm{.}$ All rearrangements are absolutely convergent by [F4]. This is exactly the stated renormalized energy.