The case of  (Coulombic lattice energy)

Williams, D. E.

doi:10.1107/97809553602060000562

International
Tables for
Crystallography
Volume B
Reciprocal space
Edited by U. Shmueli

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International Tables for Crystallography (2006). Vol. B. ch. 3.4, p. 389 | 1 | 2 |

Section 3.4.6. The case of $[{\bf{\bi n} = 1}]$ (Coulombic lattice energy )

D. E. Williams^a ^‡

^aDepartment of Chemistry, University of Louisville, Louisville, Kentucky 40292, USA

3.4.6. The case of $[{\bf{\bi n} = 1}]$ (Coulombic lattice energy )

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As taken above, the limit of the reciprocal-lattice $[{\bf h} = 0]$ term of $[S'(n, {\bf R})]$ or [S'(n, 0)] existed only if n was greater than 3. The corresponding contributions to $[V(n, {\bf R}_{j})]$ were terms (5) and (9) of Section 3.4.5. To extend the method to [n = 1] we will show in this section that these $[{\bf h} = 0]$ terms vanish if conditions of unit-cell neutrality and zero dipole moment are satisfied.

The integral representation of the term (5) is $[\eqalign{ &[1/2\Gamma (n/2)] V_{d}^{-1} \pi^{n-(3/2)} {\textstyle\sum\limits_{j \neq k}}\ Q_{jk} \textstyle\int \delta (0) |{\bf H}|^{n-3}\cr &\quad \times \Gamma [(-n/2) + (3/2), \pi w^{-2} |{\bf H}|^{2}]\cr &\quad \times \exp [2 \pi i {\bf H} \cdot ({\bf R}_{k} - {\bf R}_{j})]\; \hbox{d}{\bf H}}]$ and for term (9) is $[\eqalign{ &[1/2\Gamma (n/2)] V_{d}^{-1} \pi^{n-(3/2)} {\textstyle\sum\limits_{j}}\ Q_{jj} \textstyle\int \delta (0) |{\bf H}|^{n-3}\cr &\quad \times \Gamma [(-n/2) + (3/2), \pi w^{-2} |{\bf H}|^{2}]\; \hbox{d}{\bf H}.}]$ Combining these two sums of integrals into one integral sum gives $[\eqalign{ &[1/2\Gamma (n/2)] V_{d}^{-1} \pi^{n-(3/2)} \textstyle\int \delta (0) |{\bf H}|^{n-3}\cr &\quad \times \Gamma [(-n/2) + (3/2), \pi w^{-2} |{\bf H}|^{2}] {\textstyle\sum\limits_{j}} {\textstyle\sum\limits_{k}}\ Q_{jk}\cr &\quad \times \exp [2\pi i {\bf H}\cdot ({\bf R}_{k} - {\bf R}_{j})] \hbox{ d}{\bf H}.}]$

For [n = 1] , suppose $[q_{j}]$ are net atomic charges so that the geometric combining law holds for $[Q_{jk} = q_{j}q_{k}]$ . Then the double sum over j and k can be factored so that the limit that needs to be considered is $[\lim\limits_{|{\bf H}| \rightarrow 0} {{\left[{\textstyle\sum_{k{\phantom j}}} q_{k} \exp (2\pi i {\bf H} \cdot {\bf R}_{k})\right] \left[{\textstyle\sum_{j}}\ q_{j} \exp (-2\pi i {\bf H} \cdot {\bf R}_{j})\right]} \over |{\bf H}|^{2}}.]$ If the unit cell does not have a net charge, the sum over the q's goes to zero in the limit and this is a 0/0 indeterminate form. Let $[|{\bf H}|]$ approach zero along the polar axis so that $[{\bf H}\cdot {\bf R}_{k} = H_{3} R_{3k}]$ , where subscript 3 indicates components along the polar axis. To find the limit with L'Hospital's rule the numerator and denominator are differentiated twice with respect to $[H_{3}]$ . Represent the numerator of the limit by the product [(uv)] and note that $[{\hbox{d}^{2} (uv)\over\; \hbox{d}x^{2}} = u {\hbox{d}^{2} v\over\; \hbox{d}x^{2}} + v {\hbox{d}^{2} u\over\; \hbox{d}x^{2}} + 2 {\hbox{d}u\over\; \hbox{d}x} {\hbox{d}v\over\; \hbox{d}x}.]$ It is seen that in addition to cell neutrality the product of the first derivatives of the sums must exist. These sums are $[\left[2\pi i {\textstyle\sum\limits_{k}}\ q_{k} R_{3k} \exp (2\pi i H_{3} R_{3k})\right]]$ and $[\left[-2\pi i {\textstyle\sum\limits_{j}}\ q_{j} R_{3j} \exp (-2\pi i H_{3} R_{3j})\right],]$ which vanish if the unit cell has no dipole moment in the polar direction, that is, if $[\sum q_{j} R_{3j} = 0]$ . Since the second derivative of the denominator is a constant, the desired limit is zero under the specified conditions. Now the polar direction can be chosen arbitrarily, so the unit cell must not have a dipole moment in any direction for the limit of the numerator to be zero. Thus we have the formula for the Coulombic lattice sum $[\eqalign{ V(1, {\bf R}_{j}) &= [ 1/2\Gamma (1/2)] {\textstyle\sum\limits_{j}} {\textstyle\sum\limits_{k}}'\ Q_{jk} {\textstyle\sum\limits_{\bf d}} |{\bf R}_{k} + {\bf X(d)} - {\bf R}_{j}|^{-1}\cr &\quad \times \Gamma (1/2, \pi w^{2} |{\bf R}_{k} + {\bf X(d)} - {\bf R}_{j}|^{2})\cr &\quad + [1/2 \Gamma (1/2) V_{d}^{-1} \pi^{-1/2} {\textstyle\sum\limits_{\bf h}} |{\bf H(h)}|^{-2}\cr &\quad \times \Gamma (1/2, \pi w^{-2} |{\bf H(h)}|^{2}) {\textstyle\sum\limits_{j}} {\textstyle\sum\limits_{k}}\ Q_{jk}\cr &\quad \times \exp [2\pi i {\bf H(h)} \cdot ({\bf R}_{k} - {\bf R}_{j})]\cr &\quad - [1/\Gamma (1/2)] \pi^{1/2} w {\textstyle\sum\limits_{j}}\ q_{j}^{2},}]$ which holds on conditions that the unit cell be electrically neutral and have no dipole moment. If the unit cell has a dipole moment, the limiting value discussed above depends on the direction of H. For methods of obtaining the Coulombic lattice sum where the unit cell does have a dipole moment, the reader is referred to the literature (DeWette & Schacher, 1964; Cummins et al., 1976; Bertaut, 1978; Massidda, 1978).

References

Bertaut, E. F. (1978). The equivalent charge concept and its application to the electrostatic energy of charges and multipoles. J. Phys. (Paris), 39, 1331–1348.Google Scholar

Cummins, P. G., Dunmur, D. A., Munn, R. W. & Newham, R. J. (1976). Applications of the Ewald method. I. Calculation of multipole lattice sums. Acta Cryst. A32, 847–853.Google Scholar

DeWette, F. W. & Schacher, G. E. (1964). Internal field in general dipole lattices. Phys. Rev. 137, A78–A91.Google Scholar

Massidda, V. (1978). Electrostatic energy in ionic crystals by the planewise summation method. Physica (Utrecht), 95B, 317–334.Google Scholar

International Tables for Crystallography (2006). Vol. B. ch. 3.4, p. 389

Section 3.4.6. The case of (Coulombic lattice energy)

3.4.6. The case of (Coulombic lattice energy)

References

Section 3.4.6. The case of $[{\bf{\bi n} = 1}]$ (Coulombic lattice energy )

3.4.6. The case of $[{\bf{\bi n} = 1}]$ (Coulombic lattice energy )