Supporting Information for "MBD+C: how to include metallic character (Type C non-additivity) into atom-based dispersion energy schemes" -
Kohn-Sham electronic density response of crystalline one- and two-dimensional metals

The one-dimensional (1D) system is taken to consist of $N\rightarrow\infty$ unit cells each of length $R$ so that the total length is $L=NR\rightarrow\infty$. We treat the electrons in the 1D conductor as independent but subject to a grouandstate Kohn-Sham (KS) potential that is periodic, $V(x+R)=V(x)$. The KS eigenstates $\psi_{nk}(x)$ are labeleed by a crystal momentuum $k$ such that $\psi_{nk}(x+R)=exp(ikR)\psi_{nk}(x)$, and by an integer band index $n$. A full band carries no current. We will asume that we are dealing with a partially occupied conducton band, and will drop the index $n$ from here on. Under periodic boundary conditions across the entire length, the allowed crystal momenta are labelled by an integer $i$:

$$k_{i}=i\frac{2\pi}{L},\,\,-N/2<i<N/2$$

The number of allowed $k$ values in an interval $\Delta k$ is then

$$\Delta N=L\Delta k/(2\pi)$$

(1)

For simplicity we assume inversion symmetry so that the KS energy eigenvalue $\varepsilon(k)$ is a function of $|k-k_{0}|$ for some $k_{0}$. We also assume that for the "pocket" of condution electrons under consideration $\varepsilon(k)$ increases monotonically with $|k-k_{0}|$ up to the highest occupied energy state. These assumptions are widely valid - e.g. for free electrons in 1D, $k_{0}=0$ and $\varepsilon(k)=\hbar^{2}k^{2}/(2m)$. For electrons in the monatomic gold nanowire studied in the main paper, $k_{0}=0$ and the monotinic condition is satisfied (See Fig. 4 of Jariwala et a.l ² where the point $k=0$ is labelled $\Gamma$) . For a metallic (n,n) carbon nanotube there are two Dirac points $k_{01}$, $k_{02}$ and $\varepsilon(k)\approx v_{F}|k-k_{0i}|$ for electrons in the conducting pocket near each Dirac point $k_{0i}$. ³

The Aufbau principle implies that, in the unperturbed ground state, the Bloch orbitals are filled up to some Ferni energy $\varepsilon_{F}$. The above symmetry and monotonicity assumptions mean that the states are then filled up to a Fermi momentum $k_{F}$ where $-k_{F}<k-k_{0}<k_{F}$, Orbitals in each pocket are occupied by electrons of $N_{s}$ possible spin projections (usually $N_{s}=2$).

We consider the response of the Bloch electrons to a small, slowly-varying, long-wavelength external potential $\Delta V(x,t)$ where

$$\Delta V(x,t)=-e\phi(x,t)=\Delta Vexp(i(qx-\omega t)),qR<<1,v_{F}q<<\omega<<\varepsilon_{F}/\hbar$$

(2)

We use the semi-classical approximation (SCA) ¹, which is valid in this limit. The electron momentumn and velocity are ill-defined in a Bloch orbital because they change constantly under the action of the periodic crystal potential. In the SCA it is noted that the expectation of velocity in a Bloch state is

$$v=\hbar^{-1}\partial\varepsilon/\partial k$$

where the Bloch state extends over the entire chain of length $L$. Note that $v$ is also the group velocity - i.e. the velocity of the envelope of a spatially localized wavepacket made by superposing Bloch states with a spread of crystal momenta centered on the value $k$.

The veocity determines charge transport, and the expectation of the charge current $j$ in am extended Bloch state is

$$j=\frac{-e}{L}v=\frac{-e}{L\hbar}\partial\varepsilon/\partial k$$

(3)

Under the action of the external potential energy (2) , in the SCA the crystal momentum $k$ of each Bloch orbital changes in response to the applied force $F$ according to the deceptively classical-appearing formula

$$\hbar\frac{dk}{dt}=F=-iq\Delta Vexp(i(qx-\omega t),q\rightarrow 0$$

Each wavevector will thus be of the form

$$k_{i}^{\prime}=k_{i}+\Delta k\exp\left(-i\omega t\right)$$

where

	$\displaystyle-i\omega\hbar\Delta k$	$\displaystyle=$	$\displaystyle-iq\Delta V\$		(4)
	$\displaystyle\Delta k$	$\displaystyle=$	$\displaystyle\frac{q\Delta V}{\hbar\omega}$		(5)

In the unperturbed condition the currents from all the occupied states cancel out. In the excited state the occupied orbitals have cystal momenta $k$ lyng in the range

$$-k_{F}+\Delta k<k-k_{0}<k_{F}+\Delta k$$

Thu net current comes from from a group of electrons numbering $\Delta N=(L/2\pi)\Delta k$ with crystal momentum $k_{0}+k_{F}$ and velocity $v_{F}=\partial\varepsilon^{(0)}\partial k|_{k_{F}}$ , minus the contribution of the same number of previously-occupied states with $k=k_{0}-k_{F}$ and velocity $-v_{F}$.

The total current is thus

$$j=\Delta N\frac{-e}{L}(v_{F}-(-v_{F}))=\frac{-e}{2\pi}\frac{q\Delta V}{\hbar\omega}2v_{F}$$

(6)

where (1) and (5) were used in the last step.

This flow of current gives rise to a charge density perturbation $\Delta\rho(x,t)=-e\Delta n\exp(i(qx-\omega t))$ that must satisfy the continuity equation $0=\partial\Delta\rho/\partial t+\partial j/\partial x$ (charge conservation) so that

	$\displaystyle 0$	$\displaystyle=$	$\displaystyle-i\omega(-e\Delta n)+iqj$		(7)
	$\displaystyle\Delta n$	$\displaystyle=$	$\displaystyle-\frac{q}{e\omega}j$		(8)

Using (6) we can write (8) in the form

$$\Delta n=\chi_{0}(q,\omega)\Delta V$$

where the Kohn-sham (independent- electron) density response function for 1D Bloch electrons for $qR<<1,v_{F}q<<\omega<<\varepsilon_{F}/\hbar$ is

$$\chi_{0}(q,\omega)=\frac{q^{2}v_{F}}{\pi\hbar\omega^{2}}$$

(9)

Eq (9) applies to each "poclet" of conducting elctrons. The pockets can differ by the spin projection of the electrons (e.g. up or down). Where a Bloch band structure $\varepsilon(k)$ has multiple valleys that cross the Fermi energy, as for example in some graphenic conductors, this can also lead to multiple pockets of conduction electrons. If there are several conducting pockets then

$$\chi_{0}(q,\omega)=\frac{q^{2}}{\pi\hbar\omega^{2}}\sum_{P}v_{F}^{(P)}$$

(10)

where $P$ labels the various pockets. If there are $N_{s}$ allowed spin orientations and $N_{v}$ inequivalent valleys then the number of pockets equals $N_{s}N_{v}$ . If all pockets have the same Fermi velocity, $v_{F}^{(P)}=v_{F}$, then

$$\chi_{0}(q,\omega)=N_{s}N_{v}\frac{v_{F}q^{2}}{\pi\hbar\omega^{2}}$$

(11)

For unmagnetized metallic systems, both spin-up and spin-down bands are partially occupied so that $N_{s}=2$. For a chain of single monovalent metal atoms only one valley crosses the Fermi energy so that $N_{v}=1$. For metallic (n,n) carbon nanotubes there are valleys centered on two inequivalent Dirac ponts at the Fermi energy ³ so $N_{v}=2$.

Equation(11) is quoted as Eq (5) in the main manuscript.

As a check we can evaluate (11) for free electrons of mass $m$, for which $v_{F}=\hbar k_{F}m^{-1}$ and $N_{v}=1$. The occupied orbitals have wavenumbers in the range $-k_{F}<k<k_{F}$ and each orbital is occupied by one spin-up and one spin-down electron (i.e. $N_{s}=2$). The total number of electrons $N$ is then

$${N=N_{s}\frac{L}{2\pi}(2k_{F})\;\;\;\text{s}o}\;\;\;k_{F}=\frac{\pi N}{LN_{s}}$$

(12)

Then (11) becomes

	$\displaystyle\chi_{0}(q,\omega)$	$\displaystyle=$	$\displaystyle N_{s}N_{v}\frac{v_{F}q^{2}}{\pi\hbar\omega^{2}}$
		$\displaystyle=$	$\displaystyle N_{s}\cdot 1\cdot\frac{\hbar(\pi N)/(LN_{s}))m^{-1}q^{2}}{\pi\hbar\omega^{2}}$
		$\displaystyle=$	$\displaystyle\frac{nq^{2}}{m\omega^{2}}$

where $n=N/L$ is the total number of electrons per unit length. The last equation is the standard result for free electrons, valid for $q<<k_{F}$ and $\omega>>v_{F}q$. Eq (11) above is its generalization to Bloch electrons.

For Bloch electrons in 2D with an isotropic Bloch energy $\varepsilon(|\vec{k}|)$ we can use a similar argument to the 1D argument above, obtaining

$$\chi_{0}^{2D}\left(\vec{q},\omega\right)\approx\frac{N_{s}N_{v}v_{F}k_{F}}{4\pi\hbar}{}\frac{q^{2}}{\omega^{2}}\;\;as\,q\rightarrow 0$$

(13)

which is Eq. (6) of the main text. We will not derive this result in detail here, but will check it for the free 2D unpolarized electron gas. For ths case $N_{v}=1$, $N_{s}=2$ and $v_{F}=\hbar k_{F}/m$. The occupied orbitals of each spin species fill a circle in $k$ space of radus $k_{F}$. Thus the total number of electrons is

$$N=2\frac{A}{(2\pi)^{2}}(\pi k_{F}^{2})\>\>so\>\>k_{F}^{2}=2\pi N/A$$

and then (13) becomes

$$\chi_{0}^{2D}\left(\vec{q},\omega\right)\approx\frac{(2)(1)(\hbar k_{F}/m)k_{F}}{4\pi\hbar}\frac{q^{2}}{\omega^{2}}=\frac{k_{F}^{2}}{2m\pi}\frac{q^{2}}{\omega^{2}}=\frac{nq^{2}}{m\omega^{2}}$$

where $n=N/A$ is the total number of conduction electrons per unit area. The last formula is the correct result for free electrons in 2D when $q<<k_{F}$ and $v_{F}q<\omega<<\varepsilon_{F}$: it follows from $F=ma$ plus number conservation.