BornOppenheimer

Many-body crystal Hamiltonian

In Hartree units the crystal Hamiltonian has three many-body contributions ${\cal H} = {\cal H}_e + {\cal H}_L + {\cal H}_{eL}$, from electrons, nuclei, and electron-nucleus interactions, respectively. With $N_e$ electrons and $N$ nuclei they may be written as

$$\begin{align} {\cal H}_e & = -\sum_{i=1}^{N_e} \frac {\nabla_i^2}{2} + \frac 1 {2} \sum_{i\ne j=1}^{N_e} \frac 1{r_{ij}}\\ {\cal H}_L & = - \sum_{n=1}^{N} \frac m {M_n} \frac {\nabla_n^2} 2+ \underbrace{\frac 1 2 \sum_{n\ne m=1}^{N}\frac {Z_nZ_m}{R_{nm}} }_{\sum^\prime_{nm}V(R_{nm})}\\ {\cal H}_{eL} & = \underbrace{\sum_{i=1}^{N_e}\sum_{n=1}^{N} \frac{Z_n}{\rho_{ni}}} _{\sum_{ni} v(\rho_{ni}) }\end{align}$$

Here $r_{ij}=|\mathbf r_i-\mathbf r_j|$, $R_{nm}=|\mathbf R_n-\mathbf R_m|$, and $\rho_{ni}=|\mathbf R_n-\mathbf r_i|$. Charge neutrality in Hatree unit ($e=1$) is granted by

$$\sum_{n=1}^N Z_n -Ne=0$$

Since the lattice is periodic a vector notation may apply to lattice positions

$$\mathbf R_{n} = \underbrace{n_a \mathbf a + n_b \mathbf b + n_c \mathbf c}_{\mathbf R^0_n} + \mathbf u_{n}$$

where $\mathbf R^0_n$ is the equilibrium position and $\mathbf u_n$ the displacement from equilibrium. Equation (2) manifestly shows that the nuclear kinetic energy term scales as $m/M$, where the lightest nucleus, hydrogen, weighs 1800 times an electron and the heavier ones weigh 200 times more. Therefore, since equipartition dictates

$$\left\langle \frac {P^2}{2M} \right\rangle = \left\langle \frac {p^2}{2m} \right\rangle$$

one gets that $\sqrt {\langle{P^2}\rangle} = (m/M)^{\frac 1 2} \sqrt {\langle{P^2}\rangle}$, hence the time an electron takes to cross a lattice cell is a factor $40<(m/M)^{\frac 1 2} <600$ shorter than the vibration period of an atom. The consequence is that the electron dynamics can be calculated with the lattice frozen at any of the $\{R\}$ configurations, among which the equilibrium values will play a special rôle. This means that electrons follow adiabatically the nuclear centroids. The electron wave function around a displaced nucleus will not be exactly equal to that around an equilibrium nucleus, but the variation is very small, negligible in first approximation, as we shall discuss below We refer to it as non-adiabatic.

Born-Oppenheimer approximation

The lattice Hamiltonian, Eq. (2), contains a potential energy term evaluated at the instantaneous nuclear positions $\mathbf R_n$. It can be expressed in terms of the same potential energy at the equilibrium positions, the difference being

$$\delta V(R_{nm}) = V(R_{nm}) -V(R^0_{nm})$$

Likewise the electron-nucleus interaction $v$ appearing in Eq. (3) can be written in term of the same energy at the equilibrium positions plus the difference

$$\delta v(\rho_{ni}) = v(\rho_{ni})-v(\rho^0_{ni})$$

We can now redefine the electron, nuclear and interaction terms as

$$\begin{align} {\cal H}& = - \sum_i \frac {\nabla_i^2}{2} + \frac 1 {2} \sideset{}{^\prime}\sum_{i j} \frac 1 {r_{ij}} + \sum_{\mathbf n,i} v(\rho^0_{ni}) \\ & + \sum_n \frac m {M_n} \frac {\nabla_n^2}{2}+ \frac 1 {2} \sideset{}{^\prime}\sum_{ n m}\delta V(R_{mn}) \\ & +\frac 1 2 \sideset{}{^\prime}\sum_{n m}V(R^0_{nm}) +\sum_{n,i}\delta v(\rho_{ni}) \end{align}$$

respectively. The first term in (6) is actually the Madelung energy, the electrostatic energy of the infinite ion lattice. In the simple case of NaCl this term can be written

$$E_M =\frac {Z^2} {2 R} \alpha_M$$

with

$$\alpha_M=\sideset{}{^\prime}\sum_{m,n}\frac {(-1)^{i+j+k}}{(i^2+j^2+k^2)^{\frac 1 2}} = -1.74756\cdots$$

The sum cannot be calculated too naively. One must at least ascertain to sum over neutral shells, otherwise the convergence of the alternate sign $r^{-1}$ series is slow and erratic. A specific trick is the Ewald summation, that exploits the rapid convergence of the Fourier transform $q^{-2}$ of the potential $r^{-1}$.

All the terms, (1-3) and (4-6), could be rewritten in terms of core ions and valence electrons, instead of all electrons and nuclei.

Separation of degrees of freedom

The eigenstates of the whole crystal Hamiltonian are non factorized many-body functions of the $N_e$ electron coordinates $\{r\}$ and of the $N$ nuclear coordinates $\{R\}$

$${\cal H}|\Psi_\alpha(\{r\},\{R\})\rangle = \epsilon_\alpha|\Psi_\alpha(\{r\},\{R\})\rangle$$

Formally the comma between $\{r\}$ and $\{R\}$ indicates the non separability. This problem is not solvable exactly. In the following we omit the curly brackets.

The terms in (4) represent the electron Hamiltonian. They are the kinetic energy, the Coulomb electron-electron repulsion and the lattice potential, where the nuclei are frozen at equilibrium positions $\mathbf R^0_n$. The corresponding Schrödinger equation is a formidable many-body problem in itself and it gives rise to a wealth of surprising phenomena, at the root of condensed matter physics. We shall deal with a few of its most important features in the next pages, after electrons?. We can write this out in the following formal way, again

$${\cal H_e}|\psi_\alpha(r;R^0)\rangle = \epsilon_\alpha(R^0)|\psi_\alpha(r;R^0)\rangle$$

in short ${\cal H_e}|\psi_\alpha\rangle = \epsilon_\alpha|\psi_\alpha\rangle$. Once more $|\psi_\alpha\rangle$ is non separable into one electron factors, whereas approximate solutions are based on one electron functions. Notice that $\psi_\alpha$ represent a complete set. Leaving details for later we can plug this solution back into Eq. (4-6). The last line, (6), can be dealt with first. The Madelung constant is irrelevant for the wave function solution and the electron-nucleus difference potential $\delta v$ leads to the electron-phonon interaction. The latter may be completely neglected for the $T=0$ ground state determination, hence giving special status to the nuclear equilibrium configuration, the periodic lattice.

We are left with terms in (5 and we approach them by the following approximation

$$\begin{equation}|\Psi\rangle = |\chi(R)\rangle|\psi_\alpha(r;R^0)\rangle\end{equation}$$

The factorization is physically rooted in the Born-Oppenheimer approximation, in view of the large difference in the electron and nucleus dynamics. Plugging (7) back into the nuclear Schrödinger equation, and closing the equation on another bra $\langle \chi|\langle \psi_\beta|$ of the complete set we get

$${\cal H}_n|\chi\rangle = \left(-\sum_n \frac {\nabla_n^2} 2 + \epsilon_\beta(R^0)\right)|\chi\rangle + \langle \psi_\beta|\delta V|\psi_\beta\rangle|\chi\rangle = E |\chi\rangle$$

The last term in the middle side is the non adiabatic contribution to the nuclear energy, negligible in first approximation. The remaining terms show that in this approximation the electron energy eigenvalue $\epsilon_\beta$, a function of the lattice configuration, represents directly the potential energy for the lattice. In particular $\epsilon_0$ is the ground state potential. Each existed state represents a new sheet of the potential energy.

The nuclear dynamics may be treated classically from here on. Up to second order we have

$$\epsilon_0(R) = \epsilon_0(R^0) + \sum_{nm}\frac {\partial^2 \epsilon_0}{\partial R_n \partial R_m} u_n u_m$$

Classical virtual forces are $F_m = - \frac{\partial \epsilon_0}{\partial R_m}$ and the corresponding Newton equations are

$$M_n \ddot u_n = - \frac {\partial \epsilon_0}{\partial R_n} u_n$$

Equivalent equations may be written for the excited sheets.