Hohenberg-Kohn theorems

We now show ^8.3 that $V_{ext} [\rho_0]$, where $\rho_0$ is the ground state (GS) density, is a unique functional of $\rho_0$. With this we mean that two non-trivially different external potentials (i.e. differing by more than a constant) $v({\bf r})$ and $v'({\bf r})$ cannot produce the same GS density $\rho_0({\bf r})$. We will restrict ourselves to the case of non-degenerate GS in our proof. Consider the GS wavefunctions $\Psi_0$ and $\Psi'_0$ of the systems with potentials $v({\bf r})$ and $v'({\bf r})$ and suppose that they both produce the same GS density $\rho_0$. The GS energy $E_0$ can be written using the compact notation:

$$E_0 = \int_V d^3{\bf r}_1 \dots d^3{\bf r}_N \Psi_0^*({\bf r}_1,\... ...({\bf r}_1,\dots,{\bf r}_N) = \langle \Psi_0 \vert \hat{H} \vert \Psi_0\rangle.$$ (8.12)

Since the two wavefunctions $\Psi_0$ and $\Psi'_0$ are the GS of two different potentials they are different ^8.4, and so the variational principle implies:

$$E_0 = \langle \Psi_0 \vert \hat{H} \vert \Psi_0 \rangle < \langle... ...\Psi'_0 \rangle + \int_V d^3{\bf r} [v({\bf r}) - v'({\bf r})] \rho_0({\bf r}),$$ (8.13)

giving

$$E_0 < E'_0 + \int_V d^3{\bf r} [v({\bf r}) - v'({\bf r})]\rho_0({\bf r}).$$ (8.14)

By virtue of exchanging the primed quantities with the non-primed ones we also obtain:

$$E'_0 < E_0 + \int_V d^3{\bf r} [v'({\bf r}) - v({\bf r})] \rho_0({\bf r}),$$ (8.15)

and by summing  and  we obtain:

$$E_0 + E'_0 < E'_0 + E_0,$$ (8.16)

proving that our assumption that $\Psi_0$ and $\Psi'_0$ produce the same GS density is wrong. Therefore, there is a one-to-one correspondence not just between $V_{ext}$ and $\Psi_0$, but also between the $V_{ext}$ and $\rho_0$, and therefore a one-to-one correspondence between $\rho_0$ and $\Psi_0$, which means that we can write $\Psi_0 = \Psi_0[\rho_0]$. Since $\rho_0$ determines $\Psi_0$, it also determines all the physical properties of the system, they are functionals of the GS density. In particular, the GS energy is a functional of $\rho_0$, as well as the kinetic and electron-electron interaction contributions:

$$E_0 = E[\rho_0] = T[\rho_0] + V_{ext}[\rho_0] + V_{ee}[\rho_0].$$ (8.17)

The functional

$$F[\rho] = T[\rho] + V_{ee}[\rho]$$ (8.18)

does not depend on the external potential, and therefore it is a universal functional of the density. The unique correspondence between $\rho_0$ and $\Psi_0$ also implies the usual variational principle, which in terms of the density reads: the ground state density is the one that minimises the energy functional, subject to the constraint that the total number of electrons is constant. We can use this principle to look for $\rho_0$, and if we knew the exact form of $F[\rho]$ we could solve the electronic problem for any external potential. To find this constrained minimum we use the method of the Lagrange multipliers (see Appendix ). We define:

$$G[\rho,\mu] = E[\rho] - \mu \left ( \int_V \rho({\bf r}) d^3{\bf r} - N \right ),$$ (8.19)

and we require:

$$\delta G[\rho,\mu] = \frac{\delta G[\rho,\mu]}{\delta \rho}\delta \rho + \frac{\delta G[\rho,\mu]}{\delta \mu} \delta \mu = 0,$$ (8.20)

which implies that both variations w.r.t. $\rho$ and $\mu$ have to be zero. The latter condition of course simply defines the constraint. Unfortunately $F$ is not known, so an exact solution is still impossible. However, in the next section we will explore an approximate method, which also provides a working procedure to find the ground state density.