DensityFunctional

[This page follows loosely Grosso and Pallavicini]. The many-body ground state wavefunction is

{$$|\Psi\rangle = |\Psi(\mathbf r_1\,\sigma_1,\cdots,\mathbf r_N\,\sigma_N)\rangle$$}

and it is an eigenfunction of the Hamiltonian

{$$\hat{\cal H} = - \sum_i \frac {\hat{\nabla}_i^2} 2 + \frac 1 2 \sum_{ij} \frac 1 {\hat r_{ij}} + \sum_i v(\hat{\mathbf r}_i) $$}

where we have pedantically indicated the operators by the hat, hence it satisfies

{$$\hat {\cal H}|\Psi\rangle = E |\Psi\rangle$$}

with

{$$E =\langle \Psi|\hat {\cal H}|\Psi\rangle$$}

The Hamiltonial could be written {$\hat {\cal H}= \hat T + \hat V + \hat v$}, representing the three terms written before in the same order. The first two are universal, i.e. they have always the same form, whereas the third is the one that defines the problem, the so-called external potential. Therefore all the scalar quantities related to a given problem (to a given material), all the average value of its observables, are functionals of the external potential.

How to obtain the exact {$|\Psi\rangle$} is an unsolved problem, probably forever. If we did know the wavefunction we could compute the exact average value of any observable on the ground state. Among the others the electron density

{$$n(\mathbf r) = \langle \Psi|\sum_i \delta(\mathbf r_i-\mathbf r)|\Psi\rangle $$}

The question that HK raised is can we obtain the energy {$E$} of the ground state from the knowledge of the exact {$n(\mathbf r)$}?

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Functionals

A functional is a mapping that provides a scalar value for each function of a given manifold. Typically certain definite integrals are a functional of their kernel functions {$f$}, as in {$I=\int_{\cal D} d \mathbf r\, f(\mathbf r)$}.

So, we said, {$E=E[v(\mathbf r)] $}. But {$v(\hat{\mathbf r}) = \sum_i v(\mathbf r) \delta(\mathbf r_i-\hat{\mathbf r}) $}, hence

{$$\begin{equation}v_{ext} =\langle \Psi|\hat v(\mathbf r)|\Psi\rangle= \int \langle \Psi|\sum_i \delta(\mathbf r_i-\mathbf r)|\Psi\rangle v(\mathbf r) d \mathbf r = \int n(\mathbf r) v(\mathbf r) d \mathbf r=v_{ext}[n(\mathbf r)]\end{equation}$$}

i.e. the exact ground state energy is a functional of the external potential {$v(\mathbf r)$}, whose average value is a functional of {$n$}. Can we conclude that {$E =\langle \Psi|\hat {\cal H}|\Psi\rangle$} is a functional of {$n$}? If it is so we could compute exactly this (and other properties of the ground state) even without the knowledge of the full many-electron wavefunction. This is addressed by the Hohenberg Kohn theorem.

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Hohenberg Kohn theorem

The hypothesis of HK theorem is that there is a one-to-one correspondence between {$n(\mathbf r)$} and the external potential {$v(\mathbf r)$}. The theorem is demonstrated by contradiction, supposing that the same {$n$} could arise from two distinct external potential, say {$v_1(\mathbf r)$}, determining the Hamiltonian {$\hat{\cal H}_1$}, and {$v_2(\mathbf r)$}, determining {$\hat{\cal H}_2$}. We have therefore two distinct ground state eigenfunctions and total energy eigenvalues, that we suppose non degenerate (the degenerate case can also be demonstrated)

{$$\hat{\cal H}_1 |\Psi_1\rangle = E_1 |\Psi_1\rangle\qquad \hat{\cal H}_2 |\Psi_2\rangle = E_2 |\Psi_2\rangle$$}

For each of them we know that the ground state energy is lower that any other expectation value of the Hamiltonian, therefore

{$$\begin{equation}E_1 < \langle \Psi_2|\hat {\cal H}_1|\Psi_2\rangle = \langle \Psi_2|\hat {\cal H}_2 + \hat v_1-\hat v_2|\Psi_2\rangle = E_2 + \langle \Psi_2|\hat v_1-\hat v_2|\Psi_2\rangle\end{equation}$$}

and at the same time

{$$\begin{equation}E_2 < \langle \Psi_1|\hat {\cal H}_2|\Psi_1\rangle = \langle \Psi_1|\hat {\cal H}_1 + \hat v_2-\hat v_1|\Psi_1\rangle = E_1 + \langle \Psi_1|\hat v_2-\hat v_1|\Psi_1\rangle \end{equation}$$}

However, both

{$$\langle \Psi_2|\hat v_1-\hat v_2|\Psi_2\rangle = \int n(\mathbf r)[v_1(\mathbf r)-v_2(\mathbf r)] d\mathbf r= \langle \Psi_1|\hat v_1-\hat v_2|\Psi_1\rangle $$}

so that summing the rhs of Eq. 12 and 3, hence the lhs of the same equations, we obtain

{$$E_1+E_2< E_2 + E_1$$}

a strict inequality which is a contradiction. Therefore each density exact interacting electron density {$n$} corresponds to a unique external potential.

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Other functionals

If the ground state energy is a functional of the density, besides the expectation value of the external potential, Eq. 1, also the following contributions to the ground state energy are functionals of the same density, the Coulomb or Hartree repulsion energy

{$$V_C[n]= \frac 1 2\int d\mathbf r^\prime\int \frac{n(\mathbf r) n(\mathbf r^\prime)}{|\mathbf r-\mathbf r^\prime} d\mathbf r$$}

the kinetic energy

{$$T[n] = -\langle \Psi| \sum_i \frac {\nabla_i^2} 2 |\Psi\rangle$$}

and the exchange energy

{$$E_x[n]= V[n] - V_C[n] =\langle\Psi|\sum_i \frac 1 {\hat r_{12}}|\Psi\rangle - V_C[n]$$}

The Coulomb functional is explicitly provided above, whereas the last two are analytically unknown. Together the three form the functional {$Tn]+V[n]$}, which is universal, i.e. the same for all {$N$} interacting electron systems.

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Kohn Sham ansatz

KS assume that for each {$N$} interacting electron system a non interacting electron system exists with the same electron density {$n(\mathbf r)$}. This means that an ancilla single electron Hamiltonian {$h=-\frac {\nabla^2} 2 + V_{eff}(\mathbf r)$} must exist, for which {$h\phi_i=\epsilon_i\phi_i$}. The eigenvectors corresponding to the first {$N$} eigenvalues provide the same exact density

{$$n(\mathbf r)=\sum_i|\phi_i(\mathbf r)|^2$$}

as the interacting system. Knowing the {$\phi_i$} we could compute another functional, the total kinetic energy of the non interacting system

{$$T_0[n] = -\sum_i\int d\mathbf r \, \phi_i^*(\mathbf r)\frac {\nabla^2} 2\phi_i(\mathbf r)$$}

calculated through the ancilla wavefunctions, hopefully a good approximation to the unknown {$T[n]$}. Assuming this ansatz we can recover the form of the effective potential {$V_{eff}(\mathbf r)$} by the variational Raileigh-Ritz principle.

Notice that if we include {$T_0[n]$} in the ground state energy fucntional we have

{$$\begin{equation}E[n] = T_0[n] + V_C[n] + v[n]+ E_{xc}[n]\end{equation}$$}

with the new exchange-correlation term

{$$\begin{equation}E_{xc}[n] = V[n]-V_C[n]+T[n]-T_0[n]\end{equation}$$}

being the only unknown functional.

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Variational approach within KH theorem

We can now minimize Eq. 4 with respect to {$n(\mathbf r)=\sum_i|\phi_i(\mathbf r)|^2$} with the constraint that {$\sum_i\int |\phi_i(\mathbf r)|^2d\mathbf r=N$}, implemented by suitable Lagrange multipliers. Since {$\delta n(\mathbf r) = \sum_i \phi_i(\mathbf r) \delta \phi_i^*(\mathbf r) +c.c.$} we can use independent variations {$\delta\phi_i^*$} for each value of {$i$}.

Let us see what this means for the simplest term, {$v[n]$}. By definition

If we compute {$\delta v$} we get

{$$\delta v \equiv v[n+\delta n]-v[n] = \int d\mathbf r \frac {\delta v[n]}{\delta n} \delta n(\mathbf r) = \int v(\mathbf r)\,\delta n(\mathbf r)\, d\mathbf r$$}

This integral equation identifies the functional derivative {$\delta v[n]/ \delta n$} as the function {$v(\mathbf r)$}. With the same definition of functional derivative we identify

{$$\frac {\delta V_C[n]}{\delta n}= V_C(\mathbf r)= \int d\mathbf r^{\prime} \frac {n(\mathbf r^\prime)}{|\mathbf r^\prime -\mathbf r|}$$}

The full variational equation is

{$$\delta E = \int d\mathbf r \left(\frac {\delta T_o}{\delta n} + \frac {\delta V_C}{\delta n} + \frac {\delta E_{xc}}{\delta n} + \frac {\delta v}{\delta n}\right) \delta n(\mathbf r) - \sum_i\epsilon_i \phi_i(\mathbf r)\delta \phi_i^*(\mathbf r) + c.c. =0 $$}

Independent {$\delta\phi^*_i$} mean that their coefficent must vanish and we obtain the KS equations

{$$\begin{equation}\underbrace{\left(\frac {\delta T_o}{\delta n} + \frac {\delta V_C}{\delta n} + \frac {\delta E_{xc}}{\delta n} + \frac {\delta v}{\delta n}\right)}_{h}\phi_i(\mathbf r) - \epsilon_i \phi_i(\mathbf r)=0 \end{equation}$$}

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Kohn Sham equations

Equation (6) corresponds to the KS independent electron Schrödinger equation that we were seeking. The non interacting Hamiltonian can be rewritten as

{$$h =-\frac {\nabla^2} 2 + \underbrace{v(\mathbf r) + V_C(\mathbf r) + \epsilon_{xc}(\mathbf r)}_{V_{eff}(\mathbf r)}$$}

since

{$$\frac {\delta T_o}{\delta \phi_i^*} = -\frac {\nabla^2} 2 \phi(\mathbf r), \qquad \frac {V_{xc}[n]}{\delta n}=\epsilon_{xc} (\mathbf r)$$}