Problem: In one sentence, what is the essence of DFT?
Solution: To replace \(\Psi\mapsto n\), where the number density of a system of \(N\) identical quantum particles (usually electrons) \(n(\mathbf x)\) is:
\[n(\mathbf x):=N\int d^3\mathbf x_2…d^3\mathbf x_N |\Psi(\mathbf x,\mathbf x_2,…,\mathbf x_N)|^2\]
in terms of its \(N\)-body wavefunction \(\Psi(\mathbf x,\mathbf x_2,…,\mathbf x_N)\). Thus, \(\int d^3\mathbf x n(\mathbf x)=N\) is normalized.
Problem: Write down the non-relativistic Hamiltonian \(H\) of a molecule. Attempt to express the expected energy \(E=\langle\Psi|H|\Psi\rangle\) in many-body state \(|\Psi\rangle\in\bigwedge^NL^2(\mathbf R^3\to\mathbf C)\otimes\mathbf C^2\) in terms of \(n(\mathbf x)\) defined above; this functional \(E=E[n]\) of the density \(n\) explains the name “density functional theory“.
Solution: (the Born-Oppenheimer approximation is implicitly used whereby the electrons move on an effective/external potential energy surface \(V_{\text{ext}}(\mathbf x)\) defined by a collection of stationary nuclei)
\[H=\sum_{i=1}^N\frac{|\mathbf P_i|^2}{2m}+V_{\text{ext}}(\mathbf X_i)+\frac{1}{2}\sum_{1\leq i\neq j\leq N}\frac{\alpha\hbar c}{|\mathbf X_i-\mathbf X_j|}\]
One can check that:
\[E=\frac{\hbar^2}{2m}\sum_{i=1}^N\int d^3\mathbf x_1…d^3\mathbf x_N\biggr|\frac{\partial\Psi}{\partial\mathbf x_i}\biggr|^2+\int d^3\mathbf x n(\mathbf x)V_{\text{ext}}(\mathbf x)+\frac{\alpha\hbar c}{2}\int d^3\mathbf x d^3\mathbf x’\frac{n_2(\mathbf x,\mathbf x’)}{|\mathbf x-\mathbf x’|}\]
where \(n_2(\mathbf x,\mathbf x’)=N(N-1)\int d^3\mathbf x_3…d^3\mathbf x_N|\Psi(\mathbf x,\mathbf x’,\mathbf x_3,…,\mathbf x_N)|^2\) so in particular \((N-1)n(\mathbf x)=\int d^3\mathbf x’ n_2(\mathbf x,\mathbf x’)\).
Problem: Write down the Thomas-Fermi approximation for \(E[n]\), and explain why it sucks. Also derive the von Weizsäcker correction to the Thomas-Fermi energy functional.
Solution: The Thomas-Fermi functional treats the electrons as a locally ideal Fermi gas, thus having the usual Pauli pressure \(p\propto n^{5/3}\) in \(\mathbf R^3\) and thus (kinetic) energy density \(3p/2\) scaling likewise. Furthermore, the exchange-correlation functional is taken to vanish \(E_{\text{xc}}[n]=0\) (thus, Thomas-Fermi is often said to be semiclassical):
\[E_{\text{TF}}[n]=\frac{3\hbar^2(3\pi^2)^{2/3}}{10m}\int d^3\mathbf x n^{5/3}(\mathbf x)+\int d^3\mathbf x n(\mathbf x)V_{\text{ext}}(\mathbf x)+\frac{\alpha\hbar c}{2}\int d^3\mathbf x d^3\mathbf x’\frac{n(\mathbf x)n(\mathbf x’)}{|\mathbf x-\mathbf x’|}\]
One has:
\[\frac{\delta}{\delta n(\mathbf x)}\left(E_{\text{TF}}-\mu\left(\int d^3\mathbf xn(\mathbf x)-N\right)\right)=\frac{\hbar^2k^2_F(\mathbf x)}{2m}+V_{\text{eff}}(\mathbf x)-\mu=0\]
where \(k_F(\mathbf x):=(3\pi^2n(\mathbf x))^{1/3}\) and \(V_{\text{eff}}(\mathbf x)=V_{\text{ext}}(\mathbf x)+\alpha\hbar c\int d^3\mathbf x’\frac{n(\mathbf x’)}{|\mathbf x-\mathbf x’|}\).
(aside: although the Thomas-Fermi model is meant to be applied to molecules, a result of Teller showed that \(E_{\text{TF molecule}}>2E_{\text{TF atom}}\) so Thomas-Fermi theory predicts the molecular state is unstable with respect to dissociation. In a central external potential such as due to a single nucleus \(V_{\text{ext}}(r)=-Z\alpha\hbar c/r\), the effective potential \(V_{\text{eff}}(r)\) and electron density \(n(r)\) are both isotropic as well. Starting from Poisson’s equation:
\[\biggr|\frac{\partial}{\partial\mathbf x}\biggr|^2\frac{V_{\text{eff}}(r)}{-e}=-\frac{-en(r)}{\varepsilon_0}\]
with \(\biggr|\frac{\partial}{\partial\mathbf x}\biggr|^2f(r)=\frac{1}{r^2}\frac{d}{dr}\left(r^2\frac{df}{dr}\right)=\frac{1}{r}\frac{d^2}{dr^2}(rf(r))\) (use the latter) one can check that by defining \(f(r):=\left(\frac{8}{3\pi a_0}\right)^2rk^2_F(r)\), one recovers the Thomas-Fermi ODE:
\[\frac{d^2 f}{dr^2}=\frac{f^{3/2}(r)}{\sqrt{r}}\]
The von Weizsäcker term is a correction to the kinetic energy functional in the Thomas-Fermi model that accounts for density gradients:
\[T_{\text{vW}}[n]=\frac{\hbar^2}{8m}\int d^3\mathbf x\frac{|\partial n/\partial\mathbf x|^2}{n(\mathbf x)}\]
Heuristically, it can be justified by writing the expected kinetic energy of a single electron \(\frac{\hbar^2}{2m}\int d^3\mathbf x |\partial\psi/\partial\mathbf x|^2\) with wavefunction \(\psi(\mathbf x)\in\mathbf R\) and replacing \(\psi(\mathbf x)\mapsto\sqrt{n(\mathbf x)}\).
Problem: Prove the two Hohenberg-Kohn theorems.
Solution: First, it is useful to prove:
- Lemma: If \(V_{\text{ext}}(\mathbf x)\neq V’_{\text{ext}}(\mathbf x)\) represent \(2\) physically distinct external potentials (i.e. they differ by more than just some additive constant), then the corresponding molecular Hamiltonians \(H\neq H’\) do not share any non-zero eigenstates.
- Proof: Suppose for sake of contradiction that there exists \(|\Psi\rangle\neq 0\) such that
\[H|\Psi\rangle=E|\Psi\rangle\]
\[H’||\Psi\rangle=E’|\Psi\rangle\]
Subtracting, one obtains:
\[(H’-H)|\Psi\rangle=(E’-E)|\Psi\rangle\]
Inspecting the form of the molecular Hamiltonian \(H\) given earlier, one sees that the “universal” part (i.e. the kinetic energy and electron-electron Coulomb repulsion terms) cancel, leaving only the non-universal residue \(H’-H=\sum_{i=1}^NV’_{\text{ext}}(\mathbf X_i)-V_{\text{ext}}(\mathbf X_i)\). But the RHS \(E’-E\) is \(\mathbf X_i\)-independent for all \(i=1,…,N\); this means \(V’_{\text{ext}}(\mathbf x)-V_{\text{ext}}(\mathbf x)=\text{const}\) which is a contradiction.
With this in mind, the first HK theorem follows by specializing to the respective (distinct by the above lemma) ground states \(|\Psi_0\rangle\neq|\Psi’_0\rangle\) of \(H\) and \(H’\). Apply the variational bound both ways:
\[\]
Problem: Demonstrate the first Hohenberg-Kohn theorem for \(2\) electrons in a \(1\)-dimensional harmonic trap.
Solution:
Problem: Derive the Kohn-Sham equations.
Solution: Fictitious ideal electron gas with density \(n(\mathbf x)\).
Problem: