0
$\begingroup$

Is it possible to construct the exchange correlation functional in DFT for some exactly solvable system? My proposal is to look at a system of Harmonic oscillators. Consider two electrons moving in one dimension in the spin singlet state with Hamiltonian $$ H = \underbrace{\frac{1}{2}(p_1^2+p_2^2)}_{T} \underbrace{- \frac{\gamma^2}{2}(x_1-x_2)^2}_{V_{\mathsf{ee}}} + \underbrace{\frac{\omega^2}{2}(x_1^2+x_2^2)}_{V_{\mathsf{en}}} $$ Note that the signs in the potential have been chosen to mimic the standard electronic structure problem (which has opposite signs).

The exact ground-state energy is $E_0 =(\omega + \delta) / 2$ and the spatial part of the ground-state wave function is given by the following symmetric function $$ \psi_0(x_1,x_2) = \left(\frac{\omega \delta}{\pi^2}\right)^{1/4} \exp\left[-\frac{1}{4}(\delta + \omega)(x_1^2+x_2^2) + \frac{1}{2}(\delta-\omega)x_1x_2\right] $$ where $\delta^2 = \omega^2 - 2\gamma^2$. The ground-state density is thus $$ n_0(x) = 2\int_{\mathbb{R}}\mathsf{d}y \,|\psi_0(x,y)|^2 = \frac{2}{\sqrt{\pi}}\sqrt{\frac{2\omega\delta}{\omega + \delta}}\exp\left[-\frac{2\omega\delta}{\omega + \delta} x^2\right] $$

I presume that the next step is to compute $$ \langle \psi_0 | T + V_{\mathsf{ee}}|\psi_0\rangle = \frac{(2\delta-\omega)(\delta+\omega)}{4\delta}$$

The information above suggests that we are looking for a functional $F=F[n]$, which has the property that $$F[n_0]=\frac{(2\delta-\omega)(\delta+\omega)}{4\delta}$$

This seems like a highly under-determined problem. Is there any potential path from here to compute the full functional dependence $F=F[n]$?

Edit: The discussions in the comments inspired me to suggest the following approach. Consider the following family of Hamiltonians (the $H$ above being a special case), $$ H = \frac{1}{2}\sum_{i} p_i^2 - \frac{\gamma^2}{2}\sum_{i<j}(x_i-x_j)^2 + \frac{\omega^2}{2}\sum_{i,\alpha} Z_\alpha(x_i-X_\alpha)^2 $$ Now compute $F[n_0]$ for a a variety of difference choices of $\{(Z_\alpha, X_\alpha)\}_\alpha$ to form a dataset consisting of $(n_0,F[n_0])$ pairs. Then do some some form of interpolation to determine the ground truth $F=F[n]$. It seems likely that this interpolation can be performed analytically to capture the exact functional but I haven’t done it.

$\endgroup$
1
  • $\begingroup$ Please see the edit to my post $\endgroup$
    – Nebu
    Sep 9, 2023 at 2:45

1 Answer 1

1
$\begingroup$

I am assuming you mean the standard density functional theory definition of the exchange-correlation. I would be careful with a few things, one is that the form of the functionals is determined by the form of the interaction; this follows from the Hohenberg-Kohn theorems. Another note, there may be issues with the standard DFT constructions when the interaction is attractive, so be warned and consult with the conditions of the HK theorem.

To be a little bit pedantic, for your form of interaction we assume that we have a density functional of the form: $$ F[n] = \min_{\Psi \to n}\langle \Psi \vert \hat{T} + \hat{V}_{ee} \vert\Psi\rangle, $$ and as you have stated, we can evaluate this functional at the ground-state density of your Hamiltonian. Assuming you computed your first result correctly, $$ F[n_{0}] = \frac{(2\delta - \omega)(\delta + \omega)}{4\delta}. $$

In general, the exchange-correlation is defined to be $$ E_{xc}[n] = F[n] - T_{s}[n] - U[n], $$ where $T_{s}[n]$ is the non-interacting Kohn-Sham kinetic energy. For two electrons with spin singlet ground-state this functional is exactly known, $$ T_{s}[n] = \frac{1}{8} \int dx \, \frac{\vert \frac{\partial n}{ \partial x} \vert^{2}}{n}.$$ The Hartree contribution $U[n]$ would normally be the classical electrostatic potential energy of the density, but because of your modified form of the interaction, $$ U[n] = \int dx\int dx'\, -\frac{\gamma^{2}}{2}(x-x')^{2}n(x)n(x'). $$ Computing the rest of the energies should be doable because you have the exact density, but this itself is not THE exchange-correlation functional, but rather the value of the exchange-correlation functional at $n_{0}$, so this quantity you compute should not be treated with a lot of generality.

EDIT:

To clarify a point that seems to be getting lost in all of the comments. While you can construct some form from this scheme that you propose in your edit, this is NOT "the exact functional". It is an approximate form for this specific problem (external potential). How we know this cannot be the exact functional is that the exact $F[n]$ should yield the correct value for any valid interacting density $n$. That means if I keep the interaction fixed and solve a different external potential entirely (like a delta well instead of harmonic potential) for some $n'$, the exact functional should yield the $F[n']$ associated to that problem. Your approximate form will not be able to do this.

$\endgroup$
11
  • $\begingroup$ The interactions between the electrons are not attractive (note the minus sign in front of the inter-particle potential) $\endgroup$
    – phonon
    Sep 9, 2023 at 1:36
  • $\begingroup$ Indeed I recognize that there are few “post-processing steps” required to obtain the XC functional from F, as you explained. I’m asking how can one obtain the full functional dependence of F. This is an exactly solvable model so I feel it should be possible. $\endgroup$
    – phonon
    Sep 9, 2023 at 1:40
  • $\begingroup$ I’ll make clarifying edits to the question now. $\endgroup$
    – phonon
    Sep 9, 2023 at 1:41
  • $\begingroup$ There is no real practical sense how you would get the form of the exact functional that we know exists. Whatever the functional is it should yield the correct $F[n]$ for any valid $N$-electron interacting density (with your $V_{ee}$). If you use $T_{s}[n_{0}]$ , $U[n_{0}]$, and your current $F[n_{0}]$ the quantity you would compute is not general, as those forms work for your highly reduced case of two electrons with spin singlet ground-state. The exact functional will be much more complicated. $\endgroup$
    – Nebu
    Sep 9, 2023 at 2:12
  • 1
    $\begingroup$ Thanks this has been very helpful. $\endgroup$
    – phonon
    Sep 9, 2023 at 3:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.