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Everywhere I ready about HF or DFT the term exchange correlation functional comes up. I have a couple of fundamental questions about these:

1) Books say that the correlation energy is the difference between the exact energy (lets say we've found that somehow) and the hartree-fock energy and that the exchange term is accounted for exactly in the hartree-fock energy. While these claims depend on what exactly you include in the correlation energy, isn't the second part on getting the exact exchange totally wrong?? This is because the wave functions that are used to construct/evaluate the hartree fock energy are not the "exact" single electron wavefunctions to begin with so whatever mathematical construct we use to get the exchange energy term can only yield us a false answer. We don't know if there are things called single electron wavefunctions! Or am I wrong? I mean lets say we got the "exact" single electron wavefunctions from somehwhere. Then can we calculate the exchange energy the way its been calculated in hartree-fock method ? Is there a fundamental way to combine all the "exact" single electron wavefunctions to give us the total wavefunction?

2) Likewise, all books say that when we take into account coulombic repulsion, only using the composite wave function can only give us the same exact repulsion. If we use individual electron wavefunctions or densities, then this becomes a mean field approximation. I really do not understand this. In the end an electron is particle at least to the extent that it cannot feel repulsion from itself. So if it feels the repulsion from other charge densities (given we know what they are), why should this be a problem ?. I put forward the same proposal again - if i can get the "exact" single electronic wavefunctions from somewhere, then wouldn't it feel the repulsion of other electrons in the way this normal "averaged" coulomb term describes. To be more specific, I provide this document http://www.physics.metu.edu.tr/~hande/teaching/741-lectures/lecture-05.pdf where on page 2, Eq. (11) the real accurate term has been described for calculating the exact repulsion from the many-body density. It says there "It can be proven that this (electron electron repulsion) term cannot be written in terms of the single-particle density but instead only in terms of the two-particle density". Can someone please direct me to the proof?

3) Are we in the end saying that there is no rule of physics which can tell how one exact single many body wavefunction for N electrons can be decomposed into N exact 1-body wavefunctions or vice versa. Is this really the challenge that we haven't been able to solve in order to get to the correct exchange-correlation functional? If yes please explain a bit further. Also, has this all got to do anything with the second quantization, whatever that means.

Thanks a lot for your attention. It would be great if you could give a long answer. with sources of the mathematical proofs wherever required.

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    $\begingroup$ Perhaps you should consider breaking up this post into a few separate questions (in separate posts). $\endgroup$ – Danu Jan 14 '16 at 10:15
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    $\begingroup$ valid point. however, i was compelled to ask the exchange-correlation in respect to the interpretation of the many-body wavefunction. Also, the questions are relatively straightforward. I describe them longer so that someone who has the answer can say - "hey, you're wrong to being with because this is a false assumption" and so on. $\endgroup$ – Barry Schizzel Jan 15 '16 at 15:01
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Let's take a usual non-relativistic many body bosonic (there is not much difference in taking bosons of fermions, at least for the purpose here) Hamiltonian for $N$ particles of mass $1/2$ on $L^2_{s}(\mathbb{R}^{Nd})$ (where $s$ stands for symmetric functions, wrt exchange of particles): $$H=H_0+H_{int}=\sum_{j=1}^N -\Delta_{x_j}+\frac{1}{N}\sum_{j<i}V(x_i-x_j)\; ,$$ where the symmetric $V(\cdot)$ models the two-body interaction between particles (in the case of Coulomb, $V(x)=\pm \frac{1}{\lvert x\rvert}$).

It is pretty clear that a factorized structure for particles is not preserved by the interaction term: think of the simplest factorized symmetric function $$\Phi_N(x_1,\dotsc,x_N)=\varphi(x_1)\dotsm\varphi(x_N)\; ,$$ where $\varphi\in L^2(\mathbb{R}^d)$ is a single particle wavefunction.

While $H_0\Phi_N$ still gives a factorized structure (it suffices that $\varphi\in H^2(\mathbb{R}^d)\subset L^2(\mathbb{R}^d)$, and we get $H_0\Phi_N(x_1,\dotsc,x_N)=(-\Delta\varphi)(x_1)\dotsm(-\Delta\varphi)(x_N)$); $H_{int}\Phi_N$ is no more factorized, since in general (and specifically for Coulomb) it is not true that we can write $V(x_i-x_j)$ as $V'(x_i)V'(x_j)$.

So it is not a lack of knowledge, but a fact that systems of particles with Coulomb two-body interactions (and other potentials) are not describable by means of single-particle wavefunctions, because the interaction yields unavoidable correlations that affect the evolution of the system.

Now, said that, we would like to be able to describe the behavior of a single particle without studying the whole system, in suitable situations (because it is far more easy on a computational standpoint). This is done using the so-called mean field approximation, that is precise only in the limit $N\to\infty$ of many (infinite) particles. The huge number of particles makes possible to describe the motion of each particle by means of an effective self-interaction, that is an expression of the global action of all the other particles in the system (the mean field). The price to pay is that the mean field equation is no more linear. For the system above, it is the Hartree equation on $L^2(\mathbb{R}^d)$: $$i\partial_t\varphi=-\Delta\varphi + (V*\lvert \varphi\rvert^2)\varphi\; .$$ However its validity holds only in the limit $N\to\infty$; if you have a very large $N$ but finite, you make an error using Hartree equation. However, the error can be quantified, and depends on how large $N$ is.

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  • $\begingroup$ thanks. I'll look up the terms and try to understand your answer, as I not mathematically advanced enough to understand it. Although I do understand that the product of two "exact "single electron wave functions (even if we have them from somewhere) will never yield the correct total wavefunction and therefore we're wrong in doing this. On the other side your answer seems to be more on the side that even if we have these exact single electron wavefunctions, we'll not get the correct energetics because the formalism of calculating repulsion from these wave functions itself is wrong. $\endgroup$ – Barry Schizzel Jan 15 '16 at 15:09
  • $\begingroup$ The point is that we have two different possible dynamics, one is the "true one", that can be described only on the whole set of particles due to correlations; and the "effective one" (mean-field) that can be described for each particle. The true dynamics is never factorized, for even if the state was factorized at a certain moment, then it would be not factorized anymore after a whatever short evolution. The effective one becomes correct only in the limit of an infinite number of particles, if else you can use it, but you make an error (small if the particles are many). $\endgroup$ – yuggib Jan 16 '16 at 10:21
  • $\begingroup$ I understand this better. Now I can better explain my 3rd question "Are we in the end saying ..... decomposed into N exact 1-body wavefunctions or vice versa. " I think we do not know if there is any analytical rule (like factorization) by which the identities of single electrons can be connected to the N-body wavefunction. I mean are we sure that this is impossible to do? Don't you think that if someone found such a law of physics then it would be kind of revolutionary? It doesn't seem to me as something impossible because its hard to believe the electrons lose their individuality in a group. $\endgroup$ – Barry Schizzel Jan 18 '16 at 8:46

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