Quoting Becke [1]

Density-functional theory (DFT) is a subtle, seductive, provocative business. Its basic premise, that all the intricate motions and pair correlations in a many-electron system are somehow contained in the total electron density alone, is so compelling it can drive one mad

How do you convince yourself that all the intricate motions and pair correlations in a many-electron system are somehow contained in the total electron density alone?

PS: I am not looking for proofs for Hohenberg-Kohn theorems. More like a physical/intuitive picture


  1. Becke, Axel D. "Perspective: Fifty years of density-functional theory in chemical physics." The Journal of chemical physics 140.18 (2014): 18A301.
  • 1
    $\begingroup$ I guess one crucial point here is that we are only talking about the ground state. If you just know the density, you do not know anything about pair correlations. Only if you know that the density pertains to the ground state, then you can apply the Kohn-Sham theorem and (formally) deduce the wavefunction. $\endgroup$ – Felix May 13 '20 at 19:53
  • $\begingroup$ Once I have heard an indirect argument saying the position of nuclei (positive charges) in ground-state uniquely determines the electron distribution as well as the ground state energy/properties. Something like it is not possible to find the same electron distribution for two different sets of nuclear positions or to find two different ground-state electron density for the same nuclear positions. Similarly, we also expect that the energy of the system is determined by the position of the nuclei, therefore there should be some 1-to-1 relationship between electrons density and energy. $\endgroup$ – Greg May 13 '20 at 20:01

It's completely legitimate to want a physical intuition for something that is proved mathematically.

One very hand-wavy argument is just to say that the electron density in the ground state has to reflect the effects of all of those electron-electron correlations, interactions, exchange. After all, the electron density definitely depends on the interactions between electrons, but that doesn't mean we need to know all the details of how the electrons are actually interacting.

There are often macroscopic/thermodynamic quantities that include the effects of all sorts of more complicated microscopic interactions. For example, the Young's modulus of a solid allows us to accurately predict the stress/strain relationship without having to know anything about the microscopic rules that cause the Young's modulus to take it's specific value. the Another example would be an effective electron mass in solid state physics.

  • 1
    $\begingroup$ But unlike the example of Young's modulus in DFT the core concept is the one-to-one correspondence. $\endgroup$ – Thomas May 14 '20 at 6:43

This is a very nice question touching ongoing research. I cannot really provide a nice answer to it but I would like to share a few thoughts I had when thinking about this question.

    1. I think mentioning the Hohenberg-Kohn theorem in the question puts Becke's statement slightly into the wrong context. In this context one should first realize that there is an object even simpler than the ground-state electron density containing all details of the investigated system: The many-body Hamiltonian

    $$\hat{H} = \hat{T} + \hat{V}_\text{ee} + \hat{V}_\text{ext}$$

    consisting of the kinetic energy operator $\hat{T}$, the Coulomb interactions $\hat{V}_\text{ee}$, and the external potential $\hat{V}_\text{ext}$. You can actually write it down in every detail on a sheet of paper.

    Hohenberg and Kohn show for Hamiltonians of this form with the known expressions for $\hat{T}$ and $\hat{V}_\text{ee}$ that there is a one-to-one mapping between the ground-state density and the external potential. At least up to a constant potential shift.

    Essentially this means that the positions and charges of the atomic nuclei are encoded in the ground-state density. But more complex external potentials are also possible. The form of the external potential does not have to be known beforehand. For the rest of the Hamiltonian this is not the case. It is assumed that it has the form sketched above: There is no one-to-one mapping between the density and the form of $\hat{T}$ and $\hat{V}_\text{ee}$ which actually contain all the interactions you asked about.

    There even is an obvious counterexample for such a one-to one mapping: Kohn and Sham construct an auxiliary system of noninteracting electrons featuring the same ground-state density.

    So on this level the ground-state density only in connection with knowledge of the form of $\hat{T}$ and $\hat{V}_\text{ee}$ contains all the interesting interaction details. But as mentioned this is also encoded in the Hamiltonian.

    2. I think Becke's statement has to be seen in the context of energy functionals to extract the ground-state energy of an interacting many-electron system from the respective density.

    Such energy functionals are typically based on the Kohn-Sham system which is constructed such that the most significant contributions to the energy of the interacting many-electron system are easily accessible. This includes the Hartree energy, the energy due to the external potential, and the kinetic energies of the single-electron Kohn-Sham orbitals. Everything beyond these energy contributions is integrated into the exchange-correlation energy for which no exact expression is known.

    Fortunately this energy contribution often is not the dominating part and even simple approximations cover it with a good enough accuracy, e.g., the local density approximation that assumes the exchange-correlation energy of a homogeneous electron gas. For the homogeneous electron gas an expression for the exchange energy in terms of the density is known and the correlation energy can numerically be calculated by simulations.

    As indicated in my first sentence of this answer improving the accuracy of approximations to the exchange-correlation functional is ongoing research. One knows several properties the exact exchange-correlation functional has to fulfill. I think getting an intuitive physical picture on how the interactions are encoded in the ground-state density is strongly connected to knowing and understanding these properties. Unfortunately I cannot give you insights on this.

    Systematic improvements to the exchange-correlation functional often go beyond only using the charge density in a direct way. Since most of the exchange-correlation energy is due to exchange you see many approaches to express the exchange part exactly in terms of Kohn-Sham orbitals. This can then be used to construct a hybrid functional or to combine it with the random phase approximation to the correlation energy.

    Of course, one actually does not want to use the Kohn Sham orbitals for this because the resulting expressions are connected to significant computational demands. But the fact that it is done in this way indicates how difficult it is to directly extract the exchange-correlation energy from the density. It might also indicate that a practically usable physical picture directly connecting the density to the respecive interactions is not really available. But of course, it might also indicate that the encoding of this quantity in the rather featureless electron density is just too subtile to be usable.

  • 1
    $\begingroup$ +100. An amazing first answer! Welcome to our community, and thank you for your contribution! We hope to see much more of you !!! We are still very new (entered public beta in May 2020). $\endgroup$ – Nike Dattani Sep 20 '20 at 16:22

I really don't think that I have to be convinced in any conceptual/philosophical way.

As we are dealing with Science, and in this particular case, Exact Science, whenever the method reproduce the experimental properties, it is good enough. If also, it is capable to predict the properties/existence of new materials, is better. As soon as it fails (like "old" functionals used to simulate the interaction of graphene sheets), new methods (in this case, new functionals) are developed.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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