There is a conceptual mismatch in my personal understanding between the role of reduced density matrices in DFT and post-HF methods (possibly due to different discussions in textbooks), and I can not find arguments to bridge the gap between these two points of view.

One statement which I believe to be true is that in order to evaluate the expectation values of an $N$-electron operator, the $N$-electron reduced density matrix ($N$-RDM) is needed.

For example, within Hartree-Fock, we transform the electron-electron interaction into a Fock operator, which, being an effective one-electron operator, only needs evaluation of the 1-RDM. This means that there is no direct 2-RDM in Hartree-Fock theory (apart from an outer product of two 1-RDMs, adding no extra information), and, therefore, correlation effects are neglected. On the other hand, post-HF methods do have some expressions for their respective 2-RDMs. Moreover, since the Hamiltonian contains only terms up to two-electrons, the exact 2-RDM would be enough to evaluate the exact energy - more should be needed for operators of more electrons.

This story is seemingly not in agreement with the principle of DFT, where, starting from the Hohenberg-Kohn theorems, we start with the fact that the electron density and the Hamiltonian uniquely determine each other. If we knew the exact form of the exchange-correlation functional, we could obtain the exact density, and, from this, the exact energy. However, this seems to imply that knowing the 1-RDM is enough, since the electron density can be obtained exactly from the 1-RDM. Consequently, the 2-RDM is not an object that is even calculated (or defined) in ''standard'' DFT calculations.

Do we really need the 2-RDM to obtain the exact energy, or should the 1-RDM be enough?

  • $\begingroup$ +1. It's a well-written question. However please see this, which is the reason why I removed the second question. Perhaps you could alternatively combine the two questions by asking about $N$ for $N\ge 2$, if you want, but currently (after my edit) you have a good and suitable question. $\endgroup$ Jan 27 at 12:41
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    $\begingroup$ See this. But note that generally "in principle" is a different story than "in practice". $\endgroup$
    – Jakob
    Jan 27 at 17:18
  • $\begingroup$ That is a great answer @Jakob and maybe I should have formulated my question more around it, as it also uses the facts that I mention (1RDM is enough for one-electron expectation values, which include the density, which would also mean the external potential by Hohenberg-Kohn), but I still don't see how exactly do we resolve the fact that we do not need to use 2RDM at all. $\endgroup$
    – Szgoger
    Jan 27 at 19:54
  • $\begingroup$ What do you mean with "use the 2-RDM at all"? You can ask the same thing about the density in DFT, no? TBH I think you should phrase the question more precise, and in particular in more mathematical terms. It is unclear to me, right now, what exactly you are looking for. $\endgroup$
    – Jakob
    Jan 27 at 19:57

1 Answer 1



"Well yes, but actually no."

— the Pirate Captain (paraphrased), The Pirates! In an Adventure with Scientists!

  • The universal 2RDM functional is known (and linear!), but 2RDMs are hard to work with
  • The universal 1RDM functional is unknown and nonlinear. 1RDMs are hard to work with formally
  • The universal density functional is unknown. Densities are easy to work with! But in the usual Kohn–Sham formulation (and in finite basis sets more generally), DFT is secretly done as a 1RDM functional theory anyway.

The full version will be a little long, because you've hit on the fundamental tradeoff of DFT. My apologies in advance. But the most relevant concept, in my opinion, is the conservation of difficulty popularized by Terence Tao.

Two electrons

Clearly, you never need more than a 2RDM, because the electronic structure Hamiltonian only contains one- and two-body operators. Indeed, the energy is linear in the 2RDM $\gamma_2$, $$ E = \text{Tr}\, H_2 \gamma_2 \tag{1}. $$ $H_2$ is the two-electron reduced Hamiltonian, for which an explicit expression is known (see, e.g., Mazziotti's review in Chem. Rev.).

The trouble with this method is it's hard to tell whether a trial 2RDM could have actually been obtained by tracing out a many-body wavefunction, without actually doing the tracing directly (in which case you might as well use a higher-order method). There has been recent progress towards solving the so-called $N$-representability conditions, but I believe it's still too impractical to be worth it most of the time.

One electron, two coordinates

As you mentioned, Hartree–Fock theory is a 1RDM functional theory because the electron-electron interaction is cast as a one-electron operator. You can actually derive this another way by restricting the space of allowable ($N$-body) wavefunctions to single Slater determinants: then the 2RDM is easily expressed in terms of the 1RDM as (Eq. (2.5.2) of Parr and Yang's DFT book). $$ \gamma_2(x_1' x_2', x_1 x_2) = \frac12 \left\lvert \begin{matrix} \gamma_1(x_1', x_1) & \gamma_1(x_2', x_1) \\ \gamma_1(x_1', x_2) & \gamma_1(x_2', x_2) \end{matrix} \right\rvert. \tag{2} $$ I won't write it out here, but you can derive $V_{ee}$ in that space in terms of the 1RDM.

Anyway, it is actually possible to express the ground-state energy as a functional of the 1RDM exactly. In fact, this is done pretty much like the classic Hohenberg–Kohn theorem for DFT: but it's more general, allowing for instance nonlocal $V_{\text{ext}}$. Gilbert proved this in 1975; Levy followed up with a constrained-search version four years later; and Valone gave an ensemble version in 1980.

The universal 1RDM functional is no longer linear, and is also unknown. There's a nice comparison of the three major approaches (it's even open access!), which discusses how they trade off complexity of the functional with complexity of the domain. You see, 1RDMs have to contend with both the $v$-representability that concerned early researchers in DFT and with the $N$-representability that's so challenging for 2RDMs. (In DFT, $N$-representability is as easy as $\int dr\, \rho(r) = N$.)

One electron, one coordinate

For a local external potential, the density $\rho = \gamma_1(x_1, x_1)$ is enough to determine the ground-state energy. The universal functional is still unknown (and certainly highly nonlinear). But for 'normal' (e.g., nonmagnetic) calculations, the success of DFT certainly makes it seems like the 1RDM is overkill.

Ah, but DFT is usually 1RDM functional theory in disguise! For example, Kohn–Sham calculations are often solved rather like Hartree–Fock ones, with 1RDM methods; certainly hybrid DFT is a 1RDM functional theory rather than a pure functional of the density; and furthermore DFT in finite basis sets is really a 1RDMFT, more or less because the Dirac delta function can't be represented exactly in a finite-dimensional space and therefore don't have a truly local external potential.

Concluding thoughts

Your surprise is to be expected: the Hohenberg–Kohn theorem is highly nontrivial, even if its proof is easy. It's not at all intuitive that you can get a ground-state two-electron property from one-electron quantities.

But there is no free lunch: we trade complexity of the electronic quantity for complexity of the energy functional.

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    $\begingroup$ Valone's article is not from 2008! Other than that, good answer. BTW: That the 1-RDM suffices in DFT (if the assumption of the Hohenberg-Kohn theorem hold) simply follows from the fact that the ground state density determines the ground state, which in turn determines the ground state 1-RDM (in fact, all $p$-RDMs), which in turn determines the ground state density... Hence I was confused what OP is actually asking for... $\endgroup$
    – Jakob
    Feb 3 at 12:25
  • $\begingroup$ Whoops, you're quite right; I edited the article. There is a "2008" on JCP's website that misled me; maybe it was digitized then? I thought most likely the confusion was from the fundamental surprising-ness of HK, which I wanted to talk about a little! $\endgroup$
    – elutionary
    Feb 3 at 15:52
  • $\begingroup$ I have found a book chapter stating that due to the fact that the density can be exact in Kohn-Sham theory, any expectation value depending on only one position operator can also be exact on the 1RDM level. Based on your comment, I had the mental picture that only the energy is exact by Kohn-Sham, but I might have misunderstood it. $\endgroup$
    – Szgoger
    Feb 6 at 16:41
  • $\begingroup$ I talked about the energy because it's the main operator I work with, but Hohenberg–Kohn means that any ground-state property is a functional of the density (therefore of the 1RDM), whether it's local in position or not! Any one-position operator is included in that; it's more likely to be clear what the form of its universal functional is, though. $\endgroup$
    – elutionary
    Feb 6 at 19:41

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