I was reading on the implementation of Hubbard 'U' in Density Functional Theory. The Hubbard 'U' can be thought of as a parameter that accounts for the on-site electron-electron interactions in correlated materials.

Meanwhile, the Hund's J, I'm given to believe is an exchange effect. Many texts mention the 'J' like this one. This kind of direct exchange comes from the antisymmetry nature of the many-body electron wavefunction. How can this kind of exchange be quantified by an energy value similar to the Hubbard 'U'? I believe in literature, this 'J' value is usually taken to be less than 20% of the U value. I have not found a conclusive reason for this, but what I am more concerned with, is how can you possibly quantify such a direct exchange?

PS: I had cross-posted this question on Phys stack exch earlier but now realize that this forum might be more appropriate since the 'J' is frequently encountered in DFT+U calculations.

  • $\begingroup$ Sure Nike, have a good one. $\endgroup$
    – livars98
    Commented Jun 21, 2020 at 5:43
  • $\begingroup$ By "Hund's J" do you mean the spin superexchange interaction? $\vec S_i \cdot \vec S_j $ ? $\endgroup$ Commented Jun 26, 2020 at 3:11
  • $\begingroup$ Are you asking how it can be calculated in the context of materials modeling? $\endgroup$
    – Anyon
    Commented Jun 26, 2020 at 13:40
  • $\begingroup$ @taciteloquence No. I am referring to direct exchange, not super exchange. Super exchange can predict ground state magnetic ordering etc but direct exchange is associated only with the antisymmetric nature of the many-body fermionic wavefunction. $\endgroup$
    – livars98
    Commented Jun 27, 2020 at 19:22
  • $\begingroup$ @Anyon my original question was 'why can it be quantified' meaning why would you ascribe a physical value (of the order of few eV) to an exchange interaction that purely depends on the antisymmetric nature. I had a hard time reconciling how this could be an energy similar to coulombic repulsion. In the context of materials modeling, I believe that there is no rigorous way to calculate the Hund's J. It is taken usually to be 0-20% of the Hubbard U. So, circling back, my question is more physics oriented I believe. But I will be really happy if you had other thoughts on either of these points. $\endgroup$
    – livars98
    Commented Jun 27, 2020 at 19:24

1 Answer 1


When I first read your question, I found it somewhat puzzling. I have to admit, I still do in part. Why? Well, even your link defines $J$ as a sum of matrix elements $\langle m,m'|V_{ee}|m',m\rangle$. Mathematically, each of these matrix elements is an integral involving wave functions in the system. If we know these wave functions or know how to approximate them it is conceptually straight-forward to calculate the integrals, save for possible divergences. Leaving such issues aside, I'd say $J$ is eminently quantifiable, at least in principle. (And demonstrably also in practice for e.g. few-body systems and atoms.)

Now, I suspect the above is a kind of pedantic answer that doesn't really get to the heart of the question you wanted to ask. In the question itself and your comments below it, you make a clear distinction between the Hubbard $U$ and Hund's $J$ by saying one is linked to on-site electron-electron interactions and calling the other a pure exchange effect. You then ask how $J$ can be given a value similar to that of the Coulomb repulsion. This suggests that there might be a misconception about the nature of exchange effects at play. In fact, Hund's type exchange while magnetic in effect, is due to a combination of the Coulomb force and the Pauli exclusion principle.

A simple version

The simplest version of this physics shows up already in the treatment of two-electron atoms (notably Helium). This problem is discussed in textbooks like Sakurai's Modern Quantum Mechanics in more detail than I will go here. Thus I'll just jump to the result for the energy associated with the Coulomb repulsion between the two electrons (which is treated as a perturbation on top of a non-interacting ground state): $$ \left\langle \frac{e^2}{r_{12}} \right\rangle = I \pm J, \tag{1} $$ where $r_{12}$ is the distance between the two electrons. The upper sign goes with a spin singlet state, and the minus sign with a spin triplet state. $I$ and $J$ are called the direct and exchange integrals, given by $$ I = \int \mathrm{d}^3 \mathbf{x}_1 \int \mathrm{d}^3 \mathbf{x}_2 | \psi_{100} \left( \mathbf{x}_1 \right)|^2 | \psi_{nlm} \left( \mathbf{x}\right)|^2 \frac{e^2}{r_{12}},\tag{2} $$ $$ J = \int \mathrm{d}^3 \mathbf{x}_1 \int \mathrm{d}^3 \mathbf{x}_2 \psi_{100} \left( \mathbf{x}_1 \right) \psi_{nlm} \left( \mathbf{x}_2\right) \frac{e^2}{r_{12}} \psi_{100}^\star \left( \mathbf{x}_2 \right) \psi_{nlm}^\star \left( \mathbf{x}_1\right).\tag{3} $$ I don't want to go into detail about the notation here, but the difference in structure is clear. In $I$ we have a density-density interaction, but $J$ involves a mixing or exchange of the two wave functions at the two coordinates. While the sign of $\pm J$ is determined by the spin state and enforced by the antisymmetry of the wave function, the magnitude $J$ is set by the Coulomb repulsion, and is indeed quantifiable.

In materials

I will here loosely follow the notation of another book I like: Physics of Transition Metal Oxides by Maekawa et al. For concreteness we consider two $3d$ elecrons, labeled $1$ and $2$, with positions $r_1$, $r_2$ and wave functions $\Psi(r_1)$, $\Psi(r_2)$. In the context of this stack we can think of them either as two localized molecular orbitals (as in LCAO) or as two Wannier functions. The Coulomb interaction between them is $$ H = \frac{1}{2} \int \int \Psi^\dagger (r_1) \Psi^\dagger (r_2) \frac{e^2}{|r_1-r_2|} \Psi(r_2) \Psi(r_1) \mathrm{d}\tau_1\tau_2,\tag{4} $$ where we integrate over spatial variables and sum over spin degrees of freedom. (Both steps are hidden inside the $\tau$ symbols.) We can now decompose the wave function $$ \Psi (r) = \sum_{n,m,\sigma} c_{nm\sigma} \psi_{nm} (r) \theta_\sigma,\tag{5} $$ where $n$ denotes site, $m$ denotes orbital quantum number, and $\sigma$ the spin quantum number. $\psi_{nm}$ is the spatial wave function, and $\theta_\sigma$ is the spin wave function. In second quantization, $c_{nm\sigma}$ is the annihilation operator associated with an electron with these quantum numbers. Then, if we include the spatial integration inside the matrix elements, we can write $$ H = \frac{1}{2} \sum_{n,m,\sigma} \langle n_1 m_1, n_2m_2| \frac{e^2}{|r_1-r_2|} | n_3m_3,n_4m_4\rangle c^\dagger_{n_1 m_1 \sigma_1} c^\dagger_{n_2 m_2 \sigma_2} c_{n_4 m_4 \sigma_2} c_{n_3 m_3 \sigma_1}.\tag{6} $$

This expression includes multiple processes. First, we have the one-site ($n_1=n_2=n_3=n_4=n$) one-orbital ($m_1=m_2=m_3=m_4$) case,, $$ H_U = \frac{1}{2}\sum_\sigma \langle n,n| \frac{e^2}{|r_1-r_2|} | n,n\rangle c^\dagger_{n\sigma}c^\dagger_{n,-\sigma} c_{n,-\sigma} c_{n\sigma} = Un_{n\uparrow}n_{n\downarrow},\tag{7} $$ with $n_{n\sigma}=c_{n\sigma}^\dagger c_{n\sigma}$. This is of course nothing but our friend the Hubbard interaction, where the messy details of the matrix element is hidden behind the symbol $U$.

Two electron on the same site ($n_1=n_2=n_3=n_4=n$), but in different orbitals produce the so-called Hund coupling (c.f. Hund's rules). We get \begin{align} H_H = \frac{1}{2} \langle m_1,m_2 | \frac{e^2}{|r_1-r_2|} |m_1,m_2\rangle \sum_{\sigma_1,\sigma_2} c_{m_1\sigma_1}^\dagger c_{m_1\sigma_1}c_{m_2\sigma_2}^\dagger c_{m_2\sigma_2} \tag{8}\\ - \frac{1}{2} \langle m_1,m_2 | \frac{e^2}{|r_1-r_2|} |m_2,m_1\rangle \sum_{\sigma_1,\sigma_2} c_{m_1\sigma_1}^\dagger c_{m_1\sigma_2}c_{m_2\sigma_2}^\dagger c_{m_2\sigma_1}\tag{9}\\ \equiv K_{m_1m_2} n_{n_1}n_{n_2} - 2J_{m_1m_2} \left( \mathbf{S}_{m_1m_2}\cdot \mathbf{S}_{n_2} + \frac{1}{4} n_{m_1}m_{n_2}\right).\tag{10} \end{align} Here $K$ is a on-site inter-orbital Coulomb density-density interaction, and $J$ represents the Hund's coupling. You can see that this matches (up to coefficients) Eq. (11) in the notes you linked in the question. Note that the magnitude of $J$, like that of $U$, is related to the strength of the Coulomb force between the two electrons.

Similarly, inter-site terms like direct exchange can be derived, and ultimately quantified.

First-principles calculation

Although the Hubbard $U$ and Hund $J$ are often treated as semi-empirical parameters in materials modeling, and sometimes assigned values using rules of thumb (that I don't know the origin of), there are some first-principles frameworks that can be employed to calculate them accurately. To get reliable values, screening effects must be taken into account. This can be done while downfolding to effective Hamiltonians using e.g. the constrained Random Phase Approximation (cRPA) and constrained GW approximation, which can use DFT band structures as inputs (though there are subtleties concerning double-counting of correlation effects). For details on cRPA see e.g. Chapter 7 in this set of lecture notes. For cGW, some developments are described in this and this paper.

It's worth noting that such calculations often uncover significant non-local (i.e. intersite) Coulomb interactions, see e.g. this paper on $\alpha$-RuCl$_3$ from last year and references therein. Such non-local interactions are often overlooked, but might have important effects in strongly correlated systems.

  • $\begingroup$ +10. Very well done! $\endgroup$ Commented Jun 30, 2020 at 22:50
  • $\begingroup$ Very well-written answer. Thank you.I just wanted to clarify one thing. You say that 'J' is a combined effect of Coulombic forces and Pauli's principle - this implies that my previous impression that 'J' is purely a direct exchange effect, is erroneous, right? $\endgroup$
    – livars98
    Commented Jul 3, 2020 at 4:19
  • $\begingroup$ @livars98 Thank you. Actually, what people tend to call "direct exchange interaction" in materials is the inter-site version of $J_{m_1m_2}$ above, so the Hund's coupling and the direct exchange have essentially the same structure. From that perspective, calling $J$ an intra-site "direct exchange effect" is reasonable, although potentially confusing. However, note that the inter-site direct exchange also depends on the Coulomb force. You generally need some force/potential to induce a finite energy splitting between two otherwise degenerate states like these spin states. $\endgroup$
    – Anyon
    Commented Jul 3, 2020 at 16:25
  • $\begingroup$ As the number one user of our "model Hamiltonians" tag, I wonder if you might be able to answer this one? It's one of the oldest unanswered questions on our entire site! Also this answer is so good that it deserves a bounty, though I think my edit just now (adding the necessary numbering of the equations in case someone else wants to refer to them in their answer or journal publication) will bring you at least 50 points from all the users that joined since July who hadn't seen this yet :) $\endgroup$ Commented Dec 31, 2020 at 4:10
  • $\begingroup$ Sorry, in the last comment I meant these unanswered questions: mattermodeling.stackexchange.com/q/1548/5, mattermodeling.stackexchange.com/q/1678/5 $\endgroup$ Commented Dec 31, 2020 at 4:24

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