This term pops up through out Hartree-Fock and DFT. Does this term have any physical significance to actual(real) system or it is a mathematical by-product?


2 Answers 2


The non-local operator in Hartree Fock is the Exchange operator. Locality is a mathematical property. Look at the Coulomb operator first which acts like this: $$ \hat v_H\;\varphi_s(r) = \sum_i^{N_\textrm{occ}}\int dr' \frac{\varphi_i(r') \varphi_i(r')}{|r-r'|} \varphi_s(r).\tag{1} $$

It's value depends for the function (which I call here $\varphi_s$) on which it acts only on the value of the function at $r$.

Now look at the Exchange Operator which acts like this:

$$ \hat v_x\;\varphi_s(r) = \sum_i^{N_\textrm{occ}}\int dr' \frac{\varphi_i(r') \varphi_s(r')}{|r-r'|} \varphi_i(r).\tag{2} $$

As you can see $\varphi_s$ is now inside the integral, in order to evaluate the "effect" of the operator you will need knowledge the function $\varphi_s$ at all positions of space. Hence the designation non-local.

The physical meaning of it is that it reflects the permutation symmetry of the fermionic system. This effect is sometimes called Fermi-correlation. Just follow the derivation of Hartree Fock and it will become obvious. A book which contains this derivation is "Modern Quantum Chemistry: Introduction to Advanced Electronic Structure Theory (Dover Books on Chemistry)"

$\varphi$ is a Fock orbital. The index $i$ refers to occupied orbitals. The index $s$ refers to all orbitals. $r$ is the coordinate of a particle in space while $r'$ is the coordinate of another particle in space. $v_H$ is the Coulomb Potential, $\hat v_x$ is the exchange potential. Often in the context of Hartree Fock $\hat v_H$ is denoted as $\hat J$ and $\hat v_x$ is denoted as $\hat K$.

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    $\begingroup$ Welcome to MatterModeling! That's a great and detailed answer! Consider adding explicit definitions for $\varphi$, $\phi$, and the indices (even though they seem to have their "usual" meanings at my first glance), and adding a link to the derivation in the last sentence, or a book with it. $\endgroup$ Commented Aug 5, 2022 at 13:34
  • $\begingroup$ There is a little error. In Eq.(1),one of the $φ_i(r′)$ should be instead of its complex conjugate. And also in Eq.(2), $φ_i(r′)$ should be replaced by $φ_i(r′)^*$. $\endgroup$
    – J.G.
    Commented May 30, 2023 at 9:03
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    $\begingroup$ @J.G In this case, it would actually be better if you were to use the 'Edit' link under the answer in question and propose this change directly. If the edit is approved by community members, you would start earning the reputation points required to unlock comment privileges and so on. $\endgroup$
    – Anyon
    Commented May 30, 2023 at 14:07

It's all hollow words if no proper definition is given. The definition I know is the following: For a one-body operator $A$, define $$A(x,x^\prime):=\langle x|A|x^\prime\rangle \tag{1} \quad . $$

We call $A$ local if the following holds:

$$ A(x,x^\prime)= \delta(x-x^\prime)\, A(x),\tag{2} $$

and otherwise it's non-local. Here we employed a usual abuse of notation in equation $(2)$. An example for a local operator would be a local external potential, e.g. the external potential for an electron interacting with some nuclei in the Born-Oppenheimer approximation.

Whether or not an external potential is local becomes important in e.g. the Hohenberg-Kohn theorem.

Reference: Density-Functional Theory of Atoms and Molecules. Parr and Yang. 1989, section 2.1, p.23, equation (2.13) and (2.14) as well as the corresponding discussion.


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