I was reviewing many electronic computational methods recently and got a bit confused about relation between different approaches. In particular, what is the essential difference between Hartree–Fock (HF) and tight binding (TB)? I am not that deeply familiar with theory. And just for note I am physicist more than chemist.

I know that HF is more general method not essentially relying on LCAO approximation. But say we mainly use LCAO for HF anyway in most cases. In TB the Hamiltonian matrix elements $H_{ij}$ are mainly used from some tabulated values or fitting with experimental data. This is good for speedy calculations.

But can we just use LCAO orbitals and calculate matrix elements analytically using general many-body Hamiltonian? Will it not become HF method essentially?

  • $\begingroup$ Not too familiar with solid state stuff myself but I got an impression that TB is essentially a semi-empirical method used in solid state physics. If so, it is indeed nothing more than a simplification of HF with certain integrals precomputed 'for all cases' and included in the model. $\endgroup$ Nov 2, 2021 at 7:57
  • $\begingroup$ There is also the so-called 'density functional tight-binding' which similarly uses some precomputed elements of the Hamiltonian but within the DFT, not HF framework. A step further in some way, and considered generally better than conventional semi-empirical methods for molecules (like PM6 or AM1). $\endgroup$ Nov 2, 2021 at 8:03
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    $\begingroup$ Maybe this question could be migrated to mattermodeling.stackexchange.com. $\endgroup$
    – Antimon
    Jan 10 at 18:48

1 Answer 1


In a nutshell

The tight-binding model relies on the assumption that we can forgo the GLOBAL antisymmetry of fermions and instead purely work with a LOCALLY antisymmetric basis. In that sense, the tight-binding model is an approximation to the total wavefunction that uses site-specific antisymmetric wavefunctions to form basis states for the total wave-function.


The tight-binding Hamiltonian takes the form:

$$ H = \sum_{i} E_i | i \rangle \langle i | + \sum_{i} V_{ij} | i \rangle \langle j | $$

Where $E_i$ is the energy of the particle (usually either an electron or exciton) on each site and $V_{ij}$ is the coupling between sites. Often this model and it's parameters are selected purely empirically, but the tight-binding Hamiltonian, and the formula for it's matrix elements, can be derived from first principles.

To do so we realize what the tight binding Hamiltonian is: it is the exact many-electron Hamiltonian projected into some basis $ |i \rangle $, where each basis state $ |i \rangle $ is a tensor-product of individual site wavefunctions. To give a concrete example consider a lattice of organic molecules, each individual molecule representing a site in my tight-binding model. In such a case the basis states of my tight binding model become:

$$ |i \rangle = | \psi^{(1)}_i \rangle \prod_{j \neq i} | \psi^{(0)}_j \rangle $$

That is my basis state is a tensor-product of molecule $i$ in the $1$ state and all other molecules $j$ in the $0$ state. For electron transfer, this would mean molecule $i$ is in the charged site, while other molecules $j$ are in the neutral state. For exciton transfer $i$ is in the excited site, while $j$ are in the ground state. Note that such a basis cannot be antisymmetric, a tensor product of antisymmetric wavefunctions is NOT itself antisymmetric.

From here, with a given model for the site wavefunction (Hartree-Fock, DFT, Configuration Interaction, etc.) one can derive the matrix elements of the tight-binding Hamiltonian. The derivation in the case of CI wavefunctions was first given by Davydov in his seminal paper on molecular excitons: http://dx.doi.org/10.1070/PU1964v007n02ABEH003659

The question that remains is why such a basis is a good choice for the total material wavefunction when we know it does not even obey the Pauli Exclusion Principle. It all comes back to the idea that that the electrons must somehow be localized. Consider the HF exchange integral:

$$ K_{ij} = \int \psi^*_i (r_1) \psi_j (r_2) \frac{1}{r_{12}} \psi^*_j (r_1) \psi_i (r_2) $$

Note that this integral is 0 when there is no spatial overlap between orbitals "i" and "j", e.g. $\psi^*_i (r_1) \psi_j (r_2)=0$. When the exchange integral is 0, the Hamiltonian matrix element of the Slater Determinant and Tensor-Product wavefunctions become identical. So, in essence, when electrons can be spatially localized, they are no longer really indistinguishable, and it becomes OK to violate antisymmetry.

This also means you come come across people using tight-binding Hamiltonians for systems that really aren't tight-binding (say for atoms in an individual quantum dot), you should be extremely wary.

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    $\begingroup$ This is quite a specialist subject, and I don't know enough to judge the correctness of your answer, hence I refrain from voting; but regardless, welcome to Chemistry SE! $\endgroup$ Jan 10 at 16:18

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