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Hamiltonian written in terms of field operators: The kinetic energy (KE) part of Hamiltonian is $$ H=-\frac{\hbar^2}{2m}\int d\mathbf{r} \Psi^\dagger(\mathbf{r}) \nabla^2\Psi(\mathbf{r}) \tag{1} $$

Hamiltonian written in tight-binding form: One reaches to tight-binding form by transforming the field operators as $$ \Psi(\mathbf{r}) = \sum_i \phi_i(\mathbf{r})c_i \quad ; \quad \Psi(\mathbf{r})^\dagger = \sum_i \phi_i^*(\mathbf{r})c_i^\dagger \tag{2} $$ here $i$ goes over all the lattice sites inside the system, and $\phi_i(\mathbf{r})$ is the wavefunction of $i$-the lattice site (assume only one state per lattice site). When we put this transformation in $(1)$, we get $$ H=\sum_{ij}c_i^\dagger \left[-\frac{\hbar^2}{2m}\int d\mathbf{r} \phi_i^*(\mathbf{r}) \nabla^2 \phi_j(\mathbf{r})\right] c_j $$ $$ H=\sum_{ij}c_i^\dagger t_{ij} c_j\tag{3} $$

Question: In Hamiltonian $(1)$, the field operators create and destroy particles at $\mathbf{r}$ position. This position vector $\mathbf{r}$ includes all the lattice sites (for a discrete system). In the tight-binding Hamiltonian, the same job is done by operators ($c_i^\dagger, c_i$). So, can we say that the discrete version of the field operator Hamiltonian $(1)$ is equal to the tight-binding Hamiltonian $(3)$? What I mean is to discretize the position vector and $\nabla$ in $(1)$ $$ H=-\frac{\hbar^2}{2m}\sum_i \Psi^\dagger(\mathbf{r}_i) \nabla^2\Psi(\mathbf{r}_i)\\ H=-\frac{\hbar^2}{2m}\sum_i \Psi^\dagger(\mathbf{r}_i) \left(\frac{\Psi(\mathbf{r}_i+a) - \Psi(\mathbf{r}_i) + \Psi(\mathbf{r}_i-a)}{a^2} \right) \tag{4} $$ here $a$ is lattice constant, and $\Psi^\dagger(\mathbf{r}_i)$ creates particle at site $i$. This is exactly what $c_i^\dagger$ does. So, can replace $\Psi^\dagger(\mathbf{r}_i)$ with $c_i^\dagger$: $$ H=-\frac{\hbar^2}{2ma^2}\sum_i \left(c_i^\dagger c_{i+a} - c_i^\dagger c_{i} +c_i^\dagger c_{i-a} \right) \tag{5} $$ Is not $(5)$ similar to $(3)$? Why do we need tight-binding model when we can just discretize field operators?

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The operator $\Psi^\dagger(\mathbf{r}_i)$ creates a particle at $\mathbf{r}_i$ which is infinitely localized (i.e. it has a delta-function wave function.) In contrast, $c_i^\dagger$ creates a particle which is localized at position $\mathbf{r}_i$, but has wavefunction $\phi_i(\mathbf{r})$. You should think of this as essentially an atomic wavefunction (e.g. 1s, 2s, 2p, etc.), not a delta-function.

The choice of $\phi_i(\mathbf{r})$ depends on the problem, and the chemistry involved in your materials. Take the simplest case: Assume you are dealing with a chain of hydrogen atoms. Here, $\phi_i(\mathbf{r})$ should be taken as the 1s orbitals localized on each hydrogen atom. Of course, you could in principle include in your tight-binding model the 2s,2p,3s,3p,3d,etc. orbitals, and this should make your model more accurate. However, such complications are often unnecessary to describe the essential physics of the problem.

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  • $\begingroup$ Thank you @dan. Field operators create particles at infinitely localized point $\mathbf{r}_i$. So, this is the reason why we don't need to explicitly mention the orbitals in field operators' formalism (for example, we don't need to write $H=\sum_{\alpha\beta}\int dr \Psi_{\alpha}^\dagger \nabla^2 \Psi_\beta $). While in tight-binding modeling, if we assume that there is more than one state (orbitals) available per site, we need extra summations over all the available states. For example, $(3)$ becomes $H=\sum_{ij}\sum_{\alpha\beta} c_{i\alpha}^\dagger t_{ij}^{\alpha\beta}c_{j\beta}$ $\endgroup$
    – Sana Ullah
    Nov 27, 2022 at 20:01
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    $\begingroup$ @SanaUllah Yup! $\alpha$ and $\beta$ can also include the spin indices. Also, it might also be good to note that although you mention kinetic energy, the parameters $t_{ij}$ in a tight-binding model also has contributions from the potential energy. Indeed, if an electron on site $i$ feels the ionic potential from site $j$, there will be a nonzero integral $t_{ij}$. There is a good discussion of this in Simon's Solid State Basics, if I recall correctly. However, people typically call $t_{ij}$ the kinetic energy term anyway, even if this isn't technically correct. $\endgroup$
    – dan
    Nov 28, 2022 at 14:58
  • $\begingroup$ That's very clear and informative. Thank you very much. $\endgroup$
    – Sana Ullah
    Nov 29, 2022 at 11:44

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