# Second quantization: Hamiltonian in field operators vs Tight binding form

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?

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.
• 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 $H=\sum_{ij}\sum_{\alpha\beta} c_{i\alpha}^\dagger t_{ij}^{\alpha\beta}c_{j\beta}$ Nov 27, 2022 at 20:01
• @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.