# Variational method in the tight binding approach

I'm trying to read Professor David Tong's notes to understand the principles behind the tight-binding model - section 2.3.5 'Deriving the Tight-Binding Model'.

He first considers the Hamiltonian of one electron localised around 1 atom:

$$H_{\text{atom}} = \frac{p^2}{2m} + V(r)$$

with eigenstates $$H_{\text{atom}} \phi_n = \epsilon \phi_n$$. Then he introduces a lattice with a periodic potential:

$$V_\text{lattice} = \sum_R V(r-R)$$ where $$R$$ is a lattice vector. To solve for this Hamiltonian, he makes the Bloch wave ansatz (supposing there is only one valence electron):

$$\psi_k(r) = \frac{1}{\sqrt N} \sum_R e^{ikR} \phi(r-R)$$

and tries to find the ground state using the variational principle:

$$E(k) = \frac{ \langle \psi_k | H |\psi_k \rangle}{\langle \psi_k |\psi_k \rangle}.$$

He begins evaluating the denominator and does the following:

$$\langle \psi_k |\psi_k \rangle = \frac 1N \sum_{R,R'} e^{ik(R'-R)} \int d^3 r \phi^\ast(r -R) \psi(r-R')$$

$$= \frac 1N \sum_{R} e^{-ikR} \int d^3 r \phi^\ast(r-R) \phi(r)$$

He says 'where, in going to the second line, we’ve used the translational invariance of the lattice'. I don't understand how to find this because I don't see how the two integrals are similar. My idea was that perhaps the index of the sum could be switched because it only depends on $$R'-R$$ but I'm not able to see how the integral depends on the difference $$R'-R$$. Perhaps someone could help me with this!

$$\langle\psi_k |\psi_k\rangle=\frac{1}{N}\sum_{R, R'}e^{ik\cdot (R'-R)}\int d^3r \ \phi(r-R')\phi(r-R)$$ I would first suggest a change of variable $$R-R'=S \Longrightarrow R'=R-S$$ $$\langle\psi_k |\psi_k\rangle=\frac{1}{N}\sum_{R, S}e^{-ik\cdot S}\int d^3r \ \phi(r-R+S)\phi(r-R)=\\ =\sum_{S}e^{-ik\cdot S}\int d^3r \ \phi(r+S)\phi(r)$$ In the last step the sum over the lattice vectors R simplifies the N. After the renaming $$S\rightarrow R$$, we get $$\langle\psi_k |\psi_k\rangle=\sum_{R}e^{-ik\cdot R}\int d^3r \ \phi(r-R)\phi(r)$$
• Hey thanks for this, but I think I'm still missing something. How did you manage to sum up over the $R$'s? Are you saying that since the position argument of the eigenfunctions differs only by $S$ the sum does not depend on $R$? Mar 5, 2023 at 14:10
• @AngryPhysicist You can do another change of variable: $r\rightarrow r'=r-R$. Mar 5, 2023 at 14:38