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!


1 Answer 1


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

  • $\begingroup$ 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 $R$? $\endgroup$ Mar 5, 2023 at 14:10
  • 1
    $\begingroup$ @AngryPhysicist You can do another change of variable: $r\rightarrow r'=r-R$. $\endgroup$
    – Anyon
    Mar 5, 2023 at 14:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.