Taylor expansion of potential energy

Question

According to the book "Applied quantum Mechanics, Anthony Levi", The Hamiltonian of a monatomic linear chain is given by:

$$H=\sum_j \frac m 2\left( \frac{\text du_j}{\text dt} \right)^2+V_0(0)+\frac1 {2!}\sum_{jk} \frac{\partial^2V_0}{\partial u_j u_k}u_j u_k+ \frac 1 {3!}\sum_{jkl} \frac{\partial^3V_0}{\partial u_j u_k u_l}u_j u_k u_l+...$$

The first term on the right side comes from the sum of the kinetic energy of each particle, and the rest comes from the Taylor expansion of the multi variable potential, but I think the linear term is missing, this being:

$$\sum{_j}\frac{\partial V_0}{\partial u_j}u_j$$

Am I right or why isn't it included?

Answer 1

This is a standard expression when the energy is expanded around the equilibrium. At equilibrium, the force on each of the atoms of the chain is 0. That is, $\frac{\partial V_{0}}{\partial u_{i}}=0 \ \forall i$. For all practical purposes, you can also arbitrarily set the constant potential term $V_{0}$ to zero.

Answer 2

At equilibrium, the function(potential energy) is minimum( Morse potential curve). we know from mathematics that the first derivative of a function is always zero at minimum. that is why the linear term is missing.