A confusion about position and momentum operators in a coupled system

Question

On page 25 of "Quantum field theory for the gifted amateur" by To has written:

Consider a linear chain of $N$ atoms (see Fig. 2.5), each of mass m, and connected by springs of unstretched length a and which have spring constant $K$. The masses are normally at position $R_j = ja$, but can be displaced slightly by an amount $x_j$. The momentum of the jth mass is $p_j$. The Hamiltonian for this system is $$H=\sum_{j}\frac{p_j^2}{2m}+\frac{1}{2}K(x_{j+1}-x_j)^2.$$ In contrast to the model in the previous section, we are now dealing with a coupled problem. Each mass is strongly coupled to its neighbour by the springs and there is no way in which we can consider them as independent. However, we will show that the excitations in this system behave exactly as a set of totally independent oscillators. This works because we can Fourier transform the problem, so that even though the masses are coupled in real space, the excitations are uncoupled in reciprocal space.

Question:

I know that $[x_j,p_j]=i\hbar$. But I cannot understand for different atoms $[x_j,p_k]=i\hbar\delta_{j,k}$. I am confused that if $j\neq k, $ why $[x_j,p_k]=0$? How to explain this ?

Answer 1

The easy answer is to think of $p_k\to -i\hbar\frac{\partial}{\partial x_k}$. This makes it clear that $[x_i,p_k]=0$ unless $i=k$ since clearly $[x_i,\frac{\partial}{\partial x_k}]=0$ unless $i=j$.