What is the correct Hamiltonian for a system of coupled quantum oscillators? The Hamiltonian (see Eqn. 1 in Appendix 2 of this paper) for a system of coupled quantum oscillators is given as
$$H=\frac{1}{2}∑_{i}p^{2}_{i}+\frac{1}{2}∑_{j,k}A_{jk}q_{i}q_{k}$$
Yet, in my QM course, the Hamiltonian for such a system was given as
$$H=\frac{1}{2m}∑_{i}p^{2}_{i}+\frac{k}{2}∑_{i}x^{2}_{i}+\frac{K}{2}∑_{i}(x_{i}-x_{i+1})^{2}$$
where the third term represents the coupling between the oscillators. Why aren't these equations equivalent?
 A: They almost are. Clearly you replace $q_j$ by $x_j$ since the canonical commutation relationships are the same between these two and the $p_j$. Your QM course equation is then a special case of the one in the paper: if you then expand the last term in your QM course equation, you have equivalence if $A_{11} = A_{NN} = k+K$, $A_{jj}  = k+2 K,\;j\neq 1, N$ and $A_{j,k} = 0$ if $|j-k|>1$ but $A_{j,j+1} = k - 2 K$ for $j=1,2,\cdots N-1$ (if you have $N$ oscillators indexed by $1,2,\cdots,N$.
That the equation in the paper is "correct" follows from the fact that the system of oscillators can be diagonalised into an equivalent system of uncoupled quantum harmonic oscillators.
Note that it is stated in section 2 of the paper that $A_{jk}=A_{kj}$, a condition which must hold if the Hamiltonian is to be an Hermitian operator. This assumption is also crucial to the diagonalisation of the oscillator system into an equivalent uncoupled oscillator system.
Your QM course equation is for $N$ identical oscillators "in a row" such that there is only nearest neighbour coupling between the oscillators and furthermore such that the nearest neighbour coupling strength is the same for each pair of neighbours.
