I'm currently reading a paper which considers the (polyharmonic) periodic Schrodinger operator
$$H = (-\Delta)^l + V(x), \ \ \ x \in \mathbb{R}^d,$$
where $\Delta$ denotes the Laplace operator on $\mathbb{R}^d,$ $V$ is some smooth, periodic, real function on $\mathbb{R}^d$ acting on a function $f$ by multiplication, and $l \in \mathbb{N}.$ The paper then says the following:
Under a suitable affine change of co-ordinates, we can rewrite the operator $H$ as
$$H = H_0 + V(x),$$
where $H_0 = (\mathbf{DGD})^l,$ and $\mathbf{D}$ denotes the operator $i \nabla$ and $\mathbf{G}$ is a positive-definite $d \times d$-matrix.
The paper doesn't go into any further details here, so I have a few questions. Firstly, would I be correct to assume that, by $\mathbf{DGD},$ they mean
$$\displaystyle \mathbf{DGD} = -\sum_{\substack{1 \leqslant i \leqslant d} \\ {1 \leqslant j \leqslant d}} a_{ij}\frac{\partial^2}{\partial x_i \partial x_j},$$
where $[a_{ij}]$ is the matrix $\mathbf{G}$? Then, raising that to the power $l$ represents multiple compositions of that operator with itself?
Furthermore, could someone show me how exactly this "affine change of co-ordinates" works in this instance, to yield that expression for $\mathbf{DGD}$? That is, is there an explicit way to write such a change of co-ordinates?
I guess you are missing the point: In the article they claimed that under a change of coordinate we have $$H = H_0 + V(x),$$ with the potential $V$ being smooth and periodic with the lattice of periods of $V$ equal $(2\pi \mathbb Z)^d$ (emphasis mine).
Let $\Lambda$ be the lattice of periods of $V$. that is, $$V(x) = V(x+ 2\pi \lambda),\ \ \ \text{for all }\lambda\in \Lambda.$$ Let $A$ be the matrix (the affine change of coordinate) which sends $e_i$ to $\lambda_i$, where $\{\lambda_1, \cdots, \lambda_n\}$ spans $\Lambda$. Let $\widetilde V (y) = V(Ay)$, then $\widetilde V$ is $(2\pi \mathbb Z)^d$ periodic. Let $g(y) = f(Ay)$. If we write $x = Ay$, then $g(A^{-1} x) = f(x)$. Thus $$-\Delta f(x) + V(x) = 0$$ translates to $$-\Delta (g\circ A^{-1}) (y) + \widetilde V(y) = 0.$$ Calculating $-\Delta(g\circ A^{-1})$ gives $$-\Delta (g\circ A^{-1}) = DGD, \text{where } G = (A^{-1})^TA^{-1}.$$