# What the correct way to write the discrete kinetic energy operator?

For a bit of context, I am making simulations of a quantum algorithm that is meant to variationally find the ground state of a quantum harmonic oscillator potential. In one dimension, we know that $$\hat{p}$$ is defined (in the position basis) as:

$$\hat{p} \ = \ -\frac{\hbar}{i} \frac{\partial}{\partial x}\tag{1}$$

I am trying to discretize this, and represent it as a linear operator on a discretized state vector, in the position basis:

$$|\psi\rangle \ = \ \begin{pmatrix} \psi_{\Delta x} \\ \psi_{2 \Delta x} \\ \vdots \\ \psi_{n\Delta x} \end{pmatrix}\tag{2}$$

We logic was that one should discretize the derivative, acting on each of the components as follows:

$$\frac{ \partial \psi_{i \Delta x}}{\partial x} \ \rightarrow \ \frac{1}{2} \Big[ \frac{\psi_{i \Delta x} \ - \ \psi_{(i - 1)\Delta x}}{\Delta x} \ + \ \frac{\psi_{(i+1) \Delta x} \ - \ \psi_{(i)\Delta x}}{\Delta x} \Big] \ = \ \frac{\psi_{(i+1) \Delta x} \ - \ \psi_{(i - 1)\Delta x}}{2 \Delta x}\tag{3}$$

Essentially, taking the average of the linearized slope on each side of the point at which the derivative of being evaluated. This leads to a matrix with off-diagonal elements of $$1$$ and $$-1$$. The kinetic energy operator is defined as:

$$\hat{T} \ = \ \frac{\hat{p}^2}{2m}\tag{4}$$

So, in order to calculate the discretized matrix corresponding to this operator, one just has to square the matrix we arrived at for the momentum operator, and multiply it by some coefficient. This leads to a matrix with $$2$$s on the diagonal, $$-1$$s two columns to the left, and two columns to the right of the diagonal on every row.

This makes sense, however, I have also been reading about how the second spatial derivative on a graph can be defined using the graph Laplacian. This discretized space that I have "set-up" can be interpreted as a linear graph, $$G$$, with:

$$E(G) \ = \ \{(v_{n}, \ v_{n+1}) \ | \ v \ \in \ V(G)\}\tag{5}$$

Thus, one should be able to define the second derivative using the graph Laplacian. However, the Laplacian and the linear operator at which we arrived are not the same. They are somewhat similar, except the Laplacian has $$-1$$s one column to the left and right, rather than two columns over (evidently, as it is merely the degree matrix minus the adjacency matrix).

My question is: which of these interpretations of the "graph derivative" is correct? I have done a lot of searching, and come across conflicting answers. For instance, the Wikipedia page on quantum walks defines the second derivative with the Laplacian, while the while the top answer in this Physics SE question defines the discretized momentum operator just as I did.

• It doesn't actually matter, because as long as you adjust the overall coefficients right, both have the same continuum limit. – knzhou Mar 28 '20 at 16:38
• That does make sense, so as $n \rightarrow \infty$, they’ll be the same, due to the fact that the distance between any two points is infinitesimal, so defining the derivative based on the difference between points of distance two or one apart will be the same? – Jack Ceroni Mar 28 '20 at 16:45
• Yup! I'm sure that mathematicians can make finer distinctions here, but as a practical matter it's the same either way. – knzhou Mar 28 '20 at 16:49
• Awesome, thanks for your help! – Jack Ceroni Mar 28 '20 at 16:50
• The thing that not only mathematicians may be interested in is up to which order in $\Delta x$ your method is correct. Dependent on this you may be able to choose a coarser discretization without loosing precision. – Paul G. Mar 28 '20 at 17:17