In what sense is a separable potential maximally non-local?

Question

A (rank-one) separable potential is an operator $V$ with the coordinate representation $$ \langle \mathbf r' | V | \mathbf r \rangle = - A u(r) u(r')$$ where $u(r)$ is a real-valued function. It is clear that this is not a local potential as it is not proportional to $\delta(\mathbf r - \mathbf r')$, and hence is arguably nonphysical. Nonetheless, separable potentials have found application, particularly in many-body scattering processes in nuclear physics (with interest in their application peaking in the 1960s though they still see some use today). Approximations have been found allowing a generic potential (satisfying some weak assumptions) to be approximated by a separable one for scattering (see e.g. [1]).

This interaction is discussed in Gottfried's classic textbook [2] in section 8.2(b). There it is claimed that this interaction is "maximally non-local". This statement is not explained further. No measure of the degree of non-locality of a scattering potential is presented in the text nor am I aware of any from the literature on separable potentials. How should I interpret this statement? Is it just that the right hand side of the equation has no $\delta$ function term, or is there a quantitative sense in which this potential is more non-local than e.g. one of the form $\langle \mathbf r' | V' | \mathbf r \rangle= - A u(r) u(r') |\mathbf r - \mathbf r'|^n$ for $n > 0$ (which I would naively think is more non-local)?

References:

[1]: C. Lovelace, Phys. Rev. 135, B1225 (1964).

[2]: Gottfried, Kurt, and Tung-Mow Yan. Quantum Mechanics: Fundamentals. Springer Science Business Media, 1996.

Answer 1

This is probably more of a comment, but got a bit long and maybe this answers your question already. There is a way to see why a separable potential is in some sense 'the opposite' of a local potential if you think of them as operators acting on a wave-function: the Hilbert-Schmidt decomposition. What this is really saying is that there is a basis of states $|\psi_i\rangle$ in which you can write the operator $V$ in diagonal form

$$ V = \sum_i \lambda_i |\psi_i\rangle \langle\psi_i|.$$

In the position representation this corresponds to

$$ \langle r|V|r'\rangle = \sum_i \lambda_i \langle r|\psi_i\rangle \langle\psi_i|r'\rangle = \sum_i \lambda_i \psi(r)\psi^*(r').$$

The theorem is a generalization of the Schmidt decomposition for matrices (which is often used for example in quantum information) to continuous variables. If the theorem can be applied depends on whether the operator is compact, but I will ignore this restriction here since I only want to make a qualitative argument.

For a separable potential as defined by the OP, we see that this corresponds to a Hilbert-Schmidt decomposition with exactly one non-zero $\lambda_i$, that is a single state is enough to resolve the potential operator.

For a local potential we can see that the Hilbert-Schmidt expansion will require all states from the position basis. In formulae $$\langle r|V|r'\rangle = V(r) \delta(r-r') = \sum_{r''} V(r'') \langle r|r''\rangle \langle r''|r'\rangle.$$

So in this sense the separable potential is the opposite of the local one. If you actually want to call a one-state Hilber-Schmidt decomposition 'maximally non-local' depends on your definition of the latter term.