Locality in Quantum Mechanics

Question

We speak of locality or non-locality of an equation in QM, depending on whether it has no differential operators of order higher than two.

My question is, how could one tell from looking at the concrete solutions of the equation whether the equ. was local or not...or, to put it another way, what would it mean to say that a concrete solution was non-local?

edit: let me emphasise this grew from a problem in one-particle quantum mechanics. (Klein-Gordon eq.) Let me clarify that I am asking what is the physical meaning of saying a solution, or space of solutions, is non-local. Answers that hinge on the form of the equation are...better than nothing, but I am hoping for a more physical answer, since it is the solutions which are physical, not the way they are written down ....

This question, which I had already read, is related but the relation is unclear. Why are higher order Lagrangians called 'non-local'?

Answer 1

Presuming that there aren't nonlocal constraints, a differential operator that is polynomial in differential operators is local, it doesn't have to be quadratic. My understanding is that irrational or transcendental functions of differential operators are generally nonlocal (though that's perhaps a Question for math.SE).

A given space of solutions implies a particular nonlocal choice of boundary conditions, unless the equations are on a compact manifold (which, however, is itself a nonlocal structure). There is always an element of nonlocality when we discuss solutions in contrast to equations.

[For the anti-locality of the operator $(-\nabla^2+m^2)^\lambda$ for odd dimension and non-integer $\lambda$, one can see I.E. Segal, R.W. Goodman, J. Math. Mech. 14 (1965) 629 (for a review of this paper, see here).]

EDIT: Sorry, I should have gone straight to Hegerfeldt's theorem. Schrodinger's equation is enough like the heat equation to be nonlocal in Hegerfeldt's sense. There are two theorems, from 1974 in PRD and from 1994 in PRL, but in arXiv:quant-ph/9809030 we have, of course with references to the originals,

Theorem 1. Consider a free relativistic particle of positive or zero mass and arbitrary spin. Assume that at time $t=0$ the particle is localized with probability 1 in a bounded region V . Then there is a nonzero probability of finding the particle arbitrarily far away at any later time.

Theorem 2. Let the operator $H$ be self-adjoint and bounded from below. Let $\mathcal{O}$ be any operator satisfying $$0\le \mathcal{O} \le \mathrm{const.}$$ Let $\psi_0$ be any vector and define $$\psi_t \equiv \mathrm{e}^{-\mathrm{i}Ht}\psi_0.$$ Then one of the following two alternatives holds. (i) $\left<\psi_t,\mathcal{O}\psi_t\right>\not=0$ for almost all $t$ (and the set of such t's is dense and open) (ii) $\left<\psi_t,\mathcal{O}\psi_t\right>\equiv 0$ for all $t$.

Exactly how to understand Hegerfeldt's theorem is another question. It seems almost as if it isn't mentioned because it's so inconvenient (the second theorem, in particular, has a rather simple statement with rather general conditions), but a lot depends on how we define local and nonlocal.

I usually take Hegerfeldt's theorem to be a non-relativistic cognate of the Reeh-Schlieder theorem in axiomatic QFT, although that's perhaps heterodox, where microcausality is close to the only definition of local. Microcausality is one of the axioms that leads to the Reeh-Schlieder theorem, so, no nonlocality.

Answer 2

One way of saying this is in terms of the problem with fixed boundary conditions. If you have a differential equation with local derivative operators, and you impose a boundary condition that $\psi$ is zero on the surface of a thickened sphere of small positive width $\epsilon$ for all time, then anything you do on the interior of the sphere will not affect the exterior of the sphere.

This is true when the Hamiltonian is a polynomial, since if you make a lattice approximation, only a finite number of lattice neighbors contribute to the time derivative at any point. If you make a nonlocal Hamiltonian, like taking the square root of $\nabla^2 + m^2$ to get the positive energy Klein Gordon propagator, you will not be able to express it in terms of finite number of lattice neighbors, and the particle will be able to leak out of the interior of a sphere to influence the exterior.

I phrased it using additional boundary conditions because a nonrelativistic delta-function initial condition will spread to all space instantly even using a polynomial Hamiltonian, so it isn't local in the sense of finite maximum propagation speed. There is still a distinction between local and nonlocal operators, but it is best phrased in terms of how lattice position x depends on lattice position x' when x-x' becomes a large integer number of spacings.

Answer 3

The mathematical notion of a local operator is that if the operator $T$ is applied to a smooth function $f$, then $Tf(x)$ only depends on the behaviour of $f$ in a neighbourhood of $x$, and the neighbourhood can be arbitrarily small. In particular, the support of $Tf$ has to lie inside of the support of $f$. A local operator, in this sense, has to be a differential operator.

This is a mathematical notion which is relevant to whether you could define $T$ on an arbitrary smooth manifold in a way indepenedent of coordinates, but it does not seem very physical since, as has been pointed out in this site (See Use of Operators in Quantum Mechanics, where the question « if I simply apply the momentum operator to the wave function $−ih{d \over dx}\Psi$ Will I get an Equation that will provide the momentum for a given position? Or is that a useless mathematical thing I just did? » is answered by « The first thing you did is useless-»), just taking an observable $Q$ or a Hamiltonian $H$ and applying it to a wave function $\psi$ is not very physical. What is physical is $e^{-iHt} \psi$ or the eigenvalues of $H$ or of $Q$. For this reason the question was formulated in terms of whether you could tell from the wave function or its evolution whether things were physically « local » or not. So far as I can tell, the only concrete answer to this is whether the time evolution would violate Einstein causality or not.

It is well known that the notion of observable in Relativistic Quantum Mechanics is a tangled one (the Newton--Wigner position observables are famously non-local) and, on the other hand, the Born interpretation of the wave function becomes problematic as well (with the Klein--Gordon equation, it leads to negative probabilities). (These difficulties can be evaded in Quantum Field Theory, but then, as we all know, new difficulties arise.) Perhaps this calls into question the connection between « observables » in Relativistic Quantum Mechanics, even one-particle mechanics, and real, physical, measurements...

Thx to all who tried to help, and especially the very useful references. And correct me if I have made a mistake here.

Answer 4

Here is an experimental physicist's view.

Locality for me means that the solutions representing a particle give the particle as local, i.e. once the boundary conditions are given, all its observables and interactions depend on functions as f(x,y,z,t), (x,y,z,t) a point.

Non local means that the solutions giving the observables and interactions of a particle depend on an extended volume around each space time point (x,y,z,t).

Experimentally I would look for non locality in interactions, which would no longer be point interactions ( Feynman diagrams) but would have to be extended diagrams over the volume of non locality, and therefore the values measured for the observables would be different than the values predicted from the local theory if the non local theory holds, given enough experimental accuracy.