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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'?

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I think determining locality based on solutions is going to be difficult in any non-relativistic context, because there is no speed limit - any local perturbation of the system may spread to the entire solution in zero time. It's a lot cleaner in a relativistic context; if any given part of a solution depends only on its past light-cone, the system is local. –  Harry Johnston Dec 26 '11 at 4:31
    
Alright, let us assume we are in a relativistic context. I am thinking of the square root of the usual Klein-Gordon equation, so it is not a diff. eq. at all, but still has a propagator. –  joseph f. johnson Dec 26 '11 at 4:40
    
I'm shaky on single-particle relativistic QM. But if you have a propagator, then you should be able to apply the same criteria: if a given part of a solution depends only on the past light-cone, the theory is local. –  Harry Johnston Dec 26 '11 at 20:16
    
IC. (BTW, my impression is that everyone is shaky on fixed-finite-number-of-particles relativistic QM. this is why they hurry on the QFT). So looking at how fast a dirac delta functin spreads in time would be a good criterion. Then the non-rel. Schroedinger eq. is non-local in that sense, as is well known, but I guess any relativistic equation would have to satisfy this criterion. –  joseph f. johnson Dec 26 '11 at 20:57

4 Answers 4

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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.

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I'd like to see this answer refined a little... but +1 for the reference. a) I've never seen a def'n of 'local operator'. b) Other posters, at other places, have claimed that any order higher than 2 becomes non-local. c) If written as Fourier Integral Operators, you could not easily tell the difference between a differential operator and a square root of one.... d) if being $L^2$ (which is what is in question) is considered a non-local boundary constraint, then Schroedinger's eq. would have to be considered non-local too, right? –  joseph f. johnson Dec 25 '11 at 22:56
    
To be honest, I wrote this Answer supposing only that it might be marginally useful, but I don't know details about DEs well enough to go much further. Someone else can do better. I regret that the reference I gave is about the only definitive thing I can bring to the table. I think boundary conditions are relevant, but the Cauchy problem must be more relevant. Maybe there's no simple classification of local/nonlocal because the Cauchy problem is nontrivial? It also occurs to me that the heat equation, using the same differential operator as the Schrodinger equation, is nonlocal. –  Peter Morgan Dec 26 '11 at 0:10
    
Is there a typo in the statement of Thm. One? should it not be, 'non-relativistic particle' ? –  joseph f. johnson Dec 26 '11 at 21:13
    
From the arXiv paper, top of page 5, "relativity is not needed, as shown by the author and Ruijsenaars [11]"; in other words, the original theorem was for the relativistic case, so that's what has been repeated here, but one could equally well say relativistic or non-relativistic. –  Peter Morgan Dec 26 '11 at 21:55
    
Hey Joseph, most of the time it's the references that matter here. Off the cuff Answers rarely compete with papers that represent months or years of work, unsurprisingly. Particularly true of me, sadly. –  Peter Morgan Dec 27 '11 at 13:07

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.

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This is almost unintelligible. If there is one particle, what do you mean by 'its interactions' ? Are you saying that the concept of locality only applies to interactions between particles or between fields? If that is what you mean, perhaps you mean further that the answer to my question is that one should not speak of a one-particle QM equation as either local or non-local. ? –  joseph f. johnson Dec 26 '11 at 4:39
    
Of course there is not one particle only, though locality applies to one particle solutions too. The one particle quantum mechanical equation is local, because the solutions that give observables depend simply on (x,y,z,t) not on convolutions at each point over some non local volume.As for interactions, all experimental data come from interactions of one form or the other. –  anna v Dec 26 '11 at 5:34
    
Locality and non-locality are really used to assert on whether the very basic elementary particle interactions are local or non-local. This means the one particle equations and their solutions as well as the Quantum Field Theoretical representations of interactions of particle fields on a point and not a locus. –  anna v Dec 26 '11 at 5:47
    
you speak of « the one-particle eq.» I am thinking of three different ones, which are often discussed in this context: Schroedinger's non-relativistic eq., the Dirac equ., and the Klein--Gordon eq. Correct me if I am wrong, but your point of view has them all as being local. –  joseph f. johnson Dec 26 '11 at 5:52
    
Yes. All one particle QM equations are local. –  anna v Dec 26 '11 at 6:27

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.

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so far, there are three different answers. Could you give a reference for this one? –  joseph f. johnson Dec 26 '11 at 20:15
    
By « surface » do you mean « crust », i.e., are you including the region between the $r=1-\epsilon$ surface and the $r=1$ surface? I.e., you are requiring $\psi$ to vanish on a region of positive measure, so you are ruling out analytic solutions... –  joseph f. johnson Dec 26 '11 at 21:20
    
@joseph: I don't know a reference. Why would you trust a reference more? You have to work it out in either case, reference or not. I am requiring $\psi$ to vanish on a region of positive measure. There is no analyticity anyway. The point is that the interior and exterior are decoupled when you have a local equation. –  Ron Maimon Dec 27 '11 at 4:04
    
Well, I can't tell what you mean by « anything you do in the interior of the sphere ... influence the exterior.» Obviously after the passage of time a wave initially satisfying your conditions will stop fulfilling them. so I am hoping that either you, or a reference, can phrase your idea more clearly. What, precisely, do you mean by « anything you can do »? One thing I can do is pose a $\psi_o$ supported in the region... –  joseph f. johnson Dec 27 '11 at 7:10
    
@Joseph: I meant imposing the condition for all times, not just at t=0. If you don't impose the condition for all times, then the wavefunction will leak out instantly. –  Ron Maimon Dec 27 '11 at 7:42

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.

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