# How do we know which canonical theory the path integral is equivalent to?

So if we have a (classical) theory with fields $$\phi$$ and conjugate momenta $$\phi$$, we turn this into a (canonically quantised) quantum theory by promoting these to operators and impose some sort of commutation/anti-commutation rules, normally we inherit this from the Poisson bracket.

We then also have observables, which are functions of these field operators. Similarly, for conserved currents/charges, these are functions of these field operators. These are inspired/based off the classical expression, however, this doesn't tell us the ordering of the field operators in the observable - we can change the order classically but in QM changing the order gives changes of order $$\hbar^2$$.

However, when we use the path integral formalism, this ordering problem doesn't occur (even if we use Grassmann numbers you need contractions and hence observable quantities are bosonic).

So how do we know which canonical theory the path integral is equivalent to?

I guess a corollary question is: are the orders of operators in the observables in a canonically quantised theory unique, or are there many choices of ordering that could be valid (obviously they would be inequivalent and hence we could experimentally test all of the orderings and see what the universe actually obeys)?

• Nov 25, 2020 at 15:49

In quantum mechancs --- as opposed to quantum field theory --- the classical action in the formal expression of the path integral no information about the operator ordering, and so is inherently ambiguous. In order to get the desired quantum result which depends on operator ordering one has to regularise the path integral by discretizing the time steps and then select a way to discretize the time derivatives. Different choices that would appear to be equivalent in classical calculus then turn out to give different answers. The problem is that the Wiener measure of even a Euclidean path integral we have $$dx= O(\sqrt{dt})$$ rather than the $$dx= O(dt)$$ of ordinary calculus.

One might expect the problem to be even worse in the path integrals of a quantum field theory, but in fact it is not. This is because the process of taking a continuum limit of discrete approximation to a field theory, involves tuning the parameters to a critical point or a second-order phase transition at which the correlation length become very large compared to the lattice spacing. At such a point we have critical point universality which causes many different discretization to give the same answer. This key insight was revealed by Ken Wilson in the 1970's. The net effect that that field theories do not suffer from operator ordering ambiguites, being (in general--- there are exeptions) specified only by the space-time dimensions and the symmetries. In a sense what happens is like the central-limit theorem of statistics: the suitably weighted sum of a large number of random variables has a Gaussian distribution independently of the probability distribution of the individual variables. The fields measured at some length scale in a QFT are in the same way smoothed out local averages of many microscopic fields. The necessary weighting is the scale dependent "wave function renormalization factor"

TL;DR: This is a good question, but a general answer is unrealistic. The best we can do is to match the correspondence $$\text{Operator formalism}\qquad \longleftrightarrow \qquad \text{Path integral formalism}\tag{1}$$ for a narrow list of concrete quantum models.

1. Any textbook derivation of the correspondence (1) is just a formal derivation, which discards contributions $${\cal O}(\hbar)$$ in the process, cf. e.g. this Phys.SE post.

2. As a warning of the complexity at hand let us mention that the correspondence (1) famously becomes quite non-trivial in a curved spacetime background, cf. e.g. my Phys.SE answer here.