# Euclidean QFT and thermodynamic analogy

1. Oct 22, 2007

### OOO

I have been wondering now for quite some time about the meaning of Euclidean Quantum Field Theory.

The Wick rotation $t\to it$ allows us to transform a QFT in Minkowski space to a QFT in Euclidean space (positive definite metric). After that the expectation values of observables can be calculated as

$$<O> = \frac{1}{Z_E}\int {\cal D}\phi O[\phi] e^{-S_E[\phi]}$$

where $Z_E$ is the partition function (normalization factor) and $S_E[\phi]$ is the Euclidean action, which is positive definite in many cases. If one interprets it as a potential energy and adds a gaussian factor for an (artificial) momentum term to the path integral, then the latter factor can be integrated out, so it doesn't actually contribute. The exponential weight factor can then be written as a Boltzmann factor

$$e^{-H_E[\pi,\phi]}$$

where the formal Hamilton function is

$$H_E[\pi,\phi] = \frac{\pi^2}{2}+S_E$$.

Thus the expectation value above is indistinguishable from the expectation value for the statistical mechanics of a classical deterministic field. The deterministic field equations are just the Hamiltonian equations of motion according to $H_E[\pi,\phi]$. This point of view requires that an auxilliary time $\tau$ is introduced, along which the system evolves.

Now my questions: If the statistical mechanical equilibrium of the Euclidean deterministic system can't be distinguished from the conditions under which the Euclidean QFT produces its expectation values, isn't this a justification for saying that the Euclidean QFT is actually a deterministic system ?

And if so, doesn't the equivalence between Euclidean and Minkowskian QFT justify the claim, that all quantum behaviour is completely caused by the effects of deterministic equations and all statistical uncertainty is not much more than the usual impossibility of Maxwell's Demon ?

Last edited: Oct 22, 2007
2. Oct 22, 2007

### Haelfix

Well first off the left hand side is quite a different object. Its not the statistical ensemble of a bunch of classical variables, but rather the expectation value of an operator in some hilbert space, taken as the abstract sum of some weird paths that have no classical analogue far from the classical values.

Secondly, the measure is not the same at all. The stat mech measure is called the Wiener measure, which can be contrasted with the path integral measure which is formally ill defined.

Third of all, the euclidean path integral and the minkowski path integral are not necessarily related by Wick rotation, even when restricted to the class of physical field theories. There are additional constraints for when the procedure is well defined (ie the existence of an analytic continuation). For instance you want the Euclidean action to be bounded from below.

3. Oct 22, 2007

### OOO

I don't understand that. Are you referring to the path integral in Minkowski space or in Euclidean space now ? It seems, you mean Minkowski because in Euclidean QFT there practically is an ensemble representing the Boltzmann factor.

What you say sounds like there is mostly no relation between the Euclidean and Minkowsikian path integrals. But then, why do all textbooks just say: now let's forget about Minkowski space, we're living in Euclidean space from now on.

And don't forget about lattice field theory. A huge amount of numerical evidence is based on Euclidean field theory alone. And this again is based largely on Hybrid Monte Carlo, which is some sort of deterministic Hamiltonian approach, augmented by stochastic forces.

This does hardly make sense to me if there are few relations between Euclid and Minkowski. Then I'd say: "throw lattice field theory in the can".

4. Oct 22, 2007

### Haelfix

No I didn't say that. The Euclidean path integral can and often is equivalent to the Minkowski path integral, but not always.

An example of a field theory that has inequivalent quantizations in Euclidean vs Minkowski signature is 2+1 gravity as famously shown by Witten.

I don't follow your other objections. What we have is an analogy, not some isomorphism.

5. Oct 22, 2007

### OOO

So let's take some action for which Euclidean and Minkowskian formulations are equivalent. By what experiment can one distinguish between the corresponding deterministic system which produces statistical averages and the Minkowskian/Euclidean QFT ?

6. Oct 23, 2007

### Haelfix

Err one is a quantum mechanical thing and the other a classical thing. The objects themselves are entirely different, even though (in one and only one case) they *formally* can satisfy the same integral equation (the case where its a gaussian and the path integral is solvable analytically).

But let me ask you this, have you ever seen a perturbation series in stat mech that looks *anything* like what you would have in field theory?

And just what the hell is an operator valued functional anyway (rhetorical question). I don't even know (nor does anyone else) what mathematical space it really lives in in the first place. You better believe the path integral is not a valid Lebesgue measure, regardless of whether its in Euclidean vs Minkowski signature!

So anyway the point is its just an analogy thats useful for calculational purposes, but they really are entirely different beasts (which is a good thing, otherwise we wouldn't have spin-statistics theorems, hydrogen atoms and things like that, which have no classical analogues)

7. Oct 23, 2007

### OOO

Haelfix, I'm getting the impression that you are trying to convince me that Minkowskian and Euclidean integral measures are different things (correct me if I'm getting you wrong). You don't have to, because (I think) I know that. Let me give a elementary and discriptive example:

we agree, of course, that "sin", "cos" on the one hand and "exp" on the other, are completely different functions. If you convolute some function with "exp" you get something completely different from convoluting it with "sin" and "cos". Yet they are analytic continuations of one another. Mathematically they are the same thing expressed in different (complex) coordinate systems.

I was trying to gound my provocative statement from post #1 in the assumption, that, for some "well-formed" (i.e. no poles swept over by the wick rotation) field theory, Minkowski and Euclidean field theory are equivalent in much the same sense as "exp" is equivalent to "sin","cos".

In my opinion there is no point in sticking the label "formal" to this analogy because everything we do in theoretical physics is formal (e.g. is Hamilton's principle formal ?). If this analogy hasn't got deficiencies I see no reason why it couldn't be an alternative metaphor for reality.

So the question here is, if and to what extent this analogy is imperfect.

You got me ! I haven't seriously looked at any real world perturbation series in quantum field theory yet. That's because I'm about to do lattice qcd, and because I'm a slow and lazy guy that avoids effort that isn't likely to generate revenue related to the task at the top of the list.

Usually the path integral is approximated by a lattice first. In this approximation it becomes a finite dimensional integral which is by no means a mathematical bugbear. If you're saying that mathematicians have not yet explored the space of pathologies for which the limit "lattice spacing to zero" isn't justified or even properly defined: well that may be, but this doesn't bother me for the moment. Certainly you didn't want to say "forget about path integrals, they don't make sense at all", did you ?

My question was: if they are useful in calculations, why not take the analogy for granted, because that's what physics is for ? If it helps, you may look upon me not as a kind of renegade who wants to re-introduce determinism in physics, but someone who is still trying to figure out the underlying relationships between the two approaches, Minkowski's and Euclid's (in the QFT sense).

8. Oct 23, 2007

### Haelfix

"Haelfix, I'm getting the impression that you are trying to convince me that Minkowskian and Euclidean integral measures are different things (correct me if I'm getting you wrong). You don't have to, because (I think) I know that. Let me give a elementary and discriptive example: "

In the last post I was talking more about the difference between the expectation value of an ensemble, and the expectation value of a qm operator. Its irrelevant what signature you are in.

On one hand one measures a physical thing (like pressure), and it does it in a well defined mathematical space called a Wiener space. On the other, you have something that has to be mapped nonlinearly several times through different spaces (ranging from not defined to ill defined to somewhat under control) before you even have a chance at a physical observable.

I mean take for instance the Shroedinger equation. Its completely deterministic at face value, the problem is the thing it outputs (phi) doesn't live in a physical space at all. Now the path integral (Euclidean or Minkowski) is even worse off in this regard, b/c the very thing it outputs shouldn't even exist mathematically (in the continuum limit).

Now having said that, yes I use the PI all the time and I very much believe in it =)

9. Oct 23, 2007

### OOO

But the correspondence between operator and ensemble average is given by replacing the operator by the corresponding classical observable (if we keep out operator ordering for the moment). So it seems to me that the difference isn't that large.

Good. It's always nice to see that agreement is a feasible goal.

10. Oct 23, 2007

### jambaugh

This is correct except one need not invoke the justifying equivalence. Both classical and quantum mechanics are deterministic i.e. the wave equations are deterministic however Classical determinism and quantum determinism are categorically distinct in meaning.

What is deterministically evolving in classical theory is the state of the system (and thus the outcomes of all possible measurements). In quantum theory what is deterministically evolving is our description of the physical system.

In other words in quantum theory there is a unique future measurement of a given system which will yield the same information (correlated outcomes) as a past measurement. Thus quantum determinism says no information is lost in a closed system. But the very act of measurement "opens" the system by definition and so outcomes of arbitrary measurements and their effects on future measurements are fundamentally indeterminate.

Yes and no. Classically Maxwell's Demon is not impossible. The consideration of Maxwell's Demon is the beginning step in developing a quantum theory wherein we must pay closer attention to the operational meaning of "knowing about" a physical system.

11. Oct 23, 2007

### OOO

Thanks jambaugh, but there seems to be a lot of discussion about quantum mechanical measurement, even today. I didn't want to open Pandora's box again in this thread.

Rather I was hoping that the perspective from Euclidean QFT could shed some unusual light on the question whether classical and quantum mechanical determinism are really that different. Maybe the quantum mechanical determinism just looks so messy because it ought to be subjected to analytical continuation.

What Euclidean QFT does, at least on the lattice, is quite clear: solve a classical deterministic Hamiltonian system with some additional stochastic force (about the meaning of which there could be much speculation). This is called Hybrid Monte Carlo (HMC). After some time of thinking about this approach I'm still puzzled as to how this relates to the usual Minkowskian path amplitude

$$e^{iS}$$

and it's probabilistic interpretation (which we all know well from everyday quantum mechanics). Haelfix says the HMC approach is just an analogy, but I'm not yet convinced of this.

Last edited: Oct 23, 2007