Short version of the question: Can someone explains what physicist's 'Spin-statistic theorem' says rigorously in the context of Free Quantum fields when they are describe as the Fock space of some '1-particle space' ?

To be precise the question is not really about the spin-statistic theorem but more "Why taking a Fock space of a "1-particle" representation of the Poincarré group sometimes give a Quantum field and sometime doesn't, depending on whether the Spin of the 1-particle space and the type of Fock space under consideration are compatible with the Spin-statistic theorem."

For more details:

When reading text about Quantum field theory written by mathematicians (for eg, this), I got the impression that free Quantum fields are well understood objects and can be described relatively simply as follows (that's literally their definition in the reference mentioned):

You start from a "one particle Hilbert space" $$H$$.

For a relativistic particle, it will be a (often irreducible) unitary representation of $$\Lambda$$ the universal cover of the Poincaré group. Such representation have been classified by Wigner, and if one exclude the "non-local" and non-physical ones, they are essentially classified by a mass $$m \in \mathbb{R}$$ and a spin $$s \in \frac{1}{2} \mathbb{N}$$ (with some subtleties in the $$m=0$$ case that I'm ignoring).

You then apply 'second quantization'. That is you form either the Bosonic Fock space $$F_+$$ or the Fermionic Fock space $$F_-$$

$$F_\pm = \sum_{n=0}^\infty P_{\pm} \left( H^{\otimes n} \right)$$

where $$P_\pm$$ is either the projection on symetric or antisymetric tensor.

$$P_+ (v_1 \otimes \dots \otimes v_n )= \frac{1}{n!}\sum_{\sigma \in S_n} v_{\sigma 1} \otimes \dots \otimes v_{\sigma n }$$ $$P_- (v_1 \otimes \dots \otimes v_n )= \frac{1}{n!}\sum_{\sigma \in S_n} sg(\sigma) v_{\sigma 1} \otimes \dots \otimes v_{\sigma n }$$

These comes with a lots of structure: they have a "vacuum vector", creation and annihilation operators out of which you can define field operators, they have a "component-wise" action of $$\Lambda$$ that encodes all the usual physical concept. In particular they have a Hamiltonian obtained as the infinitesimal generator of the time translation.

If I exaggerate a bit, that is essentially all I understand of Quantum field theory (at least that is the only part I know how to make mathematically rigorous).

Now when I read instead textbook written by physicist, I got the impression that the point of view above is too general, or maybe is missing some important subtleties.

A precise point where this really appears, is with the "Spin-statistic theorem".

It claims that we can only consider the Fermionic Fock space $$F_-$$ when $$H$$ is the Hilbert space of a half-integer spin particule and only the Bosonic Fock space $$F_+$$ when $$H$$ represents an integer spin particule.

But, in terms of the description above, I see no clear reasons for this to be the case: I have no problems considering either type of Fock space of either type of representations. That obviously confirm that the previous point of view is missing something, but what ? can it be made complete ?

So I tend to assume that there is some physically significant mathematical property (or maybe structure ?) that distinguishes between the "Free Quantum fields" (in the sense above) that satisfies the Spin-statistic theorem and these that don't. And I would like to know which ones.

To rule out some obvious physical property that one expect:

• All these "attempted free quantum fields" are representation of $$\Lambda$$, so they are "Poincaré invariant", the void vector is invariant for this action in all cases.

• Unless I misunderstanding something about what it means, these are 'local' as soon as the 1 particular space $$H$$ we started from is: in the free fields the time evolution is just diagonal on $$H^{\otimes n}$$, so as soon as each single particle behave in a local way, it extends to the Quantum fields. This is the case off the representation of finite mass real mass of the $$\Lambda$$.

• The spectrum of the hamiltonian is positive as soon as this holds for the one particle space we started from.

I know that in practice physicist do not always start from an irreducible representation $$H$$ of $$\Lambda$$, for example the Dirac equation describe the sum of two irreducible representations of $$\Lambda$$ : the mass $$m$$ spin $$\frac{1}{2}$$ and the mass $$-m$$ spin $$\frac{1}{2}$$. And that might play a role in the story, but the 'how' is still unclear to me.

• You need to impose Poincare invariance and causality. Nov 18, 2020 at 20:28
• @KonstantinosKanakoglou : I mean the Bosonic Fock space of a half integer spin representation. Nov 18, 2020 at 20:31
• The wikipedia entry en.wikipedia.org/wiki/Spin-statistics_theorem for the spin-statistics theorem gives a pretty decent account of the necessary assumptions, which indeed are more specific than what you're assuming, also refuting some overly simplistic arguments along the way. Nov 18, 2020 at 21:18
• So, it's been a while and I might get this embarrassingly wrong, but I think part of the answer is that you have a Hilbert space here, not a quantum field theory. To get a QFT, you're going to start playing with creation and annihilation operators and canonical commutation relations. Pretty soon things will go wrong if you have the wrong statistics (eg, your energy states will be unbounded below). Nov 19, 2020 at 3:48
• But, I think more interesting to you might be Weinberg's version of spin-statistics where he shows (or argues at least -- it's not rigorous) that you can't have an interacting S-matrix that's consistent with causality unless you have the spin-statistics relation right. (And now I slink away embarrassed about how much QFT I've forgotten....) Nov 19, 2020 at 3:55

Streater, R. F.; Wightman, A. S., PCT, spin and statistics, and all that, New York-Amsterdam: W. A. Benjamin, Inc. VIII, 181 p. (1964). ZBL0135.44305.

In it, you find (paraphrased)

Theorem 4--10: A free quantum field $$\phi(x)$$ (corresponding to one of the allowed irreducible representations of the Lorentz group, of integer or half-integer spin) satisfying the Wightman axioms and having the "wrong" statistics is identically zero, $$\phi(x)=0$$.

Therein you will also find all the hypotheses and proofs in quite a lot of detail. The allowed representations are those classified by Wigner, excluding the non-physical ones, as the OP mentioned. The Wightman axioms include causality, relativistic invariance, positivity of energy (Hamiltonian) with respect to the vacuum state, and that the fields are represented on a Hilbert space (positive definite inner product on the states). All of these hypotheses are used and if any one is dropped then counter-examples may be found, I believe (as already mentioned in the comments, the Wikipedia page lists counter-examples).

The proof uses the analyticity properties of the 2-point function $$\langle 0| \phi(x) \phi(y) |0\rangle$$ (relies on positivity of energy), the PCT invariance (following from said analyticity and relativistic invariance), and concludes that the "wrong" statistics forces $$\phi(x)=0$$ (relying on causality and the positivity of the state inner product).

If the OP's only goal was to find a mathematically precise statement and proof of the standard spin-statistics theorem, he need look no further. In that case, it is also worth noting the publication date of the Streater-Wightman reference (1964). Many mathematical developments of these ideas have taken place since then. If one reads only material on QFT that is aimed at the level of physics graduate students, it would be hard to find this level of mathematical formalization there, since it is not needed. But the information is out there.

I think the misunderstanding at the heart of this question is about what it takes to specify a quantum field theory. A Hilbert space on which a representation of the Poincaré group acts is not enough by itself, you also need a local Hamiltonian with a positive energy spectrum. To see why that is, look at your Hilbert space of (supposedly free) multiparticle states and consider that it wouldn't change at all if you were to introduce a very weak interaction (too weak to create any bound states); but clearly the free theory and the weakly-coupled theory are different (they have different S-matrices for one), so more than the Hilbert space is needed to capture the relevant theory.

(To give a – very vague and potentially misleading, but maybe still helpful – toy analogy: in one-dimensional quantum mechanics, all theories with a purely discrete energy spectrum have Hilbert spaces isomorphic to $$\ell^2$$, but the harmonic oscillator $$V(x)=\frac{\omega^2}{2}x^2$$ is a different system from the one with potential $$V(x)=\frac{\lambda}{4!}x^4$$; the Hamiltonian makes all the difference.)

Where things go wrong if the spin-statistics relation is violated is when you try to construct a local Hamiltonian with positive energy spectrum. You either get a Hamiltonian whose energy spectrum is not bounded from below, or you find that the field operators come out to be identically zero. Neither case corresponds to a quantum field theory.

Now that the question has been edited, it seems clear that the issue is something else entirely, namely the concept of locality.

Locality in the context of quantum field theory requires that the Hamiltonian can be written in terms of products of field operators at the same spacetime point only. The single-particle Hamiltonian acting on the single-particle Hilbert space isn't written in terms of field operators at all (the field operators contain the creation and annihilation operators in Fock space, which link different $$n$$-particle Hilbert spaces, so the field operators don't even act on the single-particle Hilbert space), so the argument "these are 'local' as soon as the 1 particular space 𝐻 we started from is" doesn't work.

• The first sentence do do not make sense: A representation of the Poincarré group already defines a Hamiltonian as the generator of the time translation. Now If the representation I described in the question indeed do not have positive energy Hamiltonian when the spin-statistic theorem is violated then that would be an answer to the question. But so far I don't see that clearly. Mar 30, 2021 at 12:37
• The Hamiltonian has to be local when written in terms of the field operators $\phi(x)$. Whether that's the case is not readily apparent from its form in terms of the Fock space states.
– gmvh
Mar 30, 2021 at 12:45
• In physics texts, one usually works the other way around and starts from the local Hamiltonian of a specific free theory (Klein-Gordon or Dirac). Then one quickly finds that imposing the wrong kind of commutation relations on the creation and annihilation operators (which directly implies the wrong kind of Fock space) leads to a badly behaved Hamiltonian in terms of Fock space states. Going this way around might be tricky.
– gmvh
Mar 30, 2021 at 13:01
• I don't think Fock space reasoning alone will give you spin-statistics. The spin-statistics theorem is false in 1, 2, and 3 dimensions. There has to be some use of a fact about dim >= 4. Mar 30, 2021 at 13:08
• @user1504 : maybe you are right, but the approach I'm talking about does know about the dimension of space time : the key input is the classification of irreducible representations of the Poincarré group and this would look quite different in lower dimension. Mar 30, 2021 at 13:21