Why isn't the anticommutativity of spinors sufficient as "spin-statistics-theorem"?

Question

From the representation theory of the Lorentz algebra, we know that spinors (objects transforming under the $(\frac{1}{2}, 0)$ and $(0,\frac{1}{2})$ representation), are naturally equipped with a symplectic structure:

To get something invariant (a scalar = an object transforming according to the $(0,0)$ representation) under Lorentz transformations using two spinors $\xi, \chi$, we must use the spinor metric $\epsilon_{ij}$. For example, $ \chi_i \epsilon_{ij} \xi_j $ is a scalar.

In other words, this means that the scalar product of two spinors is antisymmetric:

\begin{align} \chi \cdot \xi &\equiv \chi_i \epsilon_{ij} \xi_j \\ &= \xi_j \epsilon_{ij} \chi_i \\ &= \xi_i \epsilon_{ji} \chi_j \\ &= - \xi_i \epsilon_{ij} \chi_j \equiv - \xi \cdot \chi\\ \end{align} where we used that in index notation we can switch all objects around freely, because, for example, $\xi_k $ is just a number.

Now, fermions are described by spinors. From the observation above, it does not seem like a big surprise that two fermions do anticommute and hence obey Fermi-Dirac statistics.

Why isn't this sufficient as a "proof" of the spin-statistics-theorem?

I've read several explanations for the various approaches to the spin-statistics-theorem, but almost all are extremely complicated and I started wondering why this is the case. It seems that the very basis observation, namely that spin $\frac{1}{2}$ particles automatically anticommute, follows directly from group theory.

Answer 1

You are putting together many different aspects.

First of all, spinors are just elements of the fundamental representation of the universal covering group of the Lorentz group, that is SL(2,$\mathbf{C}$). There are two inequivalent fundamental representations of such group, namely the defining ($(1/2,0)$; elements are $\xi^i$, complex -hence commuting- numbers) and the conjugate ($(0,1/2)$; elements are $\bar \xi^i$, complex -hence commuting- numbers). We identify the spin of this representations as $1/2$, so we would like to describe fermions with them.

Up to now no anticommutativity is introduce.

From QFT it is known that microcausality is respected if we quantize fermionic fields with anticommutation relations like \begin{equation} \{\Psi(t, \vec x),\Psi^\dagger(t, \vec y)\} \sim i\hbar\delta^3(\vec x -\vec y). \end{equation} Notice that this is really the statement of spin-statistics theorem. In the classical limit ($\hbar \to 0$) the RHS of this equation vanishes and we don't know how to make sense of it with 'usual' numbers. For this reason we introduce Grassmann numbers, ie an anticommuting algebra over the reals. Given two elements $a$, $b$ of this algebra, they are such that $ab =- ba$

Now we want to put together the two points above and hence we use spinors of anticommuting numbers to classically describe fermionic fields. For example now $\xi^i$ is a doublet of complex anticommuting numbers, that is $\xi^i \chi^j = - \chi^j \xi^i$ for two anticommuting spinors $\xi^i$ and $\chi^i$.

In other words, we use anticommuting numbers in order to have a classical analogue of the quantum anticommutator, required by spin statistics theorem.

Moreover I think that something is wrong with your inner product: a minus is missing in the first '=' sign due to the fact that you commuted two anticommuting numbers, and this product is really symmetric: \begin{align} \chi \xi \equiv \chi^i \epsilon_{ij} \xi^j = - \xi^j \epsilon_{ij} \chi^i = - \xi^i \epsilon_{ji} \chi^j = \xi^i \epsilon_{ij} \chi^j \equiv \xi \chi \end{align}

Answer 2

This is not a sufficient proof of the spin-statistics theorem because it has little to do with what the spin-statistics theorem says. For one thing, this theorem is a statement about operators, while the property discussed in the OP is purely about classical fields.

Let $a$ be an operator that transforms according to a representation $r$ of the Lorentz group $\text{Spin}(1,d-1)$. (This rep need not be irreducible; but, if reducible, it must be homogeneous with respect to $\pi_1(\text{SO}(1,d-1))=\mathbb Z_2$). We say $a$ is bosonic if $r$ lifts to a representation of $\text{SO}(1,d-1)$, and fermionic otherwise. In other words, $a$ is bosonic (resp. fermionic) if it commutes (resp. anti-commutes) with $(-1)^F\in\text{Spin}(1,d-1)$, where $(-1)^F$ denotes the image of a $2\pi$ rotation in $$ \mathbb Z_2\hookrightarrow\text{Spin}(1,d-1)\twoheadrightarrow\text{SO}(1,d-1) $$

Let also $[\,\cdot\,,\,\cdot\,]$ denote a commutator, and $\{\,\cdot\,,\,\cdot\,\}$ an anti-commutator.

With these definitions in mind, the spin-statistics theorem says the following: let $a_1(x),a_2(x)$ be two (not necessarily distinct) bosonic operators. If $\{a_1(x),a_2(y)\}=0$ for space-like $x-y$, then $a_i(x)$ must be trivial. Similarly, if $a_i(x),a_2(x)$ are fermionic operators, and $[a_1(x),a_2(y)]=0$ for space-like $x-y$, then $a_i(x)$ must be trivial.

Note that we do not say that bosonic operators must commute, and fermionic operators must anti-commute. Instead, we say that the other option leads to a trivial theory, and so is in a sense "forbidden". Of course, this does not rule out other possibilities, so the theorem is not absolutely constraining. (In any case, see this PSE post for a more detailed discussion).

The analysis in the OP does not prove this statement, and so it is not a proof of the spin-statistics theorem. That being said, it is a nice motivation for why such a theorem might hold in the first place. So it is indeed not "a big surprise" that the theorem holds, but the argument is definitely not a proof, not even at the level or rigour of physics. (And do keep in mind that the spin-statistics theorem as above holds for any $d$; but existence of a symmetric or anti-symmetric bilinear form for fermions is very much dimension-dependent, where the reality properties of the irreps of Lorentz have the well-known mod 8 (Bott) periodicity; a recent paper by Witten and Yonekura 1909.08775 does a great job at spelling out the details).