5
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ I think you have in mind a Lagrangian, hence weakly coupled, approach. But the point is that the theorem holds no matter how strongly coupled and non-perturbative effects take place. Besides, your argument says only that perturbative spin-1/2 should be fermions, but it doesn't rule out anticommuting scalars (e.g. the ghosts in YM theories). $\endgroup$ – TwoBs May 19 '17 at 9:46
4
$\begingroup$

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}

$\endgroup$
  • $\begingroup$ An element of the $(1/2,0)$ representation is not $\xi^i$, but $\xi$. The $(1/2,0)$ representation is given in terms of $2 \times 2 $ matrices, which act on two-component spinors. $\xi^i$ is just one component and hence, as you say, a complex number. However, the anticommutativity appears already hear as explained in the OP. The spinor components do commute, but the spinors themselves to not. $\endgroup$ – jak May 18 '17 at 8:47
  • $\begingroup$ Why is it necessary to talk about the commutativity/anticommutativity of the spinor components? There is no significance to them, because a fermion is always described by a Weyl spinor (a fundamental rep of the Lorentz algebra), and not by one component of it. $\endgroup$ – jak May 18 '17 at 8:49
  • $\begingroup$ The minus sign in my computation is correct. Without further assumptions the components of a spinor are just complex numbers and therefore commute. This is just what follows from group theory alone. $\endgroup$ – jak May 18 '17 at 8:50
  • $\begingroup$ Also take note that I never talked about representations of the Lorentz group, but of the Lorentz algebra (i.e. the corresponding Lie algebra) . The (complexified) Lie algebra of the Lorentz group and the Lie algebra of the double cover of the Lorentz group are isomorphic. In this sense, my statement about reps of the Lorentz algebra is correct. $\endgroup$ – jak May 18 '17 at 8:52
  • 1
    $\begingroup$ Sure it is standard, but still it is imprecise and at this point it seems very important to make this distinction. Especially: $\xi^i$ are spinor components and therefore commute, whereas the spinors themselves $\chi$ do not. $\endgroup$ – jak May 19 '17 at 11:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.