I.e., what properties of the spinors gives us a reason for using them for describing of wavefunctions of fermions?

# [Physics] Why do we use spinors for describing fermions

fermionsquantum-field-theoryspinors

#### Related Solutions

If you consider the Dirac field $\psi$ as a linear combination of the particle annihilation part $\psi^+$ and the antiparticle creation part $\psi^{-c}$, then we have the requirement that $P\psi(x)P^{-1}$ should be proportional to $\psi(Px)$, where $P$ is the parity/space inversion operator. (I commit a slight abuse of notation here - the $P$ is both the unitary operator representing parity acting on the space of states and the space inversion operator on Minkowski space sending $(t,\vec x)$ to $(t,-\vec x)$ in the argument.

In general, we have also that this should hold for $\psi^+$ and $\psi^{-c}$, i.e. \begin{align} P \psi^+(x)P^{-1} & = \eta^\ast b_u \psi^+(Px) \\ P \psi^{-c}(x)P^{-1} & = \eta^c b_v \psi^{-c}(Px) \end{align} where the $\eta^\ast$ and $\eta^c$ are the phases by which the anti-creation /annihilation operators transform and the $b_{u/v}$ are the phases by which the fundamental spinors traditionally denoted as $u_s,v_s$ transform. These formulae come about because e.g. $\psi^+(x) = \sum_s \int u_s(\vec p) \exp(-\mathrm{i}px)a_s(\vec p)\mathrm{d}^3p$ schematically, and likewise for $\psi^{-c}$. One can then show that $b_u = -b_v$, and for $P\psi(x)P^{-1}$ to be propertional to $\psi(Px)$ we must then have that $\eta^\ast = -\eta^c$ because otherwise $\psi(x) = c_1 \psi^+(x) + c_2 \psi^{-c}$ cannot be proportional because the relative sign would change under parity.

For a more detailed derivation, see *"The quantum theory of fields"* by Weinberg, volume 1, section 5.5 as suggested in a comment by TwoBs.

They are restricting their analysis to real fermions. Their dagger $\dagger$ must mean just transpose. You can see later on they break up $\Psi$ into components $\Psi=(\psi,\bar{\psi})$ and $\Psi^\dagger$ is broken up into the same two components without any notion of complex conjugation.

## Best Answer

We're using spinor fields for fermions because Nature does the same. Nature does so because She has no choice. Pauli has proven the spin-statistics theorem that says that all fields whose particles obey the Fermi-Dirac statistics (with the Pauli exclusion principle) have to carry a half-integral spin; and those with the Bose-Einstein statistics have to have an integral spin. Mixing a spin with a wrong statistics or vice versa would lead to negative prrobabilities or energies unbounded from below.

The only half-integer field that doesn't require any gauge symmetry to get rid of the negative-norm states is the spin-1/2 field, the spinor. A spinor may be viewed as an object that is more elementary than a vector and that was previously overlooked. It's also possible to build vectors and tensors out of the spinorial components – vectors and tensors may be represented by spintensors of various sorts. But the word "spinors" should be reserved for the representations with $j=1/2$.

There also exist "unphysical" field theories such as the topological ones that may violate the spin-statistics relations. Also, Faddeev-Popov ghosts used to deal with gauge symmetries in a modern way always violate the spin-statistics relationship – the rule is exactly reverted for them. $b,c$ are fermions with an integer spin for a bosonic symmetry, and $\beta,\gamma$ are bosons with a half-integral spin. They don't create physical particles.