new flip operator $\tau_\theta = F_\theta^{-1}\,\tau_0
(\Lambda) F_\theta$: $$\Delta_\theta(\Lambda) \tau_\theta = \tau_\theta \Delta_\theta (\Lambda).$$

Thus in noncommutative quantum theory, the usual fermions/bosons do not make sense, but twisted ones do. They are obtained from the projectors $S_\theta, A_\theta$: $$S_\theta~=~\frac{{\bf 1}~+~\tau_\theta}{2}, \qquad A_\theta~=~ 
\frac{{\bf 1}~-~\tau_\theta}{2}.$$

Quantum Fields
--------------

A quantum field $\chi$ on evaluation at a spacetime point (or more generally on pairing with a test function) gives an operator acting on a Hilbert space. A field at $x_1$ acting on the vacuum gives a one-particle state centered at $x_1$. When we write $\chi(x_1)\,\chi(x_2)$ we mean $(\chi\otimes\chi)(x_1,x_2)$. Acting on the vacuum we generate a two-particle state, where one particle is centered at $x_1$ and the other at $x_2$.

If $a_p$ is the annihilation operator of the free second-quantized field $\phi_\theta$ on ${\cal A}_\theta({\mathbb R}^N)$, we want, as in standard quantum field theory, $$\begin{aligned}
\langle 0 |\phi_\theta(x) a^\dagger_p |0\rangle &=& e_p(x), \\
\frac{1}{2}\langle 0 |\phi_\theta(x_1) \phi_\theta(x_2) a^\dagger_q
a^\dagger_p |0\rangle &=& \left(\frac{{\bf 1} \pm
  \tau_\theta}{2}\right)(e_p \otimes e_q)(x_1,x_2) \nonumber \\
&\equiv& (e_p \otimes_{S_\theta,A_\theta} e_q)(x_1,x_2) 
\label{tbasis} \end{aligned}$$ where $e_p(x) = e^{-i p \cdot x}$.

This compatibility between twisted statistics and Poincaré invariance has profound consequences for commutation relations. For example when the states are labeled 