How does Lorentz invariance make $(\Psi_0,J_{\mu}\Psi_0)$ vanish? Right before equation (10.4.7) in Weinberg's volume 1 on quantum field theory, he said $(\Psi_0,J_{\mu}\Psi_0)$ vanishes due to requirement of Lorentz invariance. As I understand, this term is a vector, and covariant as a Lorentz vector. But how does it vanish?
 A: For quantum field theories the vacuum state is a scalar with respect to lorentz transformations (i.e. $ M \psi = \psi $ for any boost $M$), and the angular momentum 4-vector transforms as a vector (i.e. $M J^{\mu} M^{-1} = M^{\mu}_{\nu} J^{\nu}$ where the boost $M$ sends $ x^{\mu} \to {x^{\prime}}^{\mu} = M^{\mu}_{\nu} x^{\nu} $ )
From this we see the expectation value $j$ of the angular momentum 4-vector in the ground state is equal to any boosted version of itself:
$$ j^{\mu} =( \psi, J^{\mu} \psi )=( M \psi, M J^{\mu} \psi )= ( M \psi, M J^{\mu} M^{-1} M \psi ) =( \psi,M^{\mu}_{\nu} J^{\nu} \psi) =M^{\mu}_{\nu} ( \psi, J^{\nu} \psi) = (Mj)^{\mu} $$
where $M$ is any unitary boost. But any vector that transforms into itself under any boost is necessarily the zero vector.
note: I've been lazy with respect to technicalities relating to the unitarity of general boosts. The symmetry argument is however perfectly sound.
A: The vacuum $\Psi_0$ is the only vector in the Fock representation that is Lorentz invariant. The consequence of this fact can be interpreted as "the vacuum is not polarised, so that any vector must be the zero vector, or otherwise it would determine a privileged direction in space, thus breaking its relativistic invariance".
