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?
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2 Answers
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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.
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3$\begingroup$ Actually "Lorentz invariance" usually means invariance under the action of the connected component of Lorentz group containing the identity, $SO(1,3)_+$. The transformation you consider does not belong to that subgroup. It is instead the simplest geometrical representation of the CPT action. Nevertheless, your proof can be made correct. There is no non-vanishing 4-vector $t$ satisfying $\Lambda t = t$ for every $\Lambda \in SO(1,3)_+$. $\endgroup$ Commented Jul 13, 2015 at 17:36
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$\begingroup$ Good catch. I suppose explicitly we could consider 180 degree rotations about perpendicular axes to establish the vanishing of the three spatial indices. Any finite boost would mix the time index and finish the proof, though now it seems much less pretty. For compact groups this proof requires only establishing that the operator not contain any symmetric component in its decomposition into irreducible representations, but I do not know how to prove this for a non-compact group. $\endgroup$ Commented Jul 13, 2015 at 19:24
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$\begingroup$ @creillyucla Thank you for this complete answer. I really appreciate it! $\endgroup$ Commented Jul 14, 2015 at 2:00
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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".
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$\begingroup$ Thanks for your answer. I understand that the vacuum $\Phi_0$ is Lorentz invariant. But when you say the "the vacuum is not polarised", do you mean the mean value of any vector, or tensor, should vanish if we take their expectation value in vacuum? Apparently, this is not true for a constant vector. So what is the condition for this statement. Thanks again! $\endgroup$ Commented Jul 14, 2015 at 2:05
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$\begingroup$ In creillyucla's answer the vector might as well be constant. The only tensor that can have a non-zero expectation value on the vacuum is a scalar. $\endgroup$ Commented Jul 14, 2015 at 7:19