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Good morning, I've already seen that this topic has been discussed so long, but my doubts remain unchanged. At page 61 of Peskin & Schroeder, An Introduction to QFT, there is the demonstration that $a^{\dagger}_{0}|0\rangle$ is an eigenstate of the spin operator $\Sigma^3/2$, but I don't understand some claims and passages:

  1. Peskin states that if we are in the rest-frame (that is, I think, the frame in which the particle we create by applying $\psi(x)$ to the vacuum has 3-momentum equal to zero), this means that the orbital part of $J_z$ does not contribute. How can I see this?
  2. Assuming that the first statement is correct, Peskin states that $J_z$ annihilates vacuum, but I tried to perform such a calculation, and I found $$ J_z|0\rangle=\int\text{d}^3x\,\left[{\psi}^{\dagger}(\mathbf{x})\frac{1}{2}\Sigma^{3}\psi(\mathbf{x})\right]|0\rangle $$ that is, by substituting the expressions of $\psi$ and $\psi^{\dagger}$ in terms of ladder operators and integrating on $\mathbf{x}$, $$ \int\frac{\text{d}^3p}{(2\pi)^3}\frac{1}{4E_{\mathbf{p}}}\sum_{t,r}\Big[{u^t}^{\dagger}(\mathbf{p}){a^t}^{\dagger}_{\mathbf{p}}\Sigma^3{u^r}(\mathbf{p})a^r_{\mathbf{p}}+{v^t}^{\dagger}(\mathbf{p}){a^t}^{\dagger}_{\mathbf{p}}\Sigma^3v^r(-\mathbf{p}){b^r}^{\dagger}_{-\mathbf{p}}\\ \qquad\qquad\qquad+{v^t}^{\dagger}(-\mathbf{p})b^t_{-\mathbf{p}}\Sigma^3u^r(\mathbf{p})a^r_{\mathbf{p}}+{v^t}^{\dagger}(-\mathbf{p})b^t_{-\mathbf{p}}\Sigma^3v^r(-\mathbf{p}){b^r}^{\dagger}_{-\mathbf{p}}\Big]|0\rangle. $$ Now, how can I proceed further? Do the Spin operator commute with the ladder operators?
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