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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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1 Answer 1

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It seems difficult to prove what they mentioned but we can do it.

Let me answer Question number 2: How does the angular momentum operator $\boldsymbol{J}$ annihilate the vacuum $\boldsymbol{J}\left|0\rangle\right.=0$? Of course, we can prove that fact for the specific axis, for instance, $z$ axis, but we will verify it for the general axis. You tried to check annihilation by only the spin momentum operator but it is slightly wrong. We should apply full angular momentum operator $$ \boldsymbol{J} = \int\mathrm{d}^3\boldsymbol{x} \psi^{\dagger} \left( \boldsymbol{x}\times(-i\boldsymbol{\nabla}) + \frac{1}{2}\boldsymbol{\Sigma} \right) \psi $$ to the vacuum $\left|0\rangle\right.$, and we expand the field as $$ \begin{align} \psi(\boldsymbol{x}) &= \int \frac{\mathrm{d}^3\boldsymbol{p}}{(2\pi)^3} \frac{1}{\sqrt{2E_{\boldsymbol{p}}}} \sum_{s} \left( a^s_{\boldsymbol{p}}u^s(\boldsymbol{p})e^{+i\boldsymbol{p}\cdot\boldsymbol{x}} + {b^s_{\boldsymbol{p}}}^{\dagger}v^s(\boldsymbol{p})e^{-i\boldsymbol{p}\cdot\boldsymbol{x}} \right) , \nonumber \end{align} $$ Note we will calculate at $t=0$, equivalently Schrödinger picture because the relation that we will prove follows any time.

Let's apply the angular momentum operator with orbital momentum: $$ \begin{align} \boldsymbol{J}\left|0\rangle\right. &= \int\mathrm{d}^3\boldsymbol{x}\int\frac{\mathrm{d}^3\boldsymbol{p}\mathrm{d}^3\boldsymbol{q}}{(2\pi)^6} \frac{1}{\sqrt{2E_{\boldsymbol{p}}}\sqrt{2E_{\boldsymbol{q}}}} \sum_{s,r} \left( {a^r_{\boldsymbol{q}}}^{\dagger}{u^r}^{\dagger}(\boldsymbol{q})e^{-i\boldsymbol{q}\cdot\boldsymbol{x}} + b^r_{\boldsymbol{q}}{v^r}^{\dagger}(\boldsymbol{q})e^{+i\boldsymbol{q}\cdot\boldsymbol{x}} \right) \nonumber \\ &\hspace{2cm} \times \left( \boldsymbol{x}\times(-i\boldsymbol{\nabla}) + \frac{1}{2}\boldsymbol{\Sigma} \right) {b^{s}_{\boldsymbol{p}}}^{\dagger}v^{s}(\boldsymbol{p})e^{-i\boldsymbol{p}\cdot\boldsymbol{x}}\left|0\rangle\right. \nonumber \\ &= \int\mathrm{d}^3\boldsymbol{x}\int\frac{\mathrm{d}^3\boldsymbol{p}\mathrm{d}^3\boldsymbol{q}}{(2\pi)^6} \frac{1}{\sqrt{2E_{\boldsymbol{p}}}\sqrt{2E_{\boldsymbol{q}}}} \nonumber \\ &\qquad\times \sum_{s,r} {a^r_{\boldsymbol{q}}}^{\dagger}{u^r}^{\dagger}(\boldsymbol{q}) \left( \boldsymbol{p}\times\boldsymbol{x} + \frac{1}{2}\boldsymbol{\Sigma} \right) {b^{s}_{\boldsymbol{p}}}^{\dagger}v^{s}(\boldsymbol{p})e^{-i(\boldsymbol{p}+\boldsymbol{q})\cdot\boldsymbol{x}}\left|0\rangle\right. \nonumber \\ &+ \int\mathrm{d}^3\boldsymbol{x}\int\frac{\mathrm{d}^3\boldsymbol{p}\mathrm{d}^3\boldsymbol{q}}{(2\pi)^6} \frac{1}{\sqrt{2E_{\boldsymbol{p}}}\sqrt{2E_{\boldsymbol{q}}}} \nonumber \\ &\qquad\times \sum_{s,r} b^r_{\boldsymbol{q}}{v^r}^{\dagger}(\boldsymbol{q}) \left( \boldsymbol{p}\times\boldsymbol{x} + \frac{1}{2}\boldsymbol{\Sigma} \right) {b^{s}_{\boldsymbol{p}}}^{\dagger}v^{s}(\boldsymbol{p})e^{-i(\boldsymbol{p}-\boldsymbol{q})\cdot\boldsymbol{x}}\left|0\rangle\right. \nonumber \\ &\equiv A+B . \nonumber \end{align} $$ It is expected to each term become zero since two-particles states ${a^r_{\boldsymbol{q}}}^{\dagger}{b^{s}_{\boldsymbol{p}}}^{\dagger}\left|0\rangle\right.$ and $b^r_{\boldsymbol{q}}{b^{s}_{\boldsymbol{p}}}^{\dagger}\left|0\rangle\right.$ are independent to each other, so we should show both terms vanish.

We started to evaluate the last term $B$, an easier one. Using the anticommuting relation, we get $$ b_{\boldsymbol{q}}^r{b_{\boldsymbol{p}}^s}^{\dagger}\left|0\rangle\right. = (2\pi)^3\delta^{(3)}(\boldsymbol{p}-\boldsymbol{q})\delta^{rs}\left|0\rangle\right. $$ since $b_{\boldsymbol{q}}^r$ annihilates the vaccum. Inserting this relation to the $B$, we find $$ \begin{align} B &= \int\mathrm{d}^3\boldsymbol{x}\int\frac{\mathrm{d}^3\boldsymbol{p}}{(2\pi)^3} \frac{1}{2E_{\boldsymbol{p}}} \sum_{s} {v^s}^{\dagger}(\boldsymbol{p}) \left( \boldsymbol{p}\times\boldsymbol{x} + \frac{1}{2}\boldsymbol{\Sigma} \right) v^{s}(\boldsymbol{p})\left|0\rangle\right. \nonumber \\ &= \int\mathrm{d}^3\boldsymbol{x}\int\frac{\mathrm{d}^3\boldsymbol{p}}{(2\pi)^3} \frac{1}{2E_{\boldsymbol{p}}} \sum_{s} {v^s}^{\dagger}(\boldsymbol{p}) \left( \boldsymbol{p}\times\boldsymbol{x} \right) v^{s}(\boldsymbol{p})\left|0\rangle\right. \nonumber \\ &\qquad + \int\mathrm{d}^3\boldsymbol{x}\int\frac{\mathrm{d}^3\boldsymbol{p}}{(2\pi)^3} \frac{1}{2E_{\boldsymbol{p}}} \sum_{s} {v^s}^{\dagger}(\boldsymbol{p}) \frac{1}{2}\boldsymbol{\Sigma} v^{s}(\boldsymbol{p})\left|0\rangle\right. \nonumber \\ &\equiv B_1+B_2 \nonumber . \end{align} $$ We can show that the orbital and spin angular momentum of an anti-antiparticles state vanish separately. (To be honest, I am not sure why this is the case. However, I believe that it is intuitively well understood. )

In fact, $B_1$ is zero since integrand is an odd function about the variable $x$ and canceled by integration about $x$. The second $B_2$ also vanishes though it's rather difficult to show. To prove it, we take $$ v^s(\boldsymbol{p}) = \begin{pmatrix} \sqrt{p\cdot\sigma}\eta^s \\ -\sqrt{p\cdot\bar{\sigma}}\eta^s \end{pmatrix} $$ as the representation of the $v^r(\boldsymbol{p})$. In this form, the integrand of $B_2$ becomes $$ \sum_s {v^s(\boldsymbol{p})}^{\dagger}\boldsymbol{\Sigma}\boldsymbol{v}^{s} = \sum_s \left[ {\eta^s}^{\dagger}\sqrt{p\cdot\sigma}\boldsymbol{\sigma}\sqrt{p\cdot\sigma}\eta^s + {\eta^s}^{\dagger}\sqrt{p\cdot\sigma}\boldsymbol{\sigma}\sqrt{p\cdot\sigma}\eta^s \right] $$ since $p$ and $\sigma$ are hermitian. Take $\eta^s$ as the eigenvector of $\boldsymbol{\sigma}$, i.e. $\boldsymbol{\sigma}\eta^1=+\eta^1,\boldsymbol{\sigma}\eta^2=-\eta^2$. Then, the last task of this step is to calculate the commutation relation $[\boldsymbol{\sigma},\sqrt{p\cdot\sigma}]$ and $[\boldsymbol{\sigma},\sqrt{p\cdot\bar{\sigma}}]$. Though Peskin&Scroeder didn't show, we can easily obtain the general form $$ \begin{align} \sqrt{p\cdot\sigma} &= \sqrt{E_{\boldsymbol{p}}+|\boldsymbol{\boldsymbol{p}}|} \frac{1-\hat{\boldsymbol{p}}\cdot\boldsymbol{\sigma}}{2} + \sqrt{E_{\boldsymbol{p}}-|\boldsymbol{\boldsymbol{p}}|} \frac{1+\hat{\boldsymbol{p}}\cdot\boldsymbol{\sigma}}{2} \nonumber \\ \sqrt{p\cdot\sigma} &= \sqrt{E_{\boldsymbol{p}}-|\boldsymbol{\boldsymbol{p}}|} \frac{1+\hat{\boldsymbol{p}}\cdot\boldsymbol{\sigma}}{2} + \sqrt{E_{\boldsymbol{p}}+|\boldsymbol{\boldsymbol{p}}|} \frac{1-\hat{\boldsymbol{p}}\cdot\boldsymbol{\sigma}}{2} \nonumber \end{align} $$ where $\hat{\boldsymbol{p}}$ is the unit vector of the momentum. Verifying these two formulae is left to readers since you just follow the same steps as in the text. Computing the commutation relation under this representation, we find $$ \begin{align} [\boldsymbol{\sigma},\sqrt{p\cdot\sigma}] &= \sqrt{p\cdot\sigma}\boldsymbol{\sigma} - i ( \sqrt{E_{\boldsymbol{p}}+|\boldsymbol{p}|} - \sqrt{E_{\boldsymbol{p}}-|\boldsymbol{p}|} ) \hat{\boldsymbol{p}}\times\boldsymbol{\sigma} \nonumber \\ [\boldsymbol{\sigma},\sqrt{p\cdot\bar{\sigma}}] &= \sqrt{p\cdot\bar{\sigma}}\boldsymbol{\sigma} + i ( \sqrt{E_{\boldsymbol{p}}+|\boldsymbol{p}|} - \sqrt{E_{\boldsymbol{p}}-|\boldsymbol{p}|} ) \hat{\boldsymbol{p}}\times\boldsymbol{\sigma} \nonumber . \end{align} $$ Note that $$ \boldsymbol{\sigma}(\hat{\boldsymbol{p}}\cdot\boldsymbol{\sigma}) = \sigma^i\hat{p}^j\sigma^j = \hat{p}^j\sigma^j\sigma^i + 2i\varepsilon^{ijk}\hat{p}^j\sigma^k = (\hat{\boldsymbol{p}}\cdot\boldsymbol{\sigma}) \boldsymbol{\sigma} + 2i \hat{\boldsymbol{p}} \times \boldsymbol{\sigma} . $$ There exists an extra term by commutating $\boldsymbol{\sigma}$ and others but you can easily verify that the terms are canceled in the square bracket. Therefore we get $$ \begin{align} &\quad \sum_s \left[ {\eta^s}^{\dagger}\sqrt{p\cdot\sigma}\boldsymbol{\sigma}\sqrt{p\cdot\sigma}\eta^s + {\eta^s}^{\dagger}\sqrt{p\cdot\sigma}\boldsymbol{\sigma}\sqrt{p\cdot\sigma}\eta^s \right] \nonumber \\ &= \sum_s \left[ {\eta^s}^{\dagger}\sqrt{p\cdot\sigma}\sqrt{p\cdot\sigma}\boldsymbol{\sigma}\eta^s + {\eta^s}^{\dagger}\sqrt{p\cdot\sigma}\sqrt{p\cdot\sigma}\boldsymbol{\sigma}\eta^s \right] = 0 . \nonumber \end{align} $$ The last line is easily shown, so please try it. Thus we finished proving $B_2=0$ and also get $B=0$.

Okay, let's move on to the term $A$. It was the form $$ \begin{align} A &= \int\mathrm{d}^3\boldsymbol{x}\int\frac{\mathrm{d}^3\boldsymbol{p}\mathrm{d}^3\boldsymbol{q}}{(2\pi)^6} \frac{1}{\sqrt{2E_{\boldsymbol{p}}}\sqrt{2E_{\boldsymbol{q}}}} \nonumber \\ &\qquad\times \sum_{s,r} {a^r_{\boldsymbol{q}}}^{\dagger}{u^r}^{\dagger}(\boldsymbol{q}) \left( \boldsymbol{p}\times\boldsymbol{x} + \frac{1}{2}\boldsymbol{\Sigma} \right) {b^{s}_{\boldsymbol{p}}}^{\dagger}v^{s}(\boldsymbol{p})e^{-i(\boldsymbol{p}+\boldsymbol{q})\cdot\boldsymbol{x}}\left|0\rangle\right. . \nonumber \end{align} $$ Some may try to integrate by variable $x$. Though it is a great point of view, we cannot integrate immediately by the existence of the term $\boldsymbol{p}\times\boldsymbol{x}$. However, we can remove $\boldsymbol{x}$ from that factor since it follows $$ \boldsymbol{p}\times\boldsymbol{x}\ e^{-i(\boldsymbol{p}+\boldsymbol{q})\cdot\boldsymbol{x}} = \boldsymbol{p}\times(i\boldsymbol{\nabla}_{\boldsymbol{p}}e^{-i(\boldsymbol{p}+\boldsymbol{q})\cdot\boldsymbol{x}}) $$ where $\boldsymbol{\nabla}_{\boldsymbol{p}}$ means the derivation by the variable $\boldsymbol{p}$. By using this trick, we capable to integrate by $\boldsymbol{x}$ for first and obtain $$ \begin{align} &\qquad \int\mathrm{d}^3\boldsymbol{x}\int\frac{\mathrm{d}^3\boldsymbol{p}\mathrm{d}^3\boldsymbol{q}}{(2\pi)^6} \frac{1}{\sqrt{2E_{\boldsymbol{p}}}\sqrt{2E_{\boldsymbol{q}}}} \nonumber \\ &\qquad\times \sum_{s,r} {a^r_{\boldsymbol{q}}}^{\dagger}{u^r}^{\dagger}(\boldsymbol{q}) \left( \boldsymbol{p}\times\boldsymbol{x} + \frac{1}{2}\boldsymbol{\Sigma} \right) {b^{s}_{\boldsymbol{p}}}^{\dagger}v^{s}(\boldsymbol{p})e^{-i(\boldsymbol{p}+\boldsymbol{q})\cdot\boldsymbol{x}}\left|0\rangle\right. \nonumber \\ &= \int\frac{\mathrm{d}^3\boldsymbol{p}\mathrm{d}^3\boldsymbol{q}}{(2\pi)^3} \frac{1}{\sqrt{2E_{\boldsymbol{p}}}\sqrt{2E_{\boldsymbol{q}}}} \nonumber \\ &\qquad\times \sum_{s,r} {a^r_{\boldsymbol{q}}}^{\dagger}{u^r}^{\dagger}(\boldsymbol{q}) \left( \boldsymbol{p}\times(i\boldsymbol{\nabla}_{\boldsymbol{p}}\delta^{(3)}(\boldsymbol{p}+\boldsymbol{q})) + \frac{1}{2}\boldsymbol{\Sigma}\delta^{(3)}(\boldsymbol{p}+\boldsymbol{q}) \right) {b^{s}_{\boldsymbol{p}}}^{\dagger}v^{s}(\boldsymbol{p})\left|0\rangle\right. . \nonumber \end{align} $$ In order to apply the delta function formula, we integrate the term containing derivation of $\boldsymbol{p}$ by parts. Therefore we find the relation $$ \begin{align} A &= \int\frac{\mathrm{d}^3\boldsymbol{p}}{(2\pi)^3} \frac{1}{\sqrt{2E_{\boldsymbol{p}}}} \nonumber \\ &\qquad\times \sum_{s,r} {a^r_{-\boldsymbol{p}}}^{\dagger}{u^r}^{\dagger}(-\boldsymbol{p}) \left( (i\overset{\rightarrow}{\boldsymbol{\nabla}}_{\boldsymbol{p}})\times\boldsymbol{p} + \frac{1}{2}\boldsymbol{\Sigma} \right) \frac{1}{\sqrt{2E_{\boldsymbol{p}}}} {b^{s}_{\boldsymbol{p}}}^{\dagger}v^{s}(\boldsymbol{p})\left|0\rangle\right. . \nonumber \end{align} $$ We should compute it carefully since $\overset{\rightarrow}{\boldsymbol{\nabla}}_{\boldsymbol{p}}$ acts on all $\boldsymbol{p}$ right on it. Although it seems very tough, most of it is trivial. The term $\overset{\rightarrow}{\boldsymbol{\nabla}}_{\boldsymbol{p}}$ acts on $\boldsymbol{p}$ is canceled because they are parallel to each other. (If you find this explanation difficult to understand, please do the calculation properly.) The term $$ \left(\boldsymbol{\nabla}_{\boldsymbol{p}}\frac{1}{\sqrt{2E_{\boldsymbol{p}}}}\right) \times \boldsymbol{p} $$ also vanishes by easier computation. It is probably difficult to show the remaining two terms when the derivative acts $b_{\boldsymbol{p}}^s$ or $v^s(\boldsymbol{p})$. The former is proved by using the orthogonality of $u^s$ and $v^s$: $$ {u^r}^{\dagger}(-\boldsymbol{p}) v^{s}(\boldsymbol{p}) = 0 . $$ So, the $A$ become $$ \begin{align} A &= \int\frac{\mathrm{d}^3\boldsymbol{p}}{(2\pi)^3} \frac{1}{2E_{\boldsymbol{p}}} \nonumber \\ &\qquad\times \sum_{s,r} {a^r_{-\boldsymbol{p}}}^{\dagger}{b^{s}_{\boldsymbol{p}}}^{\dagger} \underline{ {u^r}^{\dagger}(-\boldsymbol{p}) \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) v^{s}(\boldsymbol{p}) }\left|0\rangle\right. , \nonumber \end{align} $$ and what is remainded to evaluate the underlined factor $$ {u^r}^{\dagger}(-\boldsymbol{p}) \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) v^{s}(\boldsymbol{p}) $$ is vanished. There may exist several ways to prove it. We will compute it by $u^r$ and $v^s$ are eigenfunctions of the Dirac equation in the momentum space: $$ p^{\mu}\gamma_{\mu} u^s(p) = + m u^s(p) ,\ p^{\mu}\gamma_{\mu} v^s(p) = - m v^s(p) , $$ or equivalently $$ v^s(p) = - \frac{p^{\mu}\gamma_{\mu}}{m} v^s(p) ,\ {u^r}^{\dagger}(-p) = + \frac{p^{\mu}\gamma_{\mu}}{m} {u^r}^{\dagger}(-p) . $$ By using those relations, the terms are equivalent to itself multiplied by the minus sign: $$ \begin{align} &\hspace{-1cm} {u^r}^{\dagger}(-\boldsymbol{p}) \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) v^{s}(\boldsymbol{p}) \nonumber \\ &= - {u^r}^{\dagger}(-\boldsymbol{p}) \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) v^{s}(\boldsymbol{p}) . \nonumber \end{align} $$ To prove it, insert $v^s(p)=-p^{\mu}\gamma_{\mu}v^s(p)/m$ into the left hand side $$ \begin{align} &\quad {u^r}^{\dagger}(-\boldsymbol{p}) \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) v^{s}(\boldsymbol{p}) \nonumber \\ &= {u^r}^{\dagger}(-\boldsymbol{p}) \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) - \frac{p^{\mu}\gamma_{\mu}}{m} v^{s}(\boldsymbol{p}) \nonumber \\ &= -\frac{1}{m} {u^r}^{\dagger}(-\boldsymbol{p}) \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}})p^{\mu}\gamma_{\mu} v^{s}(\boldsymbol{p}) \nonumber \\ &\qquad -\frac{1}{m} {u^r}^{\dagger}(-\boldsymbol{p}) \left(\frac{\boldsymbol{\Sigma}}{2}\right)p^{\mu}\gamma_{\mu} v^{s}(\boldsymbol{p}) \nonumber . \end{align} $$ First term become $$ \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}})p^{\mu}\gamma_{\mu} v^{s}(\boldsymbol{p}) = i\boldsymbol{p}\times\boldsymbol{\gamma}v^s(p) + p^{\mu}\gamma_{\mu}\boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) v^{s}(\boldsymbol{p}) $$ where there exists an extra term but it canceled out by another extra term appearing the last term. We find $$ \begin{align} \frac{\Sigma^k}{2} p^{\mu}\gamma_{\mu} &= \frac{i}{8} \varepsilon^{kmn}[\gamma^m,\gamma^n] (p^0\gamma^0-p^i\gamma^i) \nonumber \\ &= \frac{i}{8}p^0 \varepsilon^{kmn}[\gamma^m,\gamma^n] \gamma^0 - \frac{i}{8}p^i \varepsilon^{kmn}[\gamma^m,\gamma^n] \gamma^i \nonumber \\ &= \frac{i}{8}p^0\gamma^0 \varepsilon^{kmn}[\gamma^m,\gamma^n] - \frac{i}{8}p^i\varepsilon^{kmn} \left( \gamma^i [\gamma^m,\gamma^n] + 4\delta^{im}\gamma^n - 4\delta^{in}\gamma^m \right) \nonumber \\ &= p^{\mu}\gamma_{\mu} \frac{\Sigma^k}{2} - \frac{i}{2} p^i \varepsilon^{kmn}(\delta^{im}\gamma^n-\delta^{in}\gamma^m) \nonumber \\ &= p^{\mu}\gamma_{\mu} \frac{\Sigma^k}{2} - i\varepsilon^{kim}p^i\gamma^m = p^{\mu}\gamma_{\mu} \frac{\Sigma^k}{2} - i\boldsymbol{p}\times\boldsymbol{\gamma} . \nonumber \end{align} $$ Now we notice the extra term $i\boldsymbol{p}\times\gamma$ appears with opposite sign and canceled it. So we get $$ \begin{align} &\hspace{-1cm} -\frac{1}{m} {u^r}^{\dagger}(-\boldsymbol{p}) \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) p^{\mu}\gamma_{\mu} v^{s}(\boldsymbol{p}) \nonumber \\ &= -\frac{1}{m} {u^r}^{\dagger}(-\boldsymbol{p}) p^{\mu}\gamma_{\mu} \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) v^{s}(\boldsymbol{p}) \nonumber \\ &= - {u^r}^{\dagger}(-\boldsymbol{p}) \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) v^{s}(\boldsymbol{p}) \nonumber \end{align} $$ since $u^s$ is the eigenfunction of the Dirac operator $p^{\mu}\gamma_{\mu}$. Finally, we achieved the relation $$ \begin{align} &\hspace{-1cm} {u^r}^{\dagger}(-\boldsymbol{p}) \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) v^{s}(\boldsymbol{p}) \nonumber \\ &= - {u^r}^{\dagger}(-\boldsymbol{p}) \left( \boldsymbol{p}\times(-i\boldsymbol{\nabla}_{\boldsymbol{p}}) + \frac{1}{2}\boldsymbol{\Sigma} \right) v^{s}(\boldsymbol{p}) = 0 \nonumber \end{align} $$ and the term $A$ is zero. Note that in the last computation, both the orbital term and the spin term are needed to cancel the extra term which appears when commuting the derivative and Dirac matrices. At last, the proof of $\boldsymbol{J}\left|0\rangle\right.=0$ is finished!

For Question 1, I will tell you just a hint. When you compute $J_z a_{\boldsymbol{0}}^{\dagger}\left|0\rangle\right.$, you may use the trick $J_z a_{\boldsymbol{0}}^{\dagger}\left|0\rangle\right.=[J_z,a_{\boldsymbol{0}}^{\dagger}]\left|0\rangle\right.$. You can probably understand what is happening for an orbital part by computing it for full angular momentum $J_z$. By applying the anti-commutation relation, you can easily verify $\boldsymbol{x}\times(-i\boldsymbol{\nabla})$ does not contribute to the angular momentum because of $a_{\boldsymbol{0}}^{\dagger}$.

I am sorry for the answer being too long. Thanks for your patience with my poor English.

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