【Read between lines】Peskin, Schroeder "An Introduction to Quantum Field Theory" p. 61 (Angular momentum of one particle state)

Key Words

• Angular momentum

• Spin

• One particle state

Relevant section

This is most easily done using a trick: Since $${J_z}$$ must annihilate the vacuum, $${j_za_0^{s\dagger}\ket{0}=[j_z,a_0^{s\dagger}]\ket{0}}$$. The commutator is nonzero only for the terms $${J_z}$$ that have annihilation operators at $${\bf{p}=0}$$. 
…
Taking the commutator with $${a_0^{s\dagger}}$$, the only nonzero term has the structure $${[a_{\bf{p}}^{r\dagger}a_{\bf{p}}^r, a_0^{s\dagger}]=(2π)^3\delta^{(3)}(\bf{p})a_0^{r\dagger}\delta^{rs}}$$; the other three terms in the commutator either vanish or annihilate the vacuum. Thus we find 

$$
J_za_0^{s\dagger}\ket{0}=\frac{1}{2m}\sum_r\left(u^{s\dagger}(0)\frac{\Sigma^3}{2}u^r(0)\right)a_0^{r\dagger}\ket{0}=\sum_r\left(\xi^{s\dagger}\frac{\sigma^3}{2}\xi^r\right)a_0^{r\dagger}\ket{0}
$$

where we have used the explicit form (3.47) of $${u(0)}$$ to obtain the last expression.

Questions

• Why $${J_z}$$ annihilates the vacuum

• Typo

Solution

Jz annihilation of the vacuum

Because $${\ket{0}}$$ has no special direction.

Typo

$$
\begin{array}{l}[J_z, a_0^{s\dagger}]\ket{0}\\=\displaystyle\int d^3x\int\frac{d^3pd^3p'}{(2π)^6}\frac{1}{\sqrt{2E_{\bf{p}}2E_{\bf{p}'}}}e^{i\bf{p}\cdot\bf{x}}e^{-i\bf{p}'\cdot\bf{x}}\sum_{r,r'}\left[(a_{\bf{p}'}^{r'\dagger}u^{r'\dagger}(\bf{p}')+b_{\bf{p}'}^{r'}v^{r'\dagger}(\bf{p}'))\frac{\Sigma^3}{2}(a_{\bf{p}}^ru^r(\bf{p})+b_{\bf{p}}^{r\dagger}v^r(\bf{p})),a_0^{s\dagger}\right]\ket{0}\\=\displaystyle\int\frac{d^3pd^3p'}{(2π)^6}\frac{1}{\sqrt{2E_{\bf{p}}2E_{\bf{p}'}}}(2π)^3\delta^{(3)}(\bf{p}-\bf{p}')\sum_{r,r'}\left[(a_{\bf{p}'}^{r'\dagger}u^{r'\dagger}(\bf{p}')+b_{\bf{p}'}^{r'}v^{r'\dagger}(\bf{p}'))\frac{\Sigma^3}{2}(a_{\bf{p}}^ru^r(\bf{p})+b_{\bf{p}}^{r\dagger}v^r(\bf{p})),a_0^{s\dagger}\right]\ket{0}\\=\displaystyle\int\frac{d^3p}{(2π)^3}\frac{1}{2E_{\bf{p}}}\sum_{r,r'}\left([a_{\bf{p}}^{r'\dagger}a_{\bf{p}}^r,a_0^{s\dagger}]u^{r'\dagger}(p)\frac{\Sigma^3}{2}u^r(p)+[a_{\bf{p}}^{r'\dagger}b_{\bf{p}}^{r\dagger},a_0^{s\dagger}]u^{r'\dagger}(p)\frac{\Sigma^3}{2}v^r(p)+[b_{\bf{p}}^{r'}a_{\bf{p}}^r,a_0^{s\dagger}]v^{r'\dagger}(p)\frac{\Sigma^3}{2}u^r(p)+[b_{\bf{p}}^{r'}b_{\bf{p}}^{r\dagger},a_0^{s\dagger}]v^{r'\dagger}(p)\frac{\Sigma^3}{2}v^r(p)\right)\ket{0}\\=\displaystyle\int\frac{d^3p}{(2π)^3}\frac{1}{2E_{\bf{p}}}\sum_{r,r'}\left((a_{\bf{p}}^{r'\dagger}\{a_{\bf{p}}^r,a_0^{s\dagger}\}-\{a_{\bf{p}}^{r'\dagger},a_0^{s\dagger}\}a_{\bf{p}}^r)u^{r'\dagger}(p)\frac{\Sigma^3}{2}u^r(p)+\cdots\right)\ket{0}\\=\displaystyle\int\frac{d^3p}{(2π)^3}\frac{1}{2E_{\bf{p}}}\sum_{r,r'}\left(a_{\bf{p}}^{r'\dagger}(2π)^3\delta^{(3)}(\bf{p})\delta^{rs}u^{r'\dagger}(p)\frac{\Sigma^3}{2}u^r(p)+0\right)\ket{0}\\=\displaystyle\frac{1}{2m}\sum_{r'}a_{\bf{p}}^{r'\dagger}u^{r'\dagger}(p)\frac{\Sigma^3}{2}u^s(p)\ket{0}\\=\displaystyle\frac{1}{2m}\sum_r\sqrt{m}(\begin{matrix}\xi^{r'}&\xi^{r'}\end{matrix})\left(\begin{matrix}\sigma^3/2&0\\0&\sigma^3/2\end{matrix}\right)\sqrt{m}\left(\begin{matrix}\xi^r\\\xi^r\end{matrix}\right)a_0^{r\dagger}\ket{0}\\=\displaystyle\sum_r\xi^{r'\dagger}\frac{\sigma^3}{2}\xi^ra_0^{r\dagger}\ket{0}\end{array}
$$