# Spin of single particle state of free Dirac Field

## Homework Statement

Show that the state $d^{\dagger}_{\alpha}(0)\mid 0\rangle$ describes a postrion at rest by showing that it is an eigenstate of the operators $P^{\mu}, Q, J^z$.

## Homework Equations

The fourier expansion of $\psi, \psi^{\dagger}$:
$$\psi = \int \frac{d^3k}{(2\pi)^3} \frac{m}{k_0} \sum_{s}\left(b_{s}(k)u_{s}(k)e^{-ikx} + d_s^{\dagger}(k)v_s(k)e^{ikx}\right)$$
$$\psi^{\dagger} = \int \frac{d^3k}{(2\pi)^3} \frac{m}{k_0} \sum_{s}\left(b_{s}^{\dagger}(k)u_{s}^{\dagger}(k)e^{ikx} + d_s(k)v_s^{\dagger}(k)e^{-ikx}\right)$$

## The Attempt at a Solution

I have shown without porblem that the state is an eignestate of $P^{\mu}$ and $Q$, however I am stuck at showing the same for the momentum.

I know how to get, based on the conserved noether current, that the operator $J^z$ for the Dirac Lagrangian can be written as:

$$J^z = \int d^3x \psi^{\dagger}(x\partial_y - y\partial_x + \frac{1}{2}\Sigma^3)\psi$$

However, if I expand $\psi, \psi^{\dagger}$ like I did for the momentum and charge, I get nowere since the way $\Sigma^3$ acts on $u_s(k), v_s(k)$ is not simple.

I have searching in books and have found that Peskin (page 61) provides some arguments to perform de calculation. He argues that the orbital angular moementum for the particle at rest is zero, and the only term that is left is the spin. Then he argues that since $J^z \mid 0 \rangle$ must be zero, then $J^z d^{\dagger}_{\alpha}(0) \mid 0 \rangle = [J^z, d^{\dagger}_{\alpha}(0)] \mid 0 \rangle$. Then, he continues, the only nonzero term is the commutator $[d(p) d^{\dagger}(p), d^{\dagger}(0)]$. However, I do not understand how he gets to this last step. Why doesn't $\Sigma^3$ appear between the creator and destructor operators in this commutator ? As far as I understand the spin operator shouldn't commute with the creator/annihilator operators.

Any help in understanding Peskin's arguments or any other way to achieve the same result is greatly appreciated.

Thanks,

Fedcer

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