Spin of single particle state of free Dirac Field

[Fedcer]

Homework Statement

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

Homework Equations

The Fourier expansion of [itex] \psi, \psi^{\dagger}[/itex]:
[tex]
\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)
[/tex]
[tex]
\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)
[/tex]

The Attempt at a Solution

I have shown without porblem that the state is an eignestate of [itex] P^{\mu} [/itex] and [itex] Q [/itex], 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 [itex] J^z [/itex] for the Dirac Lagrangian can be written as:

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

However, if I expand [itex] \psi, \psi^{\dagger} [/itex] like I did for the momentum and charge, I get nowere since the way [itex] \Sigma^3 [/itex] acts on [itex] u_s(k), v_s(k) [/itex] 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 [itex] J^z \mid 0 \rangle [/itex] must be zero, then [itex] J^z d^{\dagger}_{\alpha}(0) \mid 0 \rangle = [J^z, d^{\dagger}_{\alpha}(0)] \mid 0 \rangle [/itex]. Then, he continues, the only nonzero term is the commutator [itex] [d(p) d^{\dagger}(p), d^{\dagger}(0)] [/itex]. However, I do not understand how he gets to this last step. Why doesn't [itex] \Sigma^3 [/itex] 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

 

[mark726]

Dear Fedcer,

Your approach is on the right track, but there are a few things that need clarification. First, the state d^{\dagger}_{\alpha}(0) \mid 0\rangle does not describe a particle at rest, but rather a particle with no momentum. This means that the particle is allowed to have any energy, and thus the state is not an eigenstate of the energy operator H . However, it is an eigenstate of the momentum operator P^{\mu} , as you have shown.

To show that the state is also an eigenstate of the spin operator J^z , we need to consider the spin part of the current J^z . As you correctly stated, the spin operator acts on the spinor fields u_s(k), v_s(k) in a nontrivial way. However, we can use the equations of motion for these fields to simplify the expression:

\partial_{\mu}\Sigma^{\mu\nu}\psi = 0 \hspace{1cm} and \hspace{1cm} \partial_{\mu}\Sigma^{\mu\nu}\psi^{\dagger} = 0Using these equations, we can show that the spin part of the current J^z can be written as:

J^z = \int d^3x \psi^{\dagger}(x\partial_y - y\partial_x + \frac{1}{2}\Sigma^3)\psi = \int d^3x \psi^{\dagger}(x\partial_y - y\partial_x)\psiThis simplification allows us to easily calculate the commutator [J^z, d^{\dagger}(p)] , which gives only a spin operator acting on the state d^{\dagger}(p)\mid 0\rangle . Then, using the fact that the spin operator is diagonal in the spin basis, we can see that J^z d^{\dagger}_{\alpha}(0)\mid 0\rangle is indeed an eigenstate of J^z .

I hope this helps clarify the steps in Peskin's argument. Please let me know if you have any further questions.