# Interpretation of $\Psi(x)|0\rangle$

Is there an interpretation of the dirac spinor like that of the KG field? I mean, for example, does

$$\Psi(\mathbf{x}) | 0 \rangle$$

create a spin 1/2 particle and an anti-particle at a definite position $\mathbf{x}$?

• – John Rennie May 21 '18 at 4:43

does $Ψ(x)|0⟩$ create a spin 1/2 particle and an anti-particle at a definite position $x$?

That is exactly what it does, but remember that the definition of $\Psi(x)$ contains a creation operator of antiparticles and an annihilation operator for particles. So $\bar{\Psi}$ is what actually creates a particle of spin 1/2 at x.

$\bar{\Psi}(x)|0⟩ = \int \frac{d^3p}{(2\pi)^3} \frac{1}{(2E_p)^{1/2}} \sum_{s=1}^2 a_{s,p}^{\dagger} e^{ipx}\bar{u}^s(p)|0⟩ = \int \frac{d^3p}{(2\pi)^3} \frac{1}{(2E_p)^{1/2}} e^{ipx} \sum_{s=1}^2 \bar{u}^s(p)|p⟩$

is a state describing a particle with certain spin orientation (described by the spinors $\bar{u}^1(p)$ and $\bar{u}^2(p)$) and with momentum $p$. The antiparticle part does not appear here because $b_{s, p} |0⟩ = 0$ (the annihilation operator for antiparticles annihilates the vacuum).

In a similar way, $\Psi(x)|0⟩$ only creates an antiparticle with certain spin orientation at x because $a_{s, p} |0⟩ = 0$.

Note that the states are analogous with the states $\Phi|0⟩$ you get with the $KG$ equation when you apply the field operator to the vacuum. The only difference is that now the state has a spinorial structure, provided by the $\bar{u}^s(p)$ spinors.

• Usually the field operator contains both creation and annihilation operators in two separate terms. So one can write $\Psi(x)=\Psi^+(x)+\Psi^-(x)$. – flippiefanus May 25 '18 at 13:06