# Dirac fields: Do particle and antiparticle creation operators act differently on the vacuum?

Given a Dirac field $$\Psi(x):=\int\frac{d^4k}{(2\pi)^4}\delta\left(p_0-\omega(\mathbf{k})\right)\sum_s\left(a_s(k)u_s(k)e^{-ikx}+b^\dagger_s(k)v_s(k)e^{ikx}\right)$$ with the creation operators $$a^\dagger_s(k),b^\dagger_s(k)$$ for particles and antiparticles respectively, how do these operators act on the vacuum?

In particular, is it true that $$|k\rangle=a^\dagger_s(k)|0\rangle=b^\dagger_s(k)|0\rangle$$?

• physics.stackexchange.com/questions/289088/… does this help? What you've written doesn't look right: the annihilation op on the vacuum state should always give 0. Dec 22 '20 at 10:41
• Oh, that's a typo. Sorry. The post you mention does not address the difference between particle and antiparticle operators when acting on the vacuum, I'm afraid. Dec 22 '20 at 10:55

Ah I think I understand your question now and I think this is a simple notational issue. The single particle states for the particles and antiparticles should be denoted differently, i.e. trying to be as close to your notation would give something like

$$|k,s\rangle \equiv a^\dagger_s(k)|0\rangle \ \ \ \ , \ \ \ \ |\tilde{k},\tilde{s}\rangle \equiv b^\dagger_s(k)|0\rangle \ .$$ And all the usual commutation relations are the same. Perhaps more standard notation would be $$|1_{k}\rangle \equiv a^\dagger_s(k)|0\rangle$$ and $$|\bar{1}_{k}\rangle \equiv b^\dagger_s(k)|0\rangle$$, but I'm not totally sure what's most common.

It is not true that $$a^\dagger_s(k)|0\rangle=b^\dagger_s(k)|0\rangle$$. Moreover, the notation $$|k\rangle$$ is ambiguous. There is the state $$|k,s\rangle =a^\dagger_s(k)|0\rangle$$ containing one particle with momentum $$k$$ and spin state $$s$$ and the state $$|\tilde k,\tilde s\rangle =b^\dagger_s(k)|0\rangle$$ containing one antiparticle with momentum $$k$$ and spin state $$s$$. See e.g. , Section 5.4.

 G.B.Folland, Quantum field theory. A tourist guide for mathematicians, Math.Surveys & Monographs 149, AMS, 2008.

The operator $$a$$ is a particle annihilation operator, while $$b^{\dagger}$$ is an antiparticle creation operator. Acting on the vacuum, $$a_{s}(k)|0\rangle=0$$, but $$b^{\dagger}_{s}(k)|0\rangle\neq0$$. In fact, $$b^{\dagger}_{s}(k)|0\rangle$$ is a one-particle antifermion state (which is not the same as a one-particle fermion state).

The commonality between $$a$$ and $$b^{\dagger}$$ is not that they each create a particle. Rather, they each can decrease the fermion number by $$1$$. (The fermion number is the number of fermions present, minus the number of antifermions—thus zero in the vacuum.) Acting on a one-particle fermion state $$a_{s}(k)|k,s\rangle=|0\rangle$$, annihilating a fermion with momentum $$k$$ and spin $$s$$. The conjugate field $$\Psi^{\dagger}$$ (or $$\bar{\Psi}=\Psi^{\dagger}\gamma_{0}$$) involves $$a^{\dagger}$$, which creates a fermion, and $$b$$, which annihilates an antifermion. Thus, $$\Psi^{\dagger}$$ will increase the fermion number by $$1$$.

• Sorry, I forgot to add a dagger to the particle creation operator. So, the question is about the how both creation operators act differently on the vacuum. Dec 22 '20 at 10:58