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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 $|k\rangle=a^\dagger_s(k)|0\rangle=b^\dagger_s(k)|0\rangle$?

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    $\begingroup$ 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. $\endgroup$
    – Eletie
    Dec 22 '20 at 10:41
  • $\begingroup$ 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. $\endgroup$ Dec 22 '20 at 10:55
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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.

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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. [1], Section 5.4.

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

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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 $1$.

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    $\begingroup$ 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. $\endgroup$ Dec 22 '20 at 10:58

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