My first introduction to second quantization was in the context of condensed matter physics. The idea is if we have a system of $N$ indistinguishable particles then the $N$ fold tensor product of 1 - particle Hilbert spaces is unnecessarily big to describe our system because it contains states with no exchange symmetry.


Consider a free theory of bosons, the (rigged) Hilbert space of 1 - boson looks like $\mathcal{H}_1 = span_\mathbb{C}\{a^{\dagger}_{{p}}|0\rangle \space | p \in \mathbb{R}^3 \}$ and the tensor product space will be denoted $\mathcal{H} = \mathcal{H}_1 \otimes \mathcal{H}_1$.

A generic state might look like $|\Psi \rangle = a_p^{\dagger}|0 \rangle \otimes a_q^{\dagger} |0 \rangle $. This state is not symmetric under exchange of the 1st and 2nd particle.

This is dealt with in the language of second quantization. In second quantization the symmetrized version of the state is $|p,q \rangle = a_p^{\dagger} a_q^{\dagger} |0 \rangle $ as the ladder operators in this formalism commute so $|p,q \rangle = a_p^{\dagger} a_q^{\dagger} |0 \rangle = a_q^{\dagger} a_p^{\dagger} |0 \rangle = |q,p \rangle $. This makes sense to me.

The problem arises when we deal with multiple quantum fields (in either QFT or condensed matter - just whenever we have a mix of several species) and so we have several species of particles with are distinguishable from eachother. Let $a^{\dagger}_p$ create a boson of species 1 (with definite 3 momentum $p$) and $b^{\dagger}_p$ creates a boson of species 2. In QFT we would say these operators commute, and a general state is written as (a linear combination of) $|\psi \rangle = a_p^{\dagger} b_q^{\dagger} |0 \rangle $ (a). However, this formalism is not general enough to describe every possible physical state, due to the commutation of the ladder operators any state of the form (a) will be symmetric under the exchange of species 1 and 2 when it need not be. This argument extends to fermions too, where we say the fermion ladder operator commutes with all other boson ladder operators and anticommute with fermions (of a different species).

This doesn't make sense to me, so I must be misinterpreting the role of second quantization in QFT, any help would be appreciated!

  • 3
    $\begingroup$ This state $|\psi\rangle$ is not symmetric under the exchange of $1$ and $2$: it becomes $a_q^\dagger b_p^\dagger|0\rangle$. $\endgroup$
    – Meng Cheng
    Jan 7, 2022 at 21:41


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