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Suppose I have a particle of unspecified spin whose states are determined by a single quantum number $k=1,...,N$. In standard quantum mechanics notation, the state such that the particle is in a superposition of all possible states is given by $$|\varphi\rangle=\frac{1}{\sqrt N}\sum_{k=1}^{N}|k\rangle.$$ Would it make sense to use second quantization to describe the same state? So far I've only seen this formalism when dealing with many-body systems. In this case, it would perhaps look something like this: $$a_1^\dagger|0\rangle+...+a_N^\dagger|0\rangle,$$ if $a$ is the annihilation operator for the particle and $|0\rangle$ the vacuum state. There are several points of which I'm unsure:

  • Can I still use the formalism if, a priori, I don't know the spin of the particle?
  • Can I still add different amplitudes to the formal sum $a_1^\dagger|0\rangle+...+a_N^\dagger|0\rangle$, in the form $\alpha_1a_1^\dagger|0\rangle+...+\alpha_Na_N^\dagger|0\rangle$?
  • Does this make sense at all?
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  • $\begingroup$ A requirement of QFT (/second quantization) is that it be able to reduce to standard QM. As such, I think the idea behind this question should work in theory. See the definition of the field operator in QFT when expressed in ~momentum basis. $\endgroup$
    – PrawwarP
    Dec 7, 2020 at 14:47
  • $\begingroup$ Does the action of the quantised field operators not do this? $$\hat\phi(x)|0\rangle=\int d^3\tilde p\text{ } (a^{\dagger}(p)e^{-ipx}+a(p)e^{ipx})|0\rangle=\int d^3\tilde p\text{ } a^{\dagger}(p)e^{-ipx}|0\rangle.$$ $\endgroup$
    – Charlie
    Dec 7, 2020 at 14:49
  • $\begingroup$ Charlie's formula can also generalize to apply with amplitudes different from plane waves (though, if I recall correctly, they still have to be solutions of Schrodinger equation [although, I may be too restrictive in my understanding of that memory]) $\endgroup$
    – PrawwarP
    Dec 7, 2020 at 14:53
  • $\begingroup$ This proposal seems pretty different to me from the ordinary 2nd quantization scenario. Here we're taking a Hilbert space spanned by N orthogonal states, and trying to recast it as a Hilbert space spanned by 2^N orthogonal states. Unless I'm missing something, how could vector spaces with a different number of finite dimensions be isomorphic to each other? Ordinarily, we're starting with a direct product of a countably infinite number of uncountably infinite-dimensional Hilbert spaces, and just decomposing that into a direct sum of different sectors each with a well-defined particle number. $\endgroup$
    – reductionista
    Dec 10, 2021 at 3:32
  • $\begingroup$ The second one doesn't appear to change the dimensionality, while the first one does. $\endgroup$
    – reductionista
    Dec 10, 2021 at 3:37

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Yes, it is possible to use second quantization for single-particle problems in the way outlined in the question. There are a few points to pay attention to:

  • In most cases this will be redundant, since the formalism is designed expressly for treating many-particle problems, taking into account the statistics of fermions and bosons. But there are exceptions.
  • Depending on the calculation technique, one may have or may not have to pay attention to constraining the total number of particles. For example, equation-of-motion techniques, applying to a particle conserving Hamiltonian, typically pass without problems. However, all kind of statistical averages do require imposing a constraint - there is in fact a number of techniques used to impose such constrains (although normally sued in more complex settings), such as slave-boson approach, drone-fermions, Schwinger boson, etc.
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