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The states in quantum mechanics belong to some Hilbert space while the states in quantum field theory belong to a Fock space. For simplicity, let me stick to the Fock space emerging after the quantization of a real scalar field.

A Fock space is defined as a direct sum, $$\mathcal{F}=\oplus_n\mathcal{H}_n$$ of Hilbert spaces $\mathcal{H}_n$, of physical $n$-particle states.

For a real scalar field, which after quantization (which lead to only one type of particle) the states in $\mathcal{H}_n$, are in general, linear combination of $n$-particle states $\{|p_1,p_2,...,p_n\rangle\}$ of all possible momenta satisfying $p^i_{\mu }p^{\mu i}=m^2$, and $p^0_i>0$.


Questions

What is the physical interpretation of the Fock space being a direct sum of $\mathcal{H}_n$?

It looks like the Fock space has invariant subspaces of labels $n$ where $n\in \mathbb{Z}$. Does it mean that under Poincare transformation, the $n$-particle states, for a given $n$, represent an irreducible representation of the Poincare group i.e., under a Poincare transformation, the states within $\mathcal{H}_n$, for a given $n$, mix among themselves.

If the above interpretation is correct, is it also true that the states in different irreducible representations, for $n\neq m$, are labelled by different values of masses?

Does it also mean that the superposition of states belonging to two different irreducible representations (for example, superposition of a one-particle state with a two-particle state) is forbidden in nature?

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  • $\begingroup$ The said representation are only invariant, but are reducible. The structure of Fock space is the mathematical representation of the fact that states with definite and finite (but arbitrary) number of particles are possible. An infinite tensor product, barring mathematical problems in defining it, would not permit these type of states. $\endgroup$ – Valter Moretti Dec 30 '17 at 15:42
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Questions

What is the physical interpretation of the Fock space being a direct sum of $\mathcal{H}_n$?

I don't know what you are looking for here. This is the definition of a Fock space.

It looks like the Fock space has invariant subspaces of dimensions $n$ where $n\in \mathbb{Z}$. Does it mean that under Poincare transformation, the $n$-particle states, for a given $n$, represent an irreducible representation of the Poincare group i.e., under a Poincare transformation, the states within $\mathcal{H}_n$, for a given $n$, mix among themselves.

Yes, in a free (Gaussian) theory there exists a number operator $N = \sum_k a_k^\dagger a_k$ which can tell you exact number of particles in a state.

If the above interpretation is correct, is it also true that the states in different irreducible representations, for $n\neq m$, are labelled by different values of masses?

It depends on what you call the mass of a state. You could define it as $\sum_i p_i^2 = n m^2$. This mass is measured by the operator $P_1^2 \otimes P_2^2 \otimes \cdots \otimes P_n^2$ where $P_i^2$ is the momentum squared operator acting on the $i$th subspace in ${\cal H}_n$. This "mass" is of course different for $n_1 \neq n_2$.

A better definition is the invariant mass of the state, $P^2 = ( \sum P_i )^2$. The invariant mass can be the same even for $n_1 \neq n_2$.

Does it also mean that the superposition of states belonging to two different irreducible representations (for example, superposition of a one-particle state with a two-particle state) is forbidden in nature?

No, I don't see why it would be forbidden in nature. These would just not have the usual classical interpretation of a bunch of non-interacting moving particles, but in general is a perfectly good state in the quantum theory.

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  • $\begingroup$ The said representations are not irreducible if $n>1$, for this reason there is no definite mass for every such representation as it would be a Casimir operator attaining a constant value. Similarly, for the same reason there is no definite spin value, but a spectrum of possible values. Every said subspace is only invariant but not irreducible. $\endgroup$ – Valter Moretti Dec 30 '17 at 15:38
  • $\begingroup$ @ValterMoretti - That's definitely true (obviously true since for $n=2$ the representation decomposes as ${\cal H}_1 \otimes {\cal H}_1$. I think my brain just skipped the word "irreducible" everywhere in the question. $\endgroup$ – Prahar Dec 30 '17 at 15:39

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