In this question, when I write Fock space, I mean "the direct sum of the symmetric or antisymmetric tensors in the tensor powers of a single-particle Hilbert space H", as it is described by Wikipedia.

When I began with many-particle-QM in my college-course, we constructed the many-particle space as a tensorproduct of single-particle-states, and then we constructed operators that act on this many-particle-space. From that point of view, it was natural to use the fock space as the space that constains the many-particle-states.

As I understand Quantum Field Theory, everything I do is, I take a classical field theory, and try to quantize it, which means I replace the field variables (like $\phi(x,t)$) with an Operator, then I set the Operator's commutation-relations in a way that makes sense. Afterwards I write the Hamiltonian down, and express it as a sum of some operators, which I afterwards interpret as "Creation and Annihilation"-operators, the create excitation of the quantum field. We then call this excitation "one particle".

My Question: Is the space that the operators act uppon a Fock-state by definition? Or is it just "any Hilbertspace"? If it's the latter: Can we interpret the Hilbertspace to be a Fock-space, after we work with the annihilation and creation operators? For example, If I found some strange EM-Field-space in QED, can I express it as the sum of tensorproducts of "single-particle-photon"-states?

More generaly (and that is the core of my question): Is there something like a "single-particle" in QFT? While in many-particle-QM, the single-particle-QM is the basis, QFT doesn't rely on this concept, that's why I'm asking.

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    $\begingroup$ The Fock space formalism only works for very specific theories (mostly free theories). Interacting theories should not, if exactly solved, have a Fock space for their Hilbert space. $\endgroup$
    – Slereah
    Jun 23, 2016 at 12:16

1 Answer 1


It is not a postulate that the Hilbert space for QFT is a Fock space. In fact, for interacting theories is often almost surely not a Fock space. The requirements for a Hilbert space to be the space of a QFT is that the Wightman axioms are satisfied.

For free theories, a suitable Fock representation of the canonical commutation relations satisfies the Wightman axioms (in the sense that there exist a vacuum state, the unitary representation of the Poincaré group etc. in the corresponding Fock space); and this is why the Fock space is commonly introduced - also its definition is in my opinion quite intuitive. The other representations used in interacting theories, when known, are rather abstract, and it may even not be clear if they are set in a Fock space or not (almost surely not, for interesting theories).

Let me expand the ideas a bit more, related to the last part of your question. When we write the canonical commutation relations for quantum fields, it is customary to start from a suitable "one-particle" space $\mathfrak{h}$ (at least it was called like that by Irving Segal). Such space is e.g. the Hilbert space on which you build the Fock space $\Gamma_{s/a}(\mathfrak{h})$ (as direct sum of tensor products), and it encodes roughly speaking the properties of the related classical field (scalar/vector/spinor field,...). However, contrarily to what happens in quantum mechanics, there are infinitely many unitarily inequivalent irreducible representations of the canonical commutation relations $\mathrm{C}^*$-algebra built on such one-particle space $\mathfrak{h}$. One of those representations is the Fock one, the others still encode the commutation relations of the creation/annihilation operators $\{a^{\#}(h),h\in\mathfrak{h}\}$, but such operators may not behave as in the Fock representation: in particular, the intersection of their domains of definition $\bigcap_{h\in\mathfrak{h}}D(a(h))$ may fail to be a dense subset of the Hilbert space.

  • $\begingroup$ Thanks for the answer :). What exactly do you mean by an interacting theory? Would the Dirac-Theory allready be an interacting theory, since it deals with the EM-Field and the Dirac-Field? $\endgroup$ Jun 23, 2016 at 14:23
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    $\begingroup$ @Quantumwhisp By interacting theory I mean any theory representing a field (or many fields) in interaction. A simple example is the scalar $\varphi^4$ model, given by the lagrangian density $\mathcal{L}(\varphi)= \partial^\mu\varphi\partial_\mu\varphi - m^2\varphi^2-\lambda \varphi^4$. Also any theory of fermion fields interacting with a scalar or vector field are, well, interacting of course ;-) $\endgroup$
    – yuggib
    Jun 23, 2016 at 14:33
  • $\begingroup$ Shoudn't the elementary excitations of the ground state form a fock space also for interacting theories (for QED however there is a subtlety called infraparticle) ? $\endgroup$
    – jjcale
    Jun 23, 2016 at 19:06
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    $\begingroup$ 1) A non-Fock representation is e.g. a representation of the canonical commutation relations that is neither the usual one in the Fock space $\Gamma_s(\mathscr{H})$ given by $e^{i\varphi(f)}$ and $e^{i\pi(f)}$, with $\varphi(f)=a^*(f)+a(f)$ and $\pi(f)=a^*(f)-a(f)$; nor unitarily equivalent to that (and there are infinitely many inequivalent irreducible ones if $\mathscr{H}$ is infinite dimensional). It is true that every separable Hilbert space is isomorphic, but that is hardly useful when considering representations of the canonical commutation relations. $\endgroup$
    – yuggib
    Jun 25, 2016 at 7:23
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    $\begingroup$ Concerning 2), even if what you claim is correct (maybe it is, I don't know ;-) ), it is far from assured that the renormalized Hamiltonian, whatever it would be, have a non-empty discrete spectrum. $\endgroup$
    – yuggib
    Jun 25, 2016 at 7:29

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