There are at least two representations of the Hilbert spaces of quantum field theory. For a scalar field, we have

The Fock space representation, such that every state is represented as the Fock space of 'one particle' representations,

$$\mathcal H = \bigoplus_{n = 0}^\infty \text{Sym}(\mathcal H_1^{\otimes n})$$

with $\mathcal H_1^0 = \mathbb C$ and $\mathcal H_1 = \mathcal L^2(\Bbb R^3)$.

The wavefunctional representation, such that every state is represented by a functional on the space of configurations of the field on $\Bbb R^3$ which is square integrable with respect to some measure

$$\mathcal H = \mathcal L^2(\mathcal F[C(\Bbb R^3)], d\mu[\phi])$$

Since those two representations describe the same theory, I am guessing that there is some isomorphism between the two, but what would it be?

  • $\begingroup$ IIRC, there are subtelties to define the "position representation" in the QFT case, because it's not so easy to come up with a measure on an infinite-dimensional configuration space. In particular, there are no Lebesgue measure (aka translation-invariant measure) on such a space. Also, bcs it's QFT, I expect the $L_2$ spaces associated to different measures to not be unitarily equivalent (in contrast to the finite-dimensional case). $\endgroup$
    – Luzanne
    Commented May 30, 2017 at 12:56
  • $\begingroup$ That being said, using a suitable Gaussian measure (that would correspond to the Fock vacuum), maybe we can get the isomorphism you are looking for. I will look into it if nobody else does... (but I'm on my phone at the moment) $\endgroup$
    – Luzanne
    Commented May 30, 2017 at 13:00
  • $\begingroup$ The vacuum is the function 1, and the rest of the correspondence is built on the field and momentum operators, as illustrated in the Reed Simon book, Vol. II, p.228-230. $\endgroup$
    – yuggib
    Commented May 30, 2017 at 13:52
  • $\begingroup$ And yes @Luzanne , for free fields the corresponding measure is the Gaussian on $\mathbb{R}^{\infty}$ (the topological space given by the countable product of $\mathbb{R}$ with the product topology). For interacting fields, the representation is usually non-Fock, and we usually look for a representation of the type $L^2(\mathscr{S}'(\mathbb{R}^d),\mu)$, for a suitable probability measure $\mu$ on the distributions. This representation is successfully built only in very few situations, see the Glimm-Jaffe book for more details. $\endgroup$
    – yuggib
    Commented May 30, 2017 at 14:03
  • $\begingroup$ @yuggib I see. Yes, the construction you just outlined is what I suspected: with a measure that matches our vacuum, clearly the latter will be represented as the constant $1$, and we can reconstruct all other states by aplying the right operators, à la GNS construction. I presume the field conjugate momentum has to be represented as derivation + an extra divergence term to compensate for the measure being non-translation-invariant? $\endgroup$
    – Luzanne
    Commented May 30, 2017 at 14:52

1 Answer 1


The two representations are not equivalent, except for free fields, and hence in perturbation theory.

Nonperturbatively, the Fock space representation is inappropriate (because of Haag's theorem, which asserts that the Hilbert space of an interacting relativistic quantum field theory has no natural Fock space structure; see the discussions at Haag's theorem and practical QFT computations and State space of interacting theories) whereas the wave functional representation (i.e., the functional Schroedinger picture) may still work - it can capture a lot of nonperturbative information such as instantons.

In the functional Schroedinger picture, states of a QFT are treated as functionals of the field coordinates in the same way as states in QM are treated as functions of the position coordinates. A thorough discussion of the functional Schroedinger picture is in the article by Jackiw, Analysis on infinite dimensional manifolds: Schrodinger representation for quantized fields (p.78-143 of the linked document). A problem is that the infinite-dimensional measure is not really well-defined, it is given formally (i.e., nonrigorously) by a functional integral.


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