Wightman Axioms for free scalar quantum field
Article last updated: January 29, 2026
I’m writing this article so that I can learn Theorem 3.1 of Prof. Christian Brennecke’s constructive quantum field theory notes Brennecke (2026). The statement of the theorem is as follows.
Consider the strongly continuous unitary representation \(\Gamma(U)\) (gamma of U) of \(\mathcal{P}_{+}^{\uparrow}\) (P plus up) on \(\mathcal{H} = \mathcal{F}_s(L^2(S_m^+))\) (Bosonic Fock space built over \(L^2\) of the mass shell of mass \(m\)) with Fock space vacuum \(\Omega = (1,0,0,\cdots) \in \mathcal{H}\) and let \(\phi = (\phi(f))_{f \in \mathcal{S}(\mathbb R^4)}\) (family of operators indexed by Schwartz functions on \(\mathbb R^4\)) be defined as (3.14): \[\begin{equation} \phi(f) = (2\pi)^{1/2} \int_{S_m^+} \lambda_m(dp) \tilde{f}(p) a_p + (2\pi)^{1/2} \int_{S_m^+} \lambda_m(dp) \tilde{f}(p) a_p^*. \end{equation}\] Then \((\mathcal{H}, \Gamma(U), \Omega, \phi)\) satisfies the Wightman axioms and \[\begin{equation} (\square + m^2) \phi = 0 \end{equation}\] in the sense of operator-valued distributions.
The operator valued distribution (3.14)
Let me clarify (3.14) for myself. Recall that the creation and annihilation operators depend on some function \(f \in L^2(\mathcal{M})\), where in this case \(\mathcal{M} = S_m^+\). Only once you have some \(f\) can you talk about an operator from the infinite tensor product space \(\mathcal{F}_s\) to itself. That is why I can write \[\begin{equation} a(\sqrt{ 2 \pi } \tilde{f} ) : \mathcal{F}_s \to \mathcal{F}_s, \end{equation}\] and this object is defined in section 2.4 in Brennecke (2026).
Next, I want to phrase (3.14) directly in terms of rigorously defined objects. I’ll use the identities \[\begin{equation} a(f) =: \int_{\mathbb R^d} dx \overline{f}(x) a_x \end{equation}\] and \[\begin{equation} a^*(g) =: \int_{\mathbb R^d} dx g(x) a_x^* \end{equation}\] found in section 2.4 of Brennecke’s notes. It follows that \[\begin{equation} (2\pi)^{1/2} \int_{S_m^+} \lambda_m(dp) \tilde{f}(p) a_p := a(\sqrt{2\pi} \overline{\tilde{f}}). \end{equation}\] This is nice because now I can just refer back to section 2.4 to see how \(a(\sqrt{2\pi} \overline{\tilde{f}})\) acts on a tensor product on some dense subset in the Fock space.
In summary, (3.14) says \[ \phi(f) = \sqrt{2 \pi } a(\overline{\tilde{f}}) + \sqrt{2 \pi } a^*( \tilde{f}). \] (where the constant factors of pi can be taken in and out of the annihilation and creation operators because \(a\) and \(a^*\) are linear in \(f \in L^2(\mathcal{M})\).)
Let me also discuss the name “operator valued distribution.” An operator valued distribution is a family \((\phi(f))_{f \in \mathcal{S}(\mathbb R^4)}\), where for each \(f \in \mathcal{S}(\mathbb R^4)\), the symbol \(\phi(f)\) denotes a linear operator on some Hilbert space \(\mathcal{H}\). But how does this connect to another notion of a distribution, namely a map \(\Phi: \mathcal{S}(\mathbb R^4) \to \mathbb C\), with the stereotypical example being the dirac distribution \(\delta_0: \mathcal{S}(\mathbb R^4) \to \mathbb C\) centered at \(0 \in \mathbb R^4\)? To see the connection, observe that we can rewrite the quantum field \((\phi(f))_{f \in \mathcal{S}(\mathbb R^4)}\) as a map: \[ \Phi: \mathcal{S}(\mathbb R^4) \to \mathcal{L}(\mathcal{H}). \] Rewriting in this way allows me to see that a quantum field is just a distribution, but with values in \(\mathcal{L}(\mathcal{H}) = \{f: \mathcal{H} \to \mathcal{H} \mid f \text{ linear} \}\) instead of \(\mathbb C\).
On this note, one situation that happens a lot is the following. One writes down an integral \[ \int_{\mathbb R^4}dp f(p) a^*_p, \] and then says that the integral should be interpreted in a distributional sense (because the annihilation operator is not actually defined at each \(p\), and even if it were, the vector valued integral would be difficult to make sense of because the annihiliation operators are not bounded operators, so a Bochner interpretation is not immediate.) How is this connected to intepreting an analogous integral \[ \int_{\mathbb R^4} dp f(p) \delta_0(p) \] in the sense of distributions? Well, the latter integral is actually defined by \[ \delta_0(f) =: \int_{\mathbb R^4} dp f(p) \delta_0(p), \] because the delta function is not defined at the origin, and you can’t ignore the singularity at the origin (as you would for \(L^p\) functions by ignoring null sets) because the null set \(\{0\}\) actually affects the value of the integral. The main point is this, we’ve intepreted the original integral in a distributional sense, because \(\delta_0(f)\) is a well-defined complex number: it is just \(\delta_0(f) = f(0) \in \mathbb C\). To see the analogy for the operator valued distributional case, just replace the \(\delta_0\) by \(a\) on the left hand side and the \(\delta_0(p)\) inside the integral on the right hand side by \(a^*_p\). Then: \[ a(f) =: \int_{\mathbb R^4} dp f(p) a^*_p, \] so the right hand side is defined by \(a(f)\), which is no longer a complex number, but an operator. This explains how the interpretations of the identities used above from section 2.4 of Brennecke (2026) are called ‘distributional interpretations.’
The unitary representation Gamma
Consider the phrase “strongly continuous unitary representation \(\Gamma(U)\) of \(\mathcal{P}_+^\uparrow\) on \(\mathcal{H} = \mathcal{F}_s(L^2(S_m^+))\).” We understand this phrase through the following discussion. Given an abstract group \(G\) and a vector space \(V\), a unitary representation (of \(G\) on \(V\)) is a homomorphism \(\rho: G \to \mathcal{U}(V)\), where \(\mathcal{U}(V)\) denotes the set of all unitary maps from \(V\) to \(V\), equipped with composition as the multiplication operation (or matrix multiplication). What is a natural topology to put on \(\mathcal{U}(V)\)? Perhaps an operator topology?
Equipping \(G\) with a topology, we say that \(\rho\) is strongly continuous if for all \(v \in V\) \[G \ni g \mapsto \rho(g)v \in V\] is continuous. For example, if \(G = SU(2)\) and \(\iota : SU(2) \to U(2)\) is a strongly continuous unitary representation of \(SU(2)\) on \(\mathbb C^2\), then strong continuity means \[SU(2) \ni A \mapsto Av \in \mathbb C^2\] is continuous for all \(v \in \mathbb C^2\).
Thus, \(\Gamma(U): \mathcal{P}_+^\uparrow \to \mathcal{U}(\mathcal{F}_s(L^2(S_m^+)))\) should satisfy \[\mathcal{P}_+^\uparrow \ni (a,L) \mapsto \Gamma(U)((a,L))v \in \mathcal{F}_s(L^2(S_m^+))\] is continuous, where \(v\) is a fixed and arbitrary element of the Bosonic Fock space built over the mass shell. Observe that \(\Gamma(U)((a,L))\) is a unitary operator on Bosonic Fock space, so it makes sense that applying it to \(v\) gives a Bosonic function. Observe that \(\Gamma(U)\) is not a finite dimensional representation because Fock space is not finite dimensional.
It turns out that the unitary representation Gamma actually has a name: the second quantization of the unitary representation \(U= (U(a,L))_{(a,L) \in \mathcal{P}_+^{\uparrow}}\). Thus, completely understanding \(\Gamma(U)\) entails understanding the representation \(U\) of the proper Poincaré group \(\mathcal{P}_+^{\uparrow}\) on \(L^2\) of the mass shell \(S_m^+\), which can be seen in definition (3.4) in the section on the quantization of a massive relativistic particle. To this end, let me look at the following exercise, which is Probem 3.1 (a) in Brennecke (2026).
For \((a,L) \in \mathcal{P}_+^{\uparrow}\), let \(U(a,L)\) be defined as in (3.4): for \(\psi \in L^2(S_m^+, \mathcal{B}(S_m^+), \lambda_m)\), we set \[ (U(a,L)\psi)(p) = e^{ip^{\mu}a_{\mu}} \psi(L^{-1}p) \] for almost every \(p \in S_m^+\) and for all \((a,L) \in \mathcal{P}_+^{\uparrow}\). Show that \(\mathcal{P}_+^\uparrow \ni (a,L) \mapsto U(a,L) \in \mathcal{U}(L^2(S_m^+))\) defines a strongly continuous, unitary representation of \(\mathcal{P}_+^\uparrow\) on \(L^2(S_m^+) = L^2(S_m^+, \mathcal{B}(S_m^+), \lambda_m)\).
Proof. Step one is to verify the homomorphism property \[ (U((a,L)(a',L'))\psi)(p) = ((U(a,L)U(a',L'))\psi)(p) \tag{1}\] for all \((a,L), (a',L') \in \mathcal{P}_+^\uparrow\). By definition, \((a,L)(a',L') = (a+La', LL')\), so we make the appropriate substitution on the left hand side of Equation 1. Then we have by (3.4) that \[\begin{align*} (U(a+La', LL')\psi)(p) = e^{ip^\mu(a+La')_{\mu}}\psi((LL')^{-1}p). \end{align*}\] We compute for the right hand side using (3.4): \[\begin{align*} ((U(a,L)U(a',L'))\psi)(p) &= (U(a,L)(U(a',L')\psi(\cdot)))(p) && \text{i.e. composition law}\\ &= (U(a,L)e^{i(\cdot)^\mu a'_{\mu}}\psi(L'^{-1} (\cdot)) )(p) \\ &= e^{ip^\mu a_{\mu}} e^{i(L^{-1}p)^\mu a'_{\mu}} \psi(L'^{-1}L^{-1}(p)) \\ &= e^{i (p^\mu a_\mu + (L^{-1}p)^\mu a'_\mu)} \psi((LL')^{-1}p). \end{align*}\] For the homomorphism property, it remains to prove that \[ p^\mu a_\mu + (L^{-1}p)^\mu a'_\mu = p^\mu(a+La')_{\mu}. \] Note the following facts and observations about
Interlude about O(1,3)
\(O(1,3)\), the group of isometries on Minkowski space, of which \(L\) is an element by defintion of the Poincaré group. First, Minkowski space \(\mathbb R^4\) is equipped with a bilinear form \(\eta = \text{diag}({1,-1,-1,-1})\), which means that \(\eta(x,y) = x^T \eta y\), where \(x\) and \(y\) are column vectors with 4 entries. Next, an isometry \(L \in O(1,3)\) is a linear map \(\mathbb R^4 \to \mathbb R^4\) that preserves the geometry in the sense that \(\eta(Lx,Ly) = \eta(x,y)\) for all \(x,y \in \mathbb R^4\). Rewriting gives the equation \(x^TL^T\eta Ly = x^T\eta y\) for all \(x\) and \(y\), and by choosing \(x\) and \(y\) as the various combinations of basis vectors in order to single out the matrix entries of \(L^T\eta L\) on the left and \(\eta\) on the right, the isometry condition on \(L\) can be expressed equivalently as \(L^T \eta L = \eta\). Another useful thing to note is that \(a^\mu b_\mu = a^T \eta b\) for column vectors \(a\) and \(b\) in \(\mathbb R^4\).
End of Interlude, back to the proof.
Let’s rewrite the equation \(p^\mu a_\mu + (L^{-1}p)^\mu a'_\mu = p^\mu(a+La')_{\mu}\) using \(\eta\) as \[ p^T \eta a + (L^{-1}p)^T \eta a' = p^T \eta (a+La'). \] So we want to prove that this equation holds, given that \(L \in O(1,3)\) and that \(p \in S_m^+\). By linearity and cancellation, the equation becomes \[ (L^{-1}p)^T \eta a' = p^T \eta La'. \] But using that \(L^{-1} \in O(1,3)\), which implies that \((L^{-1})^T \eta L^{-1} = \eta\), which implies \((L^{-1})^T \eta = \eta L\), gives confirmation. But why is \(L^{-1} \in O(1,3)?\) Well, \(L \in O(1,3)\) implies \(L^T \eta L = \eta\), which implies that \(\eta^T = L^T \eta^T L\) because \(\eta\) is symmetric, which implies that \(\eta^T L^{-1} = L^T \eta^T\), which implies that \((L^{-1})^T \eta = \eta L\) after transposing both sides, which implies that \((L^{-1})^T \eta L^{-1}= \eta\), which implies that \(L^{-1} \in O(1,3)\). This concludes step one, the homomorphism property of the representation.
Next, I’d like to check the unitary property, which states that for all \((a,L) \in \mathcal{P}_+^\uparrow\), that \(U(a,L)\) is a unitary operator on \(L^2(S_m^+)\), which means that \[ \langle \psi, \phi \rangle_{L^2(S_m^+)} = \langle U(a,L) \psi, U(a,L) \phi \rangle_{L^2(S_m^+)} \] for all \(\psi, \phi \in L^2(S_m^+)\).
I’m not super sure if functions in \(L^2(S_m^+)\) are supposed to have a codomain of \(\mathbb C\), but I’ll guess that for now based on the imaginary \(i\) in the exponent in the formula (3.4).
\[\begin{align*} \langle U(a,L) \psi, U(a,L) \phi \rangle_{L^2(S_m^+)} &= \int_{S_m^+} d\lambda_m(p) \overline{U(a,L)\psi} (p) U(a,L) \phi (p) \\ &= \int_{S_m^+} d \lambda_m(p) \overline{e^{ip^\mu a_\mu} \psi (L^{-1}p)} e^{ip^\mu a_\mu} \phi(L^{-1}p) \\ &= \int_{S_m^+} d \lambda_m(p) e^{-ip^\mu a_\mu} \overline{ \psi (L^{-1}p)} e^{ip^\mu a_\mu} \phi(L^{-1}p) \\ &= \int_{S_m^+} d \lambda_m(p) \overline{ \psi (L^{-1}p)} \phi(L^{-1}p) \\ &= \int_{L^{-1}S_m^+} d (\lambda_m \circ L)(h) \overline {\psi(h)} \phi(h) && \text{ change of variables }\\ &= \int_{S_m^+} d \lambda_m (h) \overline {\psi(h)} \phi(h) && \text{ Lemma 3.1 of Brennecke's notes} \\ &= \langle \psi, \phi \rangle_{L^2(S_m^+)}. \end{align*}\]
I still need to learn about why \(L^{-1}S_m^+ = S_m^+\).