Do the Wightman axioms uniquely fix the representation of the Poincaré group on the one-particle states given the representation on the fields?

Question

Let $P := \mathrm{SL}(2,\mathbb{C})\ltimes \mathbb{R}^4$ be the universal cover of the connected component of the identity of the Poincaré group.

Given a classical field $\phi : \mathbb{R}^{1,3}\to V$ where $V$ carries a finite-dimensional irreducible representation $\rho : P\to\mathrm{GL}(V)$ (e.g. spinor, vector, (p,q)-tensor) and the irreducible unitary representation $U : P\to \mathrm{U}(H)$ where $H$ is the space of one-particle states associated to the field, the Wightman axioms impose that $$ \rho(\Lambda,a)\phi(\Lambda^{-1}x+a) = U(\Lambda,a)\phi(x)U(\Lambda,a)^\dagger$$ holds as an operator equation on the space of states.

Does this equation uniquely determine $U$ given $\rho$? If yes, how? If no, how do we know which one to choose? If this does not work for arbitrary fields, is there at least a recipe for free fields?

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I'll now describe what thoughts I have so far:

For the massive case I believe the answer is "natural": By Mackey's theory of induced representations, the unitary irreducible representations of $P$ are given by choosing an element of $\alpha\in\mathbb{R}^4$ and a unitary irreducible representation of the stabilizer of $\alpha$ in $\mathrm{SL}(2,\mathbb{C})$ together with a unitary character $\mathbb{R}^4\to\mathrm{U}(1)$, which I think is usually just 1 in physics. The stabilizer is known as the "little group" leaving the momentum of a particle in its rest frame invariant. For massive particles, this is $\mathrm{SU}(2)$, and since the finite-dimensional representations of $\mathrm{SO}(1,3)$ in which the fields transform are given by representations of $\mathrm{SU}(2)\times\mathrm{SU}(2)$, i.e. half-integers $(s_1,s_2)$, we choose the representation $s_1+s_2$ of $\mathrm{SU}(2)$ to induce the full representation.

For the massless case, it becomes less natural - the little group is $\mathrm{ISO}(2) = \mathrm{SO}(2)\ltimes\mathbb{R}^2$, which has one-dimensional representation (labeled by the value of "helicity") as well as infinite-dimensional irreducible representations. The latter, I believe, are the "continuous spin representations" usually not occuring in physics. But what determines which one-dimensional representation to choose? The photon usually gets the reducible representation that is a sum of the representations of helicity +1 and -1, since parity interchanges them - can this be seen simply from the finite-dimensional representation of the field and knowing it's massless?

For the tachyonic case, I'm completely at a loss. The little group is $\mathrm{SU}(1,1)\cong\mathrm{SL}(2,\mathbb{R})$, which has a plethora of rather complicated unitary irreducible representations given by Bargmann's classification. I see no way to connect the representation of the field with any of these representations.

Answer 1

Yes it does if adding the invariance requirement of the vacuum state stated in the Wightman axioms. That is a general result valid as a consequence of the GNS construction.

Let us start from the fact that the algebra of observables ${\cal A}$ is the (unital) $^*$-algebra of finite linear combinations of finite products of smeared field operators $\phi(f)$.

The action of Poincaré group induces a $^*$-algebra representation $SO(1,3)_+\times \mathbb{R}^4 \ni g \mapsto \alpha_g : {\cal A} \to {\cal A}$ completely defined by requiring that $$\alpha_g(\phi(f)) := \rho_g \phi(g(f))$$ and extending it to the whole algebra by imposing that it is linear, preserves the adjoint and the products (and the unit), where $g(f)$ is the standard action of the group on test functions.

Let us indicate the vacuum state by $\Omega$. The above representation has the property that, according to Wighhman axioms, it leaves invariant the $n$-point functions of the vacuum state $$\langle \Omega, \phi(f_1) \cdots \phi(f_n) \Omega \rangle = \langle \Omega, \alpha_g(\phi(f_1)) \cdots \alpha_g(\phi(f_n)) \Omega \rangle\:, $$ which extends to $$\langle \Omega, A\Omega \rangle = \langle \Omega, \alpha_g(A) \Omega \rangle \tag{1}$$ for a generic element $A\in {\cal A}$ and every $g \in SO(1,3)_+\times \mathbb{R}^4$.

Using the fact that $\phi({\cal A}) \Omega$ is dense in the Hilbert space (again as a consequence of Wightman axioms) and (1), it is not difficult to prove that

there is only one unitary representation of $SO(1,3)_+\times \mathbb{R}^4$ satisfying, for every $g\in SO(1,3)_+\times \mathbb{R}^4$ and every test function $f$, $$U_g A U_g^{-1} := \alpha_g(A)\:, \quad U_g \Omega = \Omega$$ where $A \in {\cal A}$, in particular $$U_g \phi(f) U^{-1}_g = \alpha_g(\phi(f))\:.$$ The representation $U$ is completely defined by requiring $$U_g A \Omega := \alpha_g(A)\Omega$$ for every $g\in SO(1,3)_+\times \mathbb{R}^4$ and $A\in {\cal A}$.

If the state $\Omega$ is Gaussian, i.e., its Wightman functions are determined by the two-point function using the Wick rule (and the $1$-function vanishes), then the Hilbert space is a Fock one and it is not difficult to prove that the above representation leaves invariant the one-particle space (and in general the $n$-particle space).

In this case, a well defined unitary representation arises in the one-particle space just by restricting $U$ to it.

However, the above result is valid also if $\Omega$ is not Gaussian.

The result is completely general, we only need a $^*$-representation of a group on the $*$-algebra of observables generated by the smeared field operators and a state determined by its Wightman functions which are invariant under the action of the group. This automatically determines a unique unitary representation of the group leaving the vector state invariant and which implements the action of the group.