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

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?

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.

• $\partial_t U(I,(t,0,0,0))\bigr\rvert_{t=0}$ is the Hamiltonian of the theory. Therefore I don't see how it can be uniquely determined by the spinor/scalar/... nature of the field and the Wightman axiom. Think of a free scalar theory, and an interacting theory with $\lambda\phi^2$ interaction (in order not to bother about well-defineteness). They both obey the axiom, but the $U$s have to be different. Jul 28, 2016 at 12:25
• @yuggib: I allow the mass as an extra input, so that specific interaction doesn't bother me, but I see your point. I'd be happy if there is a recipe for the free theories, then. Jul 28, 2016 at 12:28
• I don't understand some of your words. By the last $\phi$ you meant the field operator? Then $U(\Lambda,a)$ is not a 1-particle representation, it is the Fock representation (field operators acting on the Fock space). In this way, $\rho (\Lambda,a)$ uniquely determines $U(\Lambda,a)$ by dictating how it acts on the 1-particle states.
– chau
Jul 28, 2016 at 12:32
• @ophelia: Yes, $\phi$ is the field - and yes, the Wightman axiom is a priori for the full reducible representation on the full space of states, but it clearly must descend to the irreducible representations sitting inside that full representation, i.e. $UAU^\dagger$ cannot carry an operator $A$ restricted to an irreducible representation to an operator that acts non-trivially outside that representation. Jul 28, 2016 at 12:38
• For free theories this should hold true. I am not an expert, you should see the so-caled Shale-Stinespring theorem (this paper for free boson fields). It surely addresses existence; maybe also uniqueness. Jul 28, 2016 at 12:40

## 1 Answer

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.