17
$\begingroup$

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.

$\endgroup$
8
  • $\begingroup$ $\partial_t U(I,(t,0,0,0))\bigr\rvert_{t=0}$ 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. $\endgroup$
    – yuggib
    Jul 28 '16 at 12:25
  • $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Jul 28 '16 at 12:28
  • $\begingroup$ 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. $\endgroup$
    – chau
    Jul 28 '16 at 12:32
  • $\begingroup$ @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. $\endgroup$
    – ACuriousMind
    Jul 28 '16 at 12:38
  • $\begingroup$ 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. $\endgroup$
    – yuggib
    Jul 28 '16 at 12:40
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.