Wave Operators and Asymptotic Fields

Haag–Ruelle limits assemble into incoming and outgoing wave operators from an asymptotic Fock space to the physical Hilbert space. These maps are isometries and intertwine Poincaré transformations. Their ranges are the scattering subspaces. They become unitary onto the physical space only after asymptotic completeness is proved; calling an isometry unitary silently assumes the principal unsolved step.

Required background. Particles and one-particle subspaces defines the asymptotic one-particle space, and Haag–Ruelle scattering-state construction supplies the limits.

Helpful background. LSZ reduction: poles, residues, and stable external states gives the amplitude interpretation.

Møller maps from asymptotic Fock space

Let $\mathcal H_1$ contain all stable one-particle species under consideration and let $\mathcal F_\pm(\mathcal H_1)$ be the bosonic or fermionic Fock space. On the dense finite-particle subspace define

$$\Omega^{\mathrm{out}} (\psi_1\otimes_\pm\cdots\otimes_\pm\psi_n) =\Psi^{\mathrm{out}}(\psi_1,\ldots,\psi_n),$$

and analogously for $\Omega^{\mathrm{in}}$. The Haag–Ruelle inner-product formula gives

$$\langle\Omega^\#\Phi,\Omega^\#\Psi\rangle_{\mathcal H} =\langle\Phi,\Psi\rangle_{\mathcal F}, \qquad \#\in\{\mathrm{in},\mathrm{out}\}.$$

Hence each map extends uniquely to an isometry on all of Fock space. In particular,

$$(\Omega^\#)^*\Omega^\#=1_{\mathcal F}, \qquad \Omega^\#(\Omega^\#)^*=P^\#,$$

where $P^\#$ projects onto the closed range $\mathcal H^\#=\operatorname{Ran}\Omega^\#$. Isometry proves injectivity and closed range, not $P^\#=1_{\mathcal H}$.

Covariance of the approximants yields the intertwining relation

$$U(g)\Omega^\#=\Omega^\# U_{\mathcal F}(g)$$

for Poincaré transformations $g$. Thus the asymptotic dynamics is the second quantization of the one-particle dynamics. Haag’s and Ruelle’s constructions establish these statements under the massive local hypotheses Haag 1958, pp. 669–673 and Ruelle 1962, pp. 147–163.

Asymptotic creation operators and fields

On finite scattering vectors, define

$$a_{\mathrm{out}}^\dagger(\psi)\Omega^{\mathrm{out}}\Phi =\Omega^{\mathrm{out}}\bigl(a^\dagger(\psi)\Phi\bigr),$$

with the analogous in operator. These are unbounded operators on a specified finite-particle domain. Their canonical (anti)commutation relations follow by transporting those on asymptotic Fock space. Smearing them into positive- and negative-frequency parts produces asymptotic free fields $\phi_{\mathrm{in/out}}(f)$ satisfying the free equation with the physical particle mass.

This does not identify the interacting field with a free field at finite time. The asymptotic operators are limits on scattering domains, so Haag’s theorem is not evaded by an illicit unitary interaction picture.

Domain control is essential here. On the finite-particle core, $a^\dagger_{\mathrm{out}}(\psi)$ raises particle number by one and obeys the usual number-operator bound

$$\|a^\dagger_{\mathrm{out}}(\psi)\Phi_n\| \leq \sqrt{n+1}\,\|\psi\|\,\|\Phi_n\|.$$

This produces a closable operator after transport by the isometry, but it does not make the operator bounded or defined on an arbitrary physical vector. An asymptotic field is therefore an operator-valued distribution on a stated invariant core. Writing it as an everywhere-defined large-time operator limit would erase exactly the domain information that makes the construction rigorous.

On that core, the transported canonical relations are identities of operator-valued distributions. Their extension to larger domains requires a separate closure argument; it is not a consequence of the formal commutator alone.

The scattering operator

The comparison of the two Møller maps is

$$S=(\Omega^{\mathrm{out}})^*\Omega^{\mathrm{in}} :\mathcal F_{\mathrm{in}}\longrightarrow\mathcal F_{\mathrm{out}}.$$

Without a relation between the two ranges, $S$ is a contraction. It is unitary if the in and out ranges coincide and the corresponding wave maps are onto that common scattering space; asymptotic completeness gives the strongest case, $\mathcal H^{\mathrm{in}}=\mathcal H^{\mathrm{out}}=\mathcal H$. Matrix elements of $S$ are well-defined before any perturbative expansion, but extracting them from time-ordered functions is the separate LSZ step.

For a massive scalar model, applying the construction to every finite tensor gives isometric $\Omega^{\mathrm{in/out}}$ and the displayed $S$. This is the Hilbert-space content returned to in and out states.

An independent algebraic check is

$$\|S\Phi\| =\|(\Omega^{\mathrm{out}})^*\Omega^{\mathrm{in}}\Phi\| \leq\|\Omega^{\mathrm{in}}\Phi\|=\|\Phi\|.$$

Equality for every $\Phi$ requires the incoming range to lie in the outgoing range; surjectivity requires the converse as well.

There are consequently three logically different uses of “unitary.” Each wave map is unitary from asymptotic Fock space onto its own closed range because it is an isometry. The comparison $S$ is unitary between the in and out asymptotic spaces only when the two physical ranges coincide. A wave map is unitary onto $\mathcal H$ only after completeness. Stating the domain and codomain removes an apparent contradiction between these claims.

Failure test: the missing orthogonal complement

If only isometry is known, the physical Hilbert space decomposes as

$$\mathcal H=\mathcal H^{\mathrm{out}}\oplus (1-P^{\mathrm{out}})\mathcal H.$$

The second summand may contain bound states, topological sectors, infraparticle configurations, or simply states not controlled by the theorem. Calling $\Omega^{\mathrm{out}}$ unitary sets this summand to zero without proof. Existence of asymptotic fields likewise does not establish completeness.

Different superselection sectors may require separate asymptotic spaces and channel wave operators. A single vacuum-sector Fock space should not be declared universal.

Exercises

Prove that an isometry has closed range and identify the range projection.

Solution

If $V\psi_n$ is Cauchy, then $\|\psi_n-\psi_m\|=\|V\psi_n-V\psi_m\|$, so $\psi_n$ converges and its image is the limit; the range is closed. Since $V^*V=1$, the operator $VV^*$ is self-adjoint and idempotent. It is the identity on $\operatorname{Ran}V$ and zero on its orthogonal complement, hence is the range projection.

References

• Haag, Rudolf. 1958. “Quantum Field Theories with Composite Particles and Asymptotic Conditions.” Physical Review 112: 669–673. DOI.
• Ruelle, David. 1962. “On the Asymptotic Condition in Quantum Field Theory.” Helvetica Physica Acta 35: 147–163. Digitized article.