I am reading the book "An introduction to Symmetry and Supersymmetry in Quantum field theory" by Lopuszanski, and I have some problems understanding his argumentation about the existence of the S-matrix. On page 102 he says

'Both fields, $\varphi^{\text{in}}$ and $\varphi^{\text{out}}$, are free fields and by assumption their Fock vectors span the same Hilbert space with a common Fock vacuum. Moreover, both fields satisfy the same canonical commutation relations with the same mass. Consequently, they have to be connected by a unitary transformation

$\varphi^{\text{out}}(f) = S^{-1}\varphi^{\text{in}}(f) S$.'

In the context of this book he works with the Wightman axioms, and he assumes that the asymptotically free fields span the same Hilbert space as the original field. I do not quite see how the arguments presented imply the existence of such a unitary operator. Can someone explain what I am missing?


1 Answer 1


Well, this is a simple argument in functional analysis which neither BLT, nor Lopuszanski bother spelling out, because it is assumed known by the readers of axiomatic QFT (who are expected to know standard functional analysis).

Lopuszanski says that the axiom of asymptotic completeness means that the Hilbert-Fock space (actually rigged Hilbert space, but let us not go into finesse) of in-states $\mathcal{H}_{\text {in}}$ is isomorphic to the Hilbert-Fock space the out-states $\mathcal{H}_{\text {in}}$.

Definition: Two separable Hilbert spaces $\mathcal{H}_1 $ and $\mathcal{H}_2 $ are isomorphic, if there is a unitary transformation sending the countable orthonormal basis from $\mathcal{H}_1 $ into the countable orthonormal basis from $\mathcal{H}_2 $.

So the existence of this unitary transformation comes directly from the requirement of asymptotic completeness. It is a standard exercise to see how this unitary transformation affects the self-adjoint operators (quantum fields) from one space in order to link them with the self-adjoint operators from the other space. It is in a way similar to the calculations done to show the equivalence of the Heisenberg, Schrödinger and interaction (Dirac-Tomonaga-Schwinger) pictures of standard QM (more precisely, you transform the states by this unitary operator and then ask observables/fields to provide the same statistical/probabilistic interpretation as transformed operators in the "target space" as they had in the "original space").

And if $S$ is unitary, then of course $S^\dagger = S^{-1}$ to reach the form provided by Lopuszanski.

  • $\begingroup$ Thank you for your answer! I cannot quite see at which point in this argumentation it is necessary that the in and out fields satisfy the same commutation relations. Can you maybe elaborate on that? Also, what do you mean by the last part in parentheses? $\endgroup$
    – S.Farr
    Dec 10, 2020 at 1:36
  • $\begingroup$ Well commutators keep their form under a unitary transformation of their operators. The last part is simply how this relation you need proven comes about from the simple $ a_{out} = S b_{in}$ with S unitary. $\endgroup$
    – DanielC
    Dec 10, 2020 at 1:42
  • $\begingroup$ Yes, but how is it necessary that the in and out fields have the same commutation relations to conclude the existence of the scattering operator? I thought that maybe the reason we stress this fact is that we use representation theory of C*-algebras here. The field algebras are such algebras, and there are strong statements on the unitary equivalence of *-representations if you know that they both have a cyclic vector, in this case, the vacuum. $\endgroup$
    – S.Farr
    Dec 10, 2020 at 12:33
  • $\begingroup$ That is correct, however the reasoning by Lopuszanski doesn't use $C^*$ algebras. $\endgroup$
    – DanielC
    Dec 10, 2020 at 22:49

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