13 Non-Fock Representations
89
For the algebra generated by canonical variables of the above form, the space
translations α a are naturally defined by
α a (ψ( f i )) = ψ( f ia ), f ia (x) ≡ f i (x − a)
(13.5)
formally equivalent to α a (ψ(x)) = ψ(x + a). Clearly, the space translations define (a
one-parameter group of) ∗-automorphisms (or algebraic symmetries) of the algebra
of canonical variables. In each irreducible Fock representation the Fock state is the
unique translationally invariant state
61 and the space translations are implemented
by (strongly continuous) unitary operators U (a)
U (a)ψ(x)U (−a) = ψ(x + a).
(13.6)
We briefly sketch Haag’s theorem.
62 We consider a system described by canonical
variables of the form (13.3) and by a Hamiltonian of the form (13.1), with g = 0,
invariant under space translations. We denote by π g an irreducible representation of
the algebra of canonical variables in which H is well defined and it has a translationally invariant ground state 0g . Then, by the argument following the GNS theorem,
the space translations are implemented by (a one-parameter group of) unitary operators U g (a) leaving the ground state invariant. If π g is a Fock representation for the
a i , a
∗
i , then there exists a Fock vacuum 0 and U g (a) = U g=0 (a) ≡ U (a). In fact
U g U
−1 commutes with the a i , a
∗
i and by irreducibility it must be a multiple of the
identity exp(iθ g (a)); moreover, the group law and the continuity in a implies that
θ g (a) = θ g a and therefore by a trivial redefinition of U g (a) one may get U g = U .
This implies that 0g = 0 , i.e. the ground state is independent of the coupling constant. It is intuitively clear that it can hardly be so and actually for relativistic systems
the above coincidence of the interacting and the free ground states is compatible only
with a free theory.
63
The implications of Haag’s theorem, about the impossibility of using the Fock
representation for defining the Hamiltonian in the presence of interaction, are rather
strong. The standard Rayleigh–Schrödinger perturbative expansion in terms of eigenstates of the free Hamiltonian H 0 requires that H 0 be well defined and this is not
possible if the representation in which the total Hamiltonian is well defined is nonFock. In particular, from a mathematical point of view, Haag’s theorem excludes the
existence of the so-called interaction picture representation, which is at the basis of
the standard expansion in quantum field theory and in many-body theory. In fact,
61 In fact, since the number operator commutes with the space translations, the existence of space
translationally invariant states can be discussed in each eigenspace of N , say H K , corresponding
to the eigenvalue K , whose vectors are L 2 (R s K ) functions of s K variables. The space translation invariance would require that such a function does not depend on the sum of the variables,
incompatibly with being in L 2 .
62 R. Haag, On quantum field theories, Dan. Mat. Fys. Medd. 29 no 12 (1955); Local Quantum
Physics, Springer 1996.
63 R. F. Streater and A. S. Wightman, PC T , Spin and Statistics and All That, Benjamin-Cummings
1980.
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