Boson-fermion correspondence of type B and twisted vertex algebras

Multilocal Fermionization

“…(In deriving this equation, care must be taken of the fact that due to the different subtractions, one has : ψ(z)∂ z ψ(z) : R = : ψ(z)∂ z ψ(z) : 0 + 1 8z 2 .) Since L c= 1 2 0 has eigenvalue h R = 1 16 in the Ramond state, we conclude that ω R (β R (L curr 0 )) = 1 16 , in agreement with the ground state energy in the twisted sector of the current.…”

“…The fact that the current (3.1) j(x) = i : ψ (1) (x)ψ (2) (x) : satisfies the CCR (3.2) is purely algebraic, and therefore independent of the representation. Taking ψ (1) in the Ramond representation and ψ (2) in the vacuum representation, the isomorphism (5.1) embeds the resulting current into the Ramond algebra π R CAR(S 1 ) . The result is (in the compact picture)…”

“…where : • : R stands for the Wick product defined with the subtraction of the Ramond expectation value (which is zero in the case at hand). The right-hand side is the formula given in [1]. The new aspect here is that it arises from an underlying isomorphism of CAR algebras.…”

“…In other words, we find the current in a representation that extends anti-periodically to the circle. Indeed, this representation of the current is known as the "twisted" representation [1,9] obtained from the quasifree state with 2-point function given by (2.12). This can be established by evaluating the embedded current in the Ramond state.…”

“…It is well known that there is an algebraic isomorphism (the "split isomorphism") A(x)B(y) → A(x) ⊗ t B(y), x ∈ O 1 , y ∈ O 2 between the fields of a given QFT localized in two spacetime regions O i , and two independent copies of the same fields localized in the same regions, as long as the regions are sufficiently well spacelike separated from each other. 1 Naively, this is true because A (anti)commute with B by graded locality, while A ⊗ t 1 (anti)commute with 1 ⊗ t B by construction, so that all algebraic relations are preserved. However, this isomorphism does not preserve the vacuum state, because it eliminates all correlations between fields in O 1 and fields in O 2 : ω(A(x) ⊗ t B(y)) = ω(A(x)) • ω(B(y)) = ω(A(x)B(y)).…”