Question

Does AdS$_3$ geometry have an angular momentum?

Answer 1

In Weinberg's QFT volume 1, chapter 3.3, just below equation 3.3.19, he says P⃗ =P⃗ 0 and J⃗ =J⃗ 0 can be(easily) concluded from the definition of Møller wave operator or equivalently the Lippmann-Schwinger equations. However, I don't see how.

In virtually all known field theories, the effect of interactions is to add an interaction term V to the Hamiltonian, while leaving the momentum and angular momentum unchanged:

H=H0+V,P⃗ =P⃗ 0,J⃗ =J⃗ 0.(3.3.18)

(The only known exceptions are theories with topologically twisted fields, such as those with magnetic monopoles, where the angular momentum of states depends on the interactions.) Eq. (3.3.18) implies that the commutation relations (3.3.11), (3.3.14), and (3.3.16) are satisfied provided that the interaction commutes with the free-particle momentum and angular-momentum operators

[V,P⃗ 0]=[V,J⃗ 0]=0.(3.3.19)

It is easy to see from the Lippmann-Schwinger equation (3.1.16) or equivalently from (3.1.13) that the operators that generate translations and rotations when acting on the 'in' (and 'out') states are indeed simply P⃗ 0 and J⃗ 0.

Related commutators:

[Ji,Jj]=iϵijkJk(3.3.11)

[Ji,Pj]=iϵijkPk(3.3.14)

[Ji,H]=[Pi,H]=[Pi,Pj]=0(3.3.16)

Definition of the wave operators:

The 'in' and 'out' states can now be defined as eigenstates of H, not H0,

HΨ±α=EαΨ±α(3.1.11)

which satisfy the condition

∫dαe−iEατg(α)Ψ±α→∫dαe−iEατg(α)Φα(3.1.12)

for τ→−∞ or τ→∞, respectively. Eq. (3.1.12) can be rewritten as the requirement that:

e−iHτ∫dαg(α)Ψ±α→e−iH0τ∫dαg(α)Φα

for τ→−∞ or τ→∞, respectively. This is sometimes rewritten as a formula for the 'in' and 'out' states:

Ψ±α=Ω(±∞)Φα(3.1.13)

where

Ω(τ)=eiτHe−iτH0(3.1.14)

The Lippmann-Schwinger equations are given by:

Answer 2

the above given answer is correct