2.7 Field-operator Matching We showed in the previous section that an operator O on the conformal side corresponds to putting in a field φ on the gravity side. The example canonically presented is to consider the action of a scalar field, φ , of mass m on the AdS$_{5}$ × S 5 side: The simplest solution is obtained by assuming that φ is a function of r , the radial direction of the AdS$_{5}$ × S 5 space only. Then the solution is given by with m 2 = Δ(Δ − 4) ∗ . In fact in general, a bulk scalar φ with mass m 2 corresponds to a scalar operator on the conformal side with dimension Δ = − 2 ± √$_{4 +}$$_{m}$ 2 † . Solving ✷ φ = 0 gives two independent solutions, r Δ − 4 and r − $^{Δ}$: φ ( r ) ∼ Ar Δ − 4 + Br − $^{Δ}$. We can then identify [55–58] B as the source of the operator O and A as the VEV for the same operator. As discussed in the previous section, the most natural way [55] to combine these quantities to form a dimensionless quantity is ∫ d $^{4}$x ( A × B ) ≡ ∫ d $^{4}$x φ$_{0}$ O . We also stated in the last section that ∫ d $^{4}$xφ$_{0}$ ( x ) O ( x ) must be a singlet under all symmetries of the conformal field theory. Furthermore, it must have mass dimension 0 since it is part of an exponential, so φ$_{0}$ ( x ) O ( x ) must have mass dimension 4. Clearly such a combination of A and B satisfies these requirements. $^{∗}$This formula can be generalized to m 2 = (Δ − p )(Δ + p − 4) in the more general case of the action of a p-form [22] $^{†}$Δ refers to the full dimension of the operator (i.e. classical + anomalous) 42 S = ∫ d $^{4}$x dr d Ω$_{5}$ √ − g ( η $^{µν}$∂$_{µ}$ φ∂$_{ν}$φ − m $^{2}$φ2 ) (2.10) φ ( r ) = $_{Ar}$− (4 − Δ) + $_{Br}$− Δ (2.11)