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## 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: 

<img src='content_image/299528.jpg'>

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 ) ( x ) must be a singlet 

∫ O 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. 

<img src='content_image/299529.jpg'>

