research35 min studyLesson 795 in Physics

Boundary correlators from bulk actions

Use a source convention, a saddle and the correct functional derivatives to define the calculation.

Before you begin

On this page

The question for today

What you’ll understand

State the Euclidean generating-functional prescription with signs.
Separate a near-boundary source from an interior regularity condition.
Compute mean responses and connected correlators from a declared quadratic functional.

A dictionary entry becomes a calculational method when boundary sources determine bulk boundary-value problems and their on-shell actions generate correlators. The classical bulk approximation is a saddle approximation to a proposed dual partition function, not a universal identity between arbitrary differential equations and arbitrary quantum theories. Work here in Euclidean signature with a positive scalar kinetic normalization, standard quantization and a source coupling chosen with a plus sign in the boundary exponential. Those conventions fix the signs of the derivatives below.

Define the generating function before differentiating

For a boundary operator, introduce a source in the exponential and take the logarithm of the resulting partition function. Its first derivative gives the source-dependent mean, while its second derivative gives the connected two-point response. The logarithm matters: differentiating the partition function directly adds disconnected products. The source has the complementary scaling dimension needed to make its spacetime integral against the operator dimensionless. Internal indices and integration measures must be retained in a complete calculation.

In the semiclassical holographic prescription, evaluate the renormalized Euclidean action on the solution with that source and the chosen interior state condition. The boundary generating logarithm is minus this action in the present convention. A source sign change would change odd derivatives and must be tracked consistently. Quantum bulk loops and other saddles produce corrections; a tree-level action calculation is not the full finite-N partition function simply because its functional derivatives can be taken exactly.

Z[J]=\left\langle e^{\int d^dx\,J\mathcal O}\right\rangle,\quad W[J]=\ln Z[J]\simeq-I_{\rm ren}[\phi_{\rm cl};J],\qquad \langle\mathcal O\rangle_J=\frac{\delta W}{\delta J},\quad \langle\mathcal O\mathcal O\rangle_{c,J}=\frac{\delta^2W}{\delta J\,\delta J}

The source sign and Euclidean action sign are explicit. A zero-source normalization can divide Z by Z[0] without changing these connected response derivatives.↔ Scroll wide equations horizontally.

Reduce the scalar action only after imposing its equation

For a free bulk scalar with positive kinetic coefficient, integrating by parts expresses the action as a bulk equation-of-motion term plus a boundary term. The former vanishes on a classical solution. Near the ultraviolet cutoff, the outward normal points toward decreasing radial coordinate, which fixes a sign in the boundary flux. Boundary counterterms are then essential before removing the cutoff. Using an inward normal or an unrenormalized flux can give an incorrect sign or a divergent response.

The near-boundary expansion alone does not determine the nonlocal response coefficient: it supplies two independent asymptotic data before interior information is imposed. Euclidean regularity in a chosen state, or a separately specified Lorentzian prescription, selects their relation. Local source-dependent terms can be determined asymptotically and later appear as contact or scheme-dependent contributions. They cannot replace the global bulk solution that encodes the separated-point correlation function.

I_E=\frac{\mathcal C}{2}\int d^{d+1}x\sqrt g\left[(\nabla\phi)^2+m^2\phi^2\right],\qquad I_{E,\rm os}=\frac{\mathcal C}{2}\int_{\partial M}d^dx\sqrt\gamma\,\phi\,n^a\partial_a\phi

For a dimensionless bulk φ, C has inverse-length^(d−1) units. γ is the induced cutoff metric and n is the outward unit normal; additional boundaries and the chosen regularity conditions must be treated consistently.↔ Scroll wide equations horizontally.

Interpret simple kernels without claiming a bulk solution

A quadratic generating functional provides a transparent response benchmark. Its kernel is the connected two-point function in the declared normalization; a nonzero source produces a mean linear in that kernel. For a finite collection of integrated dimensionless source modes, these functional derivatives become ordinary matrix derivatives. Such an example teaches source versus response and connectedness, but the kernel values remain supplied data until an actual bulk boundary-value problem derives them.

Interactions lead to higher connected derivatives and, at tree level, to bulk interaction diagrams with boundary legs. Their numerical values depend on action normalization, boundary conditions, counterterms and the chosen state. A bottom-up choice of a scalar potential is therefore a model assumption, not an automatic prediction for any CFT with a similarly named operator. The function lab exactly displays one quadratic and its slope; it does not evaluate a bulk path integral or construct a boundary correlation kernel.

W[J]=\frac12\int d^dx\,d^dy\,J(x)K(x,y)J(y)\quad\Longrightarrow\quad \langle\mathcal O(x)\rangle_J=\int d^dy\,K(x,y)J(y)

K is a symmetric Euclidean connected kernel in this quadratic example. Positivity, contact terms and normalization require their own physical conditions; writing the quadratic form does not establish a holographic origin.↔ Scroll wide equations horizontally.

Work it through

Differentiate a declared source mode

For one dimensionless integrated source mode, take $W(J)=3J^2/2$ with W(0) = 0. At J = 2 find the mean response, the connected second derivative and the corresponding classical renormalized action value in the stated sign convention.

Differentiate W once: the mean response is $dW/dJ=3J$ , giving six at J = 2. It is the source-induced mean, not the connected variance itself.
Differentiate again: the connected second derivative is three, independent of J in this quadratic model. Differentiating Z instead would additionally contain the square of the mean.
At J = 2 the generating logarithm is six, so the stipulated saddle action is $I_{\rm ren}=-W=-6$ . A renormalized on-shell value need not be a positive bare integral after boundary subtraction. The kernel three was supplied, not solved from gravity.

The result

The mean is 6, the connected second derivative is 3 and the stipulated renormalized saddle action is −6.

The living notebook

Explore the model

Interactive lab

Interactive lab · function

A source quadratic and its response slope

Reset and set BOTH Coefficient a = 1.50 and Position x = 1; predict y = 1.50 and tangent slope 3. Reset and set a = 1.50 and x = 2; predict y = 6 and slope 6. Interpreting x as this dimensionless toy source illustrates W and its first derivative only; the lab does not solve a bulk field or determine a CFT kernel.

y = ax²TangentSigned integral

y at chosen x: 0.608
Tangent slope · 2ax: 1.35
Signed area · ax³/3: 0.182

Coefficient a0.75Position x0.9

t = 0 s

This lab uses only y = ax². Playback traces the curve within the fixed integration interval from 0 to x; it does not estimate numerical error or change the chosen point. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forThe derivative and the value of a generating function have different meanings.

Pause & reason

Check your understanding

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Read the checkpoint explanation

Taking the logarithm removes disconnected products. Its second source derivative is the connected correlator in the declared source convention.

Your turn

Try it yourself

For two dimensionless integrated source modes let $W=\tfrac12(2J_1^2+2J_1J_2+4J_2^2)$ . At $(J_1,J_2)=(1,-1)$ find both mean responses and the connected cross-response. Which part needs a bulk calculation if this is to become a holographic prediction?

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Differentiate the quadratic form; the off-diagonal kernel equals one because the overall factor is one half.

Compare with the solution

The means are $\partial_1W=2J_1+J_2=1$ and $\partial_2W=J_1+4J_2=-3$ . The connected cross-response is $\partial_1\partial_2W=1$ . These values follow exactly from the supplied kernel matrix, whose diagonal entries are two and four. A holographic prediction would have to derive that kernel from an appropriately normalized bulk action, source boundary conditions, an interior state prescription and renormalized boundary flux, rather than assigning it arbitrarily.

Keep the idea

Take it with you

Source sign, action sign and connected derivatives must be fixed together.
Interior conditions determine nonlocal response data.
An exactly differentiated toy kernel is not an independently derived gravitational prediction.

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