Retarded Green Functions

Why this matters

The previous page explained the real-time prescription at the level of principle: retarded correlators are computed by Lorentzian bulk fields that fall through the future horizon. This page turns that principle into a calculation.

The central boundary object is

$$G_R(t,\mathbf{x}) = -i\theta(t) \langle[\mathcal O(t,\mathbf{x}),\mathcal O(0,\mathbf{0})]\rangle .$$

It is the kernel that relates a small source to a causal response:

$$\delta\langle\mathcal O(\omega,\mathbf{k})\rangle = G_R(\omega,\mathbf{k})J(\omega,\mathbf{k}) .$$

In holography, this response is not guessed. It is extracted from the renormalized radial canonical momentum of a bulk field whose boundary value is $J$.

The short practical formula is

$$\boxed{ G_R(\omega,\mathbf{k}) = \frac{\delta \Pi_{\rm ren}(\omega,\mathbf{k})} {\delta \phi_{(0)}(\omega,\mathbf{k})} \bigg|_{\rm infalling} }$$

where $\phi_{(0)}$ is the boundary source and $\Pi_{\rm ren}$ is the finite response after adding counterterms.

Linear response on the boundary

Couple a source $J$ to an operator $\mathcal O$:

$$\delta S_{\rm QFT} = \int d^dx\,J(x)\mathcal O(x) .$$

To first order in $J$,

$$\delta\langle\mathcal O(x)\rangle = \int d^dy\,G_R(x-y)J(y) .$$

The step function in $G_R$ is not decorative: it enforces causality. A source at time $y^0$ cannot affect an expectation value at an earlier time $x^0<y^0$.

With Fourier convention

$$J(t,\mathbf{x}) = \int\frac{d\omega\,d^{d-1}k}{(2\pi)^d} e^{-i\omega t+i\mathbf{k}\cdot\mathbf{x}}J(\omega,\mathbf{k}),$$

linear response becomes multiplication:

$$\delta\langle\mathcal O(\omega,\mathbf{k})\rangle = G_R(\omega,\mathbf{k})J(\omega,\mathbf{k}) .$$

The spectral density is

$$\rho(\omega,\mathbf{k}) = -2\operatorname{Im}G_R(\omega,\mathbf{k}) .$$

For a stable thermal state and a Hermitian operator, $\rho$ measures physical spectral weight. In holography, its imaginary part is tied to absorption by the black-brane horizon.

Bulk setup

Use the planar AdS black-brane metric

$$ds^2 = \frac{L^2}{z^2} \left[-f(z)dt^2+d\mathbf{x}^{2}+\frac{dz^2}{f(z)}\right], \qquad f(z)=1-\left(\frac{z}{z_h}\right)^d,$$

with

$$T=\frac{d}{4\pi z_h} .$$

For a scalar field dual to $\mathcal O$, take

$$S_L[\phi] = -\frac{\mathcal N_\phi}{2} \int d^{d+1}x\sqrt{-g}\, \left(g^{MN}\partial_M\phi\partial_N\phi+m^2\phi^2\right) .$$

The overall normalization $\mathcal N_\phi$ depends on the ten-dimensional reduction, the five-dimensional Newton constant, and the convention used to normalize $\mathcal O$. For the logic of the prescription, it can be kept explicit.

For a Fourier mode

$$\phi(t,z,\mathbf{x}) = e^{-i\omega t+i\mathbf{k}\cdot\mathbf{x}}\phi_{\omega,\mathbf{k}}(z),$$

the wave equation is

$$\frac{1}{\sqrt{-g}}\partial_z \left(\sqrt{-g}g^{zz}\partial_z\phi\right) + \left(g^{tt}\omega^2+g^{ij}k_i k_j-m^2\right)\phi=0 .$$

The calculation has two ends:

$$\text{AdS boundary: source and response}, \qquad \text{horizon: causal interior condition}.$$

A holographic retarded two-point function is a linear-response computation. Specify the boundary source $\phi_{(0)}$, solve the bulk equation with infalling behavior at the future horizon, extract the renormalized response $\Pi_{\rm ren}$, and differentiate. Poles occur when the source coefficient can vanish while the infalling response remains nonzero.

Boundary expansion and the response coefficient

Near the boundary, the scalar behaves as

$$\phi(z;k) = z^{d-\Delta}\left[\phi_{(0)}(k)+\cdots\right] + z^\Delta\left[A(k)+\cdots\right], \qquad k=(\omega,\mathbf{k}),$$

where

$$m^2L^2=\Delta(\Delta-d) .$$

In standard quantization, $\phi_{(0)}$ is the source. The coefficient $A(k)$ is the nonlocal response data. After holographic renormalization,

$$\langle\mathcal O(k)\rangle = \mathcal N_\phi(2\Delta-d)A(k) + \text{local terms in }\phi_{(0)}(k),$$

for a scalar with no special logarithmic subtleties. The local terms are contact terms or scheme-dependent analytic pieces. They can change polynomial pieces of $G_R$, but not its nonlocal singularities.

Thus the nonlocal part of the retarded correlator is schematically

$$G_R(k) = \mathcal N_\phi(2\Delta-d) \frac{A(k)}{\phi_{(0)}(k)} + \text{contact terms},$$

where $A/\phi_{(0)}$ is computed using the infalling bulk solution.

Canonical momentum form

The more invariant way to write the prescription uses radial canonical momentum. Varying the scalar action gives a boundary term

$$\delta S_L\big|_{\text{on-shell}} = \int_{z=\epsilon}\frac{d^dk}{(2\pi)^d}\, \Pi(\epsilon;-k)\delta\phi(\epsilon;k),$$

with

$$\Pi(z;k) = -\mathcal N_\phi\sqrt{-g}\,g^{zz}\partial_z\phi(z;k),$$

up to the sign chosen for the outward normal. The renormalized momentum is

$$\Pi_{\rm ren}(k) = \lim_{\epsilon\to0} \left[ \epsilon^{\Delta-d}\Pi(\epsilon;k)+\Pi_{\rm ct}(\epsilon;k) \right],$$

where the power of $\epsilon$ shown here is the schematic conversion to the source normalization. A careful calculation uses the precise Fefferman–Graham expansion and the variation of $S_{\rm ren}$.

At linear order,

$$\Pi_{\rm ren}(k)=G_R(k)\phi_{(0)}(k).$$

Therefore

$$\boxed{ G_R(k) = \frac{\Pi_{\rm ren}(k)}{\phi_{(0)}(k)} \bigg|_{\rm linear,\ infalling} }$$

when the solution has been normalized to the desired boundary source. For multiple fields, this equation becomes a matrix relation.

The infalling solution

Near the horizon,

$$f(z)\simeq 4\pi T(z_h-z).$$

The radial equation has two independent behaviors:

$$\phi(z;k) \sim \left(1-\frac{z}{z_h}\right)^{-i\omega/(4\pi T)}, \qquad \phi(z;k) \sim \left(1-\frac{z}{z_h}\right)^{+i\omega/(4\pi T)} .$$

The retarded function uses the first behavior. In ingoing Eddington–Finkelstein time

$$v=t-z_*, \qquad z_*\sim-\frac{1}{4\pi T}\log\!\left(1-\frac{z}{z_h}\right),$$

an infalling mode is regular:

$$e^{-i\omega v} = e^{-i\omega t} \left(1-\frac{z}{z_h}\right)^{-i\omega/(4\pi T)} .$$

This condition says that the boundary perturbation may be absorbed by the future horizon. It does not allow independent radiation to emerge from behind the future horizon.

A step-by-step recipe

For a scalar two-point function in a thermal state:

1. Choose the bulk field dual to $\mathcal O$ and write its quadratic action.

2. Fourier transform along the boundary directions.

3. Solve the radial equation with infalling behavior at $z=z_h$.

4. Normalize the solution near $z=0$ so that the source coefficient is $\phi_{(0)}(k)$.

5. Compute the on-shell boundary term or radial canonical momentum.

6. Add counterterms and take $z\to0$.

7. Read off

   $$\delta\langle\mathcal O(k)\rangle = G_R(k)\phi_{(0)}(k).$$

For numerical work, a common trick is to write

$$\phi(z;k) = \left(1-\frac{z}{z_h}\right)^{-i\omega/(4\pi T)}F(z;k),$$

where $F$ is regular at the horizon. One integrates from the horizon to the boundary, then extracts the source and response coefficients.

Coupled fields and response matrices

Many important correlators are not single-field problems. Metric and gauge perturbations can mix. For example, at finite density, $a_x$ and $h_{tx}$ often couple. For stress-tensor correlators, diffeomorphism constraints mix metric components.

Suppose the independent sources are $J_a$ and the renormalized responses are $R_a$. Linear response gives

$$R_a(k)=G_{R,ab}(k)J_b(k).$$

In the bulk, one constructs a basis of infalling solutions labelled by boundary source data. In matrix notation,

$$G_R(k)=R(k)J(k)^{-1},$$

after adding counterterms and projecting onto gauge-invariant source and response variables.

This is why gauge invariance matters. A pole of a gauge-dependent component is not necessarily a physical pole. The clean objects are gauge-invariant master fields or the full renormalized response matrix.

Flux and spectral weight

For real $\omega$, the scalar equation has a conserved radial flux

$$\mathcal F_z = \frac{i\mathcal N_\phi}{2} \sqrt{-g}g^{zz} \left( \phi^*\partial_z\phi-\phi\partial_z\phi^* \right).$$

This flux is independent of $z$ when the coefficients of the radial equation are real. Evaluated near the boundary, it is related to the imaginary part of $G_R$. Evaluated at the horizon, the infalling condition turns it into an absorption flux.

This gives the geometric meaning of dissipation:

$$\operatorname{Im}G_R \quad \longleftrightarrow \quad \text{absorption by the horizon}.$$

The horizon is therefore not just a place where the bulk calculation ends. It is the bulk mechanism by which the thermal CFT loses memory of perturbations.

Hydrodynamic preview

Retarded functions encode transport. A conserved density with diffusion has a retarded correlator of the schematic form

$$G_R^{nn}(\omega,\mathbf{k}) = \frac{\chi D\mathbf{k}^2}{-i\omega+D\mathbf{k}^2} + \text{contact terms},$$

where $\chi$ is the susceptibility and $D$ is the diffusion constant. The pole

$$\omega=-iD\mathbf{k}^2$$

is a hydrodynamic relaxation mode.

Similarly, shear viscosity is extracted from a stress-tensor retarded function by a Kubo formula,

$$\eta = - \lim_{\omega\to0} \frac{1}{\omega} \operatorname{Im}G_R^{T^{xy}T^{xy}}(\omega,\mathbf{0}) .$$

The next pages develop the pole interpretation and the hydrodynamic limit more systematically.

Analytic structure

A retarded correlator is analytic in the upper half of the complex $\omega$ plane for a stable causal thermal state. Poles lie in the lower half plane:

$$\operatorname{Im}\omega_n<0 .$$

In time domain, a pole at

$$\omega_n=\Omega_n-i\Gamma_n, \qquad \Gamma_n>0,$$

contributes a late-time behavior

$$e^{-i\omega_n t}=e^{-i\Omega_n t}e^{-\Gamma_n t}, \qquad t>0 .$$

In holography, these poles are quasinormal modes of the black brane.

Dictionary checkpoint

Boundary quantity	Bulk computation
source $J=\phi_{(0)}$	non-normalizable boundary coefficient
response $\delta\langle\mathcal O\rangle$	renormalized radial canonical momentum
$G_R=\delta\langle\mathcal O\rangle/\delta J$	infalling linear solution differentiated at the boundary
spectral density $\rho=-2\operatorname{Im}G_R$	absorption flux into the horizon
transport coefficient	small-$\omega$, small-$k$ limit of $G_R$
relaxation pole	infalling, source-free bulk mode
coupled correlator matrix	matrix of renormalized responses to independent sources

The retarded prescription is a boundary-value problem with a causal horizon condition.

Common confusions

“The retarded correlator is just the ratio $A/\phi_{(0)}$.”

That is only a useful shorthand. The correct object is the renormalized variation of the on-shell action. For simple scalars, the nonlocal part is often proportional to $A/\phi_{(0)}$, but counterterms, logarithms, operator normalization, alternate quantization, and mixing can modify the precise formula.

“The imaginary part comes from the boundary.”

The boundary expansion can display $\operatorname{Im}G_R$, but its physical origin at finite temperature is horizon absorption. The conserved radial flux lets one evaluate the same imaginary part at the horizon.

“Every pole is hydrodynamic.”

Hydrodynamic poles approach $\omega=0$ as $\mathbf{k}\to0$ because they are tied to conserved quantities. Generic quasinormal poles remain at complex frequencies of order $T$ and describe nonhydrodynamic relaxation.

“Gauge fields and metric perturbations can be treated component by component.”

Not safely. Gauge and diffeomorphism constraints can mix components. Use gauge-invariant variables or the full source-response matrix.

“The infalling condition is optional.”

For the retarded correlator in a thermal state, it is the defining Lorentzian interior condition. Choosing outgoing behavior computes a different causal object.

Exercises

Exercise 1: Retarded analyticity and late-time decay

Suppose a retarded Green function has a simple pole at

$$\omega_* = \Omega-i\Gamma, \qquad \Gamma>0.$$

Show that this pole gives a decaying contribution to the response for $t>0$.

Solution

For $t>0$, the inverse Fourier transform contains

$$\int\frac{d\omega}{2\pi}\,e^{-i\omega t}G_R(\omega).$$

Closing the contour in the lower half plane picks up poles with $\operatorname{Im}\omega<0$. A simple pole at $\omega_*$ gives

$$e^{-i\omega_*t} = e^{-i\Omega t}e^{-\Gamma t} .$$

Thus the imaginary part $-\Gamma$ gives decay. A pole in the upper half plane would grow exponentially and signal an instability or an acausal prescription.

Exercise 2: Source and response from a normalized solution

Let an infalling scalar solution near the boundary behave as

$$\phi(z;k)=J(k)z^{d-\Delta}+A(k)z^\Delta+\cdots .$$

Assume standard quantization and no logarithmic subtleties. What is the nonlocal part of $G_R(k)$?

Solution

The source is $J(k)$. The nonlocal response is proportional to $A(k)$. For a scalar with normalization $\mathcal N_\phi$,

$$\langle\mathcal O(k)\rangle = \mathcal N_\phi(2\Delta-d)A(k)+\text{local terms}.$$

Therefore the nonlocal part of the retarded two-point function is

$$G_R(k) = \mathcal N_\phi(2\Delta-d)\frac{A(k)}{J(k)} + \text{contact terms}.$$

The infalling condition is what makes this the retarded correlator rather than another Green function.

Exercise 3: Why coupled fields require a matrix

Suppose two bulk fields have boundary expansions with sources $J_1,J_2$ and responses $R_1,R_2$. Explain why computing only $R_1/J_1$ with $J_2$ accidentally nonzero does not give $G_{11}$.

Solution

Linear response has the matrix form

$$\begin{pmatrix}R_1\\ R_2\end{pmatrix} = \begin{pmatrix}G_{11}&G_{12}\\G_{21}&G_{22}\end{pmatrix} \begin{pmatrix}J_1\\ J_2\end{pmatrix} .$$

If $J_2\ne0$, then

$$R_1=G_{11}J_1+G_{12}J_2.$$

The ratio $R_1/J_1$ is not $G_{11}$ unless $J_2=0$ or $G_{12}J_2$ is known and subtracted. In practice one constructs independent infalling solutions, forms the source matrix $J$, forms the response matrix $R$, and computes

$$G=RJ^{-1}.$$

Further reading

• D. T. Son and A. O. Starinets, Minkowski-space correlators in AdS/CFT correspondence: recipe and applications.
• G. Policastro, D. T. Son, and A. O. Starinets, Shear viscosity of strongly coupled $\mathcal N=4$ supersymmetric Yang–Mills plasma.
• C. P. Herzog and D. T. Son, Schwinger-Keldysh propagators from AdS/CFT correspondence.
• K. Skenderis and B. C. van Rees, Real-time gauge/gravity duality.
• D. T. Son, Viscosity, Black Holes, and Quantum Field Theory.