Green Functions and Response Preview

A weak external source can probe a quantum system without requiring a full solution of the driven dynamics. If the perturbation is

$$H_{\mathrm{pert}}(t) = -f(t)B,$$

then the first-order change in an observable $A$ has the form

$$\delta\langle A(t)\rangle = \int_{-\infty}^{\infty}dt'\, \chi_{AB}^R(t,t')f(t'),$$

where

$$\chi_{AB}^R(t,t') = \frac{i}{\hbar} \theta(t-t') \left\langle [A_H(t),B_H(t')] \right\rangle_0.$$

The step function makes the kernel retarded: a source applied at $t'$ cannot change the expectation value at an earlier time $t$. The commutator supplies the quantum dynamical content. Together they turn linear response into a Green-function problem.

The Kubo formula develops transport limits, contact terms, and many-body observables. Here a fixed source-sign convention and an oscillator example show how a commutator becomes a susceptibility.

Retarded and Advanced Response develops the paired many-body commutator kernels, adjoint relation, half-plane analyticity, spectral discontinuity, and dispersion checks.

For the complementary transition-theory dictionary—sum-over-states poles, absorption minus stimulated emission, and the golden-rule power identity—see Linear Response Preview.

Perturbing a System with a Source

Write the Hamiltonian as

$$H(t) = H_0-f(t)B.$$

Here $f(t)$ is a prescribed classical source and $B$ is the operator to which it couples. The minus sign is a convention, but it must be stated because it fixes the sign of the response kernel.

Source $f$	Coupled operator $B$	Typical measured observable $A$
force	position	position or momentum
electric field	electric dipole	polarization or current
magnetic field	magnetic moment	magnetization
electrostatic potential	minus charge density	charge density or current

The source and the measured observable need not be the same. The notation $\chi_{AB}$ means: measure $A$ after perturbing through $B$.

For the elementary stationary setup, assume:

• the unperturbed state is described by $\rho_0$ with $[H_0,\rho_0]=0$;
• the source is weak enough that terms beyond first order in $f$ can be neglected;
• the source is switched on at a specified initial time, or adiabatically from the distant past;
• operators evolve with $H_0$ inside the first-order formula.

First-order derivation sketch

In the interaction picture,

$$i\hbar\frac{d\rho_I}{dt} = [H_{\mathrm{pert},I}(t),\rho_I(t)].$$

Replacing $\rho_I$ on the right by $\rho_0$ gives

$$\begin{aligned} \rho_I(t) &= \rho_0 - \frac{i}{\hbar} \int_{t_0}^{t}dt'\, [H_{\mathrm{pert},I}(t'),\rho_0] +O(f^2)\\ &= \rho_0 + \frac{i}{\hbar} \int_{t_0}^{t}dt'\, f(t')[B_I(t'),\rho_0] +O(f^2). \end{aligned}$$

Taking the expectation value of $A_I(t)$ and using cyclicity of the trace yields

$$\begin{aligned} \delta\langle A(t)\rangle &= \frac{i}{\hbar} \int_{t_0}^{t}dt'\, f(t') \operatorname{Tr} \left( \rho_0[A_I(t),B_I(t')] \right). \end{aligned}$$

Take $f(t')=0$ before $t_0$ (or switch it on adiabatically from the past). The integral can then be extended over all $t'$ using $\theta(t-t')$, giving the retarded kernel stated above. This is the first-order response identity. The Interaction Picture and Dyson Expansion as Formal Evolution derive the underlying evolution operators.

Response Functions

Operationally, the susceptibility is a functional derivative evaluated at zero source:

$$\chi_{AB}^R(t,t') = \left. \frac{\delta\langle A(t)\rangle_f} {\delta f(t')} \right|_{f=0}.$$

It answers a precise question: how much does $\langle A(t)\rangle$ change per infinitesimal source impulse at $t'$? In the time domain, $\chi$ includes the response per unit time; after Fourier transformation it has the ordinary units of $A/f$.

If $H_0$ and $\rho_0$ are time-translation invariant, then

$$\chi_{AB}^R(t,t') = \chi_{AB}^R(t-t').$$

Use the Fourier convention

$$\chi_{AB}^R(\omega) = \int_{-\infty}^{\infty}dt\, e^{i\omega t} \chi_{AB}^R(t).$$

The convolution becomes multiplication:

$$\delta\langle A(\omega)\rangle = \chi_{AB}^R(\omega)f(\omega).$$

The real and imaginary parts describe in-phase and out-of-phase response, with the precise assignment depending on the Fourier and driving conventions. Causality makes $\chi^R(\omega)$ analytic in the upper half of the complex $\omega$ plane. Under suitable decay assumptions, its real and imaginary parts obey dispersion relations such as

$$\operatorname{Re}\chi^R(\omega) = \frac{1}{\pi} \operatorname{PV} \int_{-\infty}^{\infty} d\omega'\, \frac{\operatorname{Im}\chi^R(\omega')} {\omega'-\omega}.$$

This Kramers–Kronig structure is not an extra dynamical law. It follows from retarded support, analyticity, and sufficiently controlled high-frequency behavior.

Retarded Correlators

Several objects called Green functions appear near linear response, and their normalizations vary. With the convention used on Retarded and Advanced Green Functions, define

$$G_{AB}^R(t,t') = -\frac{i}{\hbar} \theta(t-t') \langle[A_H(t),B_H(t')]\rangle_0.$$

For the coupling $H_{\mathrm{pert}}=-fB$ used here,

$$\chi_{AB}^R(t,t') = -G_{AB}^R(t,t').$$

Some texts instead call $\chi^R$ itself the retarded Green function. Others omit $1/\hbar$. Comparing symbols without comparing definitions is therefore unsafe.

Object	Ordering or support	Main question
$\chi_{AB}^R$	retarded commutator	How does $A$ respond to a source coupled to $B$?
$\langle A(t)B(t')\rangle$	fixed operator order	What fluctuations or transition weights occur in this order?
time-ordered correlator	chronological ordering	What object appears in perturbative amplitudes or generating functionals?
symmetrized correlator	anticommutator average	What fluctuation level is insensitive to operator order?

These functions can share spectral data without being interchangeable. The Correlation Functions in Path Integrals page develops time-ordered insertions, while Noise Spectra explains ordered and symmetrized noise conventions.

Spectral form

Let

$$H_0\lvert n\rangle = E_n\lvert n\rangle, \qquad \rho_0 = \sum_n p_n\lvert n\rangle\langle n\rvert.$$

For $A_{nm}=\langle n\rvert A\lvert m\rangle$ and the Fourier convention above,

$$\begin{aligned} \chi_{AB}^R(\omega) &= \sum_{n,m} \frac{ (p_m-p_n)A_{nm}B_{mn} }{ \hbar\omega+E_n-E_m+i0 }. \end{aligned}$$

The poles occur at transition frequencies, while the numerators combine matrix elements with population differences. Response therefore knows not only the energy spectrum but also which transitions the source can excite and which observable can detect them. The broader pole-and-weight organization is developed in Spectral Representation of Green Functions.

The infinitesimal $i0$ is the frequency-domain record of retarded support. Reversing the time-support prescription reverses the boundary value. Adding a finite linewidth is a physical or phenomenological modification, not merely another notation for $i0$.

Linear Response Preview

Consider a harmonic oscillator driven by a weak force:

$$H(t) = \frac{p^2}{2m} + \frac12m\Omega^2x^2 - f(t)x.$$

Here $A=B=x$. The unperturbed commutator gives

$$\chi_{xx}^R(t) = \theta(t) \frac{\sin(\Omega t)}{m\Omega}.$$

Its Fourier transform is

$$\chi_{xx}^R(\omega) = \frac{1}{ m\left[ \Omega^2-(\omega+i0)^2 \right] }.$$

The same function is the retarded inverse of the differential operator in

$$m\frac{d^2}{dt^2}\delta\langle x(t)\rangle + m\Omega^2\delta\langle x(t)\rangle = f(t).$$

This equivalence exposes the central bridge:

1. as an inverse kernel, $\chi^R$ solves a driven equation;
2. as a correlator, $\chi^R$ is a retarded commutator;
3. as a frequency-domain Green function, it has poles at the oscillator transitions;
4. as a susceptibility, it maps an applied force to a measured displacement.

For a static force $f_0$ reached by adiabatic switching, the zero-frequency response is

$$\delta\langle x\rangle = \chi_{xx}^R(0)f_0 = \frac{f_0}{m\Omega^2},$$

which is the equilibrium displacement. An isolated oscillator with a suddenly applied constant force instead has $\delta\langle x(t)\rangle=f_0[1-\cos(\Omega t)]/(m\Omega^2)$ for zero initial displacement; without damping it does not settle. At resonance, the ideal isolated oscillator has infinitely sharp poles. Real damping, finite observation time, and nonlinear response regulate the idealization in different ways and should not be collapsed into one ad hoc prescription.

Equilibrium and resonant limits

Linear response can be formulated about nonequilibrium states, but a thermal fluctuation–dissipation relation needs an equilibrium ensemble and matching ordering conventions. Near a sharp resonance, small drive amplitude alone does not ensure small response for arbitrarily long times. Switching, damping, observation time, and saturation answer different physical questions.

Common Mistakes

• Writing a susceptibility without stating whether the perturbation is $-fB$ or $+fB$.
• Forgetting that $B$ is the operator coupled to the source while $A$ is the measured observable.
• Calling an ordinary, symmetrized, or time-ordered correlator a causal response function.
• Omitting the step function or $i0$ prescription that enforces retarded support.
• Assuming time-translation invariance when the initial state is not stationary or the background Hamiltonian is driven.
• Treating a finite broadening as identical to the infinitesimal retarded prescription.
• Applying a thermal fluctuation–dissipation relation to an arbitrary nonequilibrium state.
• Ignoring contact terms when the measured observable depends explicitly on the source.
• Taking $\omega\to0$ and $\mathbf q\to0$ limits without specifying their order.
• Trusting first-order response when the induced change is no longer small.

References

• H. B. Callen and T. A. Welton, “Irreversibility and generalized noise,” Physical Review 83, 34–40 (1951).
• A. L. Fetter and J. D. Walecka, Quantum Theory of Many-Particle Systems, Dover, 2003.
• R. Kubo, “Statistical-mechanical theory of irreversible processes. I,” Journal of the Physical Society of Japan 12, 570–586 (1957).
• R. Kubo, “The fluctuation-dissipation theorem,” Reports on Progress in Physics 29, 255–284 (1966).
• G. D. Mahan, Many-Particle Physics, 3rd ed., Kluwer Academic/Plenum, 2000.
• J. J. Sakurai and J. Napolitano, Modern Quantum Mechanics, 3rd ed., Cambridge University Press, 2020.

Exercises

1. Starting from the interaction-picture density operator, verify the sign of the response kernel for $H_{\mathrm{pert}}(t)=-f(t)B$.

Solution

To first order,

$$\delta\rho_I(t) = -\frac{i}{\hbar} \int_{t_0}^{t}dt'\, [H_{\mathrm{pert},I}(t'),\rho_0].$$

Substituting $H_{\mathrm{pert},I}=-fB_I$ gives

$$\delta\rho_I(t) = \frac{i}{\hbar} \int_{t_0}^{t}dt'\, f(t')[B_I(t'),\rho_0].$$

Therefore

$$\begin{aligned} \delta\langle A(t)\rangle &= \operatorname{Tr} \left( \delta\rho_I(t)A_I(t) \right)\\ &= \frac{i}{\hbar} \int_{t_0}^{t}dt'\, f(t') \langle[A_I(t),B_I(t')]\rangle_0. \end{aligned}$$

Thus the susceptibility carries $+i/\hbar$ for the stated minus-sign coupling. A coupling $+fB$ would reverse this sign.

2. Explain why retarded support implies that an impulse applied after the measurement time contributes nothing to the response.

Solution

For an impulse $f(t')=f_0\delta(t'-t_s)$,

$$\delta\langle A(t)\rangle = f_0\chi_{AB}^R(t,t_s).$$

The kernel contains $\theta(t-t_s)$. If $t_s\gt t$, then

$$\theta(t-t_s)=0,$$

so the impulse makes no contribution. This is temporal causality of the response kernel. It does not by itself assert relativistic light-cone support in space.

3. Derive the harmonic-oscillator susceptibility from the commutator.

Solution

The unperturbed Heisenberg operator is

$$x_H(t) = x_H(0)\cos\Omega t + \frac{p_H(0)}{m\Omega}\sin\Omega t.$$

Using $[x_H(0),p_H(0)]=i\hbar$ gives

$$[x_H(t),x_H(0)] = -\frac{i\hbar}{m\Omega} \sin\Omega t.$$

Hence

$$\chi_{xx}^R(t) = \frac{i}{\hbar} \theta(t) [x_H(t),x_H(0)] = \theta(t) \frac{\sin\Omega t}{m\Omega}.$$

This commutator is a multiple of the identity, so the result is independent of the oscillator state.

4. For $H_0=(\hbar\omega_0/2)\sigma_z$, take $\omega_0>0$, the ground state, and $A=B=\sigma_x$. Find $\chi_{xx}^R(t)$ and its static value.

Solution

The Heisenberg operator is

$$\sigma_x(t) = \sigma_x\cos\omega_0t - \sigma_y\sin\omega_0t.$$

Therefore

$$[\sigma_x(t),\sigma_x] = 2i\sigma_z\sin\omega_0t.$$

In the ground state, $\langle\sigma_z\rangle=-1$, so

$$\chi_{xx}^R(t) = \frac{2}{\hbar} \theta(t) \sin\omega_0t.$$

At zero frequency,

$$\chi_{xx}^R(0) = \int_0^\infty dt\, \frac{2}{\hbar} e^{-0^+t} \sin\omega_0t = \frac{2}{\hbar\omega_0}.$$

This agrees with the first-order static polarization produced by a perturbation $-f\sigma_x$.

5. Which assumptions are added when a fluctuation–dissipation relation is inferred from a retarded susceptibility?

Solution

Causal linear response supplies the retarded commutator, but a fluctuation–dissipation relation additionally requires an equilibrium stationary ensemble, usually a Gibbs state, together with the corresponding Kubo–Martin–Schwinger condition. Fourier, ordering, and sign conventions must also match. A generic driven, aging, or technical-noise environment can possess a response function and fluctuation spectrum without obeying the equilibrium relation.