Linear Response Formula Sheet

Linear response predicts the first change in a detector observable caused by a weak source. The formulas are compact only after the source sign, coupled operator, detector, reference state, Fourier transform, contact terms, and limiting protocol have been fixed.

This sheet uses one convention throughout:

$$\begin{aligned} H_{\mathrm{pert}}(t) &= -f(t)B, \\ \chi_{AB}^{\mathrm R}(t) &= \frac{i}{\hbar} \theta(t) \times \langle[A(t),B(0)]\rangle_0. \end{aligned}$$

With the transform $e^{+i\omega t}$, a passive Hermitian autochannel has

$$\operatorname{Im} \chi_{AA}^{\mathrm R}(\omega) \geq 0 \qquad (\omega>0).$$

Changing the source sign or retarded-kernel sign changes intermediate formulas. It must not change measurable absorbed power after all definitions are transformed consistently.

Required background. Correlation function definitions fix the ordering and spectrum conventions; the Kubo formula derives the source response.

The ordered label $\chi_{AB}$ means the response of $A$ to a source coupled through $B$. The measured operator may also depend explicitly on that source, producing a contact term. Equilibrated thermodynamic derivatives and isolated dynamical responses are distinct protocols.

Master Kubo Formula

Let a multicomponent source couple as

$$H_{\mathrm{pert}}(t) = -\sum_b \int d^d r\, f_b(\mathbf r,t) B_b(\mathbf r).$$

For a detector $A_a(\mathbf r,t)$ with no explicit source dependence,

$$\delta\langle A_a(\mathbf r,t) \rangle ={} \sum_b \int_{-\infty}^{\infty} dt' \int d^d r' \times \chi_{ab}^{\mathrm R} (\mathbf r,t;\mathbf r',t') \times f_b(\mathbf r',t') + O(f^2),$$

where

$$\chi_{ab}^{\mathrm R} (\mathbf r,t;\mathbf r',t') ={} \frac{i}{\hbar} \theta(t-t') \times \bigl\langle \bigl[ A_a(\mathbf r,t), B_b(\mathbf r',t') \bigr] \bigr\rangle_0.$$

The operators evolve with the unperturbed generator. The expectation value is evaluated in the unperturbed state.

The operational definition is

$$\chi_{ab}^{\mathrm R} = \left. \frac{ \delta\langle A_a\rangle_f }{ \delta f_b } \right|_{f=0}$$

when $A_a$ has no explicit dependence on $f_b$. Consequently,

$$[\chi_{ab}(\mathbf r,t;\mathbf r',t')] =\frac{[A_a]}{[f_b]\,\mathrm{time}\,\mathrm{length}^{d}}.$$

The Fourier-transformed susceptibility has units $[A_a]/[f_b]$ because the time and space integrals have been performed.

One-line sign check

In the interaction picture,

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

Cyclicity of the trace gives

$$\delta\langle A(t)\rangle = \frac{i}{\hbar} \int_{t_0}^{t} dt'\, f(t') \times \langle[A_I(t),B_I(t')]\rangle_0.$$

This is the quickest way to recover the response sign from $H_{\mathrm{pert}}=-fB$ instead of relying on memory.

Explicit Source Dependence and Contact Terms

If the measured operator depends on the source, the commutator is not the full functional derivative. Write

$$A_a[f] = A_a^{(0)} + \sum_b \mathcal C_{ab}f_b + O(f^2).$$

For a local instantaneous dependence,

$$K_{ab}^{\mathrm R} (\mathbf r,t;\mathbf r',t') ={} \chi_{A_a^{(0)}B_b}^{\mathrm R} (\mathbf r,t;\mathbf r',t') + \langle\mathcal C_{ab}\rangle_0 \delta^{(d)}(\mathbf r-\mathbf r') {}\times \delta(t-t').$$

The full response is

$$\delta\langle A_a\rangle = \sum_b K_{ab}^{\mathrm R} *f_b.$$

Depending on context, $\mathcal C$ is called a contact, diamagnetic, stress, or seagull term. Derive it from the source-dependent Hamiltonian or observable. Omitting it can violate gauge invariance, continuity constraints, and sum rules.

Stationarity and Fourier Dictionary

If

$$[H_0,\rho_0] = 0,$$

then

$$\chi^{\mathrm R}(t,t') = \chi^{\mathrm R}(t-t').$$

For a translation-invariant state and geometry,

$$\chi^{\mathrm R} (\mathbf r,t;\mathbf r',t') = \chi^{\mathrm R} (\mathbf r-\mathbf r',t-t').$$

This sheet uses

$$\begin{aligned} X(\mathbf q,\omega) &={} \int_{-\infty}^{\infty} dt \int d^d r\, e^{i\omega t-i\mathbf q\cdot\mathbf r} \times X(\mathbf r,t), \\ X(\mathbf r,t) &={} \int\frac{d\omega}{2\pi} \int\frac{d^d q}{(2\pi)^d} \times e^{-i\omega t+i\mathbf q\cdot\mathbf r} X(\mathbf q,\omega). \end{aligned}$$

The convolution becomes

$$\delta\langle A_a(\mathbf q,\omega) \rangle = \sum_b K_{ab}^{\mathrm R} (\mathbf q,\omega) f_b(\mathbf q,\omega).$$

For

$$f(t) = \operatorname{Re} \left[ f_0e^{-i\omega t} \right],$$

the steady first-order response is

$$\delta\langle A(t)\rangle = \operatorname{Re} \left[ \chi_{AB}^{\mathrm R}(\omega) f_0e^{-i\omega t} \right].$$

Under this transform:

• retarded kernels are analytic for $\operatorname{Im}z>0$;
• undamped finite-system poles lie on the real boundary with the retarded $-i0$ displacement; damped stable poles lie below it;
• the retarded Lehmann denominator carries $+i0$;
• $E=\hbar\omega$ converts angular frequency to energy.

Lehmann Lookup

For

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

define

$$A_{nm} = \langle n\lvert A\rvert m\rangle, \qquad B_{mn} = \langle m\lvert B\rvert n\rangle.$$

Then

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

The four ingredients are transition energies, detector matrix elements, source matrix elements, and population differences. A state can exist in the spectrum yet remain absent from a chosen response because either operator matrix element vanishes.

For numerical work, replace $i0$ by a finite width only after declaring whether it represents instrumental resolution, physical damping, a finite observation window, or visualization. A broadening parameter is not generated by the exact finite-system formula.

Response Spectral Density

Define

$$\rho_{AB}(\omega) \equiv{} \frac{1}{2\pi\hbar} \int_{-\infty}^{\infty} dt\, e^{i\omega t} \times \langle[A(t),B(0)]\rangle_0.$$

For $\operatorname{Im}z\ne0$ and adequate high-frequency behavior,

$$\chi_{AB}(z) = \int_{-\infty}^{\infty} d\omega'\, \frac{ \rho_{AB}(\omega') }{ \omega'-z }.$$

The retarded boundary value is

$$\chi_{AB}^{\mathrm R}(\omega) ={} \operatorname{PV} \int_{-\infty}^{\infty} d\omega'\, \frac{ \rho_{AB}(\omega') }{ \omega'-\omega } + i\pi\rho_{AB}(\omega).$$

Therefore define the general discontinuity

$$\chi''_{AB}(\omega)=\frac{\chi^{\mathrm R}_{AB}(\omega)-\chi^{\mathrm A}_{AB}(\omega)}{2i} =\pi\rho_{AB}(\omega).$$

For a matrix of Hermitian channels, $\chi^{\mathrm A}=\chi^{\mathrm R\dagger}$. Only a diagonal Hermitian channel reduces this automatically to $\operatorname{Im}\chi^{\mathrm R}_{AA}$; a complex cross-channel spectrum need not equal the elementwise imaginary part.

For a Hermitian autochannel in a stationary passive equilibrium state,

$$\begin{aligned} \rho_{AA}(-\omega) &= -\rho_{AA}(\omega), \\ \operatorname{Re} \chi_{AA}^{\mathrm R}(-\omega) &= \operatorname{Re} \chi_{AA}^{\mathrm R}(\omega), \\ \operatorname{Im} \chi_{AA}^{\mathrm R}(-\omega) &= -\operatorname{Im} \chi_{AA}^{\mathrm R}(\omega), \end{aligned}$$

and

$$\operatorname{Im} \chi_{AA}^{\mathrm R}(\omega) \geq 0 \qquad (\omega>0).$$

These parity and positivity statements do not apply component by component to a generic cross-susceptibility.

Absorption and Passivity

For a Hermitian source channel driven harmonically,

$$\overline P = \frac{\omega}{2} \lvert f_0\rvert^2 \operatorname{Im} \chi_{BB}^{\mathrm R}(\omega).$$

Thus the imaginary part is absorptive under the conventions of this sheet. The real part is reactive.

For several source components, define the dissipative matrix

$$\boldsymbol{\chi}'' \equiv \frac{ \boldsymbol{\chi}^{\mathrm R} - \boldsymbol{\chi}^{\mathrm R\dagger} }{ 2i }.$$

Then

$$\overline P = \frac{\omega}{2} \mathbf f_0^\dagger \boldsymbol{\chi}'' \mathbf f_0.$$

Passivity requires $\boldsymbol{\chi}''$ to be positive semidefinite for $\omega>0$. An individual off-diagonal entry can have either sign.

Causality and Dispersion

Causal support,

$$\chi_{AB}^{\mathrm R}(t) = 0 \qquad (t<0),$$

implies upper-half-plane analyticity. When no subtraction is needed,

$$\operatorname{Re} \chi_{AB}^{\mathrm R}(\omega) = \frac1\pi \operatorname{PV} \times \int_{-\infty}^{\infty} d\omega'\, \frac{ \operatorname{Im} \chi_{AB}^{\mathrm R}(\omega') }{ \omega'-\omega },$$

and

$$\operatorname{Im} \chi_{AB}^{\mathrm R}(\omega) = -\frac1\pi \operatorname{PV} \times \int_{-\infty}^{\infty} d\omega'\, \frac{ \operatorname{Re} \chi_{AB}^{\mathrm R}(\omega') }{ \omega'-\omega }.$$

Slow high-frequency decay requires subtracted dispersion relations. A finite experimental frequency window does not generally support an accurate direct Hilbert transform without tail modeling.

The advanced kernel is

$$\chi_{AB}^{\mathrm A}(t) = -\frac{i}{\hbar} \theta(-t) \langle[A(t),B(0)]\rangle_0.$$

For a channel matrix on the real axis,

$$\boldsymbol{\chi}^{\mathrm A}(\omega) = \boldsymbol{\chi}^{\mathrm R\dagger}(\omega).$$

Equilibrium Fluctuation–Dissipation Dictionary

Let

$$\delta A = A-\langle A\rangle_\beta, \qquad \delta B = B-\langle B\rangle_\beta.$$

Using angular-frequency spectra without a $2\pi$ prefactor, define

$$\begin{aligned} S_{AB}^{>}(\omega) &\equiv \int_{-\infty}^{\infty} dt\, e^{i\omega t} \langle\delta A(t)\delta B(0)\rangle_\beta, \\ S_{AB}^{<}(\omega) &\equiv \int_{-\infty}^{\infty} dt\, e^{i\omega t} \langle\delta B(0)\delta A(t)\rangle_\beta. \end{aligned}$$

For one equilibrium thermal generator and a compatible operator pair,

$$S_{AB}^{<}(\omega) = e^{-\beta\hbar\omega} S_{AB}^{>}(\omega).$$

The absorptive response is

$$\chi''_{AB}(\omega) = \frac{ S_{AB}^{>}(\omega) - S_{AB}^{<}(\omega) }{ 2\hbar }.$$

Equivalently,

$$S_{AB}^{>}(\omega) = \frac{ 2\hbar }{ 1-e^{-\beta\hbar\omega} } \chi''_{AB}(\omega).$$

For a Hermitian autochannel,

$$S_{AA}^{\mathrm{sym}}(\omega) \equiv \frac12 \int_{-\infty}^{\infty} dt\, e^{i\omega t} \times \bigl\langle \bigl\{ \delta A(t), \delta A(0) \bigr\} \bigr\rangle_\beta.$$

For nonzero-frequency spectral weight, the fluctuation–dissipation theorem is

$$S_{AA}^{\mathrm{sym}}(\omega) = \hbar \coth\left( \frac{\beta\hbar\omega}{2} \right) \operatorname{Im} \chi_{AA}^{\mathrm R}(\omega).$$

In the classical low-frequency regime,

$$\beta\hbar\lvert\omega\rvert \ll 1,$$

this becomes

$$S_{AA}^{\mathrm{sym}}(\omega) \simeq \frac{ 2k_{\mathrm B}T }{ \omega } \operatorname{Im} \chi_{AA}^{\mathrm R}(\omega).$$

An exactly static connected component $C_0$ contributes $2\pi C_0\delta(\omega)$ to the fluctuation spectrum but no commutator response. Add that component separately: the singular thermal multiplier cannot reconstruct it from zero. The theorem requires thermal equilibrium, not merely stationarity. For a grand-canonical ensemble, number-changing operators must use a consistent generator so that the detailed-balance exponent includes chemical work correctly.

Static Thermodynamic Response

For

$$H_f = H_0-fB,$$

the isothermal self-susceptibility is

$$\chi_T^{BB} \equiv \left. \frac{ \partial\langle B\rangle_f }{ \partial f } \right|_{f=0}.$$

In the canonical ensemble, for a Hermitian $B$ preserving any sector fixed by the ensemble,

$$\chi_T^{BB} = \int_0^\beta d\lambda\, \left\langle \delta B(-i\hbar\lambda) \delta B(0) \right\rangle_\beta,$$

where

$$B(-i\hbar\lambda) = e^{\lambda H_0} B e^{-\lambda H_0}.$$

Equivalently,

$$\chi_T^{BB} = \frac1\hbar \int_0^{\beta\hbar} d\tau\, \langle\delta B(\tau)\delta B(0)\rangle_\beta.$$

If

$$[B,H_0] = 0,$$

then

$$\chi_T^{BB} = \beta \left( \langle B^2\rangle_\beta - \langle B\rangle_\beta^2 \right).$$

For a nondegenerate zero-temperature ground state,

$$\chi_{T=0}^{BB} = 2 \sum_{n>0} \frac{ \lvert\langle n\lvert B\rvert0\rangle\rvert^2 }{ E_n-E_0 }.$$

The thermodynamic derivative need not equal the zero-frequency isolated retarded response. In particular, if $B$ is conserved,

$$\chi_{BB}^{\mathrm R}(t) = 0,$$

while $\chi_T^{BB}$ can remain nonzero because equilibration changes statistical weights among sectors.

Order of Limits

For a spatial response, distinguish

$$\chi_{\mathrm{static}} = \lim_{\mathbf q\to0} \lim_{\omega\to0} \chi(\mathbf q,\omega)$$

from

$$\chi_{\mathrm{uniform}} = \lim_{\omega\to0} \lim_{\mathbf q\to0} \chi(\mathbf q,\omega).$$

Hydrodynamic poles and conservation laws can make these limits unequal. Transport may additionally require an explicit order for

$$\begin{aligned} L&\to\infty, & \eta&\to0^+, \\ \mathbf q&\to0, & \omega&\to0. \end{aligned}$$

A finite isolated spectrum usually gives delta functions, recurrences, or a vanishing uniform commutator response rather than a bulk dc coefficient.

Common Source and Detector Channels

Physical source	Coupled operator	Typical detector	Essential convention
force $F$	position $x$	position or velocity	$H_{\mathrm{pert}}=-Fx$
spin field $h^\beta$ in energy units	spin $S^\beta$	spin $S^\alpha$	add $g\mu_{\mathrm B}$ when converting to magnetic-field units
local chemical potential $\delta\mu$	number density $n$	density $n$	distinguish $\partial n/\partial\mu$ from $n^{-2}\partial n/\partial\mu$
vector potential $\mathbf A$	paramagnetic current $\mathbf J^p$	physical current $\mathbf j[\mathbf A]$	include the diamagnetic or stress term
pairing source $\eta$	pair operator and its adjoint	pair amplitude	declare charge, phase, and Nambu normalization
strain or metric source	stress tensor	stress tensor	include instantaneous elastic terms

The source should be defined by its Hamiltonian coupling, not only by a verbal label.

Magnetic and Density Lookups

For a spin source in energy units and dimensionless spin operators $S_j^\beta$,

$$H_{\mathrm{pert}}(t) = -\sum_{j,\beta} h_j^\beta(t) S_j^\beta,$$

the response is

$$\delta\langle S_i^\alpha(\omega) \rangle = \sum_{j,\beta} \chi_{S_i^\alpha S_j^\beta}^{\mathrm R}(\omega) h_j^\beta(\omega).$$

Magnetic susceptibility quoted per tesla acquires the appropriate powers of $g\mu_{\mathrm B}$ and any volume or molar normalization.

For a local chemical-potential source,

$$H_{\mathrm{pert}}(t) = -\int d^d r\, \delta\mu(\mathbf r,t) n(\mathbf r),$$

one has

$$\delta n(\mathbf q,\omega) = \chi_{nn}^{\mathrm R}(\mathbf q,\omega) \delta\mu(\mathbf q,\omega).$$

Two common compressibilities are

$$\kappa_n = \left( \frac{\partial n}{\partial\mu} \right)_T, \qquad \kappa_T = \frac1{n^2} \left( \frac{\partial n}{\partial\mu} \right)_T.$$

At exactly $\mathbf q=0$, a conserved total number has no finite-frequency isolated self-response. The thermodynamic compressibility is obtained from the static equilibrated protocol.

Conductivity Lookup

For continuum particles of signed charge $Q$ in a uniform vector potential,

$$H[\mathbf A] = \sum_i \frac{ (\mathbf p_i-Q\mathbf A)^2 }{ 2m } + V.$$

Expanding gives

$$H[\mathbf A] = H_0 - \mathbf A\cdot\mathbf J^p + \frac{NQ^2}{2m} \mathbf A^2.$$

The physical current density is

$$\mathbf j[\mathbf A] = -\frac1V \frac{\partial H}{\partial\mathbf A} = \mathbf j^p - \frac{nQ^2}{m} \mathbf A.$$

Define

$$\begin{aligned} \Pi_{\alpha\beta}^{\mathrm R}(\omega) &\equiv \frac1V \chi_{J_\alpha^pJ_\beta^p}^{\mathrm R}(\omega), \\ D_{\alpha\beta} &\equiv \frac{nQ^2}{m} \delta_{\alpha\beta}. \end{aligned}$$

Since

$$\mathbf E(\omega) = i\omega\mathbf A(\omega),$$

the optical conductivity is

$$\sigma_{\alpha\beta}(\omega) = \frac{i}{\omega+i0} \left[ D_{\alpha\beta} - \Pi_{\alpha\beta}^{\mathrm R}(\omega) \right].$$

On a lattice, $D_{\alpha\beta}$ is the appropriate kinetic or stress expectation obtained by differentiating the gauged Hamiltonian. It is not generally $nQ^2/m$.

A standard decomposition is

$$\operatorname{Re} \sigma(\omega) = \pi D_{\mathrm{Dr}} \delta(\omega) + \sigma_{\mathrm{reg}}(\omega).$$

The Drude weight $D_{\mathrm{Dr}}$ is the surviving zero-frequency weight after paramagnetic and diamagnetic contributions are combined. It is not the bare tensor $D_{\alpha\beta}$.

Diffusion and Einstein Lookup

For one conserved density in a diffusive regime,

$$\begin{aligned} \chi_{nn}^{\mathrm R}(\mathbf q,\omega) &= \chi \frac{ Dq^2 }{ Dq^2-i\omega }, \\ \chi &= \left( \frac{\partial n}{\partial\mu} \right)_T. \end{aligned}$$

The pole is

$$\omega = -iDq^2.$$

It reproduces the noncommuting limits

$$\begin{aligned} \lim_{\mathbf q\to0} \lim_{\omega\to0} \chi_{nn}^{\mathrm R} &= \chi, \\ \lim_{\omega\to0} \lim_{\mathbf q\to0} \chi_{nn}^{\mathrm R} &= 0. \end{aligned}$$

For carriers of signed charge $Q$, consistent number-current normalization gives the Einstein relation

$$\sigma = Q^2\chi D.$$

These formulas require a genuine diffusive window, local equilibrium, and the same conserved density in $\chi$, $D$, and $\sigma$.

Reciprocity Lookup

Adjoint relations follow from definitions. Onsager–Casimir reciprocity requires equilibrium microscopic reversibility and correctly transformed time-reversal-odd controls.

If $A$ and $B$ have time-reversal parities $\epsilon_A$ and $\epsilon_B$, a common convention gives

$$\chi_{AB}^{\mathrm R} (\omega;\mathcal B) = \epsilon_A\epsilon_B \chi_{BA}^{\mathrm R} (\omega;-\mathcal B).$$

Momentum, current, pseudovector, and complex-basis indices can add component transformations. Reversing operator labels while leaving a magnetic field or rotation unchanged is not a reciprocity test.

Matsubara Route

For an equilibrium bosonic observable channel, a Matsubara calculation samples an analytic response function at

$$\nu_n = \frac{2\pi n}{\beta}, \qquad n\in\mathbb Z.$$

The retarded response is the upper real-axis boundary of the same analytic function after all overall signs, factors of $\hbar$, static zero-mode terms, and contact terms have been matched.

Symbolic replacement

$$i\nu_n \longrightarrow \hbar\omega+i0$$

is meaningful only after that analytic function has been identified. Recovering a real-frequency spectrum from finitely many noisy Matsubara values is an ill-posed inverse problem, not a direct substitution.

Thermal Green Functions defines imaginary-time ordering. Spectral Representation explains the continuation bridge and static bosonic terms. Analytic Continuation explains numerical reconstruction.

Exact Benchmark: Harmonic Oscillator

Let

$$\begin{aligned} H_0 &= \frac{p^2}{2m} + \frac12m\Omega^2x^2, \\ H_{\mathrm{pert}}(t) &= -F(t)x. \end{aligned}$$

Since

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

the retarded response is

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

Its frequency-domain form is

$$\chi_{xx}^{\mathrm R}(\omega) = \frac1m \frac1{ \Omega^2-(\omega+i0)^2 }.$$

The static response is

$$\chi_{xx}^{\mathrm R}(0) = \frac1{m\Omega^2},$$

and the positive-frequency absorptive line is

$$\operatorname{Im} \chi_{xx}^{\mathrm R}(\omega) = \frac{\pi}{2m\Omega} \delta(\omega-\Omega).$$

This benchmark checks the source sign, pole side, static spring compliance, spectral weight, and positive absorption simultaneously.

Exact Benchmark: Two-Level System

Take

$$H_0 = \frac{\hbar\Omega}{2} \sigma_z, \qquad H_{\mathrm{pert}}(t) = -f(t)\sigma_x.$$

At inverse temperature $\beta$, define the population difference

$$\Delta p = \tanh\left( \frac{\beta\hbar\Omega}{2} \right).$$

The transverse response is

$$\begin{aligned} \chi_{xx}^{\mathrm R}(\omega) ={}& \frac{\Delta p}{\hbar} \biggl[ \frac1{\omega+\Omega+i0} - \frac1{\omega-\Omega+i0} \biggr] \\ ={}& \frac{2\Omega\Delta p}{\hbar} \frac1{ \Omega^2-(\omega+i0)^2 }. \end{aligned}$$

Thus

$$\chi_{xx}^{\mathrm R}(0) = \frac{ 2\Delta p }{ \hbar\Omega },$$

and, for $\omega>0$,

$$\operatorname{Im} \chi_{xx}^{\mathrm R}(\omega) = \frac{\pi\Delta p}{\hbar} \delta(\omega-\Omega).$$

Positive temperature gives $\Delta p>0$ and absorption. Population inversion gives $\Delta p<0$ and gain.

Calculation Routes

Route	Direct output	Main caution
exact diagonalization	discrete Lehmann weights	finite spectra require declared broadening and limit order
real-time evolution	causal kernel or ordinary correlator	finite time causes windowing and frequency convolution
equilibrium fluctuations	ordered or symmetrized spectra	fluctuation–dissipation conversion requires thermal equilibrium
Matsubara calculation	imaginary-axis samples	continuation is ill posed with finite noisy data
perturbation theory	approximate analytic kernel	preserve causality, symmetries, and contact terms consistently
experiment	convolved detector response	de-embed source calibration, resolution, backgrounds, and units

Exercises

Exercise 1: Oscillator static and absorptive checks

For the oscillator benchmark, verify the static limit and the sign of the positive-frequency spectral line.

Solution

Setting $\omega=0$ in

$$\chi_{xx}^{\mathrm R}(\omega) = \frac1m \frac1{ \Omega^2-(\omega+i0)^2 }$$

gives

$$\chi_{xx}^{\mathrm R}(0) = \frac1{m\Omega^2}.$$

This is the classical displacement per applied static force. Using

$$\frac1{x+i0} = \operatorname{PV}\frac1x - i\pi\delta(x),$$

the pole at $\omega=\Omega$ contributes

$$\operatorname{Im} \chi_{xx}^{\mathrm R}(\omega) = \frac{\pi}{2m\Omega} \delta(\omega-\Omega)$$

for positive frequency. Its coefficient is positive, as passivity requires.

Exercise 2: Conserved quantity paradox

Suppose $[B,H_0]=0$. Show why the isolated retarded self-response vanishes while the equilibrium isothermal susceptibility can be nonzero.

Solution

Conservation gives

$$B(t) = B,$$

so

$$\chi_{BB}^{\mathrm R}(t) = \frac{i}{\hbar} \theta(t) \langle[B,B]\rangle_0 = 0.$$

In a canonical equilibrium family with $H_f=H_0-fB$,

$$\chi_T^{BB} = \beta \left( \langle B^2\rangle - \langle B\rangle^2 \right),$$

which is nonzero whenever different conserved sectors have fluctuating thermal weights. The retarded protocol evolves one isolated state; the thermodynamic protocol compares re-equilibrated states.

Exercise 3: Diffusive order of limits

Evaluate both iterated limits of

$$\chi_{nn}^{\mathrm R}(\mathbf q,\omega) = \chi \frac{ Dq^2 }{ Dq^2-i\omega }.$$

Solution

At fixed nonzero $\mathbf q$,

$$\lim_{\omega\to0} \chi_{nn}^{\mathrm R}(\mathbf q,\omega) = \chi.$$

Taking $\mathbf q\to0$ afterward leaves $\chi$. At fixed nonzero $\omega$,

$$\lim_{\mathbf q\to0} \chi_{nn}^{\mathrm R}(\mathbf q,\omega) = 0.$$

Taking $\omega\to0$ afterward leaves zero. The pole at $\omega=-iDq^2$ makes the origin path dependent and encodes conservation of the uniform mode.

Exercise 4: Diamagnetic contact term

For

$$H[\mathbf A] = \sum_i \frac{ (\mathbf p_i-Q\mathbf A)^2 }{ 2m } + V,$$

derive the source-dependent physical current and identify the contact term.

Solution

Expanding the kinetic energy gives

$$H[\mathbf A] = H_0 - \mathbf A\cdot\mathbf J^p + \frac{NQ^2}{2m} \mathbf A^2.$$

The physical current density follows from differentiating the Hamiltonian:

$$\begin{aligned} \mathbf j[\mathbf A] &= -\frac1V \frac{\partial H}{\partial\mathbf A} \\ &= \mathbf j^p - \frac{nQ^2}{m} \mathbf A. \end{aligned}$$

The second term is instantaneous and therefore contributes

$$-\frac{nQ^2}{m} \delta_{\alpha\beta} \delta(t-t')$$

to the current response to $\mathbf A$. Combining it with the paramagnetic current commutator yields the gauge-consistent conductivity kernel.

References

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