Kubo Formula

The Kubo formula gives the first change in a many-body observable produced by a weak external source. If the perturbation is

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

then the response of an observable $A$ is

$$\delta\langle A(t)\rangle = \int_{-\infty}^{\infty} dt'\, \chi^R_{AB}(t,t') f(t') + O(f^2),$$

with retarded susceptibility

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

The sign is tied to the explicit choice $H_{\mathrm{pert}}=-fB$. The step function enforces causality. The commutator measures the difference between the two operator orders. The expectation value is taken in the unperturbed reference state, and the Heisenberg operators evolve with the unperturbed Hamiltonian.

This identity is exact to first order in the source. The hard work in a many-body application is specifying the correct source, detector, ensemble, current or density operators, contact terms, Fourier convention, boundary conditions, and order of limits.

Helpful background. The interaction picture separates unperturbed evolution from the source; thermal density operators supplies equilibrium averages. The derivation itself also applies to a specified nonthermal initial state.

Source and detector

The source and measured observable play different roles:

$$\underbrace{f(t)}_{\text{source}} \quad\longleftrightarrow\quad \underbrace{B}_{\text{coupled operator}}, \qquad \underbrace{A}_{\text{detector}}.$$

The ordered label $\chi_{AB}$ means “response of $A$ to a source coupled through $B$.” In general,

$$\chi^R_{AB} \ne \chi^R_{BA}.$$

Representative choices are:

Source	Coupled operator	Detector	Response
magnetic field	magnetization	magnetization	magnetic susceptibility
local chemical potential	density	density	density response or compressibility
electric vector potential	paramagnetic current	physical current	conductivity
force	position	position	mechanical susceptibility
pairing field	pair creation and annihilation	pair amplitude	pairing susceptibility

We assume the source-dependent Hamiltonian is Hermitian and the required correlation functions exist. A non-Hermitian pair field needs both it and its Hermitian conjugate in the coupling. Write $H_{\mathrm{pert}}$ explicitly. The phrase “apply a field” is insufficient because the sign, units, spatial profile, and coupled operator determine the response convention.

Kubo response separates the source $f$, coupled operator $B$, and measured observable $A$. The retarded commutator gives the state-mediated response. If $A$ depends explicitly on the source, its derivative contributes an additional instantaneous contact term.

General Space-Time Form

Let the source have components $f_b(\mathbf r,t)$ and write

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

The induced change in $A_a(\mathbf r,t)$ is

$$\delta\langle A_a(\mathbf r,t) \rangle ={} \sum_b \int_{-\infty}^{\infty} dt' \int d^d r' \times \chi^R_{ab} (\mathbf r,t;\mathbf r',t') f_b(\mathbf r',t').$$

The retarded kernel is

$$\chi^R_{ab} (\mathbf r,t;\mathbf r',t') ={} \frac{i}{\hbar} \theta(t-t') \times \left\langle \left[ A_{a,H}(\mathbf r,t), B_{b,H}(\mathbf r',t') \right] \right\rangle_0.$$

On a lattice, replace the spatial integrals by site sums. The formula remains valid for a nonstationary reference state, but the kernel then depends on $t$ and $t'$ separately.

Derivation from the Density Operator

Write

$$H(t) = H_0+\lambda H_{\mathrm{pert}}(t),$$

where $\lambda$ tracks source order. In the interaction picture,

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

To first order, replace $\rho_I(t)$ on the right by the unperturbed state $\rho_0$:

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

For $H_{\mathrm{pert},I}(t')=-f(t')B_I(t')$,

$$\delta\rho_I(t) ={} \frac{i\lambda}{\hbar} \int_{t_0}^{t} dt'\, f(t') [B_I(t'),\rho_0] + O(\lambda^2).$$

Taking the trace with $A_I(t)$ and using cyclicity,

$$\delta\langle A(t)\rangle ={} \frac{i\lambda}{\hbar} \int_{t_0}^{t} dt'\, f(t') \times \left\langle [A_I(t),B_I(t')] \right\rangle_0 + O(\lambda^2).$$

For a source that vanishes before $t_0$, extending the integral over all $t'$ inserts $\theta(t-t')$. Setting $\lambda=1$ after identifying the first-order coefficient gives the Kubo formula.

The derivation assumes that the initial state is unmodified at the lower integration limit. Adiabatic switching from the distant past, a finite preparation time, and a quench protocol are physically different prescriptions and can select different steady or transient responses.

Functional-Derivative Definition

Operationally,

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

when $A$ has no explicit source dependence. The time-domain kernel has units $[A]/([f]\,\text{time})$, because the response integrates over $dt\prime$. After a time Fourier transform, $\chi(\omega)$ has units $[A]/[f]$. A spatially resolved kernel also carries the inverse volume appropriate to its integration measure.

For multiple components,

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

The response can be anisotropic, nonlocal, and nondiagonal in internal indices even when the source is scalar.

Stationarity and Translation Invariance

Suppose

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

Then the reference state is stationary and

$$\chi^R_{AB}(t,t') = \chi^R_{AB}(t-t').$$

If the Hamiltonian, state, and geometry are also translation invariant,

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

Use the Fourier convention

$$X(\mathbf q,\omega) = \int_{-\infty}^{\infty} dt \int d^d r\, e^{i\omega t-i\mathbf q\cdot\mathbf r} X(\mathbf r,t).$$

The convolution becomes

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

With this $e^{+i\omega t}$ transform and a stable kernel of at most polynomial growth, a retarded function is analytic for $\operatorname{Im}\omega\gt0$. The retarded boundary value carries $+i0$ in denominators.

Lehmann Representation

Choose a common eigenbasis of $H_0$ and a stationary $\rho_0$:

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

Define

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

The frequency-domain susceptibility is

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

This formula exposes four independent ingredients:

1. transition energies $E_m-E_n$;
2. source matrix elements $B_{mn}$;
3. detector matrix elements $A_{nm}$;
4. population differences $p_n-p_m$.

A pole in the spectrum is invisible if either the source or detector matrix element vanishes. Population differences encode absorption minus stimulated emission. In an inverted state they can reverse the sign of the absorptive response.

The infinitesimal $i0$ enforces the retarded boundary condition. Replacing it by a finite $\Gamma$ is a physical or phenomenological broadening assumption and must be justified separately.

Absorptive Response

For a Hermitian operator $A=B$,

$$\operatorname{Im} \chi^R_{AA}(\omega) ={} \frac{\pi}{\hbar} \sum_{n,m} (p_n-p_m) \lvert A_{nm}\rvert^2 \times \delta\!\left( \omega-\frac{E_m-E_n}{\hbar} \right).$$

For a positive-temperature thermal state and $\omega\gt0$, upward transitions have $p_n\gt p_m$, so this convention gives

$$\operatorname{Im} \chi^R_{AA}(\omega) \ge 0.$$

Let a real harmonic source be

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

The cycle-averaged work done on the system is

$$\overline P = \frac{\omega}{2} \lvert f_0\rvert^2 \operatorname{Im} \chi^R_{BB}(\omega)$$

for the coupling $H_{\mathrm{pert}}=-fB$. The imaginary part is therefore the absorptive or dissipative component under these conventions. The real part is reactive: it changes the in-phase response and stores rather than irreversibly absorbs energy over a cycle.

This positivity statement applies to a passive equilibrium state and a diagonal source–detector choice. A generic cross-susceptibility need not have a sign-definite imaginary part.

Causality and Dispersion Relations

Because

$$\chi^R_{AB}(t) = 0 \qquad \text{for }t\lt0,$$

its Fourier–Laplace transform is analytic in the upper half-plane when the kernel is tempered (in particular, not exponentially growing). Under suitable decay or subtraction conditions,

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

A companion relation reconstructs the imaginary part from the real part. These Kramers–Kronig relations follow from causality and analyticity, not from a particular microscopic model.

Slow high-frequency decay may require subtracted dispersion relations. Truncating a measured frequency interval can also make a direct numerical transform unreliable.

Static Response Is Not One Limit

Several quantities are casually called “the static susceptibility”:

• the thermodynamic derivative after the system re-equilibrates;
• $\chi^R(\mathbf q,\omega=0)$ at finite $\mathbf q$;
• the limit $\omega\to0$ after the thermodynamic limit;
• the uniform limit $\mathbf q\to0$;
• the response of an isolated finite system under adiabatic switching.

They need not agree.

For an equilibrium Hamiltonian

$$H_f = H_0-fB,$$

the isothermal susceptibility is

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

In a canonical ensemble it can be written

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

where

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

If $[B,H_0]=0$, this reduces to

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

Fluctuations and Susceptibilities develops this equilibrium identity, the general Kubo–Mori covariance, and the energy, number, and magnetization applications. The real-time commutator probes a different protocol.

Yet the isolated retarded self-response of a conserved $B$ is zero because $[B(t),B]=0$. Re-equilibration changes the statistical weights of conserved sectors; closed-system unitary response cannot do so by itself. This distinction is essential for uniform magnetization, total particle number, and other conserved quantities.

Order of Limits

For a spatial response $\chi(\mathbf q,\omega)$, compare

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

with

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

The first lets an arbitrarily slowly varying spatial perturbation equilibrate before making it uniform. The second makes the perturbation uniform first, so conservation laws can block relaxation. Hydrodynamic poles often make the limits noncommuting.

A simple diffusive density response makes this concrete. For diffusion constant $D>0$ and static susceptibility $\chi_0>0$,

$$\chi_{nn}^R(q,\omega)=\chi_0\frac{Dq^2}{Dq^2-i\omega}.$$

At fixed nonzero $q$, setting $\omega=0$ gives $\chi_0$. At fixed nonzero $\omega$, sending $q\to0$ gives zero. The relaxation time $1/(Dq^2)$ diverges as the modulation becomes uniform; there is no single continuous value at $(q,\omega)=(0,0)$. This form assumes a regime of ordinary diffusion; it does not apply to ballistic density dynamics.

Transport adds further limits:

$$L\to\infty, \qquad \omega\to0, \qquad \mathbf q\to0, \qquad \eta\to0^+.$$

A defensible result states their order. Taking $\omega\to0$ in a finite isolated spectrum generally produces delta functions or zero response rather than a bulk dc coefficient.

Explicit Source Dependence and Contact Terms

The commutator formula is not always the whole derivative. Suppose the measured operator itself depends on the source:

$$A[f] = A^{(0)} + fD + O(f^2).$$

Then

$$\delta\langle A(t)\rangle ={} \int dt'\, \chi^R_{A^{(0)}B}(t,t') f(t') + \langle D\rangle_0 f(t).$$

Equivalently, the full response kernel contains

$$K_{AB}(t,t') = \chi^R_{A^{(0)}B}(t,t') + \langle D\rangle_0 \delta(t-t').$$

The instantaneous term is often called a contact, diamagnetic, or seagull term depending on context. Omitting it can violate gauge invariance, conservation-law sum rules, or the correct static limit.

If $A[f]$ is defined as a derivative of the Hamiltonian, the contact term comes from a second source derivative of $H[f]$. Its sign must be derived from that definition rather than memorized.

Magnetic Susceptibility

Let a field in energy units couple to local spins:

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

The response is

$$\delta\langle s_i^\alpha(t) \rangle ={} \sum_{j,\beta} \int dt'\, \times \chi^R_{s_i^\alpha s_j^\beta}(t,t') h_j^\beta(t').$$

For a homogeneous system, Fourier transformation gives a tensor

$$\chi_{\alpha\beta}^R(\mathbf q,\omega).$$

The uniform static susceptibility is related to derivatives of total magnetization after specifying equilibration and volume normalization. If $M^z$ is conserved,

$$\chi_{M^zM^z}^R(\mathbf q=0,\omega) = 0$$

for the isolated commutator response, while the equilibrium isothermal susceptibility can be

$$\chi_T = \frac{\beta}{V} \left[ \langle(M^z)^2\rangle - \langle M^z\rangle^2 \right].$$

There is no contradiction: these quantities implement different protocols.

Density Response and Compressibility

Use a local chemical-potential source,

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

Then

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

The equilibrium compressibility probes a static, long-wavelength density response. Two common conventions are

$$\kappa_n = \frac{\partial n}{\partial\mu}$$

and

$$\kappa_T = \frac{1}{n^2} \frac{\partial n}{\partial\mu}.$$

They differ by $n^{-2}$, so the symbol $\kappa$ is incomplete without a definition.

At exactly $\mathbf q=0$, the density operator is total particle number. If it is conserved, its finite-frequency retarded self-response vanishes. The thermodynamic compressibility instead compares equilibrium states with different chemical-potential weights. The static and uniform limits must therefore be distinguished.

Electrical Conductivity

Conductivity is the standard example in which a contact term is indispensable. For particles of charge $q$ in a spatially uniform vector potential,

$$H[\mathbf A] = \sum_{i=1}^{N} \frac{ \left( \mathbf p_i-q\mathbf A \right)^2 }{2m} + V.$$

Expanding to second order,

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

where $\mathbf J^p$ is the total paramagnetic current. The physical current density is

$$\mathbf j[\mathbf A] = -\frac{1}{V} \frac{\partial H}{\partial\mathbf A} = \mathbf j^p - \frac{nq^2}{m} \mathbf A.$$

Define the paramagnetic current-response tensor per volume,

$$\Pi_{\alpha\beta}^R(\omega) = \frac{1}{V} \chi^R_{J_\alpha^pJ_\beta^p}(\omega),$$

and the diamagnetic tensor

$$D_{\alpha\beta} = \frac{nq^2}{m} \delta_{\alpha\beta}.$$

Then

$$\delta j_\alpha(\omega) = \sum_\beta \left[ \Pi_{\alpha\beta}^R(\omega) - D_{\alpha\beta} \right] A_\beta(\omega).$$

With vanishing scalar potential and the Fourier convention used here,

$$E_\beta(\omega) = i\omega A_\beta(\omega).$$

The optical conductivity is therefore

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

The signs follow from three stated choices: charge enters as $\mathbf p-q\mathbf A$, current is $-\partial H/\partial\mathbf A$, and Fourier transforms use $e^{+i\omega t}$. Other conventions can rearrange signs while leaving measurable response unchanged.

On a lattice, $D_{\alpha\beta}$ is an expectation of the appropriate kinetic-energy or stress tensor rather than simply $nq^2/m$. It must be derived by coupling the lattice Hamiltonian to a gauge field, usually through Peierls phases.

Drude and regular weight

A common decomposition is

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

$D_{\mathrm{Dr}}\gt0$ signals a ballistic zero-frequency contribution under the stated order of limits. It is not the same symbol as the bare diamagnetic tensor before paramagnetic cancellation.

For a clean one-species continuum with parabolic dispersion, translation-invariant position-dependent interactions, and conserved total momentum,

$$\Pi^R(\omega) = 0$$

at zero wavevector, giving

$$\sigma(\omega) = \frac{i}{\omega+i0} \frac{nq^2}{m}.$$

Its real part is a delta function, not a finite dc conductivity. A finite resistivity requires relaxation of the component of current protected by conservation. In this Galilean-invariant example, current is proportional to total momentum, so interactions among the same particles cannot relax it. A lattice can allow Umklapp momentum transfer to the crystal; disorder, boundaries, or an equilibrating environment can also remove momentum. Merely coupling electrons to a freely moving closed phonon fluid does not guarantee relaxation of the combined conserved momentum.

Gauge and Conservation Checks

Electromagnetic response must be gauge consistent. A scalar potential and a longitudinal vector potential can represent the same electric field after a gauge transformation. Density and current correlators are therefore linked by the continuity equation and associated Ward identities.

Practical checks include:

• include both paramagnetic and contact contributions;
• use a current operator derived from the same gauged Hamiltonian;
• verify charge conservation and the continuity equation;
• check the optical identities developed on Sum Rules;
• distinguish a uniform vector potential that is pure gauge from flux through a periodic ring;
• state whether the thermodynamic limit is taken before removing the vector potential or frequency regulator.

A violation often signals a missing contact term, inconsistent truncation, nonconserving approximation, or incompatible boundary convention.

Finite Systems and Numerical Spectra

For a finite-dimensional isolated system, or a confined system with a discrete spectrum, the Lehmann representation consists of discrete poles. Its absorptive part is a sum of delta functions:

$$\chi''(\omega) = \sum_\ell w_\ell \delta(\omega-\omega_\ell).$$

Replacing each delta function by

$$\delta_\eta(\omega-\omega_\ell) = \frac{1}{\pi} \frac{ \eta }{ (\omega-\omega_\ell)^2+\eta^2 }$$

is a plotting or regularization choice unless $\eta$ is derived from a physical decay process. Report it explicitly.

For bulk response:

1. resolve exact symmetry sectors and matrix elements;
2. include contact terms and conserved contributions;
3. check sum rules at every size;
4. vary the broadening independently of system size;
5. take a controlled thermodynamic limit;
6. only then analyze zero-frequency or long-time behavior.

A smooth curve from one broadened cluster does not establish a finite lifetime or bulk dc transport coefficient.

Imaginary-Time Calculations

Equilibrium Monte Carlo and Matsubara methods often compute an imaginary-time correlator rather than $\chi^R(\omega)$ directly. The two share spectral information, but obtaining real-frequency response requires analytic continuation.

That inverse problem is ill-conditioned: many real-frequency spectra can fit noisy imaginary-time data within error bars. A trustworthy continuation reports priors, regularization, covariance of the input data, resolution limits, and tests on synthetic spectra.

The Kubo formula does not make analytic continuation automatic. It identifies the target retarded object; a separate inference problem remains.

Analytic Continuation develops that inference problem, including covariance, channel-valid constraints, static terms, method limitations, and feature-level resolution.

Validity of Linear Response

The response expansion requires the induced change to remain first order over the observation window. It can fail when:

• the source is not small relative to intrinsic scales;
• a resonant drive acts long enough for secular growth or saturation;
• the source changes populations appreciably;
• the system crosses a phase boundary;
• heating changes the reference state;
• nonlinear selection rules or harmonics are the observable of interest;
• the reference state itself changes appreciably under the added probe.

A small instantaneous amplitude does not guarantee linear behavior for arbitrarily long time. Near a narrow resonance, a dimensionless accumulated transition amplitude can become order one.

A strongly driven or aging background can still have a valid linear response to an additional weak probe. Use its actual two-time or Floquet response kernel, and check the size of the probe-induced change.

Common pitfalls

Small amplitude does not mean short-time perturbation theory lasts forever. A resonant source can build an order-one change over a sufficiently long interval. Compare the accumulated response with the unperturbed populations and scales.

The current operator may change with the source. In electromagnetic response, differentiating $H[\mathbf A]$ gives both the paramagnetic current and its explicit vector-potential term. Their combination is needed before taking a static or uniform limit.

A nonstationary reference state does not invalidate linear response. It invalidates replacing a two-time kernel by an equilibrium function of one frequency. The weak-probe expansion still follows from the interaction-picture derivation.

Exercises

Sign of the retarded kernel

Starting from the interaction-picture density operator, derive the sign of $\chi^R_{AB}$ for

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

How does the result change for $H_{\mathrm{pert}}=+fB$?

Solution

To first order,

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

Substitution 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

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

Changing the perturbation to $+fB$ reverses the overall sign.

Lehmann weights and absorption

For a thermal state and Hermitian $A$, show that

$$\operatorname{Im} \chi^R_{AA}(\omega) \ge 0$$

for $\omega\gt0$ under this page’s conventions.

Solution

At positive frequency, the delta function selects

$$E_m-E_n = \hbar\omega \gt 0.$$

Thermal probabilities obey

$$p_n = \frac{e^{-\beta E_n}}{Z} \gt \frac{e^{-\beta E_m}}{Z} = p_m.$$

Every selected term in

$$\operatorname{Im} \chi^R_{AA}(\omega) = \frac{\pi}{\hbar} \sum_{n,m} (p_n-p_m) \lvert A_{nm}\rvert^2 \delta(\omega-\omega_{mn})$$

is therefore nonnegative. This is the spectral origin of positive absorbed power for a passive thermal state.

Transverse spin susceptibility

Let

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

For a thermal state, compute $\chi^R_{xx}(t)$ and $\chi^R_{xx}(0)$ in frequency space.

Solution

The Heisenberg operator is

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

Hence

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

The thermal polarization is

$$\langle\sigma_z\rangle_\beta = -\tanh\!\left( \frac{\beta\hbar\Omega}{2} \right).$$

Therefore

$$\chi^R_{xx}(t) = \frac{2}{\hbar} \tanh\!\left( \frac{\beta\hbar\Omega}{2} \right) \theta(t) \sin(\Omega t).$$

Fourier transformation gives

$$\chi^R_{xx}(\omega) = \frac{ 2\Omega }{ \hbar \left[ \Omega^2-(\omega+i0)^2 \right] } \tanh\!\left( \frac{\beta\hbar\Omega}{2} \right).$$

Thus

$$\chi^R_{xx}(0) = \frac{2}{\hbar\Omega} \tanh\!\left( \frac{\beta\hbar\Omega}{2} \right).$$

This agrees with the small static tilt of the equilibrium spin.

Conserved quantity paradox

Suppose $[B,H_0]=0$. Show that the isolated retarded self-response $\chi^R_{BB}(t)$ vanishes, while the canonical isothermal susceptibility can be $\beta\operatorname{Var}(B)$.

Solution

Conservation gives

$$B_H(t) = B.$$

Therefore

$$[B_H(t),B] = 0$$

and

$$\chi^R_{BB}(t) = 0.$$

For the equilibrium Hamiltonian $H_f=H_0-fB$, commuting $B$ permits direct differentiation of the Boltzmann weights:

$$\left. \frac{\partial\langle B\rangle_f} {\partial f} \right|_{f=0} = \beta \left( \langle B^2\rangle - \langle B\rangle^2 \right).$$

The thermodynamic protocol reweights sectors with different $B$ after equilibration. The isolated unitary protocol preserves their populations. The two susceptibilities answer different physical questions.

Diamagnetic term for free particles

For

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

derive the physical current and the zero-wavevector conductivity when the total paramagnetic current commutes with $H_0$.

Solution

Differentiating the Hamiltonian,

$$\mathbf j = -\frac{1}{V} \frac{\partial H}{\partial\mathbf A} = \frac{q}{mV} \sum_i\mathbf p_i - \frac{nq^2}{m} \mathbf A.$$

The first term is $\mathbf j^p$. Because total momentum is conserved,

$$\Pi^R(\omega) = 0$$

at zero wavevector. Hence

$$\sigma(\omega) = \frac{i}{\omega+i0} \frac{nq^2}{m}.$$

Using

$$\frac{i}{\omega+i0} = i\operatorname{PV} \frac{1}{\omega} + \pi\delta(\omega),$$

gives

$$\operatorname{Re}\sigma(\omega) = \pi \frac{nq^2}{m} \delta(\omega).$$

The contact term produces ballistic Drude weight. Dropping it would incorrectly give zero electromagnetic response.

Static and uniform density limits

Explain why

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

can differ from

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

Solution

At nonzero but small $\mathbf q$, density can redistribute over long distances. Taking $\omega\to0$ first allows the spatial modulation to equilibrate; sending $\mathbf q\to0$ afterward probes compressibility.

Taking $\mathbf q\to0$ first turns the density operator into total particle number. If total number is conserved, its finite-frequency retarded self-response vanishes. The subsequent $\omega\to0$ limit remains constrained by that conservation law.

The two orders describe different protocols, so their disagreement is physical rather than an algebraic inconsistency.

Related reading

• Retarded and Advanced Response derives adjoint identities, analyticity, and dispersion relations.
• Transport Coefficients Preview applies response kernels to diffusion, thermoelectricity, and viscosity.
• Driven Many-Body Systems examines absorption, heating, and nonlinear breakdown.
• Transport Measurements relates conductivity to terminal wiring, geometry, and low-frequency protocols.
• Chern Numbers in Band Theory specializes clean transverse response to occupied Bloch bands.

References

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