In linear response theory, the Kubo formula describes the change in $\langle A \rangle$ for some observable $A$, due to applying a time-dependent perturbation $V(t)$ to the Hamiltonian $H_0$. It is given by:

\begin{align*} \langle A\rangle (t) = \langle A \rangle_0 - i \int_{-\infty}^{t} dt' \langle [\bar{A}(t), \bar{V}(t')] \rangle_0 \end{align*} where $\bar{A}(t) = \exp(iH_0t)A\exp(-iH_0t)$.

The expectation of the commutator has the subscript $0$, indicating that it is carried out over the equilibrium density matrix, typically $\rho_0 \sim \exp(-\beta H_0)$.

My question concerns the case(s) in which either $A$ or $V$ commute with $H_0$, and the response term vanishes. To illustrate, assume $A$ commutes with $H_0$. Then the expectation of the commutator can be written as:

$$\langle [\bar{A}(t), \bar{V}(t')] \rangle_0 = \mathrm{Tr}[\rho_0\bar{A}(t)\bar{V}(t') - \rho_0\bar{V}(t')\bar{A}(t)]$$ $$= \mathrm{Tr}[\rho_0 A\bar{V}(t') - \rho_0 \bar{V}(t')A]$$ $$= \mathrm{Tr}[\rho_0 A\bar{V}(t') - A\rho_0 \bar{V}(t')] \qquad(\text{Cyclic trace rule})$$ $$= \mathrm{Tr}[\rho_0 A\bar{V}(t') - \rho_0 A\bar{V}(t')] = 0 \qquad (A\text{ commutes with } H_0)$$

This result of zero confuses me, because I can think of a few experimental probes matching the above description that ordinarily show a finite response. For example, take $H_0$ to be an Ising Hamiltonian $-J\sum_{\langle ij \rangle}S^z_i S^z_j$ and $V(t)$ to be a constant field $-h\sum_i S^z_i$. Here, choosing $A$ to be the magnetization $\sum_i S^z_i$ is supposed to give a response that indicates the magnetic susceptibility.

The Ising model has non-zero susceptibility, yet the above calculation yields zero owing to $A$ commuting with $H_0$.

What's the resolution to this apparent discrepancy? My guess is that the corrections show up only at higher orders in this case, but I can't quite see it.


1 Answer 1


What we mean when we say that the Ising model has non-zero susceptibility is that $$ \chi_0(T)=\frac{\partial^2 \mathcal{F}(T,H)}{\partial H^2}|_{H=0}\neq 0. $$ What is implicit here is that, when the temperature or magnetic field change, the occupation numbers of different microstates change as well. As is usual in statistical physics (and indeed fundamental to it) the residual interactions responsible for the change of the occupation of the microstates are neglected. We simply assume that such interactions exist and they will eventually result in the system relaxing to an equilibrium state - this may take a very long time, but, as long as we are interested in the equilibrium state only, these are not essential.

Once we consider a non-equilibrium situation, even if a linear response regime, we are essentially being interested in the dynamics of relaxation to equilibrium, and these interactions cannot be neglected anymore. I suspect that they would still drop from the final result in the zero frequency limit (which is what we get from thermodynamics/stat.phys.), but using Kubo formula in absence of such interactions is not possible (in my opinion - I still might be missing some nuance.)

Kubo himself seems to be introducing explicitly a transition probability to calculate dynamics of Ising model: Dynamics of the Ising Model near the Critical Point. I

A simple dynamical model of interacting Ising spins is discussed. Each spin is assumed to flip spontaneously with a transition probability which depends on the temperature and the configuration of surrounding spins, but its functional form is assumed to be the simplest. The frequency-wave number dependent susceptibility χ( q , ω) is given exactly in the one-dimensional case. In two-and three-dimensional cases the model is treated in the molecular field and the generalized approximations.

(emphasis mine) Unfortunately I do not have access to this paper:

  • $\begingroup$ Thanks for the answer and reference. From what I understand, 'residual interactions' refer to the unknown microscopics of a system in contact with a heat bath. Let's say I apply the Kubo formula to the following situation: 1. System ($H_0$) equilibrates at temperature T. 2. It is removed from contact with the heat bath, while still retaining the same density matrix $\rho_0$. 3. The perturbation $V(t)$ is applied. $\endgroup$
    – Crisco
    Commented May 12, 2023 at 6:49
  • $\begingroup$ In this way of looking at it, it follows that the response vanishes in the Ising case, because the density matrix remains unchanged by $V(t)$ to all orders. However, I think the broader question about zero response remains. If $V(t) = -h\sum_i S^x_i$ instead, $\rho_0$ does evolve, but the Kubo result still vanishes if $A$ is taken to be the magnetization. $\endgroup$
    – Crisco
    Commented May 12, 2023 at 6:54
  • $\begingroup$ @Crisco Residual interactions are not necessarily interactions with the bath - we expect them to exist even in a microcanonical ensemble - e.g., collisions between the molecules in an ideal gas. In the setting that you describe the thermodynamic susceptibility also becomes meaningless, since the system does not equilibrate in response to the perturbation. $\endgroup$
    – Roger V.
    Commented May 12, 2023 at 7:17
  • $\begingroup$ @Crisco Overall, I think this is an interesting question - how to get the static susceptibility from the Kubo formula. It requires some calculation, even for simple models. $\endgroup$
    – Roger V.
    Commented May 12, 2023 at 7:59
  • 1
    $\begingroup$ @Crisco What I would do is take a Hamiltonian with a bath, write the expression for the linear response function, and then expand it in the coupling to bath (not forgetting the denominator of the density matrix.) My guess is that the coupling to bath would cancel out and one ends with the same susceptibility as the thermodynamic one. Doing it for a single spin would suffice. $\endgroup$
    – Roger V.
    Commented May 12, 2023 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.