How imaginary part of susceptibility is measure of dissipation?

Question

In linear response theory, we focus only imaginary part of the generalized susceptibility and consider it a measure of dissipation in the system. Can someone throw some light at it that what is meant by dissipation in the generalized sense?

Answer 1

Fluctuation-Dissipation theorem

When we want to know the property of a system with Hamiltonian $H _0$, we can give it a perturbation, (i.e. apply an external field) and research the response in term of such perturbation. Namely, the external field should couple with an operator A (physical quantity) of original Hamiltonian $H_0$: $$H=H_0-f(t)A$$ What we concerned is the change of the expectation value of such physical quantity $\langle A(t)\rangle$, which couple with the external field, so that it must be the function in term of external field $\langle A(f)\rangle$ . If the external field $f(t)$ is weak, we can expand such $\langle A(f)\rangle$ to the linear term of $f$: $$\langle A(t)\rangle \simeq\langle A\rangle_{0}+\int_{-\infty}^{+\infty} \chi\left(t-t^{\prime}\right) f\left(t^{\prime}\right)+O\left(f^{2}\right)\\\langle A(\boldsymbol{q}, \omega)\rangle \simeq \chi(\boldsymbol{q}, \omega) f(\boldsymbol{q}, \omega)+O\left(f^{2}\right)$$where $\chi$ is called "response function", which is the key of linear response function, and it is just the "susceptibility" in your question.

Then, we will express the correlation function in real time: $$S\left(t-t^{\prime}\right)=\left\langle A(t) A\left(t^{\prime}\right)\right\rangle\sim\left\langle (A(t)-\langle A(t)\rangle)(A(t')-\langle A(t')\rangle)\right\rangle$$from this equation, it is natural to note that $S(t-t')$ actually describes the fluctuation of system.

Now, we have two elements of story:

• susceptibility, which measures the response in term of external field;
• structure factor $S(t-t')$, which measures the fluctuation

After spectrum expansion, we can obtain the relation between such two physical quantities: $$S(\omega)=2 \hbar\left[1+n_{B}(\hbar \omega)\right] \text{Im} \chi(\omega)$$ which is so-called fluctuation-dissipation theorem. The right side of equation is the "dissipation" term, i.e. imaginary part of response function.

To understand why such term represent the "dissipation", we can recall the classical damped forced oscillation.

Damped forced harmonic oscillation

We begin with the forced oscillation without damped term, or say "no dissipation": $$m\frac{d^2x}{dt^2}=-kx + F_0\cos(\omega t )$$ and the displacement will be : $$x=\frac{F_0/m}{\omega_0^2-\omega^2}\cos(\omega t )\sim AF_0\cos(\omega t )$$ which means the particle will oscillate following external field oscillation. Also, such two oscillation is the synchronization, i.e. the phase of displacement and external field are the same.

Then, we know the for forced oscillation without damped term $-\mu \frac{dx}{dt}$: $$m\frac{d^2x}{dt^2}=-kx -\mu \frac{dx}{dt}+ F_0\cos(\omega t )$$ the displacement is: $$x=\frac{F_0/m}{\sqrt{(\omega_0^2-\omega^2)^2+(\gamma\omega)^2}}\cos(\omega t +\Delta)\sim BF_0\cos(\omega t +\Delta)$$ where the phase shift $\Delta=\tan^{-1}(\frac{-\omega\mu}{k-m\omega^2})$ origin from the damped term, or say "dissipation term". As the result, we find that one of key effect of dissipation is to shift the oscillation phase relative to the external field. Also, we can regard the $\frac{x}{F_0\cos(\omega t)}$ as the response of oscillation in term of external field. Thus, the imaginary part of response comes from the dissipation term.

"Dissipation" in many-body system

Now we can back to your question, the right side of Fluctuation-Dissipation theorem, i.e. imaginary part of response function, actually measures the response of $A$ which is out of phase with the applied field $f(t)$. As in the damped forced harmonic oscillation, it is the out-of-phase component which measures the energy absorbed by the system from external field, thus justifying the meaning of imaginary part of response function.