# Kubo formula to compute the energy dissipated in heat bath

In the case of a perturbed Hamiltonian $$H_0$$

$$$$H=H_0 +\theta(t-t_0)W(t)$$$$

at $$t=t_0$$ the Hamiltonian admits eigenvalues $$E_n(t_0)$$ and for positive $$t-t_0$$ then the eigenvalues are $$E_n(t)$$. The Kubo formula then states that, up to linear order in $$W(t)$$, the expectation value of an operator $$A$$ is $$$$\langle A(t)\rangle = \langle A\rangle_0 -i \int_{t_0}^t ds\langle\left[A(t),W(s)\right] \rangle_0$$$$ where $$\langle\rangle_0$$ means that the expectation value is taken on the eigenstates of $$H_0$$.

If one wants to compute the energy dissipation of a system $$H_0 = \frac{p^2}{2M}+V(q)$$, then one can couple it with an infinite set of harmonic oscillators, $$$$H= H_0 + H_{HO} + H_C$$$$ where $$H_{HO}$$ is a set of free harmonic oscillators, and $$H_C$$ is a coupling term.

My question is, does the application of the Kubo formula in this case to the Hamiltonian gives the energy disspation of the system? \begin{align} \langle H(t)\rangle &= \langle H\rangle_0 -i \int_{t_0}^t ds\langle\left[H(t),W(s)\right] \rangle_0\\ &= \langle H\rangle_0 -i \int_{t_0}^t ds\langle\left[H_0(t),W(s)\right] \rangle_0 \end{align}

If not, how can one compute the energy dissipated in a heat bath modeled by a set of the harmonic oscillators?

• Your notation is potentially confusing. $V(q)$ does not play the same role as the $V$ in your first example. Is this $V$ the small term you want to compute linear response from? Why should $V(s)$ commute with $H_C$ (your last equation)? Or did I miss something?
– lcv
Commented Jun 8, 2020 at 13:20
• @lcv So I just changed the notation to clear the confusion. $W$ is now the perturbation. To answer your second question, I wrote $H = H_0 +H_C+H_{HO}=H_0+\theta(t-t_0)W(t)$. So I considered $W(t)$ to be the $H_{HO}+H_C$, that's why it should commute with itself. Commented Jun 8, 2020 at 14:13
• Then the answer is yes. That is precisely Kubo formula. But there is a little detail, Kubo formula assumes that the perturbation is time dependent (it's based on the Taylor series of the evolution operator). However in your first formula there's no time dependence. By the way the switching does not need to be instantaneous (like a Heaviside theta).
– lcv
Commented Jun 9, 2020 at 2:33
• @lcv Ok. Then how about this to include time dependency. The equation of motion can be written as a Langevin equation: $$M\ddot{q} + M\int_0^t\gamma(t-s) \dot{q}(s) +V'(q) = \xi(t)$$ So I can rewrite it as $$M\ddot{q} =-V'(q)-f'(q)$$ where $f(t)=M\int_0^t ds\gamma(t-s)-\xi(t)$ and in this case the Hamiltonian is $$H = H_S - \hat{W}(t)\theta(t)$$ where $$\hat{W}(t)= \int_0^tdsf(s)\dot{q}(s)$$ Commented Jun 9, 2020 at 7:41
• I don't understand. What is $H(t)$ in your post? How does it depend on time? If $H_C$ and $H_{HO}$ are time independent leave them as such. In this case you compute the variation of expectation values not with the dynamical susceptibility (Kubo formula) but with the isothermal susceptibility.
– lcv
Commented Jun 9, 2020 at 18:10

While the logic is sound, let me note that it is more meaningful to look not at the energy dissipated by the system, but at the rate of dissipation, which can be defined as $$R = \dot{H}_0 = \frac{1}{i\hbar}[H_0, H]_-$$ The energy and the rate defined above are in the same relation as the charge and the current. Although deriving Kubo formula for electric charge is possible, it is usually impractical.
• I have one small issue. When computing the energy dissipated, I get this$$\Delta E = -\hbar\frac{p(t)}{M}\bigg\langle\sum c_kq_k-2q(t)\sum\frac{c_k^2}{2m_k\omega_k}\bigg\rangle_0$$ this sounds to be completely independant of the potential of the original system (because $\hat{W}$ has no dependance on $p$), which suggest that all systems dissipate in the same way? How come? (or where is my mistake). i think this would apply also to the rate of dissipation you mentioned. Commented Jun 12, 2020 at 10:26
• It is hard to comment, since the form of coupling and the notation are not specified in your question (perhaps, you could add more details there). What strikes me however is: a) why $p(t)$ appears outside of averaging, and b) what happened to the time variable $s$? - the commutator in Kubo formula contains the operators at two different times, and evaluating it directly is possible only after solving for the the time evolution of all the operators. The time evolution of $p(t)$ is dependent on the potential: $i\hbar \dot{p}(t) = [p(t), H_0]_-\Rightarrow \dot{p} = -V'[x(t)], \dot{x}(t)=p(t)/m$. Commented Jun 12, 2020 at 11:26
• a) you are right, $p(t)$ should appear inside the average, b) because I had a commutator of the form $[p(t),q(s)]$. This will result in a delta function $\delta(t-s)$. With the integral, $s$ will be replaced by $t$. Commented Jun 12, 2020 at 11:58
• Look at the example in my previous comment: $[p(t), q(s)] = [p(s), q(s)] = -i\hbar$ - where does the delta-function come from? Or in the other direction: $[p(t),q(s)]=[p(t), q(t) + \frac{p(t)}{m}(s-t)] = [p(t),q(t)]=-i\hbar$. You might be confusing particle momentum and position with field operators. Commented Jun 12, 2020 at 12:33