I have a problem understanding what exactly is non-equilibrium steady state. Let's say the system starts with an equilibrium state at remote past. We should have the density operator given by: $$\hat{\rho}_{eqm}=\frac{\prod_{i}e^{-\beta_i\hat{K}_{0,i}}}{\mathrm{Tr}(\prod_{i}e^{-\beta_{i}\hat{K}_{0,i}})}$$ where $$\hat{K}_{0,i}$$ is the unperturbed grand canonical hamiltonian for the $$i$$-th sub-system. Usually, we have $$[\hat{K}_{0,i},\hat{H}_0]=0$$, so the density operator is identical at any time in all the Schrödinger's picture, Heisenburg picture and interaction picture. Therefore, treating it as an operator expressed in the interaction picture at $$t=0$$, the perturbed density operator in Schrödinger's picture at time $$t$$ should be given by: $$\hat{\rho}(t)=S(t,-\infty)\hat{\rho}_{eqm}S(-\infty,t)$$ where $$S(t,t_0)=\lim_{\eta\to 0^{+}}\mathcal{T}\exp\left[-i\int_{t_0}^{t}\mathrm{d}t'e^{\eta t'}\hat{V}_{I}(t')\right]$$ The perturbation is turned on adiabatically. When you calculate the expectation value of any observable in the non-equilibrium state, you use $$\left\langle{\hat{O}}\right\rangle(t)=\mathrm{Tr}\left(\hat{\rho}(t)\hat{O}\right)$$ Here comes my problem, how do you define a steady state? If $$\left\langle{\hat{O}}\right\rangle$$ is independent of time, its time integral will obviously increases linearly with time since the remote past, i.e. it diverges! If the expectation value describes a particle current, it would mean one side of the system is taking in infinite number of particles and it keeps increasing! If the expectation value describes a energy flow, one side is gonna receive infinite amount of energy!

How can you even "fix" the chemical potential and temperature of the sub-systems? They can only be fixed in the remote past via $$\hat{\rho}_{eqm}$$, aren't they? The density operator contains everything about the system and the Hamiltonian governs the time evolution of it. These are fundamental laws of quantum physics. So shouldn't $$\hat{\rho}(t)$$ already define what the current state is? Just like energy spectrum can be altered after perturbation, temperature and chemical potential of the sub-systems should also be changed. I don't see how a non-equilibrium steady state can exist.

As you noted in your question, $$\langle \hat{O} \rangle$$ increases linearly with time... which means that its rate is constant! E.g., if $$\langle \hat{O} \rangle$$ is the electric charge, it gives us a situation with a constant current.

I think conceptually the difficulty is that a steady state is more of a theoretical/modeling concept rather than a kind of a situation actually existing in nature. What I mean is that a steady state is actually a transient state, observed during the time period much shorter than the time required for the system to actually reach the equilibrium, but quite long to ignore the fast relaxation processes that might have happened when the system was first driven out equilibrium.

To model this situation mathematically one often artificially imposes a kind of boundary conditions, such that the system can never equilibrate. For example, one can impose a constant potential difference between two regions, which drives a current. If we were waiting long enough, then a significant amount of charge would move from one region to the other and and screen the potential driving the current - the system would then reach the equilibrium. However, we would often model it neglecting this screening potential, as well as the potentially limited amount of the electric charge.

Thus, steady state is an approximation that is made even before we write the equations describing our system (unlike more obvious mathematical approximations, such as perturbation theory, adiabatic approximation, etc.)

• I fail to see how "...steady state is an approximation that is made even before we write the equations describing our system..., etc." is any different from any other problem we solve in physics or in sciences in general. After all, in a long enough time scale we are all dead. Oct 7, 2020 at 15:25
• @hyportnex the point that I make in my answer is that it is an implicit approximation, as opposed to the explicit ones made in mathematical derivations. Have you ever worked on non-equilibrium problems? Oct 7, 2020 at 16:49
• Do you mean that in order to describe a system that can achieve equilibrium, you actually need more terms in the initial Hamiltonian? So by using a simple Hamiltonian itself already make an assumption that the system does not know how to achieve detailed balance? Oct 7, 2020 at 17:17
• @WongHarry steady non-equilibrium is not necessarily achieved through the Hamiltonian - e.g., one may impose different chemical potentials or temperatures in parts of the system. Oct 7, 2020 at 19:07

I will expose a couple of ideas that maybe can help you:

-I understand non-equilibrium steady states like those steady states that cannot be predicted by Statistical Mechanics, where your steady state cannot be described by the microcanonical, canonical, etc... ensembles. An example of this is the many-body localization, where local observables of closed quantum systems with interacting particles reach stationary values that depend on the initial conditions (failing to thermalize). See for example this experiment with ion traps: https://www.nature.com/articles/nphys3783. In this phenomenon, the emergence of quasi-local conserved quantities is the responsible of the memory effect in the system. If you are interested in this phenomenon, check this review: https://journals.aps.org/rmp/abstract/10.1103/RevModPhys.91.021001

-Respect to the definition of steady states of observables, you can write the temporal average of an observable for long times. Let's define a time evolution of our system (as a pure state, but this can be generalized to density matrices): $$|\Psi(t)\rangle=\sum_mC_me^{-iE_mt}|m\rangle,$$ where $$E_m$$ are the eigenvalues of the Hamiltonian and $$C_m$$ the coefficients that codify the initial condition. The temporal average of an observable in the limit of a long time is: $$\overline{\langle\hat{O}\rangle}=\lim_{T\rightarrow\infty}\frac{1}{T}\int^T_0dt\sum_{m,n}C^*_mC_n e^{i(E_m-E_n)t}O_{mn}=\lim_{T\rightarrow\infty}\frac{1}{T}\int^T_0dt\sum_{m}|C_m|^2O_{mm}$$ $$+\lim_{T\rightarrow\infty}\frac{1}{T}\int^T_0dt\sum_{m,n\neq m}C^*_mC_n e^{i(E_m-E_n)t}O_{mn}$$ Here $$C^*_m$$ is the complex conjugate and $$O_{mn}$$ are the matrix elements of the observable in the eigenstate basis. Applying the limit, the equation simplifies to: $$\overline{\langle\hat{O}\rangle}=\sum_{m}|C_m|^2O_{mm}$$ where the second term vanishes. We have to remember that $$\overline{\langle\hat{O}\rangle}$$ is just the temporal average, so it does not mean that $$\langle\hat{O}\rangle(t)$$ will be close to a fixed value at any instant of time after the initial transient. Now, the value of your observables in the stationary state will depend on the statistics of your eigenstates and the properties of your system. For example, if your system fulfills the eigenstate thermalization hypothesis, $$\overline{\langle\hat{O}\rangle}$$ will coincide with the microcanonical prediction. If you are interested in this topic of thermalization of closed systems and the eigenstate thermalization hypothesis, check this review: https://www.tandfonline.com/doi/full/10.1080/00018732.2016.1198134.

-Short comment respect to the discussion about the "existence" of steady states in nature: you can observe experimentally that systems (classical and quantum) show what we call steady states. This is a really complicated discussion but something that we must take into account is the temporal scales. In closed quantum systems, one can see thermalization of local observables, for examples with 10 spins (see the reference of the ion trap experiment). On the other hand, the Poincaré recurrence theorem says that after a sufficiently long but finite time, our system will become close to the initial state. However, for systems with large number of degrees of freedom, this time where we see the recurrence could happen at a really long times, like thousands of years or more. Then, for practical porposes, we consider them stationary.

I know that I'm introducing some concepts that could be unfamiliar, and I just gave a short mention to them, but I hope this can help you!