Can an arbitrary quantum system of finite size be made to reliably relax to its ground state? Is there a physical principle prohibiting this? I am talking about the possibility of reliably cooling an arbitrary quantum system of FINITE size (for example, localized on earth), to its ground state through any means, like exposure to a special kind of environment etc. By reliable, I mean that for the given system, can we calculate a finite time T after which it has a high probability, lets say 75%, of relaxing to the ground state?
Can this always be done? If not, is there a physical argument prohibiting it?
More Details for context: I read a [paper][1] that gives a physical argument why this may not be possible. It asserts that the ground state of carbon is known to be graphite, yet tiny pieces of diamond have been found to survive in meteorites for billions of years.
I tried to set up the GKSL equation for the relaxation of a two state system exposed to some environment at temperature T. I found that a quantum cooling analogous to newton cooling in the sense that there are exponential decay terms for the off diagonal elements and so on, can be incorporated as the lindblad equation.
So the catch may be that the system may get caught somewhere in a local minima for an indefinite amount of time, prohibiting it from reaching the ground state.
[1]: http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.85.6029&rep=rep1&type=pdf “ Smith, Warren D. "Three counterexamples refuting Kieu’s plan for “quantum adiabatic hypercomputation”; and some uncomputable quantum mechanical tasks." Applied Mathematics and Computation 178.1 (2006): 184-193.”
Note: to get to the relevant paragraph of the paper above, search for the term 'cool' in the pdf.
 A: Yes, in principle at least. A number of points

*

*Cooling down corresponds to a time-evolution in imaginary time $\tau$ i.e. $\hat{\rho}(t)=e^{-\hat{H}\tau}\hat{\rho}_0 e^{-\hat{H}\tau}$, and it is easy to see that this leads to the ground state.


*The whole notion of cooling down relies on the Second Law of Thermodynamics: basically, any (equilibrium) system will tend towards a Boltzmann-Gibbs state at infinite time, and the corresponding temperature will go to the one of its environment. So if you put the system in a $T=0$ bath, eventually it goes to its ground state. The assumption is here that thermodynamics is valid, meaning we are relying on ergodicity (for any microstate, there exist some incoherent process connecting it to the other microstates), but this is widely accepted except for some very specifically engineered cases. Thermodynamics doesn't tell anything about the timescales involved, though.


*Consider as a practical setup, that the system couples with empty space, a bath at $T=0$. Cooling down occurs by spontaneous emission, single-photon losses in the simplest case. We can describe this process with a Lindblad equation, and an associated Liouvillian superoperator $\mathcal{L}$. In case where where the quantum-dynamical semigroup is irreducible (basically, this is the same assumption as ergodicity), $\mathcal{L}$ has a unique lowest eigenvector, the steady state (also known as Spohn's theorem). If the emission process can only decrease energy, the steady state is the same as the ground state. The spectrum of $\mathcal{L}$ has also a gap, and the inverse of this Liouvillian gap corresponds to the slowest timescale associated with cooling.
Of course, metastabilities exist, but that's what they are, metastable so at some point, possibly longer than the age of the universe, the system will decay further. A good example of long-living metastabilities are glasses. Typically, slow cooling or thermal annealing where the temperature is decreased slowly, overcomes metastabilities faster than connecting it to a $T=0$ bath as described above.
