Can a quantum measurement breach the third law of thermodynamics? The third law of thermodynamics states, in one rather intuitive formulation, that:

It is impossible to refrigerate a system to a temperature of exactly $0$ $K$ using a finite process (finite energy or finite lapse of time).

The question is, though, is this actually true? While it seems hard to see how the first and second laws are always considered to hold, I've been a little more bugged by this third law, for the following reason: namely, all the derivations I've heard of it seem to rely on classical, not quantum, mechanics, yet we know the real Universe is quantum.
Hence, consider the following. Let's consider building up a crystal which will be our system to "cool to zero kelvin". Now I believe that, in quantum mechanics, $0$ $K$ indicates the crystal ground state (ignoring overall kinetic stuff).
Consider a "crystal" composed of 2 metal atoms bonded together. This is quantum-mechanically a diatomic molecule - there's no real difference between a metallic bond holding 2 atoms and a covalent bond holding 2 atoms; it's all just various degrees of interfering single-atom electronic wavefunctions. Hence, it will roughly be described, at least for the first few energy levels, as a quantum harmonic oscillator, and thus will have a ground state and first excited state with separated discrete energies, i.e. there is a minimum energy which is required to transition from one to the other. But more importantly, if I make a quantum measurement of energy with something better, but which need not be infinitely better, than this finite resolution, then it is possible to yield the ground state as the result of said measurement, and the system will be assured to be exactly in that ground state afterward.
Now, of course, 2 atoms do not make a thermodynamic system. So add another atom, to make it 3. Now you have a 3-atom cluster, but the same logic will still apply. There will be a ground state and a first excited state, though the latter will be considerably closer in energy. Still, it's a finite gap. Use a finer measurer. Now add another atom. 4 atoms. Same logic. Another atom. 5. Same logic. ... Up to $\approx 6.022 \times 10^{23}$ atoms. We now have something we can call "thermodynamic", and a procedure we can use to get exactly the ground state: cool it down real close so the probability of measuring it in the ground state is maximized. Measure. If not in the ground state, cool some more. Measure. Cool. Measure. Cool. Measure. While this is probabilistic and the worst-case number of steps is unbounded, "with probability zero" the actual steps should still be finite, no?
Does this mean that yes, it turns out we can cool things to exactly $0$ $K$ if we have access to a suitably discriminating quantum measurer? Note that I don't know if such is possible or not - given that I'm not sure exactly how the gap between the ground and the first excited level of the crystal shrinks as atoms are added (much less as to when you include also the inner electrons and the like to make a truly thorough non-kinetic "ground state" model but still it seems it should "intuitively" still apply), in particular, whether it is linear or exponential in the number of atoms in the crystal, or somewhere in between, for if it's exponential then it's quite possible that your device collapses into a black hole before it is able to achieve enough resolution to perform the necessary measurement, thus rendering your attempt futile, but if it's linear, then "only" some amount more than 23-digit precision would suffice to render the system exactly in the ground state, making the Avogadro-crystal exactly $0$ $K$ in temperature, in a finite - even if very sophisticated - process.
Is there something wrong with this reasoning or scheme? If so, what?
 A: The uncertainty principle states
$\Delta E$ $\geq$ $\dfrac{\hbar}{(2\Delta t)}$.
A temperature of exactly $0 \ \text{K}$ implies $\Delta E = 0$. But this would mean $\Delta t = \infty$ or an infinite lapse of time. This is how the statement of the third law you gave is compatible with quantum mechanics.
A: Third law of thermodynamics is meant as limitation on macroscopic processes using heat transfers and work to do refrigeration on macroscopic bodies.

I make a quantum measurement of energy with something better, but which need not be infinitely better, than this finite resolution, then it is possible to yield the ground state as the result of said measurement, and the system will be assured to be exactly in that ground state afterward.

This can be done on atoms with magnetic moment in external magnetic field, when they make a record on a screen (the Stern-Gerlach experiment); part of atoms will end up in the minimum magnetic energy possible and thus we can say we "cooled them to 0K" in the sense that their magnetic energy is the minimum possible value.
I don't see how we can make similar thing happen to total energy of many-particle interacting system. There are other kinds of energies there than just energy in external magnetic field which we can minimize in the SG-manner. Energy levels of macroscopic bodies are very close to each other due to inter-particle interaction removing degeneracy. And there is always interaction with other bodies, and with background radiation, which transfers energy from and to the system.
