Is it theoretically possible to reach $0$ Kelvin? I'm having a discussion with someone.
I said that it is -even theoretically- impossible to reach $0$ K, because that would imply that all molecules in the substance would stand perfectly still.
He said that this isn't true, because my theory violates energy-time uncertainty principle.
He also told me to look up the Schrödinger equation and solve it for an oscillator approximating a molecule. See that it's lowest energy state is still non-zero.
Is he right in saying this and if so, can you explain me a bit better what he is talking about.
 A: from WP-negative temperature 

In physics, certain systems can achieve negative temperature; that is,
  their thermodynamic temperature can be expressed as a negative
  quantity on the kelvin scale.
A substance with a negative temperature is not colder than absolute
  zero, but rather it is hotter than infinite temperature. As Kittel and
  Kroemer (p. 462) put it, "The temperature scale from cold to hot runs:
  +0 K, . . . , +300 K, . . . , +∞ K, −∞ K, . . . , −300 K, . . . , −0 K."
. The inverse temperature β = 1/kT (where k is Boltzmann's constant)
  scale runs continuously from low energy to high as +∞, . . . , −∞.
from Positive and negative picokelvin temperatures :
  ... of the procedure for cooling an assembly of silver or rhodium nuclei
  to negative nanokelvin temperatures.

A: For a temperature to be definable and measurable the distribution of the kinetic energies of the molecules in the medium under discussion  should be known.

The process of cooling involves removing thermal energy from a system. When no more energy can be removed, the system is at absolute zero, which cannot be achieved experimentally. Absolute zero is the null point of the thermodynamic temperature scale, also called absolute temperature. If it were possible to cool a system to absolute zero, all motion of the particles comprising matter would cease and they would be at complete rest in this classical sense. Microscopically in the description of quantum mechanics, however, matter still has zero-point energy even at absolute zero, because of the uncertainty principle.

The uncertainty principle assures that molecules cannot stay perfectly still and continue being in a certain position , i.e. in the material under study. Certainly not all molecules of the material, this would be necessary to define a 0K temperature.
The solution with the vibrational degrees of freedom that molecules may have is not conclusive , though sufficient as proof for that the  specific material that displays these vibrational modes cannot go to 0K. It is the HUP that is general for all materials.
A: By the third law of thermodynamics, a quantum system has temperature absolute zero if and only if its entropy is zero, i.e., if it is in a pure state. 
Because of the unavoidable interaction with the environment this is impossible to achieve. 
But it has nothing to do with all molecules standing still, which is impossible for a quantum system as the mean square velocity in any normalized state is positive.
A: I think you are both wrong.
"The lowest energy state still has non-zero energy" does not mean that the temperature cannot be zero. If the system is in the ground state with 100% probability, then the temperature is zero. It doesn't matter what the ground state energy is.
It's true that all molecules in the substance would stand perfectly still at absolute zero [well, they don't have exact positions by the uncertainty principle, but the probability distribution of position would be perfectly stationary]. But so what? Why would that make absolute zero impossible? [see update below]
Nevertheless, there is no process that can get a system all the way to absolute zero in a finite amount of time or a finite number of steps. There's just no way to get that last little bit of energy out. This is one aspect of the third law of thermodynamics, as discussed in some (but not all) thermodynamics textbooks.
-- UPDATE --
It seems likely that I misunderstood. By "stand perfectly still", I guess you meant "have a fixed and definite position, and a fixed and definite velocity equal to 0". If that's what you meant, then "standing perfectly still" is indeed impossible (because of the Heisenberg Uncertainty Principle). But "standing perfectly still" is not expected or required to happen at absolute zero. Again, a harmonic oscillator which is in the ground state with 100% probability is at absolute zero, but does not have fixed and definite position or velocity.
A: I wonder why the measurement postulate has not been mentioned so far. Consider a cubical microcrystal of sodium chloride containing 64 atoms (4 on each side). If we cool it off so it is as close to absolute zero as possible, then we can represent its state as a superposition of pure states. One of those states is the ground state. If we then measure its energy, is there not some finite probability that it will be found in its ground state?
The atoms will not be stationary. They still have their zero-point energy. But in the ground state the temperature of the crystal is absolute zero.
A: This is what my Science teacher said on the matter. Nothing can reach absolute zero because Energy is linked to Mass, in the sense that if there is no energy, there is no mass. It would disappear. That can't happen due to other laws, so 0K can't be reached.
A: What people don't understand is that the laws of thermodynamics are not exact in the same way as, for example, energy conservation is. They are only quite probable, meaning that for a finite system there always exists a non-zero propability to break them.
So, even though it's quite improbable to reach $T= 0 \ \textrm{K}$, in principle it is possible. With quite some huge effort, e.g. immense heat baths and heat pumps working for thousands of years, the propability $p$ for the system to be at zero temperature within a finite time $\Delta t$ could reach $p \approx 1$. Then it's only a matter of waiting for chance.
The argument about the harmonic oscillator not reaching $E=0$ in the ground state is no argument against zero temperature, since temperature is more or less the mean excitation energy per degree of freedom. This also corresponds to the fact that potential energy is always only know up to a constant. If $H$ ist the Hamiltonian for the oscillator then $H + \textrm{const.}$ is just as good.
A: Literally saying, I like common sense more than physics, because that is more applicable in daily life.
Theoretically, I can say anything but in real life, it might not be so. You see, to make anything colder I must introduce an object more colder than that to transfer the energy it already has. Now since that is not possible, (even theoretically it can't be so) nothing can be colder than 0K.
