# I Extremely cold crystal

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1. Nov 23, 2017

### Petr Matas

Is it theoretically possible to cool a macroscopic crystal (for example NaCl 1×1×1 cm) to an extremely low temperature, like 1 nK?
Will it retain its microscopic structure?
Why?
Does it have anything to do with zero point energy (I mean something like the lowest energy level of a quantum oscillator) or quantum uncertainty?
And how about a microscopic crystal, like 10×10×10 atoms?

2. Nov 23, 2017

### hilbert2

The low-temperature behavior of the heat capacities of solids is very important in the statement of the 3rd law of thermodynamics, and also in the experimental testing of the Einstein and Debye crystal models. I'm not sure about temperatures of the nanokelvin range, though.

If a crystal is cooled quickly enough, it can be "frozen" in a state that is not the thermodynamically most stable one at low temperatures.

3. Nov 23, 2017

### Staff: Mentor

Theoretically, there is no limit to how close to 0 K one can cool down to (without ever reaching it, per the 3rd law of thermodynamics). Practically, macroscopic crystals have been cooled below 1 K, using for instance adiabatic magnetization, see https://en.wikipedia.org/wiki/Magnetic_refrigeration.

Yes. Why not?

4. Nov 23, 2017

### Petr Matas

Because at extremely low temperatures the energy of individual atoms and thus their momentum is also extremely low and therefore very precisely specified. This should result in high uncertainty of their position. Shoudn't this lead to the crystal becoming liquid?
However, from another point of view, the crystal is just a large molecule. De-exciting it to its ground state makes it lose all energy and its temperature should be 0 K. And we know that molecules usually do retain their structure in the ground state.
The microcrystal's energy levels will be spaced coarsely and thus its temperature should be quantized.
Am I right?

5. Nov 23, 2017

### hilbert2

If you have a harmonic oscillator in the ground state (a simple model of a single atom in a crystal), the expectation value of the magnitude of its momentum vector is not zero. That's because of the ground state energy.

Temperature is not quantized because it depends on the statistical probabilities of the excitation states of parts of the system. If you have a set of atoms that can only have energies $E_1$ or $E_2$ because of quantization, the probability of a randomly picked atom from the set having energy $E_2$ can still be any real number between 0 and 1.

The crystal does not de-excite to ground state in a realistical situation because it's constantly receiving thermal radiation and molecular collisions from its environment.

6. Nov 23, 2017

### Staff: Mentor

I don't see why low momentum means "more precisely specified."

7. Nov 23, 2017

### Petr Matas

@hilbert2: Thank you for your clear explanation.

Oh yes, that did not come to my mind. And any system is always entangled with its environment. Is it even possible to prepare a single atom with a 100% probability of it being in its ground state then? ΔE⋅Δt ≥ ½ħ comes to my mind as well, so maybe the answer is no, unless you have infinite time?

This is not because the momentum's expectation value is very close to a zero vector (that would permit high uncertainty of the momentum vector), but because its magnitude is expected to be very low, which means that the momentum probability distribution has a sharp peak around the zero vector. However, that applies to a free particle. As the atom is bound in our case, its momentum uncertainty is high even in the ground state.

Last edited: Nov 23, 2017
8. Nov 24, 2017

### hilbert2

I don't think a 100% probability is possible, but if you prepare a Bose condensate from a dilute alkali metal vapor and then pick a single atom from it and measure its energy as accurately as possible, it could be very close to ground state after that (not sure how that would be done in practice).

9. Nov 24, 2017

### DrDu

That's a very good question. But what is relevant here is the momentum uncertainty of the whole crystal, not that of it's compound atoms. For a crystal or molecule of 1000 atoms, free to move in space, the uncertainty in position at a given temperature is given by the De Broglie wavelength, which is a consequence of Heisenberg's uncertainty relation: https://en.wikipedia.org/wiki/Thermal_de_Broglie_wavelength
As this length is also the uncertainty with which we can know the position of an atom in a crystal, it is better to say that in a crystal, the difference of the atomic positions (or the spatial correlation function) has a sharp value, rather than the positions itself.