# Could the uncertainty principle theoretically be violated at 0 K? [duplicate]

Ok so please excuse me if the following mental argument is completely ridiculous or obviously flawed :P I was reading about how, even at 0 K (assuming we could experimentally reach such a temperature), there must be some slight particulate motion, because of the uncertainty principle. What about if we have some particles in a closed box that is in a dark vacuum (assuming a complete absence of any energy or light). If we then proceed to cool the particles to 0 K, then wouldn't we have succeeded at violating the uncertainty principle? We know that the speed of the particles is exactly 0 m/s, and we know with 100% certainty that they are in the box; we do not need to observe them to know so.

## marked as duplicate by Martin, Kyle Kanos, JamalS, ACuriousMind♦, John RennieJan 23 '15 at 14:33

• possible duplicate of Absolute zero and Heisenberg uncertainty principle – Martin Jan 23 '15 at 12:57
• you should check the answers to the question linked above^ @marm_96 – Hritik Narayan Jan 23 '15 at 13:01
• Absolute Zero represents a minimum amount of thermal energy, but it does NOT correspond to zero atomic motion. – Sean Jan 23 '15 at 13:05
• Ok, for argument's sake, let's say it did correspond to zero atomic motion; then would the uncertainty principle be violated? Maybe the uncertainty principle only holds when there is a minimum particle motion? – marrm_96 Jan 23 '15 at 13:09
• No, for example, removing energy from ('cooling') a quantum harmonic oscillator until it is in the minimum energy (ground) state means that no further energy can be extracted from the system - it is as 'cold' as it can get. The system has minimum energy and the state saturates the lower bound of uncertainty. – Alfred Centauri Jan 23 '15 at 13:27

You say that the particle is closed in the box. By the time you limit the linear dimensions of the particle, you already fall under the reign of the uncertainty principle. If the box is, for simplicity, a cube of linear dimension $L$,
$\Delta p_i \ge \hbar/L$, where $i$ stands for $x, y, z$.
From now on, you have no control on $p_x, p_y$, and $p_z$ ? You can cool the box, so what?