# Vibrations at Absolute Zero

Is matter still vibrating at absolute zero temperature? If it isn't, wouldn't we then be able to know both its position and momentum?

A system at absolute zero will still have nonzero uncertainty in position and momentum.

As an example, consider the simplest vibrating system, the one dimensional harmonic oscillator with (angular) frequency $\omega$ and mass $m$. In it's ground state (i.e. the lowest possible energy for the system) the position probability amplitude is a Gaussian ('bell curve'): $$\psi(x)\propto e^{\frac{-m \omega x^2}{2 \hbar}}$$. The finite extent of that curve reflects the position uncertainty: $\Delta x = \sqrt{\frac{\hbar}{2 m \omega}}$. One can also show that in this state the momentum uncertainty is $\Delta p= \sqrt{\frac{ m \omega \hbar}{2}}$, and it is a minimal uncertainty state ($\Delta p \Delta x = \hbar/2$).

This example is easily generalized to 3D and is actually quite relevant since vibrating systems can often be approximated quite well by a simple harmonic oscillator, especially at low energies.

• Are you able to say why they are still vibrating? – Shookster May 23 '14 at 3:30
• Not sure how to answer that. Are they "vibrating" if the system is in the lowest stationary state? I would say not, but it is perhaps a matter of terminology. The wavefunction has finite extent so that there is uncertainty in position/momentum, but it is not 'moving' in any observable sense. – DrEntropy May 23 '14 at 20:48

There will still be vibrations due to zero-point energy. So it's still impossible to know both its position and momentum.

• Is there any state below this zero point where the vibrations stop? – Shookster May 20 '14 at 2:57
• No. That's why it's called zero-point. – velut luna May 20 '14 at 3:04