# Why doesn't a monoatomic particle have 6 degrees of freedom? [duplicate]

A monoatomic particle can move in three directions: $$x$$, $$y$$, and $$z$$. So the number of degrees of freedom (DOF) for translation is 3. The particle can also rotate around three axes. So the number of DOF for rotation is 3.

But it is written everywhere that the number of DOF of a monoatomic particle is 3.

Why aren't the DOF (caused by rotational motion) counted?

• Feb 14 '19 at 19:32
• How do you know if a PARTICLE is rotating? it is a perfect sphere or pint-like. Feb 14 '19 at 23:57

The energy levels for rotation are too far apart for rotational energy to be given to (or taken from) monatomic molecules when they collide in a gas at 'normal' temperatures. So effectively the molecule has no degrees of rotational freedom. The gaps between levels are inversely proportional to the moment of inertia, $$I,$$ of the molecule about any axis through its centre, and $$I$$ is very small for a monatomic molecule (or for a diatomic molecule about an axis through both its nuclei).

In quantum theory, if the atom is in a spherically symmetric state, as typically it will be in the ground state, then nothing happens to the wavefunction when you rotate the coordinate system.

Here is an example wavefunction for an atom in such a state, in some given inertial frame, expressed in spherical polar coordinates: $$\psi(r,\theta,\phi) = N e^{-r/a} .$$ As you see, there is no dependence on $$\theta$$ and $$\phi$$. Now let's see what form this wavefunction takes when we change to a reference frame rotating at one zillion revolutions per second about the centre of the atom. In the new frame the wavefunction is $$\psi(r,\theta',\phi') = N e^{-r/a}$$ It's the same! Similarly, if you apply to the atom an active transformation that would induce rotation if the atom were not spherically symmetric, then all you get is $$\psi = N e^{-r/a} .$$

No change.

The upshot of all this is that there is no such thing as rotation for a physical entity which is truly spherically symmetric right down to its quantum details. The state of affairs that you might want to say is 'rotating' is not in fact a different state from the one you have when it is not rotating. So there really isn't any degree of freedom there. It's not just hard to get it to rotate. It's a mathematical impossibility.

Having said that, in the case of atoms one can always disturb the electron's wavefunction into some non-spherically symmetric form, and then rotation is possible. That would typically require a collision or something that supplied to the internal motion several electron-volts of energy.

• One can increase the angular momentum of an atom, also when the electron density remains symmetric. For example hydrogen with its electron in a $Y_{1,1}$ orbital. But it is about 10 eV above the ground state.
– user137289
Feb 15 '19 at 0:12
• @Pieter $Y_{1,1} \sim \sin(\theta) e^{i\phi}$ which is not spherically symmetric. Did you have something else in mind? Feb 15 '19 at 9:40
• The density is axially symmetric ("circular orbital"), but it has angular momentum, the phase is rotating with time.
– user137289
Feb 15 '19 at 9:48