# Indistinguishability versus Lack of energy for degrees of freedom of a symmetrical atom

Consider the degrees of freedom (thermodynamic) for an Argon atom. It has 3 translational degrees of freedom. Everyone seems to agree that at normal temperatures it has no rotational degrees of freedom. However the reasoning given is slightly different.

My thermodynamics professor said that these rotational degrees of freedom aren't counted because it is meaningless to rotate a symmetrical object like an argon atom, because you can't distinguish the rotations of the atom. He mentions:

Rotation doesn't mean anything. The reason is that classical objects you can imagine making a mark on it and seeing that mark rotate around. In Quantum Mechanics its impossible to make marks on the atom. There's no structure associated with the surfaces. In QM, rotation for an argon is meaningless.

In other places however, (such as this question and answer), it seems as though rotation of the argon atom isn't counted because it isn't noticed at normal temperatures. There just isn't enough energy to excite that degree of freedom. At high enough temperatures, the argon can start to 'meaningfully' rotate, and thats A - OK.

To me, these two explanations seem incompatible. The first seems to assert that no matter what temperature I am at, rotations for objects such as an argon atom simply don't mean anything in the realm of Quantum Mechanics. The latter seems to assert that they only don't matter at low temps, and at high temperatures there is meaning in rotation of an argon atom.

Is one interpretation a simplification? Am I misunderstanding something, perhaps they are actually mutually compatible interpretations? Thanks

• Take a look at the theory of the quantum mechanical rigid rotor: en.wikipedia.org/wiki/…. As you can see, the energy eigenvalues are inverse proportional to the moments of inertia. If the moments of inertia become small, then the energy needed to excite the rotor is large. – CuriousOne Feb 16 '16 at 11:41