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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

Community
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QCD_IS_GOOD
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2 Answers2

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If you take the case of a point particle, it won't contribute any rotational degrees of freedom. In your case, at low temperatures, the rotational degrees of freedom of the Argon atom don't contribute significantly, (simply because of the fact mentioned in the comment by CuriousOne, that the energy required to excite the rotor is high, that even the lower excited states are not easily excitable) so it can be approximated as a point particle having no rotational degrees of freedom. But at higher temperatures, such levels can be excited with ease, and as a result, the rotational degrees of freedom come into play, and Argon cannot be regarded as a point particle anymore. Your professor is not entirely correct in saying that symmetrical objects don't possess any rotational degrees of freedom, they do and can be modelled by the rigid rotor approximation to get the required energy spectrum, although these may or may not come into play at room temperature (as it doesn't in case of Argon). The rotational spectra of symmetrical molecules/atoms has been long known and observed. Thats the resolution of the contradiction.

Bruce Lee
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Note that your linked question is about the moments of inertia for diatomic molecules, but the ground state for argon at ordinary pressures and temperatures is a monatomic.

The angular momentum stored in an isolated atom is the sum of the intrinsic spins and orbital angular momenta of its constituent protons, neutrons, and electrons. If you were to reach into this population and stir it into a higher-angular-momentum state, like you might reach into a pot of soup and stir, you would do so by exciting electric or nuclear transitions. The lowest-energy electronic transitions typically have energies of a few electron-volts, though it’s higher in argon because argon has a closed valence shell; the lowest-energy nuclear transitions have typical energies of a few million electron-volts. These excited states are not populated at equilibrium in an environment where the temperature is tens of milli-eV, which we call “room temperature.”

That is to say: the term symbol for the ground state of an argon atom, ${}^1S_0$, is a shorthand for “this ensemble of electrons doesn’t have any angular momentum.” If you wanted to “rotate” an argon atom, you’d have to change that.

rob
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