Blackbody radiation in quantum mechanics

[Malamala]

Hello! If I place a particle with more energy levels (of the order of kT) in a well defined state, in a thermal bath at temperature T, how will the blackbody radiation affect the internal state of the particle i.e. will the distribution be classical or QM? Basically, if I prepare that particle in that state, let it there a long enough time and measure it and I repeat this many times, I expect the populated energy levels to be given by a Boltzmann distribution. In the end, did each particle have a well defined energy, but different particles had different energies. Or all the particles ended up in the same superposition of different energies with weights given by the Boltzmann distribution, and when I measure them I make the energy collapse according to that weight? So, is the particle after that interaction with the thermal bath in a well defined state or in a superposition? Thank you!

 

[Twigg]

So, is the particle after that interaction with the thermal bath in a well defined state or in a superposition?

As a (very very good) rule of thumb, the final state will be an incoherent state (a mixed state where the off-diagonals of the density matrix, i.e. the coherences, tend to zero). However, if you do this with atoms in cavities, things get a little weird. I'm not an expert, but there are more complications to consider in that case.

In general, interactions with random, macroscopic noise sources tends to destroy coherence.

To put your question in a practical context, blackbody radiation is a major source of instability in atomic clocks. The interaction of the blackbody photons with the atoms results in dephasing, which shows up in Rabi or Ramsey sequences as decoherence (loss of contrast). This is why blackbody shielding is a key ingredient in any precision spectroscopy experiment.

 

[Malamala]

As a (very very good) rule of thumb, the final state will be an incoherent state (a mixed state where the off-diagonals of the density matrix, i.e. the coherences, tend to zero). However, if you do this with atoms in cavities, things get a little weird. I'm not an expert, but there are more complications to consider in that case.

In general, interactions with random, macroscopic noise sources tends to destroy coherence.

To put your question in a practical context, blackbody radiation is a major source of instability in atomic clocks. The interaction of the blackbody photons with the atoms results in dephasing, which shows up in Rabi or Ramsey sequences as decoherence (loss of contrast). This is why blackbody shielding is a key ingredient in any precision spectroscopy experiment.

To give an actual example, say I place a molecule in a Penning trap, in the ground vibrational state (by vibrational I mean CM vibrations), and in the ground internal vibrational and electronic state and J = 5. If the walls of the trap have a temperature T and I somehow check the J state after a time long enough for BB radiation to matter but shorter than the decay lifetime of the state, and say I find J=4, is the molecule fully in that state, as it was in J=5 initially (e.g. by stimulated emission due to the BB photons), or it is somehow a linear combination of J=4, J=5 and eventually other J values, and I just happened to make the wavefunction collapse in J=4?

 

[Twigg]

it is somehow a linear combination of J=4, J=5 and eventually other J values, and I just happened to make the wavefunction collapse in J=4?

It is theoretically possible but extremely statistically unlikely for blackbody radiation to cause a coherent superposition after a long interaction time. You will most likely find your single molecule classically in one J state or the other. The reason for this is that broadband, incoherent light tends to scramble superpositions (or any coherent information, for that matter). There is a way to prove this, but I'm a bit fuzzy on the details. Let me know if you're interested in a mathematical argument and I'll try to dig it up.

 

[Malamala]

It is theoretically possible but extremely statistically unlikely for blackbody radiation to cause a coherent superposition after a long interaction time. You will most likely find your single molecule classically in one J state or the other. The reason for this is that broadband, incoherent light tends to scramble superpositions (or any coherent information, for that matter). There is a way to prove this, but I'm a bit fuzzy on the details. Let me know if you're interested in a mathematical argument and I'll try to dig it up.

Thanks a lot, that makes sense. And sure, if you can I would like to see a derivation of that.