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If we drive a quantum harmonic oscillator (e.g. starting from its ground state) at its resonance frequency then it will not just create an excitation in the first excited state but will create a coherent state superposition over all number states. This is why, for instance, we don't use quantum harmonic oscillators as qubits, as we cannot isolate the ground and first excited states from the higher order excitations.

Is there a good classical analogy for why this might be the case? Something similar to how if you try to pluck a guitar string then you won't just excite the fundamental, but will instead also get various contributions from the higher order harmonics? With the guitar string this is a consequence of the way a string is plucked necessarily being a mixture of stationary states, so this isn't an appropriate analogy in this case.

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  • $\begingroup$ I would say that the harmonic oscillator is more like a spring than a string. It's only composed of a single frequency. It doesn't have a harmonic series. The coherent states appear because they are the states that can be most phase-locked (while still satisfying Heisenberg uncertainty) with the drive. The occupation number of the one frequency is like the classical amplitude (and this is how to understand laser light, for instance). $\endgroup$ – Ryan Thorngren May 2 '18 at 7:31
  • $\begingroup$ Thanks for the comments. Agreed the string analogy is not appropriate here because, as you rightly point out, a QHO only has a single frequency. I guess then the classical analogy I'm looking (if a decent one actually exists) for is to explain why multiple oscillations, each of the same frequency, but different amplitudes would be excited when you drive the oscillator. I should have mentioned that my goal is to explain superconducting qubits to non-physicists, so any analogy that can help them understand why it is not easy to isolate the ground and first excited state of a QHO is welcome. $\endgroup$ – Jon A May 2 '18 at 8:22

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