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Consider the following formal implications:

A quantum particle moving on the circle performs periodic motion even in the absence of an external potential. It can therefore be considered as a simple harmonic oscillator. It should then in principle be possible to define an annihilation operator, even though it is well-known that operators on the circle are troublesome. One can then obtain coherent states on the circle as eigenstates of an annihilation operator associated to a free particle, in significant contrast to quantum mechanics on the line.

Is this line of reasoning correct?

Edit: For inspiration, here is the motion of an initial wavefunction $\psi \in \mathcal{L}_2(S^1)$. I think it can be solved analytically, but I did it numerically. What is left is to construct the annhilation operator.

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QuantumBrick
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    "A quantum particle moving on the circle performs periodic motion even in the absence of an external potential" -- says who? – Emilio Pisanty Jul 14 '21 at 12:14
  • @EmilioPisanty Isn't this a direct implication of periodic boundary conditions? – QuantumBrick Jul 14 '21 at 12:15
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    Not really (or at least, not directly). Wavepacket spreading is rather unlikely to be undone by simple propagation in the absence of a quadratic potential. – Emilio Pisanty Jul 14 '21 at 12:16
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    (That said, the energy eigenstates do have commesurate energy differences, so it's not implausible, but it needs to be argued in detail.) – Emilio Pisanty Jul 14 '21 at 12:17

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