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From what I understand, the physical relevance and interest of a coherent state is that its dynamics closely resembles the one of its classical analogue.

For example, for a quantum SHO $\langle x \rangle \sim \cos(\omega t)$ and $\langle p \rangle \sim \sin(\omega t)$ just like in the classical case.

Mathematically, a coherent state $|\alpha\rangle$ is defined to be the eigenstate of the annihilation operator $a$, such that $$ a|\alpha\rangle = \alpha|\alpha\rangle.$$ Question: is there a relation between being the eigenstate of the annihilation operator and having a classical-resembling dynamics, or is it just a pure coincidence?

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  • $\begingroup$ I'm not sure how to formalize it, but one handwavey way that I think about this correspondence for the EM field specifically is that a classical macroscopic EM field should have the property that it is changed as little as possible when a photon is added or removed from it, and this is what the condition of being an eigenstate of $\hat{a}$ says. See also: physics.stackexchange.com/questions/89018/… . $\endgroup$ – Rococo Sep 12 '16 at 4:51
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A maximally classical state should have minimum and equally distributed uncertainty in $X$ and $P$. In other words the uncertain in $X$ should equal the uncertainty in $P$ and this uncertainty should be as small as possible. This leads to $$ (X-\langle X\rangle)|\alpha\rangle=-i(P-\langle P\rangle)|\alpha\rangle $$ or if we rearrange the equation $$ \alpha|\alpha\rangle=\langle X+iP\rangle|\alpha\rangle=(X+iP)|\alpha\rangle=a|\alpha\rangle $$ where $a=X+iP$ is the lowering operator. It is important to realize that $|\alpha\rangle$ is still a quantum state. As you have pointed out, $\langle X\rangle$ and $\langle P\rangle$ follow the classical trajectory, but if you calculate the variance of $X$ and the variance of $P$ you will find they are non-zero. The uncertainty principle must be satisfied.

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    $\begingroup$ Thanks! And is there a relation between being maximally classical and minimum uncertainty? $\endgroup$ – SuperCiocia Sep 11 '16 at 20:57
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    $\begingroup$ Also why do you have a $-i$ in front of the momentum in the first equation? $\endgroup$ – SuperCiocia Sep 11 '16 at 21:51
  • $\begingroup$ Why should the uncertainties be equal? I thought that $Δx Δp>=0.5$ is all it satisfies wrt the uncertainty relation. $\endgroup$ – TheQuantumMan Jun 29 '17 at 9:53
  • $\begingroup$ @TheQuantumMan physics.stackexchange.com/questions/90034/… $\endgroup$ – Vladimir F Jun 29 '17 at 12:39
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Thanks, both of these are great explanations. I am offering a heuristic explanation: Any state of the electromagnetic field is characterized by a probability distribution of occupancy of photon number states |m>. Suppose that the state we are looking for has an expected number of photons greater than zero. During absorption, photons are removed from every state that has photons in it. There is also, in general, a finite probability that the electromagnetic field is in the state with no photons |0>. When we remove a photon, owing to the distributed probability of the state of the EM field we have a finite probability of reducing the number of photons in any number state including the state with zero photons. If the EM field happens to have no photons, nothing happens, we do not get a photon and this is not a problem: the expected number of photons in the state is still reduced by one owing to the distributed probability of population in states with number of photons greater than zero. Hence, there is no reason why the resulting state could not have a continuous probability distribution albeit corresponding to one photon less, namely, the state should be an eigenstate of the annihilation operator. However, if one tries adding a photon to such a state, it will be impossible to ever alter or contribute to the probability of eigenstate |0> and therefore it will be impossible, but adding a photon, to reconstruct a probability distribution we started from, namely, the state we are looking for is not an eigenstate of the creation operator.

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