# Coherent state being the eigenstate of the annihilation operator

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?

• 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/… . – Rococo Sep 12 '16 at 4:51

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.
• Also why do you have a $-i$ in front of the momentum in the first equation? – SuperCiocia Sep 11 '16 at 21:51
• Why should the uncertainties be equal? I thought that $Δx Δp>=0.5$ is all it satisfies wrt the uncertainty relation. – TheQuantumMan Jun 29 '17 at 9:53