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I'm asking because I got introduced to the state $|0\rangle$ as a fock-state. Nevertheless: $$ \hat{a} |0\rangle = 0 |0 \rangle $$ It is an eigenstate of $\hat{a}$ with eigenvalue $0$, and it can be obtained the same way any other coherent states are obtained via the displacement operator with parameter 0: $$ \hat{D}(\alpha=0)|0 \rangle = e^{0 \hat{a}^\dagger - 0 \hat{a}}|0\rangle = |0\rangle $$

Would one consider the vacuum state a coherent state?

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  • $\begingroup$ Yes, the vacuum is the unique state which can be considered both a Fock state and a coherent state. $\endgroup$ – Rococo Sep 8 '17 at 22:02
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The coherent state $\vert \alpha\rangle$ is just a vacuum state $\vert 0\rangle$ translated in $x$ and $p$ space so $\alpha=x_0+ip_0$. Thus the vacuum state is a coherent state that has not been displaced, i.e. $x_0=p_0=0$.

In fact, a nice way to see this is in the Wigner function formalism. The vacuum state is just a Gaussian sitting at the centre of $(x,p)$ space whereas a coherent state is the same state displaced to another point. This is illustrated in the figures below, taken from this site: on the left is the Wigner function of the vacuum state, and on the right that of a coherent state.

enter image description here enter image description here

Note also that the Wigner function for the coherent state is everywhere positive, and positivity of the Wigner function is sometimes taken as a marker of classicality so in this sense coherent states (and the vacuum state) are "classical states".

A short movie illustrating the time evolution of the Wigner function of a coherent state can be found on the coherent state wikipage; it shows the Wigner function does not deform and remains non-negative at all times Of course since the vacuum state is an eigenstate of the Hamiltonian and lies at the centre of $(x,p)$, its Wigner function would actually remain there at all times.

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Coherent state is a superpostition of states with different particle number with a weight of Poisson distribution. In other words, $$|\alpha\rangle =e^{-{|\alpha|^2\over2}}\sum_{n=0}^{\infty}{\alpha^n\over\sqrt{n!}}|n\rangle $$ where $|n\rangle$ is a state with $n$ number of particles.

Thus when you say that my system is in coherent state, you mean that your system does not have a definite particle number. This is why if you remove a particle from that state with $$a|\alpha\rangle=\alpha|\alpha\rangle$$ you dont change the state. That's why coherent state is an eigenstate of annihilation operator.

But the reason that vacuum is an eigenstate of annihilation op is; it does not have any particle so there is nothing to annihilate. that's why it does not change. So in this sense, I would not say vacuum is a coherent state because it has a definite particle number $0$. and the reason of this similarity under the act of annihilation operator is due to different reasons.

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