# Is the vacuum state a coherent state?

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?

• Yes, the vacuum is the unique state which can be considered both a Fock state and a coherent state. – Rococo Sep 8 '17 at 22:02

## 2 Answers

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.

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.

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.