Does the expectation value of an operator with respect to Quantum Vacuum State has experimental meaning? I'm learning about the coherent state in Quantum Optics. And I'm confused with the statement in the book, which states that the coherent state is the minimum uncertainty state. And since the vacuum state is a special kind of coherent state, it is also a minimum uncertainty state.
My question is: I can understand the calculation of why the coherent state is the minimum uncertainty state, but what does the mean of an operator with respect to vacuum state mean in the language of measurement, since there is no photon in the vacuum state, and let alone a measurement with vacuum state?
 A: To help answering your question perhaps it's in order to remember ourselves why people use vacuum expectation values in the first place.
Mathematically, the vacuum state is just another pure state. You could think of it as a basis element of $\mathcal{H}$ for a suitable basis, like the computational basis. It also has the property that when acting with its associated ladder operators (which are necessary to define the vacuum state in the first place) you get a $0$ if you try to lower it further or eventually exhaust the spectrum of possible states  by rising it and combinations thereof.
Thus it has a clear physical/computational advantage: a priori, from the vacuum state you can get any state of physical relevance. So you only care about vacuum expectation values, as you can compute every expectation value from the vacuum state, at least in principle.
For completeness, your coherent state is no exception, quoting Wikipedia
$$
|\alpha\rangle=e^{-{|\alpha|^2\over2}}\sum_{n=0}^{\infty}{\alpha^n\over\sqrt{n!}}|n\rangle = e^{-{|\alpha|^2\over2}}e^{\alpha\hat a^\dagger}e^{-{\alpha^* \hat a}}|0\rangle.
$$

Edit:
In a nutshell, my argument is for any $|\psi\rangle = \mathcal{O}(a, a^\dagger )|0\rangle$, where $\mathcal{O}$ is an operator constructed from $a$ and $a^\dagger$’s. An example is the coherent state above with $\mathcal{O} \sim \exp(-\alpha a^\dagger)$. So all I’m saying is that if you know what all operators expectation value on $|0\rangle$ are, then you know all expectation values on any state. I believe this argument is in Weinberg’s vol. I of his QFT books, in the cluster decomposition principle chapter.
