which is the state if I add a photon to a coherent state of the light? Imagine a situation in which I introduce a photon of frequency w, described by the state $|1\rangle$ in a cavity previously filled by radiation in a coherent state $|\alpha\rangle$. My question is, which the final state of the system?
My problem to answer it, is because the $|\alpha\rangle$ state has an uncertainty in the number of photons equal to the sqrt(average of photon number of alpha). If the uncertainty is very large, adding a photon to the cavity will not change the state, then $|1\rangle|\alpha\rangle=|\alpha\rangle$? In some sense, the $|\alpha\rangle$ state would not notice of the new photon.
 A: $\newcommand{\ket}[1]{|#1\rangle} \newcommand{\bra}[1]{\langle #1|}$
These states are is known by the (admittedly unimaginative) name of single photon added coherent states. They are for example described in a 2004 paper by Zavatta, Viciani and Bellini (pdf). They have been (together with photon subtracted coherent states) a major object of research of continuous variable quantum information over the last decade, because they are essentially the only experimentally realistic non-Gaussian states and evade as such many no-go theorems (in computing, Bell inequalities, entanglement distillation, etc.)
To go back to your initial question, these state are intermediate between a coherent states and a Fock states. If I call this state $|\alpha+1\rangle$, 
the distinguishibility between $\ket\alpha$ and $\ket{\alpha+1}$ is characterized by the scalar product $\langle\alpha|\alpha+1\rangle$, which is $0$ if the states are orthogonal (perfectly distinguishable) and $1$ if they are identical.
We have
$$|\alpha+1\rangle = \frac{a^\dagger\ket{\alpha}}{\sqrt{1+|\alpha|^2}} $$
and therefore
$$\langle\alpha|\alpha+1\rangle 
  = \frac{\bra{\alpha} a^\dagger\ket{\alpha}}{\sqrt{1+|\alpha|^2}} 
  = \frac{\bra{\alpha} \alpha^* \ket{\alpha}}{\sqrt{1+|\alpha|^2}}
  =  \frac{\alpha^*}{\sqrt{1+|\alpha|^2}}
$$
The photon added state is orthogonal to the corresponding coherent state if and only if $\alpha=0$, but this the boring case, where we compare the vacuum and the single photon state. The brighter the coherent state is, the closer the scaler product is to 1. When $|\alpha|\gg1$, we have
$$|\langle\alpha|\alpha+1\rangle|=\frac{1}{\sqrt{\frac{1}{|\alpha|²}+1}}\simeq 1 - \frac{1}{2|\alpha|²}$$
which corresponds to your intuition: if $\ket{\alpha}$ is bright enough, it almost does not feel the additinnal photon.
