# 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.

• Welcome to Physics.SE! Please note that MathJaX is supported on this site, meaning that you can (and should) use simple LaTeX markup for mathematical notation in your question. Please edit your question accordingly, because as it stands it's hard to decipher what you're asking. – Michael Seifert Feb 22 '18 at 17:11

$\newcommand{\ket}{|#1\rangle} \newcommand{\bra}{\langle #1|}$ 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.
• The photon-added thermal states has been investigated too: see e.g. this paper (arxiv:0704.019/PRA 75 052106). The state is almost the one you describe, $∝a^{†}ρa$, see eq (3) (You’ve put the dagger in the wrong $a$, describing the photon-subtracted thermal state). The behaviour is then similar as the thermal state: if the average photon number is big, the final state is almost identical to the initial state, but: 1. It’s not the case if the number of photon is not big, 2. they are never absolutely identical – Frédéric Grosshans Feb 28 '18 at 11:58
• Wait! I did some calculations to confirm the above comment and they proved me wrong. Thermal states don’t behave like coherent states! Bosonic statistics favours high $n$, leading to a final average photon number $2μ+1$ when starting with $μ$, and states which are significantly different. Adding a single photons indeed changes the average $n$ 100 photons to 201, manly because you cannot add a singe photon deterministically, and the corresponding operation has a probability of success $∝(n+1)$, no matter how you do it, skewing your probability distribution towards high numbers. – Frédéric Grosshans Feb 28 '18 at 14:06