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.
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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
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$\newcommand{\ket}[1]{|#1\rangle} \newcommand{\bra}[1]{\langle #1|}$ 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.
Frédéric Grosshans
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Thanks a lot for your clear answer! From the answer, should I induce that the final quantum state of a system in which I add a photon to a thermal state of light (inside a cavity), characterized by the usual density operator rho is: a rho a(dagger)/tr(a rho a(dagger))? Similarly, in this case, the final state did not notice the new added photon, isn’t? – P. Salas Feb 27 '18 at 16:48
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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
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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
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Thanks again! I see, in the case of thermal states there is a significant difference if I add a photon or not. – P. Salas Mar 05 '18 at 15:58
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...Coming back to add a photon to a coherent state (CS), reading some papers about single photon-added coherent states (SPACS), it seems necessary some interaction to get a SPACS: for instance sending an excited atom through a cavity containing a CS. But, just introducing a photon (literally) in the cavity previously filled by a CS, that interaction is missing, in this situation, do I still have a SPACS as a final state of the radiation? – P. Salas Mar 05 '18 at 15:59
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The big question is "how would you physically introduce a single photon (literally) in a cavity containing a CS?". If you don't want to do it via an atom, you should assume that one of the mirror is is slightly transparent and you try to send the photon through it. It will be probabilistic, like the atom. There is litterally no way to deterministically add a photon in a cavity containing a CS – Frédéric Grosshans Mar 05 '18 at 16:22
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In fact, it occurred to me several possibilities to reach the goal: 1) I have a cavity with a photon (for instance, dropped by an excited atom crossing a micromaser cavity, and this is a deterministic process if the transition is resonant with the cavity) and, after that, I would introduce a CS inside it; 2) I have a cavity filled by a CS and, I send one of the photons (e.g. signal) created by parametric down conversion through it. If I detect an idler photon in a detector outside, this would means the signal is inside the cavity. – P. Salas Mar 06 '18 at 10:41
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In fact, from the point of view of quantum states, my question is so simple as, is the equality 1> alpha> = a(dagger) alpha> fulfilled? – P. Salas Mar 06 '18 at 10:41
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To the last question: no, because (1) $|1\rangle|\alpha\rangle$ lives in two modes and $a^\dagger|\alpha\rangle$ lives in one and, more importantly, (2) the 1st state is normalized while the second has norm $|\alpha|^2+1$. The condition (2) is the reason why photon added coherent states cannot be prepared deterministically by “just adding a photon”. – Frédéric Grosshans Mar 06 '18 at 13:23
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For your possibilities, 1) is adding a coherent state to a photon, not adding a photon to a coherent state. This can be done through a partially reflective mirror, but the end state will show entanglement between the light inside the cavity and the reflected light. Some computation has to be done to check whether the limit of highly reflective mirror is indeed a photon added coherent state (but I doubt it). 2) No, the photon might be reflected out of the cavity. You have to check that the photon is not reflected, and the probability of reflection depends on the # of photons inside the cavity – Frédéric Grosshans Mar 06 '18 at 13:52