Lasers work by stimulated emission of atomic transitions. Stimulated emission produces two photons which, because the particle number is well-defined, projects the field into a Fock state. However, it is a known fact that lasers emit light in a coherent state. How does the field evolve from a particle-state to a superposition of particle-states? Omitting normalization:

$$ | n \rangle \rightarrow \sum_{n=0}^{\infty}\frac{\alpha^n }{\sqrt{n!}}| n \rangle $$

I guess one way of looking at it is that the field shifts according to $\Delta n \Delta \phi \geq 1$ from certain particle number to certain phase but it feels like a superficial answer to me. What I want to understand is the mechanism that allows this to happen. Is it the reflection with the mirror? Is it the imposed boundaries of the resonating cavity? Pumping method?

  • 1
    $\begingroup$ @Steven The derivation you're looking for, as specified in the bounty banner's final paragraph, does not exist -- it is provably impossible. A Fock state is a time-translation-invariant state; coherent states, in contrast, are not. It is impossible to go from the one to the other using any time-translation-invariant interaction with any external system. The creation of an optical coherent state requires a nontrivial symmetry-breaking step. This has been challenged and deeply discussed in the literature (see the references in Wouter's answer and my comment there). $\endgroup$ Jan 15, 2021 at 16:08
  • $\begingroup$ @EmilioPisanty, thanks - when I get to a computer that has access to these journals I'll check it out. Intuitively I don't know how it makes sense to claim that laser light is poissonian light when homodyne tomography and quantum state estimation will reconstruct a coherent state with nonzero offdiagonal elements in the density matrix (and scrambling the phase will cause this purity to be lost and these terms will go to zero). But when I have access to the paper I'll read it and try to learn the arguement. $\endgroup$ Jan 16, 2021 at 20:46

3 Answers 3


You are making an incorrect assumption in your question: There is no physical evolution from a number state (aka Fock state). This evolution happened purely inside physicists' heads as it was realized that laser light is not properly described by number states. The problem is your assumption that the particle number ever is well-defined.

Lasing action is an inherently quantum-mechanical process: A photon interacts with a two-level system in its upper state. Unlike the simplified description you seem to be using, this does not always result in two photons and the two-level system in its lower state. What really happens is that a superposition between that result and the boring one, with no interaction at all, is created. Hence you have a superposition between a light field with one and with two photons. Continue this to the (theoretical, but sensible) limit of infinitely many such interactions (with interaction strength tuned to give your desired mean photon number), and you get coherent states.

  • 1
    $\begingroup$ Hi. This answer is quite enlightening to me. I was wondering where I could obtain a more detailed discussion. How is the Poisson-like superposition formula is actually derived out? I'd appreciate any suggestion of reading material. Thanks. $\endgroup$
    – Zheng Liu
    Jul 24, 2016 at 6:11
  • $\begingroup$ @ZhengLiu What you are looking for are textbooks or good lecture notes on coherent states (or quantum optics in general). Often, the guy who invented the concept is a great source (which you can easily consult: nobelprize.org/nobel_prizes/physics/laureates/2005/…). But in this case, you may be better off finding something less historically influenced. $\endgroup$
    – user73762
    Jul 30, 2016 at 10:42
  • $\begingroup$ It's not correct to say that the described process produces coherent states. The described process creates states like $a|1e\rangle + b|2g\rangle$, but this is an entangled state between the atom and the light field. Because of this entanglement, the light field does not have a definite phase, like coherent states do. See Wouter's answer or physics.stackexchange.com/questions/695993/…. $\endgroup$
    – user34722
    Mar 23, 2022 at 0:43

I'm going to stir things a bit and say that laser light is actually not a coherent state.

Because the emission events are random and independent to good approximation, this leads to a Poisson process. Consequently, the laser light will be in a classical mixture of Fock states with Poissonian number statistics (as is the number statistics for coherent states, but without a well-defined phase) . I don't think that this part is really controversial, I believe standard quantum optics books (e.g. Walls-Millburn) mention it. The common explanation to describe them with coherent states later on is spontaneous symmetry breaking: the mixed states interact weakly with an environment, and since the coherent states are pointer states, $U(1)$ phase symmetry is broken and the photon field assumes a pure coherent state. This is not so different from the onset of Bose-Einstein condensation, I believe.

There has also been an alternative claim, in the paper

"Optical coherence: A convenient fiction", Klaus Mølmer, Phys. Rev. A 55, 3195 (1997)

which to my understanding says that the symmetry breaking never really occurs, and everything we think to know about laser light having a well-defined phase is merely an illusion because of a circularity in reasoning about interference experiments.

I'm not enough of an expert to really say that I can completely agree with the latter claim, but based on the number of citations and not being aware of anyone really debunking it, it's tempting to believe that it might hold some truth.

  • $\begingroup$ Just to make sure this is not misunderstood: random and independent in this context means the emission times, but not the phase. If the phases of the emitted photons were uncorrelated the random interference would lead to a thermal state, as explained in this answer. $\endgroup$
    – A. P.
    Jan 15, 2021 at 11:15
  • 1
    $\begingroup$ @A.P. I don't really agree with you. The point is that number and phase are conjugate variables, satisfying Heisenberg uncertainty. So questioning the phase of an individual photon is as meaningless as questioning the position of a plane wave. The difference between both situations is made clear for the case of a dye-microcavity photon condensate for example iopscience.iop.org/article/10.1088/1361-6455/aad409/meta . There, there is a clear crossover between the canonical regime (Poisson statistics) and the grandcanonical regime (Bose-Einstein=thermal statistics). $\endgroup$
    – Wouter
    Jan 15, 2021 at 15:43
  • 3
    $\begingroup$ The Mølmer paper is an important landmark but it is really the start of a conversation on the literature. For a good review of where it went after that, see Dialogue Concerning Two Views on Quantum Coherence: Factist and Fictionist. Among other things, it's not really tenable to say that Mølmer's claims have never been challenged. $\endgroup$ Jan 15, 2021 at 15:58
  • $\begingroup$ @EmilioPisanty thanks a lot! $\endgroup$
    – Wouter
    Jan 15, 2021 at 16:00
  • 1
    $\begingroup$ @EmilioPisanty and Wouter: I'm confused by this topic and you understand it well. Would you please care to have a quick look at a follow-up question of mine? $\endgroup$
    – The Vee
    Jan 17, 2021 at 8:39

It is true that the light coming from a laser is not precisely a coherent state. One can measure the photon statistics to see that it only approximates the Poisson statistics. However, the OP is not concerned with the precise modeling of the light coming form a laser. Instead, the question is how a coherent state can come about as a result of the stimulated emission that occurs in a laser. To address this aspect, I present here a simplistic view of the process. It neglects the possibility that the excited atom remains excited and not radiate.

When the laser is switch on, it starts with the creation of a population inversion. Then one of the excited atoms decays spontaneously. The photon that is produced by the spontaneous decay stimulates other atoms to decay producing more photons. However, one can also see the subsequent stimulated decays together with the initial spontaneous decay as multiple spontaneous decays together with a normalization process.

Each spontaneous decay will effectively produce a superposition $$ |\psi_{spon}\rangle = |\text{vac}\rangle\beta + |1\rangle\zeta , $$ where $|\text{vac}\rangle$ is the vacuum state, $|1\rangle$ is a single photon state with the correct cavity mode, and $\beta$ and $\zeta$ are coefficients. The reason for the vacuum state is not because the atom did not radiated, which I exclude here, but because some of the photons are produced in modes that would not survive in the long run. These I remove and replace by the vacuum state.

For $n$ such spontaneous decays, one gets $$ |\psi_{n-spon}\rangle = \mathcal{N}\left(|\text{vac}\rangle\beta + |1\rangle\zeta\right)^n = \mathcal{N}\sum_{p=0}^n \frac{n!}{p! (n-p)!} \left(|1\rangle\right)^p \frac{\zeta^p}{\beta^p}, $$ where $\mathcal{N}$ is a normalization constant. In the last expression I absorbed $\beta^n$ in $\mathcal{N}$ and removed the tensor products with the vacuum states. Note that a tensor product of $p$ single photon states becomes a $p$-photon Fock state $$\left(|1\rangle\right)^p= \sqrt{p!} |p\rangle . $$

In the limit $n\rightarrow\infty$, we have $$ \frac{n!}{(n-p)!}\rightarrow 1 . $$ So, the state becomes $$ |\psi\rangle = \mathcal{N}\sum_{p=0}^{\infty} \frac{\alpha^p}{\sqrt{p!}} |p\rangle , $$ where we defined $$ \frac{\zeta}{\beta} \equiv \alpha . $$ What remains is to compute the normalization constant of the state, which would then lead to the well-known expression for the coherent state.

  • 3
    $\begingroup$ This is incorrect. Spontaneous/stimulated emission generates a state like $\alpha |e\rangle|n\rangle + \beta |g\rangle|n+1\rangle$, where $|e\rangle$ and $|g\rangle$ are atomic excited and ground states, $|n\rangle$ is an n-photon field state and $\alpha,\beta$ are probability amplitudes. Trace out the atoms and you will immediately see that there is no local field coherence. As Emilio has stated already, one can prove in a couple of lines of trivial algebra that coherent states can never be produced by atom-field interactions at optical frequencies. See Wouter's answer and comments. $\endgroup$ Jan 20, 2021 at 14:05
  • $\begingroup$ As I said, it is a simplistic view of the process. It excludes the probability that the atom remains in the excited state. In that case the radiated photon is not entangled with the atom and would therefore not lead to a mixed state when you trace out the atoms. $\endgroup$ Jan 21, 2021 at 3:02
  • $\begingroup$ This is what I was originally looking for. (+1) But it seems as though it's apparently controversial to write something like this...I guess I have to go through the literature to understand the argument - although I suspect there is a misunderstanding somewhere. $\endgroup$ Jan 22, 2021 at 17:57
  • $\begingroup$ see @The Vee's question/answer for more of the discussion of Mølmer's work. $\endgroup$ Jan 22, 2021 at 19:48
  • $\begingroup$ Thanks Steven. I thought this was what you were looking for. The issue about the controversy is not so significant in my view. There are papers that defend the approximate formation of a coherent state in a laser. I guess that is what your new bounty is all about. $\endgroup$ Jan 25, 2021 at 5:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.