# Laser output

1. Nov 6, 2011

### phonon44145

Are photons in the laser output beam mutually independent and can the multi-photon output be described as the tensor product of individual photon states?

2. Nov 7, 2011

### Cthugha

The photons in the laser output are statistically independent. That means that one detection event does not give you any new information about when the next detection event is likely to occur, so there is no statistical tendency for photons to be close to or far from each other.

The multi-photon output can be described as a Poisson-weighted sum over Fock states if that was your question.

3. Nov 14, 2011

### phonon44145

That's not quite what I meant but it's conceptually close. I just don't understand how the photons can be statistically independent. If it's a sum over Fock states, then the exact number of photons in the output must be indeterminate. But if it's indeterminate, doesn't that imply that we cannot write the output state as a product of photon states?

4. Nov 14, 2011

### Cthugha

No, it does not imply that. The result of a single measurement of the photon number is indeterminate in the sense that it is not predictable. The mean result after many repeated measurements - the mean photon number - is instead well defined. Now in a coherent state the weights of each Fock state in the coherent state are given by a Poissonian distribution around a mean photon number. The result of a single measurement is indeterminate, but the probability to get a certain result is well known and given by a Poissonian distribution.

Now for photons to be statistically independent you need to fulfill the condition that the detection of a photon now must not change the probability to detect another one. While that sounds trivial, it is in fact absolutely not. Each detection event destroys one of your photons and should therefore decrease the probability of detecting another one. It turns out that for coherent states the photon number uncertainty exactly cancels this term that is introduced by the destruction of a photon.

Or if you are more math-oriented statistical independence means that the photon pair detection rate $$\langle n^2 \rangle$$ factorizes into the mean photon count rates $$\langle n \rangle^2.$$

Now you can incorporate the fluctuating photon number as the momentary state of the light field being the mean photon number $$n$$ with some added fluctuation $$\delta n.$$

So you have
$$\langle n^2 \rangle=\langle (\langle n \rangle+\delta n) (\langle n\rangle +\delta n -1) \rangle.$$
The -1 comes from the destruction of a photon during the detection process. Now this is
$$\langle \langle n \rangle^2 -\langle n \rangle +(\delta n)^2 \rangle.$$
All the terms linear in $$\delta n$$ vanish of course. As you see, this value depends on the variance $$(\delta n)^2$$ of the photon number distribution. If $$(\delta n)^2=\langle n \rangle$$ is fulfilled, you get factorization of the photon pair count rate into the product of the mean count rates. This is fulfilled for every distribution where the variance equals the mean. The best known one is the Poissonian distribution which brings us back to the beginning and to why a coherent state has Poissonian statistics.

5. Nov 14, 2011

### phonon44145

I think I understand the math above. So basically, the probability distribution is Poissonian, and the laser output is a superposition of Fock states with amplitudes equal to square roots of respective probabilities? However, does that assume the absence of noise, or will statistics remain Poissonian even in the presence of spontaneous emission?

6. Nov 15, 2011

### Cthugha

Well, spontaneous emission is usually associated with a Bose-Einstein distribution of the emitted light which has high variance. The light field is therefore noisier and the photons are not independent. There is a tendency for them to clump together, the so-called bunching. So the photon statistics for spontaneous emission are also Bose-Einstein-like for timescales shorter than the coherence time of the light.

In the presence of both stimulated and spontaneous emission, the resulting statistics are somewhere in between. The exact outcome depends on the relative contributions of stimulated and spontaneous emission. This can be seen for example in the output of a laser operated around threshold.

7. Nov 15, 2011

### phonon44145

So if I understood correctly, in a real laser the photons are not completely independent since one cannot suppress spontaneous emission? If that's the case, should we describe the multi-photon output as an entangled state, or does entanglement only apply when we have a fixed number of photons?

8. Nov 15, 2011

### Cthugha

I never said that. Of course you cannot completely suppress spontaneous emission, but for common lasers the relative contribution of stimulated emission to spontaneous emission is roughly the inverse of the laser beta factor which can be as large as 1000000:1 or above, depending on the kind of laser used. Under these circumstances spontaneous emission is pretty much negligible. It is of course more critical for lasers using a small number of emitters.

Why should it be an entangled state? You get entangled states from parametric processes like parametric down conversion. In this case you will just get a mixture of coherent and thermal light or in other terms light with partial higher-order coherence.

9. Nov 15, 2011

### phonon44145

"the inverse of the laser beta factor which can be as large as 1000000:1 or above"

That's true. My question was if mutual dependence between output photons exists in principle.

"It is of course more critical for lasers using a small number of emitters."

Suppose this number is small (let's say, N=1 emitter to take the extreme case). How would we then write the resulting output state?

10. Nov 15, 2011

### Cthugha

Well, from a purist point of view there will be some mutual dependence. However, you will most likely not find it in the photon pair detection rate I mentioned above. However, you can go to higher orders. So you can also check whether the three-photon, four-photon or n-photon coincidence rate will factorize. Any real light source will show some deviations from complete factorization at some order which will often be very large. Basically this is the same as comparing the emission photon number statistics to Poissonian statistics order by order. You start with the mean and go on with the variance. Then you will compare skewness and kurtosis. In some order you will find a difference. However, for practical purposes the difference usually does not matter.

This is highly nontrivial as the exact behavior depends on a lot of parameters. Even if you have just one emitter you can have several relations between spontaneous and stimulated emission depending on the quality of the laser cavity mirrors. It also depends strongly on the exact kind of laser used and the pumping strength. In principle all most common kinds of light (non-classical, coherent, thermal) are possible in such devices. This is still an active area of research today. See for example C. Gies et al., "The single quantum dot-laser: lasing and strong coupling in the high-excitation regime", Optics Express 19, 14370 (2011). There are many other articles on that topic, but this one is freely available here: http://www.opticsinfobase.org/oe/abstract.cfm?uri=oe-19-15-14370"

Last edited by a moderator: Apr 26, 2017
11. Nov 15, 2011

### phonon44145

May I take a specific example? Interaction Hamiltonian between an atom and the incoming photon is given by

H = g Σμ∙ϵ_s σ− a+s + h.c.

where μ is transition dipole moment, ϵ1 or ϵ2 - two possible polarization states of the incoming photon, σ− is Pauli spin-flip operator and a+s is the creation/annihilation operator, and h.c. is apparently Hermitian conjugate. If we Taylor-expand the evolution operator as e^iHt = 1 - iHt neglecting higher-order terms, will the output be an entangled state like in parametric down-conversion? As far as I understand the resulting math, the probability to get a pair of identical photons (both in the incoming state ϵ1) and the probability of a pair of different (one ϵ1 and the other ϵ2) photons will be related as 2:1. Does that mean there is only 1/3 chance of stimulated vs. 2/3 chance of spontaneous emission in this case?

12. Nov 16, 2011

### Cthugha

I do not get your Hamiltonian. Where is the atomic inversion operator? Is it supposed to be sigma? Is a+s a bosonic operator representing photon creation and annihilation? Also, I do not see stimulated emission incorporated at all. Is it supposed to be some version of the Jaynes-Cummings hamiltonian?

However, you usually do never get entangled states from spontaneous emission. How should you get the entanglement? There are rather special cases where it becomes possible. In such cases you need an excited state which has two decay channels leading to the states with the same energy. In such cases you may get an entangled state of the atom ending up in the up state and the photon being sigma+ polarized and the atom ending up in the down state and the photon being sigma- polarized. However, that is atom-photon entanglement.

Entanglement between two photons is even harder to create. You could use a decay cascade where the energy level after two consecutive photon emission events are equal. For example there are possibilities to use the biexciton cascade decay in quantum dots. Again this is rather complicated.

So where exactly do you think entanglement between emitted photons might enter? Or how should it be created?

13. Dec 2, 2011

### phonon44145

This is indeed Jaynes-Cummings hamiltonian, in the rotating wave approximation. Sigma is the atomic transition operator (sigma- means the excited atom falls to the ground level, emitting a photon). The idea is that the photon output is

|1,0> ----> sqrt(2) |2,0> + |1,1>

if |a,b> is a state with a vertically polarized and b horizontally polarized photons. So the likelihood of stimulated emission is 2 times greater that of spontaneous emission. Suppose now I pick a photon from the output beam, measure its polarization, and find it horizontally polarized. That allows me to predict with certainty that the other photon will be polarized vertically. On the other hand, if my photon is polarized vertically, then the other photon could have any polarization. That sounds to me like a case of weak entanglement between the two output photons, is that right?