Stimulated photons direction: book suggestion.

[provolus]

Hi everyone! I will very thankful to whom can suggest me some texts or webpages where I can find a demonstrated explanation of why photons emitted by stimulated
atoms travels in same direction of the photons which interact with the atoms. Every
author says photons are coherent, same phase, energy and momentum but none with an equation.

Thanks in advance!

 

[mathman]

http://en.wikipedia.org/wiki/Stimulated_emission

Will this help?

 

[provolus]

Thx. But I need something more specific. Like 99 out of 100 books, even your link from wikipedia states

The notable characteristic of stimulated emission compared to everyday light sources (which depend on spontaneous emission) is that the emitted photons have the same frequency, phase, polarization, and direction of propagation as the incident photons.

I'm looking for a full quantum treatment (or at least semiclassical/time dependent perturbation theory/harmonic perturbation, Fermi GD and so on...).
I've already searched a demonstration on several books among them also Foundamental of Photonics (2009),the article in wikipedia (FoP 1991) is taken from.

 

[Cthugha]

Almost any text on quantum optics should cover this. Try "Optical coherence and quantum optics" by Mandel and Wolf. "Introductory quantum optics" by Gerry and Knight should cover it, too, but I am not completely sure about that.

However, the reason you will find is quite trivial. The emission rate into some mode of interest is proportional to the number of photons already present in this state. As a photon moving in a different direction also means that it is in a state of different momentum, the stimulated emission probability for the state of the light field with momentum in the same direction as the intracavity field will be highest.

If you want a detailed comparison of fully quantized and semiclassical approaches, you are looking for the Jaynes-Cummings model. Jaynes and Cummings compared the differences between semiclassical and quantum approaches. Their most helpful papers should be "Stimulated Emission of Radiation in a Single Mode" (Phys. Rev. 140, A1051–A1056 (1965)) and "Comparison of quantum and semiclassical radiation theories with application to the beam maser" (Proc. IEEE 51: 89–109 (1963)).

 

[provolus]

Thx Cthugha! I've already taken read chapter 15 of Mandel and Wolf but not specific answer. I will try to take a look on your suggested Cummings papers. Thx a lot.

The emission rate into some mode of interest is proportional to the number of photons already present in this state

All semiclassical treatments (without involving Einstein coefficients but perturbation theory) I found, are about one photon and one excited-two level atom. They say the emitted photon will have the same momentum (momentum conservation) and then direction of the initial one. They don't consider gas of photons into cavity or whatever.

Using field quantization operators the explanation is quite straightforward, but I want to see how they derive the "same direction" using just 1 photon and 1 atom.

 

[Cthugha]

Using field quantization operators the explanation is quite straightforward, but I want to see how they derive the "same direction" using just 1 photon and 1 atom.

Ok, I am not quite sure what your texts say exactly, but having one photon present is not too different from having many photons present in the quantized description. The stimulated emission always works only for photons ending up in a state which is already populated, so momentum and direction of the emitted photon are determined by the state of the already populated mode. What changes are the relative probabilities of having spontaneous or stimulated emission happen. However, I guess the spontaneous emission is neglected in most introductory texts as the authors are only interested in explaining stimulated emission. In semiclassical pictures the explanation of course becomes a bit handwaving.

All semiclassical treatments (without involving Einstein coefficients but perturbation theory) I found, are about one photon and one excited-two level atom.

This is a bit puzzling to me as a semiclassical treatment usually means that only the atom is treated in a quantized manner while the electromagnetic field is assumed to be classical.

 

[provolus]

The stimulated emission always works only for photons ending up in a state which is already populated, so momentum and direction of the emitted photon are determined by the state of the already populated mode.

How do you demonstrate this? By momentum conservation of photon?

This is a bit puzzling to me as a semiclassical treatment usually means that only the atom is treated in a quantized manner while the electromagnetic field is assumed to be classical.

Try for example Bransedn and joachain Physics of Atoms and Molecules page 190.

 

[Cthugha]

How do you demonstrate this? By momentum conservation of photon?

No, this is a statistical interference effect relying on the bosonic nature of photons. It is sometimes called bosonic final state simulation. Basically, the probability for indistinguishable bosons to end up in the same final state is twice as high as for distinguishable particles. A short summary is given in these lecture notes from Rutgers university:http://www.physics.rutgers.edu/~steves/273/notes/Lec18_Stimulated_Emission.pdf" . However, this short introduction is very brief. The whole story can be found for sure somewhere in the Mandel/Wolf. Most probably near the section, where the Hong-Ou-Mandel dip is discussed.

Try for example Bransedn and joachain Physics of Atoms and Molecules page 190.

Hmm, sorry, we do not have a copy of that book at the library of our chair. But I will try to get it elsewhere.

 

[provolus]

Thx for the link! I've just downloaded the lecture. Now I'm reading also "the quantum theory of light" by R. Loudon. But just as final question, what would you answer to the simple question "why a photon coming from stimulated emission must have the same vector momentum of the initial photon?"

 

[Cthugha]

I am afraid you will not like the answer, but I would say: by definition.

To be a bit more clear, if I had to explain stimulated emission in a didactic style, I would always start the other way round. Indistinguishability of particles is an important concept in quantum mechanics. I would start by explaining emission processes where the emission is indistinguishable from the external applied light field (no matter whether there is one photon or many) and then consider emission processes where the emission is distinguishable from the external applied light field. The result of the first calculation will lead to stimulated emission. The other case leads to spontaneous emission.

So I would say it is not the process of stimulated emission which causes photons to be emitted with the same momentum. It is the high emission rate of photons into modes which are already populated by photons indistinguishable from the emission - as follows from statistical arguments - which is then termed stimulated emission.

 

[provolus]

Thx Cthugha for the explanation but still I can't get your point about the reversed picture. Using the Fermi Golden Rule I can calculate the emission rate and I can SEE that the emitted photon has the same energy of the difference between the two atomic levels, but I can't see from where comes the same photon momentum vector, let say K.
If I use full quantized matrix element, books I considered (like loudon, schwabl, etc) state: with an initial Fock or number state "n" in the mode "K", so |n_K>, after the interaction with the atom (with interaction hamiltonian H) the final state can be |n_{K}+1> (emission) or |n_{K}-1> (absorption), that is a different photon number state n (plus or minus 1) but in the same mode K.
So my question is:
which is the physical principle that forbids to have, after the interaction, a different final mode K', not equal to K? In other words: why the matrix element <n_{K'}+1|H|n_{K}> is nosense?
Einstein argued the "same direction" without using quantum language but some kind of "drag" force...

 

[Cthugha]

You always have the spontaneous emission rate. For stimulated emission it is multiplied by (n+1). This enhancement can be explained in terms of interference of the indistinguishable probability amplitudes leading to the same final result. If the emitted photon is in principle indistinguishable from the external field (that means you cannot tell afterwards which was the emitted photon and which photon was already present in the field) you have to sum up all probability amplitudes and square the result to get the emission probability. If the emission is distinguishable from the external field, you have to square all the probability amplitudes first and then sum them up. This is pretty similar to Feynman's explanation of the double slit. The enhancement in the indistinguishable boson case gives you the factor of (n+1) which is characteristic for stimulated emission. However, they can only be indistinguishable if they share the same quantum state. This requires them to have the same energy and also momentum.

edit: Maybe I should be a bit more clear. I assume the matrix element <n_{K'}+1|H|n_{K}> means that you start with an excited atom and a photon in state k and end up with a ground state atom, a photon in state k and a photon in state k' which has been in the vacuum state before. This is not forbidden. It is spontaneous emission.
Or to use a different line of reasoning: For Fermi's golden rule you need the density of final states. In the final state you have an atom in the ground state and two photons (let me call them a and b) leaving and you have an initial photon (called i) and an emitted photon (called e). So you can have the final states a=i,b=e or a=e,b=i. If the initial end emitted photon are distinguishable, these are indeed two separate final states, so you can apply Fermi's golden rule to calculate the transition probability for each of them. If they are indistinguishable, the states a=i,b=e and a=e,b=i are the same state. You have to add all indistinguishable processes in Fermi's golden rule before squaring them.

So you get either:
$$\frac{2 \pi}{\hbar}|\langle a=i,b=e|H|i\rangle^2 +\frac{2 \pi}{\hbar}\langle a=e,b=i|H|i\rangle|^2$$
or
$$\frac{2 \pi}{\hbar}|\langle a=i,b=e|H|i\rangle+\langle a=e,b=i|H|i\rangle|^2$$

Which makes a huge difference. This is somewhat like a bosonic exchange interaction or an interference effect. In my opinion it should not really be pictured as a photon hitting the atom and knocking out a photon or something similar (at least at an advanced level) as this can be misleading sometimes.

 

[provolus]

First of all thanks for the detailed explanation. Second, I think I get the point now: reasoning about the photon bosonic nature and the interference of wave function based on their indistinguishableness (so enhanced probability). Like for the explanation of the Hanbury-Brown and Twiss experiment.

 

[Cthugha]

Ah, sorry. If I had known that you are familiar with HBT experiments, I would have pointed you in that direction earlier as I spent some time doing similar stuff. However, most people interested in stimulated emission never stumble across that topic, so I assumed the parallels would only confuse you.