What is the nature of the photons?

[Haorong Wu]

TL;DR Summary

photons; Fock states; Coherent states;

Hi. I am reading quantum optics by Marlan Scully currently. Although the book has devoted a whole section to explain the photon concept, I am still confused about some nature of photons.

First, how do photons distribute in space? For example, there are $n$ photons for a Fock state $\left | n \right >$, or $\left | \alpha \right | ^2$ photons for a coherent state $\left | \alpha \right >$. Let a light beam be emited from point A and be detected in point B. Then how do these photons distribute between point A and point B?

Personally, I visualize that in a small interval in space, there are $n$ or $\left | \alpha \right | ^2$ photons for a Fock state or a coherent state, respectively. Each of these photons behaves like a wave packet which occupies the whole space of the interval. And these photons superpose together in the interval.

Is this image correct? If so, then what is the length of the interval? If not, what is the correct picture?

Second, how to create a Fock state or a coherent state. The fock states are easy to make by letting $n$ excited atoms to decay simoutaneously. But how to create a coherent state $\left | \alpha \right > \propto \sum_0^ \inf \left | n \right >$? I can not figure it out, since it is a superposition of many different numbers of photons.

Thanks!

 

[Paul Colby]

Is this image correct? If so, then what is the length of the interval? If not, what is the correct picture?

No. The electromagnetic field is written as a sum of normal modes. Each mode amplitude is replaced with creation and annihilation operators. Photons are energy intervals of each normal mode.

The fock states are easy to make by letting n excited atoms to decay simoutaneously.

I completely disagree. Eigenstates of occupation number for photons are difficult to create. The only way I've seen is by using correlated atomic decays that produce a secondary photon that can be used in correlation with a primary photon.

In fact, the vast majority of light sources are, on a mode by mode basis, thermal. An ideal laser which produces just one mode, the photon number is described by an $\alpha$-state. Photons in an $\alpha$-state follow a black body distribution.

 

[PeterDonis]

The fock states are easy to make by letting excited atoms to decay simoutaneously. But how to create a coherent state

As @Paul Colby has pointed out, you have this backwards. It is the coherent states that are easy to create: just turn on your laser. Fock states are the ones that are hard to create; "just let excited atoms decay simultaneously" sounds easy to you, perhaps, but you can't control when excited atoms decay or which decay mode they decay into, so in practice this is a very poor and unreliable method for producing Fock states.

 

[Haorong Wu]

Thanks, @Paul Colby . But I still do not understant this sentence:

Photons are energy intervals of each normal mode.

So, could I say that, for a quantum electric field, there are different numbers of photons for different normal mode and the mode with a greater number of photons has more energy?

If so, then photons are not particles at all. They just represent some quantization properties for the electric field.

Meanwhile, I have not reached the chapter of quantum theory of laser. In the semiclassical laser theory, I do not find the relation between laser and coherent state, although it gives a expression about the number of photons in a laser. I guess I would understand better your words after I learn the quantum theory of laser.

Thanks!

 

[Haorong Wu]

Thanks, @PeterDonis . I am sorry I expressed myself poorly. By saying easy, I mean it is easy to picture a possible way to create a Fock state although I understand it is very hard to require those excited atoms to decay at the same time.

Since I have not read the chapter of the quantum theory of the laser, I did not know that a coherent state can be created by a laser amplifier. Thanks for point out that for me. I just finded it in the book. How exciting!

 

[vanhees71]

Another much more efficient way to create Fock states of photons is the use of parametric downconversion creating entangled photon pairs. You can use it as single-photon source using one photon of the pair as "trigger", indicating that there's one and only one other photon ("heralded-photon source").

There's a very recent review by Zeilinger at al in Nature Reviews about how to create fancy entangled multi-photon Fock states of various kinds,

https://www.nature.com/articles/s42254-020-0193-5
https://arxiv.org/abs/1911.10006

I'd also define "a photon" just as a one-photon Fock state. A real single photon must of course be described by a true normalizable state, i.e., it's not a monochromatic plane wave state.

BTW, the book by Scully and Zubairy is very good in carefully explaning in the first few sections what a photon really is!

 

[Paul Colby]

If so, then photons are not particles at all. They just represent some quantization properties for the electric field.

Well, particle here is a technical term that gets blurred by common usage. It's really very complicated. That's why there is nearly endless discussions of wave-particle duality. A mode of the EM field has momentum and energy but no location. It's a plane wave spread over all of space (ideally). To get a localized field one must superimpose many modes including modes of more than one frequency. That is why I say ideal laser. A real laser excites many modes, a continuum in fact. These modes are tightly grouped but they have a finite spread or width in frequency.

Now on top of there being many modes, one must also add in the quantum mechanical state of these modes which may or may not be entangled. The end result is hopeless to understand or visualize in terms of a classical particle picture. You'd best be served purging yourself of this view.

 

[PeterDonis]

could I say that, for a quantum electric field, there are different numbers of photons for different normal mode and the mode with a greater number of photons has more energy?

If "normal mode" means "Fock state" and you only consider Fock states, not any other states, something like this can be a reasonable heuristic.

However, as soon as you try to talk about any other states besides Fock states (such as coherent states), you don't even have a well-defined photon number (since non-Fock states are not eigenstates of the photon number operator) or a well-defined energy (since non-Fock states are not eigenstates of the Hamiltonian), so this heuristic no longer works.