I am reading some introductory materials on modern optics, in which they mention two-photon processes everywhere. I know fundamental optics and a bit on quantum mechanics. Can anyone explain in a plain way what is a two-photon process or any simple introduction to a related topic.

  • $\begingroup$ have you looked at the wiki article? en.wikipedia.org/wiki/Two-photon_excitation_microscopy . This gives the diagrams en.wikipedia.org/wiki/Two-photon_physics for high energy but similar should hold for chemical processes too with interactions on the molecules. $\endgroup$
    – anna v
    Commented Feb 27, 2013 at 5:29
  • $\begingroup$ Yes, I did. But that's too technical. There mentioned something on high-energy physics which is too hard for me to understand. $\endgroup$ Commented Feb 27, 2013 at 5:40
  • $\begingroup$ My answer to Is energy exchange quantized? includes a discussion of one process that is related to Oli's answer below. There is a limit to how "plain" a discussion of these processes can be before they become simply wrong, as quantum mechanics is required to understand these processes. $\endgroup$ Commented Feb 27, 2013 at 16:53
  • $\begingroup$ Classical physics goes a long way to explain the concept of second harmonic generation in a nonlinear crystal. It is just one two-photon process. Other effects cannot be introduced that easily and need a full quantum treatment from the start. $\endgroup$
    – Oli
    Commented Feb 27, 2013 at 22:00

2 Answers 2


Here is a very basic explanation (hopefully not too much), which involves only basic physics.

Let's consider light passing through a piece of transparent glass. Under the influence of the oscillating electric field, the electric dipoles formed for instance by pairs of nucleus-electrons (i.e. atoms) will oscillate. These oscillating dipoles will in turn behave as antennas, which will radiate light, at the frequency at which they are vibrating. Macroscopically, the way the dipoles vibrate affects the optical index and the losses of the material.

In a linear medium, this effect depends linearly on the electric field. This means for instance that if you send twice the optical intensity on your piece of glass, you expect twice the intensity exiting it. It also means that if you send light composed of several colours, they will be well behaved and ignore each other.

In a nonlinear medium, the polarisation of some dipoles will behave differently. These dipoles will also have some response that depends on the square of the electric field. Let's consider an electric field at a frequency $\omega$ : $$E(t)=E_0\cos (\omega t) $$

Through trigonometry, the square of this field can be rewritten as $$E^2(t) = \frac{E_0^2}{2}(1+\cos(2\omega t))$$

This means that under the influence of a field at a frequency $\omega$, our second order non-linear dipole will also start to oscillate at a frequency $2\omega$. It will then radiate light at this frequency. This effect is called second harmonic generation.

Looking at it another way, in this process 2 photons at a frequency $\omega$ now produce a single photon at a frequency $2 \omega$. This is an example of a two-photon process! More generally, photons at frequencies $\omega_1$ and $\omega_2$ could produce photons at frequencies $\omega_1+\omega_2$ (thus conserving energy). Other effect are also possible, for instance, starting from a photon at a frequency $\omega_1+\omega_2$ and creating a pair of photons at $\omega_1$ and $\omega_2$.

These effects need strongly nonlinear materials and/or very high intensities to be significant so they are not usually witnessed in everyday's life. However, one example would be those fancy green laser pointers. In them, there is a small laser producing infrared light which is then frequency doubled in a non-linear crystal, producing green visible light. The infrared laser is pulsed so that for the same average power, the electric field can reach much higher (peak) intensities, enhancing the non-linear effects greatly.

Amazing technology at a few bucks' cost!

P.S. I simplified things a bit too much, so to clear things up: Individual atomic dipoles can't pick up a second order non-linear polarisation (for symmetry reasons). In that case, it can be the result of dipoles formed by ions in some asymmetric crystals.

  • $\begingroup$ hi Oli, thank you so the simple explanation. I have few questions. 1) as my understanding, two-photon process is one photon at $\omega_1$ and one at $\omega_2$ can produce one photon $\omega_1+\omega_2$ or one photon at $\omega_1+\omega_2$ can generate two photons each at $\omega_1$ and $\omega_2$, but why this requires strong nonlinear effect? 2) If I have a photon at frequency $\omega$, could it be possible to generate two photons at $a\omega$ and $b\omega$ where $a<1, b<1, a+b=1$, so there could be many different photons at different frequencies generated? $\endgroup$ Commented Feb 28, 2013 at 0:02
  • $\begingroup$ 1) You need a non linear effect, which is usually week. Since the efficiency of second harmonic generation increases with the power, high power allows the effect to produce bright beams. Note that the efficiency of the opposite effect (creation of pairs of photons) doesn't increase with higher power of the pump beam, so you usually just create pairs of photons and not a bright two-color beam. See en.wikipedia.org/wiki/Parametric_down_conversion. You can then put your crystal between mirrors to form a en.wikipedia.org/wiki/Optical_parametric_oscillator $\endgroup$
    – Oli
    Commented Feb 28, 2013 at 0:39
  • $\begingroup$ 2) This statement is correct. $\endgroup$
    – Oli
    Commented Feb 28, 2013 at 0:47

When we speak of photons we speak of elementary particles, that means that we are in the realm of quantum mechanics . Quantum mechanical problems can be solved using the Feynman Diagrams,. These are a pictorial way to allow one to write down the integrals that will describe the probability of a reaction to take place, either high energy or low energy.

They are simple conceptually but complicated in mathematical formalism.

Take this two photon to two photons diagram:

enter image description here

The curly lines are the photons, the dark straigth ones can be electrons, quarks etc, charged particles, the photons are neutral. There exists a one to one correspondence for every line and vertex in the diagram to a specific formula which multiplied will give a long formula with integrals and functions within them, and the whole thing squared will give the probability of a photon hitting a photon, exchanging energy and two photons leaving the interaction. Probability means if I have N_1 photons on N_2 photons , how many of them will have interacted to produce N_3 +N_4 photons.

Now in the chemical/molecular two photon language, one can draw such Feynman diagrams.

Two-photon excitation employs a concept first described by Maria Goeppert-Mayer (1906–1972) in her doctoral dissertation in 1931, and first observed in 1961 in a CaF2:Eu2+ crystal using laser excitation by Wolfgang Kaiser.3 Isaac Abella showed in 1962 in cesium vapor that two-photon excitation of single atoms is possible.

Usually an atom can be excited by an electron being hit by one photon and going into a higher energy orbital. One in principle could write a Feynman diagram where the two photons are incoming and hit an electron in its orbital as it is shown here on the left. The right side needs some thought, as it would need an outgoing excited atom but you should get the idea. Here is a whole thesis on the problem, being treated with the Hamiltonian formalism. They find in section 5.2.1 that there is a probability for two photons to excite one atomic transition.

In this book they have developed specific Feynman diagrams for molecular processes, so this a subject that needs study, if one is serious in understanding it.

  • $\begingroup$ This is a good answer. A pure two-photon process at quantum level. $\endgroup$
    – JKL
    Commented Feb 27, 2013 at 11:54
  • $\begingroup$ Thanks anna. Honestly, I don't fully understanding all but I am trying to read the related materials you give here. Hopefully, I can go a bit more deeper. $\endgroup$ Commented Feb 28, 2013 at 0:15

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