# What's the difference between Raman scattering and second harmonic generation in crystal?

As far as I know, the Raman scattering is from the stokes and anti-stokes scattering that the laser light interact with molecular vibrations. So we know that ""laser light interact with molecular vibrations"" in crystal can cause Raman scattering.

As for second harmonic generation, the pump waves with frequency w go through the crystal creating waves with 2w. I've searched information from internet and there are all explained this phenomenon by math methods, but I want to know exactly the physical model that what's happening in the crystal which create second harmonic. Is it the same as Raman scattering (incident waves interact with molecular vibrations)?

• If you're looking for a classical/visual picture of what's happening at the molecular level when SHG occurs, then yes, there is one. The basic analogy is that SHG is like a full-wave rectifier, whereas Raman is like FM radio. The molecular picture can be found in some nonlinear optics books. I'll post a semi-classical picture in a bit. Commented Apr 9, 2014 at 12:57

I want to know exactly the physical model that what's happening in the crystal which create second harmonic

One physically intuitive model for thinking about light-matter interactions is in terms of an energy level picture. In this picture, light propagating through a material can be thought of as series of absorptions and emissions. In one of these cycles the material absorbs a photon, which will excite the material into a virtual state (i.e. not a true energy state of the material). Since this state is virtual it will almost immediately decay back down into a real state emitting a photon in the process.

Below is an image which shows 3 different situations. The Rayleigh scattering does not change the energy of the light because the decay is back into the original state. Raman scattering though will decay into a different state (the different energy levels due vibrational modes of the molecules) with the difference in the energy resulting in a change in color/frequency of the scattered light.

In second harmonic generation a photon is absorbed and before the material decays into a real energy level, a second photon is absorbed. When the material re-radiates the energy it only emits a single photon, so effectively 2 photons were combined to create a single photon with twice the energy/frequency, and the final state of the material is in the same level it was to begin with (unlike the Raman scattering cases).

There are several ways to look at these two effects. Their mechanisms are very similar, but their experimental realizations are rather distinct from one another.

Here's one of the physical models, a classical one. Since you ask for physical intuition, I think the classical picture works best. Quantum mechanical models use a different language and tool set. There are other classical models. It's helpful, ultimately, to understand all of the models. (Even in the classical model there is one bit of quantum mechanics that we can't get around. You'll see it below.)

Warm up - linear transmission: For practice, and to set the stage, lets first consider simple transmission of light through a solid. (I'll focus on solids as an example.) A light wave of frequency $f$ is present in the crystal. The electrons in the crystal are initially "stationary", by which I mean that the electron density is not changing with time. The electrons respond to the wave, and "slosh" in sync with the light, both in time and space, resulting in a polarization wave. (There could be a phase shift between the light wave and the polarization wave. We'll ignore that.) The polarization wave acts as a source of a new electromagnetic wave at the same frequence $f$. The new wave interferes with the original wave. This is the mechanism of linear transmission. It can explain the reduced wavelength of the light in the medium, and various other phenomena like reflection and refraction. It explains all the "linear" effects, where the frequency of the light doesn't change.

Second harmonic generation (SHG): Things start the same as linear transmission. The EM wave in the solid induces a polarization wave oscillating at $f$. But now we can go one step further (second order) and recognize that the original wave is now in the presence not of a stationary electron density, but one that is oscillating at $f$. The EM wave interacts with the oscillating charge distribution, resulting in beating at the sum and difference frequencies, $2f$ and $0$. The polarization wave at $2f$ generates an EM wave at $2f$: the second harmonic. The $0f$ component also exists, but is usually of no interest.

Before moving on to the Raman effect, we need an aside on the structure of solids (same applies to molecules). A simple model of a solid (or molecule) has fixed ions producig a potential where the electrons live. A model one step closer to reality allows the ions to move, and we have phonons (vibrations in a molecule), but we don't allow the electrons and phonons to interact: they are uncoupled. One step further allows electron-phonon interactions. In this model we can have infrared absorption: an incident EM wave causes a polarization wave at $f$ as usual. But now, due to electron-phonon interaction, the sloshing electrons drag the ions along a little bit. If the frequency $f$ matches a vibration frequency (and perhaps a few other conditions), that phonon (vibration mode) will be excited: energy moves from the EM wave to the vibration mode. This happens at frequencies much lower than visible light frequecies. In the infrared.

Now the Raman effect: The incident light at $f$ causes an electron polarization at $f$. But we've allowed electron-phonon interactions. Vibration modes are also causing an oscillating polarization, this one at the phonon (vibration) frequency, $\Omega$. As before, the electrons are experiencing two sources of oscillation, and the result is beat frequencies at $f-\Omega$ and $f + \Omega$, and new EM waves are generated at those frequencies. The former is called the Stokes frequency, and is what is ususally observed in Raman scattering. The second is called the anti-Stokes frequency, and it can be observed, also. But now we have a problem. The Stokes frequency is always observed. Even if there was no vibrational excitation in the first place to cause the polarization at $\Omega$, light at $f-\Omega$ is observed. How can that be?

This is where we have to reach out just slightly to quantum mechanics. Each phonon (vibration mode) has a zero point energy. They vibrations are never truly at rest. There is always some vibration in every mode. It is the zero-point vibrations that couple to the electrons, causing a polarization at $\Omega$ and ultimately new light at $f-\Omega$.

Zero-point fluctuations induce other effects. For example, spontaneous emission from an excited atom is induced by zero-point EM field fluctuations. As zero-point fluctuations induce the conventional Raman effect, it is more properly called the spontaneous Raman effect.

There is a lot more to the complete story. There is a stimulated Raman effect, and other effects, and there is a discussion to be had concerning the coherence properties of the light. And, of course, there is the quantum mechanical picture.

• Also, here're some question about SHG. Is it related to the dipole in the crystal? I've search several papers that researched about surface effect, and they discussed SHG difference between monolayers and bulks(inverse symmetry or not ). So I'm wondering that if the vibration of the dipole moment and the symmetry are related? Commented Apr 9, 2014 at 23:52

The basic analogy to consider is that SHG is like full-wave rectification in electronics, whereas Raman scattering is like FM radio modulation. While Punk_Physicist gave an energy-level diagram that is helpful, you may also be interested in a purely classical/mechanical picture of the process.

## Picture of a molecule

Think of a molecule as a pair of charges $q$ and $-q$ attached by a spring (covalent bonds), like so:

When you apply an $\mathbf{E}$-field, the charges will attempt to separate. The dipole moment becomes $$\mu(\mathbf{E})=qr(\mathbf{E})$$ where $r(\mathbf{E})$ is the distance separation as a function of applied field $\mathbf{E}$. For simplicity, let's assume $r(\mathbf{0})=0$.

If the bond is harmonic, then Hooke's law is valid, and $\mu\propto\mathbf{E}$. Thus, the macroscopic polarization caused by millions of these "molecules" is going to be a field linear in $\mathbf{E}$. Remember that for later.

Otherwise, if the bond is not harmonic, then $d(\mathbf{E})$ is not linear in $\mathbf{E}$, so let's expand to second order: $$d(\mathbf{E})=a_1|\mathbf{E}|+a_2|\mathbf{E}|^2.$$ In this case, if you have millions of molecules reacting to a field $\mathbf{E}$, the polarization has a quadratic dependency on $\mathbf{E}$. Remember that for later.

## SHG:

Now that you have that microscopic picture of molecules, let's recall some macroscopic E&M, and try to connect the two pictures together.

Inside matter, an externally-applied electric field $\mathbf{E}$ causes matter to rearrange and separate slightly into dipoles, which amplifies the field. This total field inside the matter is called $\mathbf{D}$, and since it depends on $\mathbf{E}$, it can be Taylor-expanded as $$\mathbf{D}(\mathbf{E})=\sum_{n=0}^\infty\boldsymbol{\epsilon}^{(n)}\cdot\mathbf{E}^n$$ where $\boldsymbol{\epsilon}^{(n)}$ is a rank $n+1$ tensor (and I've absorbed the factors of $1/n!$ into the $\boldsymbol{\epsilon}^{(n)}$) and where $\cdot$ represents $n$-fold tensor contraction.

Unless you're dealing with an electret material, $\boldsymbol{\epsilon}^{(0)}$ vanishes, and we can also leave out the order-3 and higher terms for simplicity, leaving $$\mathbf{D}(\mathbf{E})=\boldsymbol{\epsilon}^{(1)}\cdot\mathbf{E}+\boldsymbol{\epsilon}^{(2)}:\left(\mathbf{E}\otimes\mathbf{E}\right).$$

The linear response term $\boldsymbol{\epsilon}^{(1)}\cdot\mathbf{E}$ represents the Hooke's law (linear spring) behavior in the preceding "picture of a molecule". The quadratic response term $\boldsymbol{\epsilon}^{(2)}:\left(\mathbf{E}\otimes\mathbf{E}\right)$ represents the anharmonicity of the spring-molecule.

So now we have a visual connection between our "spring-molecule" and the induced field $\mathbf{D}(\mathbf{E})$. How does the quadratic term give rise to SHG?

Let $\mathbf{E}=\mathbf{E}_0\sin\left(\omega t\right)$. Then $$\mathbf{D}(\mathbf{E})=\boldsymbol{\epsilon}^{(1)}\cdot\mathbf{E}_0\sin\left(\omega t\right)+\boldsymbol{\epsilon}^{(2)}:\left(\mathbf{E}_0\sin\left(\omega t\right)\otimes\mathbf{E}_0\sin\left(\omega t\right)\right) \\ =\boldsymbol{\epsilon}^{(1)}\cdot\mathbf{E}_0\sin\left(\omega t\right)+\frac{1}{2} (1-\cos (2 \omega t ))\boldsymbol{\epsilon}^{(2)}:\left(\mathbf{E}\otimes\mathbf{E}\right) \\ =\mathbf{D}_\text{DC}+\mathbf{D}_\text{1st}\sin\left(\omega t\right)+\mathbf{D}_{\text{2nd}}\cos\left(2\omega t\right)$$ where $\mathbf{D}_\text{DC}$ represents the DC component of the field, $\mathbf{D}_\text{1st}\sin\left(\omega t\right)$ is a component which oscillates at the same frequency, and $\mathbf{D}_{\text{2nd}}\cos\left(2\omega t\right)$ is a component which oscillates at twice the input frequency. This last component is where 2nd harmonic generation comes from. Note there is also a $90^\circ$ phase shift.

second harmonic generation is a non-linear optic effect, this mean that is the dielectric polarization of the crystal responds non-linearly to the electric field of the light -in this case laser pulses.the photons emitted by the laser (monochromatic light) interact with the non-linear media (crystal), and because of the constructed properties of crystals the refractive index depends on the polarization and propagation of the light. and Because of the refractive index of the crystal there will be phase matching which will be present as a new photon with twice the energy. Raman scattering on the other hand is the observation of the inelastic scattering of photons -let's say in a gas- caused by energy transitions of the gas molecules. source - wikipedia

• Whoever said "normally in a gas" is very misinformed. The Raman effect has been and continues to be an important tool in liquid solutions, solids, surfaces, and interfaces. Commented Apr 9, 2014 at 12:09