# Nonlinear susceptibility in second harmonic generation

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1. Nov 27, 2015

### Konte

I have read a book that demonstrate the origin of electrical susceptibility of high order in harmonic generation: (in Robert Boyd's book : "Nonlinear optics").

For example, he show clearly for the case of second harmonic generation, how $\chi^{(2)}$ depends on matrix element of electric dipole operator on the basis of eigenvector of electronic states.
$\chi^{(2)}_{ijk}(\omega_{\sigma},\omega_q,\omega_p)=\frac{N}{\hbar^2} P_F\sum_{mn} \frac{\mu_{gn}^i\mu_{nm}^j\mu_{mg}^k}{(\omega_{ng}-\omega_{\sigma})(\omega_{mg}-\omega_p)}$

Where $\mu_{nm}=\langle{\phi_n} |\hat{\mu} |{\phi_m}\rangle$ and $\phi_n$, $\phi_m$ are electronic wavefunctions.

It assumes that the transitions mentionned in the $\chi^{(2)}$ expression are between eletronic level.

My question:
Is it possible to make second or third harmonic generation with transition taking place in vibrational level or rotationnal level of a molecules?

2. Nov 28, 2015

### jfizzix

I don't think there's any special reason why the electronic levels couldn't correspond to rotational or vibrational levels in Second Harmonic Generation (SHG).
However..

-the final energy state will have to be at twice the energy of the initial state (so you couldn't ordinarily do SHG between adjacent vibrational/rotational energy levels, since the energy difference between these levels is (usually) just too small to account for a whole factor of two.
If we're talking about doing SHG between the vibrational/rotational levels adjacent to the ground state, this would not be a problem

Considering SHG between different vibrational/rotational levels near the ground state..
-the dipole transition elements may end up being quite small compared to the standard electronic energy levels..
-and with energy levels so close together, you may need to consider (Boltzmann) thermal statistics, as you won't be able to assume that almost all of the atoms are in their respective ground states, so the perturbation theory calculation of this susceptibility will require added treatment.

Also, you will want to make sure that phase matching is possible for this process (i.e., that the index of refraction at the low frequency is identical to the index of refraction at the SHG frequency). Given that the levels are so close together, I expect this to actually be easier than in the standard case, as you may get at least partial phase matching for free.

That being said, I don't think it's impossible, but the process may be a lot less efficient than standard SHG.

3. Nov 29, 2015

### Konte

Real thanks for your answer, it helps me to understand many things, however, I still have some questions. And forgive me for my English because I begin to learn it now.

If I consider the case of purely rotational level, for example a quantum planar rotor, the energy of $j$-th level is $E_j = \frac{\hbar^2 j^2}{2I}$.
This case show us that difference between rotational leves increase with $j$. So, I want ask you if in a such case, SHG between adjacent rotational energy levels is more possible?

Why doing SHG becomes possible if we consider rotational levels near the ground state?

Could you give me some reference (books or articles) about mixing thermal statistics and perturbation theory, please?

Thanks.

4. Nov 29, 2015

### jfizzix

in the case of the rigid rotor, there are no two levels where the energy of one is twice the energy of the second, so I don't think you could get SHG as far as I know. The reason for this can be seen by taking the ratio of two energy levels of the rigid rotor. If
$\frac{E_{i}}{E_{j}}=\frac{i^{2}}{j^{2}}=2$
then
$\frac{i}{j}=\sqrt{2}$,
which can't happen when $i$ and $j$ have to be integers.

Now, the vibrational levels have equal spacing, so that if you take the vibrational ground state to have energy zero, then the second excited vibrational energy level will have twice the energy of the first excited vibrational energy level, and SHG might be possible (since then you can generate an output photon at twice the frequency (and energy) of each of a pair of input photons).

Boyd's book it my go-to guide for everything in nonlinear optics. I don't know of any better references regarding second harmonic generation.

As far as needing thermal statistics with perturbation theory, that's really just my educated guess. Any reference on perturbation theory that doesn't automatically assume you start in the ground state would be your best bet for seeing how it would work.

5. Nov 30, 2015

### Daz

Interesting discussion! I’ve got a couple of questions and a comment.

Firstly the comment: I think there may be a good reason why rotational modes of a molecule cannot mediate SHG. It is well known that the second-order susceptibility may be non-vanishing only if the dielectric lacks a centre of inversion symmetry. If you are considering the oscillator strength of rotational modes of molecules, then presumably the molecules must be free to rotate. Therefore, in thermal equilibrium they will do just that: rotate - but randomly. An ensemble of randomly rotating molecules will appear isotropic and therefore chi(2) vanishes.

Now a question. Are you discussing resonant or non-resonant SHG. I can’t see anything in the expression for chi(2) in post 1 that looks like an absorption term, so presumably you are considering non-resonant SHG. In this case, wouldn’t that expression be summed over (virtual) intermediate states? Then why the requirement that the energy levels to be spaced according to the photon energies of the participating fields?

Or are you considering resonant SHG? In which case, where does the absorption part of chi(2) come from in that expression? And wouldn’t that be a hindrance to efficient SHG as the dielectric would simply absorb the light?

6. Nov 30, 2015

### Konte

Why do you specify on the need of "two levels where the energy of one is twice the energy of the second"? I was understanding that SHG is possible through a two virtual levels that ensure the absorption of the two incident $\omega$ even if there is no real levels of the system satisfying this? Have I misunderstand something?
How to know in advance that the dipole transition elements will be small?
Is there a way to know which dipole transition matrix elements will be close to zero? Does this have anything to do with the fact that the matrix element was evaluated between two very distant states?
example : $\langle n_{=2}| \mu |n_{=100} \rangle$

I think, our friend jfizzix can answer you better than me. I am a novice on this matter, so before your question, I never knew that there is another subtleties like resonant or non-resonant...

Thanks a lot for all of you comments.

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7. Nov 30, 2015

### Daz

Actually, I think it may be more helpful to talk in terms of a summation of virtual transitions between real states. You could think of it in terms of virtual states, but that doesn’t help with the calculations where the only thing you have to work with are the matrix elements between the actual states of the molecule. Sorry of that was confusing in my previous post.

You diagram shows a non-resonant interaction (the fields do not resonate with any real transitions in the dielectric.) i.e. The photon energies do not coincide with any allowed transitions. In this case the SHG is a purely parametric interaction and there are no real transitions. In particular, there is no energy or momentum exchanged between the field and the dielectric. This picture also corresponds with your expression for chi(2) in post 1, and this is the conventional configuration for optical SHG.

Conversely, if the fields do resonate with the dipole moment of allowed transitions then you have the resonant situation. In this case chi(2) may be very big as it is enhanced by the resonance but you also have to contend with the fact that real transitions between states of the dielectric also occur. In other words: the dielectric absorbs the incident light. I believe this situation is what jfizzix is describing.

8. Nov 30, 2015

### Konte

Ok, I see now !

So in the case of "non-resonant" way, am I right when I think that for any spectrum of energy (even a continuum) there is always a chance to obtain $\chi^{(2)} \ne 0$ (since $\mu_{ij} \ne 0$) ?

Thanks.

9. Nov 30, 2015

### Daz

Generally no: chi(2) vanishes identically if the dielectric possess inversion symmetry. That precludes all gases, liquids and amorphous solids, plus a good number of the crystallographic groups as candidates for SHG media.

This is why I questioned (above) whether you could obtain SHG mediated by molecular dynamics. In order to talk meaningfully about molecular dynamics, the molecules must be “free to do their own thing.” If they are free in that sense, then presumably each degree of freedom will contain 1/2 KT of energy. Wouldn’t this make the medium appear isotropic, at least in the sense of its macroscopic constitutive relations, such as dielectric susceptibility?

In the absence of inversion symmetry, then yes, there is a possibility to have a non-vanishing chi(2), providing (as you noted) the selection rules don’t make the matrix elements vanish.

10. Nov 30, 2015

### Konte

Thanks for your answer,

For the case of non centro-symmetric media
In your opinion, is there a particular characteristic or special criterion on the spectrum that make $\chi^{(2)}$ higher or smaller?
For example, does dense spectrum (like continuum) give higher or smaller $\chi^{(2)}$?

11. Nov 30, 2015

### Daz

In the lossless (non-resonant) regime, I don’t think that the distribution of energy states is particularly important. I say that because it is conventional to take the susceptibility tensor as being independent of frequency and this is generally held to be a good approximation. Amnon Yariv discusses this point in some detail in his (excellent) book: Quantum Electronics, Third Edition (J. Wiley & Sons, New York, 1988)

Yariv’s book also contains an abridged version of an analysis originally due to N. Bloembergen in his book Nonlinear Optics (W. A. Benjamin, New York, 1965) which is quite illuminating.

Bloembergen derived an expression for chi(2) semi-classically by considering the dielectric to be an ensemble of electrons constrained to move in anharmonic potential wells. In this formalism it is easy to see that the more anharmonic the potential (i.e. the less symmetrical) the greater chi(2). This view also agrees well with the experimental observation that the least symmetrical crystals tend to have the highest chi(2).

For example lithium niobate (LiNbO3) has trigonal structure and its chi(2) tensor contains 18 non-vanishing elements, the largest of which is an order of magnitude higher than the usual suspects (KTP, KDP, KD*P, etc.)

In more recent years highly-asymmetric polymers have also been found to exhibit huge value of chi(2). [No reference - I just got that titbit from Google!]

So I would say: you are looking for the least symmetrical dielectric and the precise distribution of energy levels is not of great importance.

One problem with such “exotic” dielectrics is that it is rarely possible to achieve phase-matching using birefringence alone. You might want to look into quasi-phase-matching as one technique used to overcome this obstacle.

12. Dec 1, 2015

### Konte

I have just read the Yariv's book and I found the part that demonstrate the dependance of nonlinear susceptibility on the anharmonicity of the potential well. Excellent.

You imply here that symmetry of potential where the electrons move is linked to the symmetry of the crystal? If it is the case how you link those two symmetries?

13. Dec 1, 2015

### Daz

That’s a good question!

I guess that the answer to your question is that in a crystal all the electronic states have the symmetry of the bravais lattice - that’s guaranteed by Bloch’s theorem.

I suppose we should really note that the anharmonic potential well is a semi-classical phenomenology. Despite that, it has some merits. For example one can see why the distribution of energy levels is of less importance in the calculation of chi(2): there are no (real) transitions. I guess the distribution of energy levels does play a part in determining the shape of that potential well, however.

Despite the intuitive appeal of the anharmonic potential well, I think that for a crystalline solid the approach in your post no.1 would be the way to go because you can get the matrix elements from the irreducible symmetry representations.

I don’t honestly know how one would approach the calculation in a more general case, such as polymers and the like. Hopefully, there may be others more knowledgeable than me in this area.

If anyone reading this has any further insights, please don’t hesitate to chip-in.

14. Dec 1, 2015

### Konte

Thank you for your answer.
Other than the general case (as you noted about polymers and the like), could you suggest me some readings (books or articles) that deal with " matrix elements from the irreducible symmetry representations" for crystalline solids case please?

15. Dec 2, 2015

### Daz

16. Dec 2, 2015

### Konte

Thank you very much. I begin the reading and I will comment when I finish.

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