A crystal in 422 needs two resonances

Two classes go dark far from absorption counted what Kleinman’s rule does to the second-harmonic coefficients of the thirty-two crystal classes. Far from any absorption the three fields in the process can be exchanged, the tensor becomes symmetric in all three indices, and eighty-three coefficients across the classes become forty-eight. The part removed transforms as the traceless gyration tensor plus a polar vector. Two classes, $422$ and $622$, lose everything: they have no cubic invariant left, and every second-harmonic coefficient they are permitted is one Kleinman’s rule forbids.

That essay ended with a question it could not answer by counting. The forbidden coefficients exist only through dispersion, the difference between a material’s response at the fundamental and at the harmonic. How do they grow as the frequencies approach an absorption band, and is the growth the same in every class, or does it depend on which representation the forbidden part belongs to?

The obvious model of dispersion answers it, and the first thing it says is unexpected. In a model with one resonance, $422$ and $622$ are not merely dark far from absorption. They are dark at every frequency. Dispersion with a single resonance cannot light them. They need a second resonance, and $432$ needs a third. Which classes need how many is a count of invariants, and the number of resonances a coefficient needs decides how fast it can grow: as the square of the frequency with two, as the sixth power with three.

The oscillator model #

The classical model of optical nonlinearity goes back to Lorentz for the linear part and to Bloembergen and Miller in the 1960s for the rest. A charge bound in a crystal is a damped oscillator. Its displacement $x$ responds to the field, and its potential energy has a harmonic part, which gives the refractive index and its dispersion, and an anharmonic part, which makes the response depend on the square of the field. The polarisation is the charge times the displacement.

Here the model is allowed $n$ oscillators, each a displacement that transforms under the crystal’s operations as a polar vector. The harmonic potential $\tfrac12 X^T K X$ and the cubic potential $\tfrac16 \sum T_{\alpha\beta\gamma} X_\alpha X_\beta X_\gamma$, in all $3n$ displacements, are both required to be invariant under the class. The field drives each oscillator with its own charge. In the transparent region, with no damping, the linear response at frequency $\omega$ is $X = L(\omega) Q E$ with $L(\omega) = (K - \omega^2)^{-1}$, and the second-order polarisation at the sum frequency is the cubic coupling evaluated on two linear responses and passed back through $L$ at the sum frequency:

$$\chi_{ijk}(\omega_1 + \omega_2;\, \omega_1, \omega_2) = -\sum_{\alpha\beta\gamma} \bigl(Q^T L(\omega_1+\omega_2)\bigr)_{i\alpha}\, T_{\alpha\beta\gamma}\, \bigl(L(\omega_1) Q\bigr)_{\beta j}\, \bigl(L(\omega_2) Q\bigr)_{\gamma k}.$$

$T$ is symmetric in its three indices, because it comes from a potential energy. That is what makes the static limit obey Kleinman’s rule exactly: when every $L$ is the same matrix, the whole tensor inherits $T$’s symmetry. Away from that limit the three matrices $L$ differ, and the part of $\chi$ that is not fully symmetric, the part Kleinman’s rule forbids, is whatever the difference between them makes. The model is simple and exactly solvable, and its invariant potentials are built by averaging random ones over the class, so every number below is computed rather than assumed.

How many oscillators a class needs #

With one oscillator the cubic potential is a cubic form in one polar vector, an element of $\mathrm{Sym}^3 V$, and the class must keep it. The number of such invariants is the third coefficient of the class’s Molien series, which the earlier essay computed: zero for $422$, $622$ and $432$. So with one oscillator those three classes have no anharmonic potential at all, and no second-order response at any frequency. Dispersion cannot help, because there is nothing to disperse.

With two oscillators $x$ and $y$ the potential can couple them, through terms like $x_i x_j y_k$. Those live in $\mathrm{Sym}^2 V \otimes V$, which splits into the fully symmetric part again and a mixed part of dimension eight, the same eight-dimensional piece that Kleinman’s rule removes. $422$ and $622$ each keep exactly one invariant in it. For $422$ it is $x_z (x_x y_y - x_y y_x)$: the vertical displacement of one oscillator times the vertical component of the cross product of the two. $432$ keeps none. With three oscillators the potential can contain $x \cdot (y \times z)$, which transforms as the determinant and is kept by every class without improper operations, $432$ among them.

The picture at the head of this essay sorts the classes accordingly, by the counts of three characters averaged over each class, the averaging that Neumann’s principle, as one sum uses for every property tensor and that a character does not know its basis shows can be done in any coordinates. Each count is a coefficient of the kind how many invariants of each degree reads from a Molien series. The eleven centrosymmetric classes have no second-order response at all. Eighteen classes need only one oscillator. $422$ and $622$ need two. $432$ needs three. Four of the eighteen are shaded: $23$, $\bar43m$, $\bar6$ and $\bar62m$ have no mixed invariant, so whatever the model, their second-order tensor stays fully symmetric at every frequency. Those are the four classes the earlier essay found Kleinman’s rule leaves untouched, now seen from the other side. Kleinman’s rule constrains them by nothing, because their symmetry already does.

The couplings that light them are handedness #

The invariants that rescue $422$, $622$ and $432$ are not arbitrary couplings. The earlier essay found that the eight-dimensional mixed part, the part of a second-order tensor Kleinman’s rule removes, transforms as a polar vector plus the traceless gyration tensor, the tensor that carries optical activity. Which half a class keeps depends on whether it is polar or handed. The ten polar classes keep the vector. The classes that may rotate the plane of polarised light, fifteen of them, keep gyration components. $422$ and $622$ are handed and not polar, so their one mixed invariant is a gyration invariant.

So the coupling that lets two oscillators in $422$ radiate a second harmonic has the symmetry of optical activity. That is visible in its form. The cross product $x \times y$ of two polar displacements is an axial vector, and its vertical component times a vertical polar displacement is a pseudoscalar along the four-fold axis, the quantity that changes sign in a mirror and that a handed crystal is allowed to carry. A mirror-image crystal would have the opposite coupling, and its second harmonic would have the opposite sign of $d_{14}$.

$432$ is the extreme case. Its one cubic coupling, $x \cdot (y \times z)$, is the determinant, the pseudoscalar itself, and it is the same invariant that makes $432$ optically active at all: a cubic crystal can rotate light only through the trace of its gyration tensor, which is a pseudoscalar. The only way a $432$ crystal can mix two frequencies is through its handedness. It is also why the two cases need more than one resonance. A pseudoscalar built from vectors needs vectors that are independent, and one oscillator’s response to one field supplies only one direction. Each permits what the other forbids traced the same pseudoscalar through the linear properties, where it separates optical activity from piezoelectricity. Here it separates the classes that need one resonance from those that need several.

With one oscillator, the departure is linear optics #

The eighteen classes that one oscillator can serve do depart from Kleinman’s rule as the frequency rises, and the model says by how much. A single oscillator in an anisotropic crystal has different resonance frequencies along different axes. A tetragonal crystal resonates at one frequency for displacements in the basal plane and another along the four-fold axis. Then $L(\omega)$ and $L(2\omega)$ are diagonal in the crystal axes with different entries, and the coefficients that Kleinman’s rule makes equal pick up different products of them.

In $4mm$ the two coefficients are $d_{15}$, in which the polarisation is along a basal axis and one field along the four-fold axis, and $d_{31}$, in which the polarisation is along the four-fold axis and both fields in the plane. Kleinman’s rule says they are equal. The upper panel shows their ratio, one at low frequency and rising to 1.19 as the harmonic approaches the lowest resonance. The lower panel divides each coefficient by the three linear susceptibilities it involves, at the harmonic for the output direction and at the fundamental for each input direction. What is left is Miller’s $\delta$, and with one oscillator it is exactly constant and exactly equal for the two coefficients, to the last digit the arithmetic carries.

With one oscillator, the whole departure from Kleinman’s rule is predicted by the linear optics. The refractive indices at $\omega$ and $2\omega$, along each axis, fix how far $d_{15}$ and $d_{31}$ part company, and no nonlinear measurement is needed to say so. Miller noticed the empirical version of this in 1964, that $\chi^{(2)}$ divided by three linear susceptibilities varies far less between materials than $\chi^{(2)}$ itself. In the one-oscillator model it is not an empirical regularity but an identity. That is the answer the earlier essay’s question was reaching for in the eighteen classes: the departure has the same form in every one of them, because it is the anisotropy of the refractive index divided out of a constant.

Two resonances light 422 #

$422$ is outside that argument, because with one oscillator it has no constant to divide. Give it two oscillators, with different resonances, coupled through its one mixed invariant, and it radiates.

The response needs the two resonances to differ, and the figure shows it vanishing exactly where they coincide. The reason is visible in the invariant. $x_z(x_x y_y - x_y y_x)$ is antisymmetric in the two oscillators’ basal displacements. If the two oscillators respond identically to the field, their basal displacements are parallel, and the force each exerts on the other through this coupling cancels against the force it receives. Writing the result out for the coefficient $d_{14}$, which couples a field along one basal axis and one along the four-fold axis into a polarisation along the other basal axis, gives

$$d_{14} \propto L_y(2\omega)\, L_x(\omega) - L_x(2\omega)\, L_y(\omega),$$

with $L_x$ and $L_y$ the two oscillators’ basal responses. It is a two-by-two determinant of the two dispersions at the two frequencies. It is zero in the static limit, where both columns are equal, and zero when the two oscillators are the same, where both rows are. That is Kleinman’s rule and its failure in one line: the static limit is dark because a determinant with equal columns vanishes, and the response appears only when the frequencies separate and the oscillators differ.

A real crystal has many resonances, electronic and vibrational, and a $422$ crystal such as paratellurite, $\mathrm{TeO_2}$, will have several to couple. What the model says is that its second-harmonic coefficient is not set by any one of them. It is set by how differently two of them disperse, through a coupling constant that no linear measurement can see. Unlike the eighteen classes, $422$ and $622$ get no help from Miller’s rule.

How fast the forbidden part grows #

At low frequency, expanding the determinant gives $d_{14} \propto 3\omega^2 \bigl(1/\omega_y^2 - 1/\omega_x^2\bigr)$, with $\omega_x$ and $\omega_y$ the two resonances. The response grows as the square of the frequency, in proportion to the difference of the inverse squared resonances. The forbidden part of the one-oscillator classes grows the same way, since there the two “oscillators” are the same charge’s motion along two axes with two resonance frequencies, and the same two-by-two determinant appears.

The computed slopes are 2.00 for $4mm$, $\bar42m$ and $3m$ with one oscillator, and 2.00 for $422$ with two. That answers the earlier question directly. The rate is the same whether the forbidden part belongs to the polar vector, as in $4mm$, or to the gyration tensor, as in $\bar42m$ and $422$. The representation decides which coefficients are forbidden. It does not decide how fast they grow. What decides that is how many resonances the coefficient has to set against each other.

$432$ shows the rule’s other case. With three oscillators coupled through $x \cdot (y \times z)$, the response is a three-by-three determinant of three dispersions at three frequencies. Two of the frequencies must differ, because the coupling is alternating, and a second harmonic, with both input frequencies equal, makes two columns of the determinant identical. So $432$ never doubles a frequency, even with three resonances, which is the earlier essay’s sum-frequency exception reached from the model. Mixing two different frequencies, it does respond, and its response vanishes like a Vandermonde determinant in the three squared frequencies, as the product of their three differences. That is the sixth power of the frequency, and the computed slope is 6.00. A forbidden coefficient that needs $k$ distinct resonances vanishes at low frequency as the power $k(k-1)$: two for $422$, six for $432$.

What the model leaves out #

One kind of nonlinearity. The model’s nonlinearity is a potential energy, an anharmonic restoring force. Real second-order response also comes from processes that are not of this form: electrons moving between bands, magnetic-dipole and electric-quadrupole contributions, and the motion of the ions at frequencies below the vibrational resonances. Some of those do not reduce to a symmetric cubic potential, and they could make $422$ radiate with one resonance. The statement here is about the anharmonic-oscillator picture, which is the one Miller’s rule comes from, and within it the counts are exact.

No absorption. The model is solved without damping, in the transparent region, where Kleinman’s rule is supposed to apply. Near a resonance the responses become complex and the determinant’s growth is overtaken by the resonance itself.

Random couplings. The invariant potentials are random elements of the right invariant spaces, averaged over the class, so the sizes in the figures are illustrations. The zeros, the exponents and Miller’s constancy do not depend on the draw. Each is either an exact cancellation or the leading power of an expansion, and each was checked to the precision the arithmetic carries.

Permitted is not present is the general warning that applies here too. A coefficient a class permits may still be small in a material, and the model says that in $422$ and $622$ it will be, far from absorption, by a factor of the frequency squared over the resonance squared and of the difference between two resonances.

Who worked out the pieces #

The anharmonic-oscillator model of nonlinear susceptibility was developed by Bloembergen and his colleagues and by Garrett and Robinson through the 1960s. Kleinman’s symmetry was stated by Kleinman in 1962 and Miller’s rule by Miller in 1964. That the departure from Kleinman’s rule follows linear dispersion in the one-oscillator model is implicit in Miller’s rule, and the observation that $422$, $622$ and $432$ need more than one oscillator follows from the cubic-invariant counts of the earlier essay, where the third Molien coefficient decides whether an order parameter may change smoothly. The tiers here are those counts read as a statement about resonances.

What the model must refuse #

The two refusals are the ones a quick reading of the model would get wrong. Two oscillators are not enough by themselves to light $422$: they must have different resonances, and when they are given identical ones the response is zero to sixteen decimal places. And three oscillators are not enough for $432$ to double a frequency, because the alternating coupling is cancelled by the two identical input fields. Each refusal is a determinant with two equal rows or columns, which is the whole mechanism.

Still open: a material that shows the rate #

The model predicts a measurable signature. In a $422$ or $622$ crystal the second-harmonic coefficient should rise with frequency as the square of the frequency, from zero, while in a polar crystal like a $4mm$ one the coefficients rise from a non-zero static value and their ratio departs from one as the square. Paratellurite is the obvious $422$ crystal to test it on, and $\beta$-quartz, hexagonal $622$ above 573 degrees Celsius, the obvious $622$ one, where the coefficients $\alpha$-quartz owes to its one-oscillator class $32$ are forbidden and the second harmonic should drop sharply at the transition. Whether measured dispersion of $d_{14}$ in either follows the square of the frequency, and whether it can be fitted by two resonances read off the absorption spectrum, is a question for data the model cannot supply.