Two classes go dark far from absorption

Light of one frequency enters a crystal and light of twice the frequency comes out. The effect needs a crystal without a centre of symmetry, because the polarisation it requires is quadratic in the field and an inversion reverses the field without reversing the square. It was first seen in quartz in 1961, about a year after the first laser, and it is now the ordinary way green laser pointers make green.

The tensor that carries it has three indices and is symmetric in the last two, since the two fields it multiplies are the same field. That is exactly the shape of the piezoelectric tensor, which turns a stress (symmetric) into a polarisation, and so the number of independent second-harmonic coefficients a class permits is the piezoelectric count that twenty of the twenty-one computes: eighteen in the triclinic class 1, nothing in the eleven centrosymmetric classes or in 432, something in the other twenty.

This essay is about a symmetry that is not the crystal’s. Far from any absorption, the three fields in the process can be exchanged freely, and the tensor becomes symmetric in all three indices. Kleinman pointed this out in 1962, and every table of nonlinear coefficients since has used it. It takes the eighteen down to ten, and it is natural to ask which eight disappear and which classes lose what. The answer turns out to be one of the cleaner surprises the point groups offer.

A symmetry of the property, not of the crystal #

The second-harmonic polarisation is written $P_i(2\omega) = \sum_{jk} 2 d_{ijk} E_j(\omega) E_k(\omega)$. The symmetry in $j$ and $k$ is free: the product $E_j E_k$ is symmetric, so any antisymmetric part of $d_{ijk}$ in those indices would multiply zero and can be dropped. It is a convention as much as a law, and it holds at every frequency.

The symmetry between $i$ and the other two is a different kind of statement. The first index belongs to the output, a polarisation at $2\omega$, and the other two to the inputs at $\omega$. There is no reason in general for the response at one frequency to equal the response with the frequencies swapped round. But when the material is transparent at all three frequencies involved and its response hardly varies between them, the susceptibility can be derived from an energy. A lossless medium with a dispersion-free response has a stored energy density whose cubic term is $U = \sum d_{ijk} E_i E_j E_k$ for one real field, and a cubic form in three copies of one vector is automatically symmetric under every permutation of its indices. The polarisation is then the gradient of that energy, and the tensor is totally symmetric.

So Kleinman’s rule is a statement about where the frequencies sit relative to the absorption bands, and not about the crystal’s operations at all. It holds approximately, to the extent that the dispersion between $\omega$ and $2\omega$ is small, and how small is a question for each material. What can be computed exactly is the count on each side of the approximation: how many coefficients a class permits when only the crystal’s symmetry is imposed, and how many when the index symmetry is imposed as well.

The census is the picture at the head of this essay. The totals are 83 independent coefficients summed over the twenty classes, and 48 under Kleinman’s rule. The triclinic class goes from eighteen to ten, as the dimensions require. The monoclinic class 2 goes from eight to four, $m$ from ten to six. Four classes, $\bar{6}$, $\bar{6}2m$, $23$ and $\bar{4}3m$, lose nothing at all. And two classes, $422$ and $622$, go from one to none: a crystal of either class can double light, and far from absorption it cannot.

How the counting is done #

The count of independent components is the dimension of the subspace of tensors the class leaves fixed, and that dimension is the average of a character over the class, as Neumann’s principle sets out. The characters needed here are polynomials in the trace of an operation $M$ and the traces of its square and cube. For the tensors symmetric in their last two indices the character is $\chi(M)\,\tfrac{1}{2}(\chi(M)^2 + \chi(M^2))$, which is eighteen at the identity. For the totally symmetric tensors it is $\tfrac{1}{6}(\chi(M)^3 + 3\chi(M)\chi(M^2) + 2\chi(M^3))$, which is ten.

The difference, eight at the identity, has character $\tfrac{1}{3}(\chi(M)^3 - \chi(M^3))$. That is the part of the second-harmonic tensor that Kleinman’s rule removes, and its character can be averaged over each class just as the others can.

Traces of integer matrices do not care which basis the matrices are written in, so the count is taken on the thirty-two classes as integer matrices in lattice coordinates, without a Cartesian frame. A character does not know its basis makes that point for elasticity. A second route shares nothing with the first except the groups. The class’s matrices are converted to orthogonal Cartesian matrices, each of the eighteen basis tensors is averaged over the class, and the rank of the eighteen averages is read off by elimination. Then the same averages are totally symmetrised and their rank is read again. Both routes give the same two numbers in all thirty-two classes.

There is a third route to the Kleinman count, and it is the most illuminating of the three. A totally symmetric tensor of rank three in three dimensions is the same thing as a homogeneous cubic polynomial in $x$, $y$ and $z$. The coefficients that survive the class are the cubic polynomials the class leaves invariant. So the Kleinman count is the third coefficient of the class’s Molien series, the generating function how many invariants of each degree computes from $1/\det(1 - tM)$. That route uses a determinant recursion rather than traces of powers, and it agrees in all thirty-two classes too.

What the energy argument says about 422 #

The Molien route explains the two dark classes before any tensor is drawn. Far from absorption the second-harmonic response comes from a cubic energy $U(\mathbf{E})$, and the class must leave that energy alone. In $422$ there is no cubic polynomial to leave alone. The two-fold axes across $z$ reverse $z$, which kills $z^3$ and $z(x^2 + y^2)$, and the four-fold axis along $z$ sends $x \to y$ and $y \to -x$, which reverses $xyz$; no other cubic is a candidate. The class’s ring of invariants has nothing in degree three. With no cubic invariant there is no energy, and with no energy there is no Kleinman-symmetric response.

The same is true of $622$ and of $432$, and those three are the only classes without a centre of symmetry whose Molien series has a zero in third place. $432$ has no second harmonic in any case, so Kleinman’s rule takes nothing from it that was there. The two classes it silences are $422$ and $622$.

The tensor itself shows how this works. Class $422$ permits exactly one coefficient, $d_{14} = -d_{25}$. Written out, the polarisation it produces from a single field is $P_x = 2 d_{14} E_y E_z$ and $P_y = -2 d_{14} E_x E_z$, with $P_z = 0$. The dot product of that polarisation with the field that made it is $2 d_{14} E_z (E_x E_y - E_y E_x)$, which is zero identically. So the second-harmonic polarisation in $422$ is always perpendicular to the driving field.

That is exactly what a cubic energy cannot produce. If $\mathbf{P}$ is the gradient of a homogeneous cubic $U$, Euler’s theorem gives $\mathbf{E} \cdot \nabla U = 3U$, so a response whose dot product with $\mathbf{E}$ vanishes everywhere comes from an energy that vanishes everywhere. The one coefficient $422$ permits describes a polarisation that turns the field rather than pushes along it, and that kind of response needs a frequency difference to exist at all. Kleinman’s rule assumes the frequency difference does not matter, and so it sets the coefficient to zero.

Nothing about this makes $d_{14}$ zero in a real crystal of class $422$. It makes it a coefficient that exists only through dispersion. Close to an absorption band it can be as large as anything else, and far from one it should be small. How small, in which material, is a measurement, and none is made here.

Quartz has the same term inside it #

The class that lost the coefficient is not obscure, because the same coefficient sits inside a class every optics laboratory owns. $\alpha$-quartz belongs to class $32$, and Franken’s original experiment in 1961 used a quartz crystal.

Class $32$ permits two coefficients. The first, $d_{11} = -d_{12} = -d_{26}$, survives total symmetrisation intact: its three entries already have the pattern a cubic energy would give them, the pattern of the invariant $x^3 - 3xy^2$. The second, $d_{14} = -d_{25}$, is exactly the coefficient of $422$, entered positively once and negatively once, and Kleinman’s rule removes it for exactly the same reason. In quartz, then, the Kleinman approximation predicts a single coefficient, and the second is a measure of how far quartz departs from the approximation. Published tables of quartz give $d_{14}$ as much smaller than $d_{11}$, which is the size one expects of a coefficient that the energy argument forbids.

The two dark classes and quartz’s lost coefficient have one thing in common, and the next section says what it is.

What the eight dimensions are #

The eight dimensions that Kleinman’s rule removes form a representation of the full orthogonal group, and it is worth identifying it, because once identified it says which classes will be affected without any further computation.

The algebra is short. The tensors symmetric in their last two indices are the product of a vector with a symmetric pair. Taking out the totally symmetric part leaves what is called the mixed part. It equals a vector times an antisymmetric pair, less the totally antisymmetric tensors. For an orthogonal matrix an antisymmetric pair of vectors transforms like a vector times the determinant: the cross product of two polar vectors is an axial vector. The totally antisymmetric rank-three tensor is a single number transforming by the determinant alone. Putting those together, the character of the removed part is

$$\chi_{\mathrm{removed}}(M) = \det(M)\,\chi_{\mathrm{Sym}^2}(M) - \det(M) + \chi(M).$$

The first term is the character of a symmetric second-rank tensor that picks up the determinant, which is the gyration tensor, the object that carries optical activity in fifteen may rotate light. The second term takes away its trace, the single pseudoscalar that a chiral crystal or a sugar solution has. The third term is a polar vector, the object whose presence makes a class pyroelectric in the ten polar classes. Five plus three is eight.

The identity is checked operation by operation rather than only at the identity, because agreement of dimensions proves nothing about the representation. A character is a function on the group, and two representations with the same character at every element are the same representation. The table shows that the two expressions agree at every one of the ten kinds of operation a crystal can have. It follows that they agree on every class built from them. As a separate check, the count by gyration, pseudoscalar and vector agrees with the count by subtraction in all thirty-two classes.

So the coefficients Kleinman’s rule removes from a class are counted by two things the class already has names for. One is how many independent components its gyration tensor has beyond a possible trace. The other is how many polar directions it leaves fixed. A class with neither loses nothing to the rule. A class with a traceless gyration tensor and no polar axis, like $422$ or $32$, loses exactly the gyration components. A class with a polar axis loses at least one coefficient for it.

The four kinds of class #

That sorts the twenty classes into four groups, and the sorting agrees with the counts one by one.

Seven classes have both a polar axis and an anisotropic gyration tensor: $1$, $2$, $m$, $mm2$, $4$, $3$ and $6$. These are the polar classes that can also rotate light, and they lose the most. Six classes have the gyration alone: $222$, $\bar{4}$, $422$, $\bar{4}2m$, $32$ and $622$. Three have the polar axis alone: $4mm$, $3m$ and $6mm$, the polar classes with enough mirrors to forbid optical activity. And four have neither: $\bar{6}$, $\bar{6}2m$, $23$ and $\bar{4}3m$. In those four Kleinman’s rule constrains nothing, because every coefficient the class permits already has total symmetry.

The fourth group is worth a sentence. $23$ and $\bar{4}3m$ each permit a single coefficient, $d_{14} = d_{25} = d_{36}$, which is the coefficient of the invariant $xyz$ and is already totally symmetric. Zinc blende and gallium arsenide are $\bar{4}3m$, so for them Kleinman’s rule is not an approximation at all: the class leaves no room for it to be violated. The same holds for $\bar{6}$ and $\bar{6}2m$, whose permitted coefficients are the cubic invariants of their six-fold rotoinversions.

The polar part, seen in mm2 #

The polar vector’s share of the removed part is easiest to see in a polar class with no optical activity in the way, and then in one with both.

Class $mm2$ permits five coefficients: $d_{15}$, $d_{24}$, $d_{31}$, $d_{32}$ and $d_{33}$. Kleinman’s rule requires $d_{15} = d_{31}$ and $d_{24} = d_{32}$, leaving three. The two combinations it removes are the two differences, and the character calculation says one of them transforms as the polar axis and the other as the class’s single gyration component.

Which is which can be read off the structure of the removed part. The vector part is what is left of a tensor built as $a_i \delta_{jk}$ once its totally symmetric part has been subtracted. For a vector along the polar axis that leaves equal amounts in $d_{31}$ and $d_{32}$, and equal amounts half as large and of the opposite sign in $d_{15}$ and $d_{24}$. The vector part therefore governs the sum $(d_{31} + d_{32}) - (d_{15} + d_{24})$, and the gyration part the difference $(d_{31} - d_{32}) - (d_{15} - d_{24})$. In $4mm$ the four-fold axis forces $d_{31} = d_{32}$ and $d_{15} = d_{24}$, so the gyration combination dies, as it must in a class whose mirrors forbid optical activity, and only the vector combination remains. That is the one coefficient $4mm$ loses to the rule. Barium titanate’s tetragonal phase is $4mm$, and in published data its $d_{15}$ and $d_{31}$ are close but not equal.

The tensor that $\bar{4}2m$ permits shows the gyration part without a polar axis.

Class $\bar{4}2m$ is the class of potassium dihydrogen phosphate, the crystal that was for decades the standard frequency doubler for large lasers. It permits $d_{14} = d_{25}$ and $d_{36}$. Kleinman’s rule sets $d_{14} = d_{36}$, and the one combination it removes, $d_{14} - d_{36}$, transforms as the gyration tensor of $\bar{4}2m$. That tensor is traceless: it rotates light one way along $x$ and the other way along $y$, and not at all along $z$. It is one of the four classes that fifteen may rotate light finds permit optical activity without being chiral.

So the statement that KDP’s $d_{14}$ and $d_{36}$ differ is, as far as symmetry is concerned, the statement that a quantity transforming like its optical activity is non-zero. That does not make the two effects the same. Optical activity is a first-order response to the gradient of a field, and the Kleinman violation is a second-order response at different frequencies. But they belong to the same representation of the same group, and so any class that forbids one forbids the other.

The surprising connection, stated plainly #

Here is the connection in one sentence. The second-harmonic coefficients a crystal has only because of dispersion are, as far as symmetry can tell, its optical activity and its pyroelectric axis.

It is surprising because none of the three effects has an obvious relation to the others. Pyroelectricity is a static polarisation and a property of the crystal at rest. Optical activity is a linear optical effect that depends on the wavevector. Kleinman violation is a nonlinear effect that depends on the frequency difference between input and output. Each is computed here from a different character, and the three characters satisfy one identity, operation by operation. A crystallographer who wants to know whether a material can show a measurable departure from Kleinman’s rule can read the answer from two columns already in the table the property table computes. One is whether the class is polar. The other is whether its gyration tensor has a traceless part.

The identity also explains why the two dark classes are where they are. A class goes dark under Kleinman’s rule when everything it permits is in the removed part, and the removed part of a non-polar class is its traceless gyration. In $422$ and $622$ the gyration tensor has two components, one along the axis and one across it, and one combination of them is the trace. What remains is a single traceless component, and that single component is the class’s entire second-harmonic tensor. $432$ is cubic, so its gyration tensor is pure trace and its traceless part is empty, which is the representation-theoretic form of twenty of the twenty-one: there is nothing for a second harmonic to be.

Two different fields, and 432 comes back #

The argument so far assumed one input field. If two beams of different frequencies are mixed to make their sum, the two input indices are no longer interchangeable. The tensor then has twenty-seven components, the counts rise, and something worth noticing happens in $432$.

The extra nine dimensions are the tensors antisymmetric in the two input indices, which by the same cross-product argument transform as a vector times the determinant, times a vector. Among them is the totally antisymmetric tensor $\epsilon_{ijk}$, which survives every proper rotation. So every chiral class gains at least one sum-frequency coefficient that its second-harmonic tensor lacks, and in $432$ that coefficient is the only one: a sum-frequency polarisation $\mathbf{P} \propto \mathbf{E}_1 \times \mathbf{E}_2$, perpendicular to both input fields, which vanishes the moment the two beams have the same frequency and direction. The count confirms it, one sum-frequency coefficient in $432$ against none for the second harmonic, and between one and nine extra in each of the other ten chiral classes.

The same term is allowed in a liquid of molecules of one hand, which has no lattice at all and the full rotation group as its symmetry. Giordmaine pointed this out in 1965, and the effect has since been measured in chiral solutions, where it is one of the few nonlinear signals an isotropic liquid can give. In the language of this essay it is the pseudoscalar that the removed part took away, returned when the two input fields stop being the same field.

What the counting cannot show #

Every number on this page is a count of permitted coefficients, and none is a measurement. That distinction is the one permitted is not present is about, and it bites twice here. A class that permits a coefficient does not make it large. And a coefficient that Kleinman’s rule removes is not thereby zero. It is a coefficient whose size depends on how far the frequencies sit from the material’s absorption. Its smallness is a prediction about dispersion, not about symmetry, and nothing computed here says how small it is in any material. The remarks above about quartz, barium titanate and KDP are about the shape of published data, and the data were not reproduced here.

The convention the counts depend on should be named. The coefficients are $d_{ijk} = \chi^{(2)}_{ijk}/2$, and the contracted $d_{il}$ uses $l = 1$ to $6$ for the field pairs $xx$, $yy$, $zz$, $yz$, $xz$, $xy$. Neither factor changes a count. The axis convention matters more: the matrices drawn above put the principal axis along $z$ and, in $32$, a two-fold axis along $x$, and in another setting the same coefficients appear in other cells of the matrix. What the setting cannot change is the number of independent coefficients and which of them the rule removes, because those are characters, and a character does not depend on the axes.

There is also something the pictures cannot show at all. A three-by-six matrix of letters looks like a tensor, but the claim about the removed part is a claim about a representation, and no single picture of a single class displays that. The matrices show which combinations are tied, and the table of characters shows why. The link between them is the argument in the prose.

The second refusal is there because it is the natural mistake. The removed part has a symmetric second-rank tensor in it, and a symmetric second-rank tensor without its determinant is the dielectric tensor, which every class permits. Counting with that tensor in place of the gyration tensor would give the cubic class $m\bar{3}m$ five removed coefficients from a second-harmonic tensor that is identically zero. The determinant is what makes the removed part vanish in every centrosymmetric class, and it is what makes the identification with optical activity rather than with birefringence.

Who found it and when #

Franken, Hill, Peters and Weinreich reported second-harmonic generation in quartz in 1961. The full permutation symmetry of the nonlinear susceptibility in a lossless medium was set out by Armstrong, Bloembergen, Ducuing and Pershan in 1962, and in the same year Kleinman stated the dispersion-free form that bears his name. The identification of the removed part with the gyration tensor and a polar vector is a consequence of standard representation theory of the orthogonal group, applied here to the thirty-two classes. It is assembled for this essay rather than quoted, so the reader should treat the identity in the table as a computed result with its checks beside it rather than as a citation.

Still open: how far a real crystal departs #

The census says which coefficients exist only through dispersion. It does not say how they grow as the frequencies approach an absorption band, and that growth is not a symmetry question. A simple model of the dispersion, one resonance with a known strength, would give the ratio of a Kleinman-violating coefficient to a Kleinman-allowed one as a function of how close the doubled frequency sits to the resonance. Whether that ratio has the same form in every class, so that the departure from Kleinman’s rule in $422$ and in $\bar{4}2m$ grows at the same rate, or depends on which representation the violating part belongs to, is the next question. The cubic term that forbids a continuous change is the other place the third Molien coefficient decides something. There it decides whether an order parameter may change smoothly. Whether the classes with no cubic invariant for their vector representation share anything else is open.