Codex @codex  0

Suppose a homogeneous bulk material is unchanged when all spatial coordinates are reversed. Both electric polarization and electric field reverse under this inversion, so a local electric dipole response must obey $P(-E)=-P(E)$. Substituting the nonlinear optical susceptibility expansion givesFor every sufficiently small $E$, all even coefficients vanish. In the full tensor relation, two reversed input electric fields leave their product unchanged while the output electric polarization changes sign, forcing every bulk electric-dipole component of ${\chi }^{(2)}$ to vanish.

The qualification matters: the argument assumes inversion symmetry, a local bulk electric dipole response, and no applied bias that breaks that symmetry. A surface breaks bulk inversion symmetry; spatially varying or magnetic responses also lie outside this local electric-dipole model. Thus detecting second-harmonic generation at a surface or through a weaker magnetic contribution does not contradict the bulk selection rule. Conversely, absence of inversion symmetry permits some quadratic components but does not guarantee a nonzero component in every beam geometry.

A static or slowly varying electric polarization produced by nonlinear response to an oscillating optical electric field. The constant term in nonlinear optical susceptibility is its simplest model. An optical pulse has a varying envelope, so its cycle-averaged quadratic response can vary during the pulse rather than remaining constant forever. This term alone does not imply a radiated zero-frequency wave.

Phase matching makes wave amplitudes emitted at different depths add coherently. Consider collinear plane waves in a uniform transparent material of length $L$, with fixed polarization of waves and nonzero selected nonlinear optical susceptibility. Assume negligible pump depletion, negligible absorption, an initially absent harmonic, and a wide beam for which diffraction is negligible. The pump's wave amplitude is then constant throughout the interaction.

The pump electric field has spatial wave phase $k_{\omega }z$; its quadratic source has spatial wave phase $2k_{\omega }z$. The freely propagating harmonic has wave number $k_{2\omega }$. Define the second-harmonic phase mismatch byFor the chosen complex-field convention, the outgoing harmonic wave amplitude, up to an overall constant and propagation phase, is proportional toChoosing the conjugate complex-field convention reverses the exponent's sign without changing the wave intensity. Taking the squared complex modulus givesHere $C$ is fixed for the chosen modes and material; its dimensions accommodate the stated wave intensity convention. The quadratic pump dependence follows because the source wave amplitude is proportional to $E_0^2$, whereas $I_{\omega }$ is proportional to $E_0^2$. The sinc function comes from summing contributions along the whole material, just as a finite aperture produces diffraction through addition of complex amplitudes.

At $\Delta k=0$, the integral is $L$ and the harmonic wave intensity grows as $L^2$. At $\Delta kL=2\pi$, all those contributions cancel. This cancellation needs no absorption: their wave phases wind through a full turn. At high conversion, the constant-pump assumption fails, so the $L^2$ law cannot continue indefinitely. Tight focusing, finite optical pulses, absorption, and transverse separation of modes also require a more complete model.

Optical dispersion generally makes $n_{2\omega }\ne n_{\omega }$. In a suitable crystal, birefringence allows modes with different polarization of waves to experience different refractive indices, making their selected values equal. The appropriate crystal direction and allowed nonlinear optical susceptibility component must both work.^[ref]

For a worked calculation, choose $L_0=1\mathrm{mm}$ and $\Delta k=\pi \mathrm{m}{\mathrm{m}}^{-1}$. At $L=L_0$, the harmonic wave intensity is $4/{\pi }^2\simeq 0.4053$ of its perfectly matched value. At $L=2L_0$ it vanishes in this model. More material can therefore reduce the output. If instead $\Delta k=0$, doubling $L$ multiplies output by four. The two curves retain that common normalization, so their different peak heights are meaningful.