Past exam of the laser science and quantum information processing course of the University of Oxford / 2015 / 4 / c / Solution

Type-I phase matching requires $n_{e,2\omega }(\theta )=n_{o,\omega }$, because both pump photons are ordinary and the harmonic is extraordinary. Apply the direction-dependent index equation to the harmonic, not to the pump. Solving givesFor the ruby-laser data this is $0.626210$, whereas the shorter-wavelength data give $0.958873$. ThereforeThe second calculation uses the supplied $257\mathrm{nm}$ indices as rounded harmonic data: the exact half-wavelength of $515\mathrm{nm}$ is $257.5\mathrm{nm}$. The index dispersion between those two ultraviolet values was not supplied.

For an extraordinary wave, its Poynting vector makes an angle ${\theta }_S$ with the optic axis satisfying $\tan {\theta }_S=(n_o^2/n_e^2)\tan \theta$. Thus the spatial walk-off angle relative to the wave vector isBoth angles refer to the harmonic. The ordinary pump does not have this angular walk-off. For beam radius $w_b$, the approximate overlap length is $w_b/\tan \rho$, so the shorter-wavelength geometry has about $2.19$ times the walk-off-limited interaction length.^[ref]

To compare coupling, take propagation azimuth $\phi$ and ordinary unit polarization $(-\sin \phi ,\cos \phi ,0)$. Its quadratic source is $P_z=-{ϵ}_0{d}_{36}{E}_0^2\sin 2\phi$. Projection onto an extraordinary displacement direction with $z$ component proportional to $\sin \theta$ gives $|d_{\mathrm{eff}}|\propto |d_{36}\sin \theta \sin 2\phi |$, up to the small anisotropic normalization correction. The optimal azimuth is $45^{\circ }$. At equal beam area, interaction length, input power and approximately equal nonlinear coefficients, the shorter pump has both stronger angular coupling and the favorable frequency-squared factor:Here $\eta =P_{2\omega }/P_{\omega }$ in the small-conversion, equal-area model, with harmonic index phase-matched to the pump. Its smaller spatial walk-off further favors the $515\mathrm{nm}$ pump in a walk-off-limited geometry. Equal input power alone does not determine an absolute efficiency ranking: ultraviolet absorption, wavelength-dependent $d_{36}$, beam size, crystal length and coatings are unspecified. For example, a sufficiently strongly absorbing $257.5\mathrm{nm}$ output can reverse the ideal ranking. The stated preference is conditional on comparable transparent crystals and comparable focusing, not a prediction independent of those parameters.