Michael Scalora U.S. Army Research, Development, and Engineering Center

OPTICS BY THE NUMBERS L’Ottica Attraverso i Numeri Michael Scalora U.S. Army Research, Development, and Engineering Center Redstone Arsenal, Alabama, 35898-5000 & Universita' di Roma "La Sapienza" Dipartimento di Energetica Rome, April-May 2004

The Lorentz Oscillator Polarization Intrinsic Optical Bistability The FFT-Beam Propagation Method

E e- x Classical Theory of Matter Lorentz Atom: Electron on a Spring Simple Harmonic Oscillator Under the action Of a driving force Nucleus: ~2000 times electron mass. In the language of theorists, this means infinite mass Electron position is perturbed periodically and predictably

E x Restoring Force Driving Force Damping Damped Harmonic Oscillator P=Nex Average number of dipoles per unit volume e- Perform Fourier Transform:

E x Damped Harmonic Oscillator e- c(w): Dielectric Susceptibility

Im(c): Absorption Re(c): Dispersion

Away from any resonances, in regions of flat dispersion… We will assume propagation occurs in a uniform medium with constant (i.e., dispersioness) e

Practical Example Resonance is that way. Index of refraction of GaN Index of refraction of AlN

Component that oscillates at frequency w is E x |b| << k Nonlinear Oscillator

|Et|2 |Pt|2

|Et|2 |Pt|2 Optical Bistability: Two Stable Output States Exits for the Same Input Intensity

Expanding the denominator:

is of the form… Then…

Substituting and retaining terms of lowest order:

Many Nonlinear Optical Effects Are of This Type and can Explain Everything From Optical Bistability to Fiber Solitons. Continuing the expansion in a perturbative manner, one can show that…

In General For Nonlinear FrequencyConversion, i.e., Harmonic Generation,& Sum-difference, and Nearly all Nonlinear Optical Effects of Interest are Described Well by the First Two Terms of the Nonlinear Oscillator Potential

Away from sharp resonances and absorption lines, The index of refraction can be taken to be nearly constant as a function of frequency. Constitutive Relations: Assumptions as to how matter interacts with the propagating fields It follows that once the suceptibility has been determined, an index of refraction can be assigned:

Im(c): Absorption Re(c): Dispersion In general, n is complex. But far from Absorption lines the imaginary part Is small and can usually be neglected.

Beam Propagation In The Presence Of Matter As we saw earlier using the nonlinear oscillator model, the total polarization P is composed of two parts: a linear and a nonlinear response. Assuming only a third order Nonlinear potential, then, with… Assuming a single vector component, dropping the vector notation, And substituting above one finds:

Once again, assuming CW operation, no boundaries or interfaces in the longitudinal direction (z), we make the SVEA approximation, i.e., drop second order spatial (z) derivatives: For A Uniform Medium, The Choice k=(w/c)n is Appropriate. The result is:

Using the scalings… We can simplify the equation and rewrite it in simple form: and… where…

Equation is of the form: Formal Solution:

Let’s assume that H varies slowly inside the interval. For small Intervals:

We must expand the operator in order to evaluate it! D and V generally DO NOT commute.

D and V generally DO NOT commute... However, using the same sort of expansion of the operators, It can be shown that…

The single mixed integration step (D+V) has been split into three parts: • Free space propagation by half of the spatial step • Interaction with the medium by the full propagation step • Account for remaining half free space propagation step Split-Step Beam Propagation Method, or more commonly known as FFT-BPM

1. 2. 3. By Construction, each free space propagation step requires two FFTs, for a total of four per interval. But there Is some good news.

Then, given the symmetric disposition of each term, it must be true that… Which can be verified by direct substitution or by the simple transformation D->V V->D Clearly this algorithm requires half as many FFTs per step, and so it is more efficient

1. 2. 3. Each free space propagation step requires only two FFTs per interval with the same kind of accuracy.

As usual, improved accuracy requires more work at the expense of efficiciency, and may not always be worth it. C===========================================C a=0.05361185 b=0.62337932451322 C===========================================C c=0.89277629949778 d=-0.12337932451322 C===========================================C e=-0.1203850412143 f=-0.12337932451322 C===========================================C g=0.17399689146541 h=0.62337932451322 C===========================================C

Actual Implementation of each step: free space Is the solution of the equation Using spectral methods: Solve numerically as follows:

Which gives a stable, third order accurate solution. Then…

Actual Implementation of each step: medium Is the solution of the equation Solve numerically as usual:

Michael Scalora U.S. Army Research, Development, and Engineering Center