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5. GAUSSIAN BEAMS 5.1. Solution to the Wave Equation in Cartesian Coordinates

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5. GAUSSIAN BEAMS 5.1. Solution to the Wave Equation in Cartesian Coordinates 5. GAUSSIAN BEAMS 5.1. Solution to the wave equation in Cartesian coordinates • Recall the Helmholtz equation for a scalar field U in rectangular coordinates ∇+22UU(r,ωβ) ( rr , ω )( , ω) = 0, (5.1) • β is the wavenumber, defined as β22(rrr,,, ω) = ω µε( ω) − i ωµσ( ω ) ω 2 (5.2) = ni2 (rr,,ω) − ωµσ( ω ) c2 • Assuming lossless medium (σ = 0) and decoupling the vacuum contribution (n =1) from βω(r, ), we re-write Eq. 2 to explicitly show the driving term, ∇+2 ω2 ω =−−22 ωω U(r,,) kU00( r) k n( rr,) 1 U( ,,) (5.3) • k = ω . 0 c 1 • Eq. 3 preserves the generality of the Helmholtz equation. Green’s function, h, (an impulse response) is obtained by setting the driving term to a delta function, ∇+2h(rr,,ω) kh2 ( ωδ) =−(3) ( r) 0 (5.4) = −δδδ( xyz) ( ) ( ) • To solve this equation, we take the Fourier transform with respect to r, 22 −+kh(kk,ωω) k0 h( ,1) =−, (5.5) • k 2 =kk ⋅ . • This equation breaks into three identical equations, for each spatial coordinate, 1 hk( x ,ω) = 22 kkx − 0 (5.6) 11 1 = − 2k000xx kk−+ kk x 2 • To calculate the Fourier transform of Eq. 6, we invoke the shift theorem and the Fourier transform of function 1 , k fk( −→ a) eiax ⋅ fx ( ) 1 (5.7) → ixsign( ) k • Thus we obtain as the final solution 1 hx( ,ω) =⋅> eik0 x ,0 x (5.8) ik0 • The procedure applies to all three dimensions, such that the 3D solution reads ⋅ he(r,ω) ∝ ik0 kr, (5.9) • k is the unit vector, kk = / k . 3 • Equation 9 describes the well known plane wave solution, which is characterized by the absence of amplitude modulation upon propagation. This is an infinitely broad wavefront that propagates along direction kˆ (Figure 1). y r k x Figure 1. Plane wave. 4 5.2. Solution of the wave equation in spherical coordinates • For spherical symmetry, such as emission from a point source in free space, the problem becomes one dimensional, with the radial coordinate as the only variable, h(r,,ωω) = Ur( ) h(k,,ωω) = Uk ( ) 1 (1) (5.10) (3)δδ(r) = (r) 2π r 2 (3)δ (r) →1 • Recall that the Fourier transform pair is defined in this case as ∞ h( r) ∝⋅∫ h( k) sinc( kr) k2 dk 0 ∞ (5.11) h( k) ∝⋅∫ h( r) sinc( kr) r2 dr 0 5 • The Fourier properties of (3)δ (r) and ∇2 extend naturally to the spherically symmetric case as (3)δ (r) →1 (5.12) ∇22 →−k • Thus, by Fourier transforming Eq. 4, we obtain the frequency domain solution, 1 hk( ,ω) = 22 (5.13) kk− 0 • Not surprisingly, the frequency domain solutions for the Cartesian and spherical coordinates (Eqs. 6 and 13, respectively) look quite similar, except that the former depends on one component of the wave vector and the latter on the modulus of the wave vector. 6 • The solution in the spatial domain becomes ∞ 1 sin(kr) h r,ω = ⋅⋅k2 dk ( ) ∫ 22 0 k− k0 kr 11∞ 1eeikr− − ikr = ∫+ dk (5.14) 22r0 kk−+00 kk i 1 ∞ eikr = ∫ dk. 4ir0 k− k0 • We recognize in Eq. 15 the Fourier transform of a shifted 1 function, which k we encountered earlier (Eqs. 7). Evaluating this Fourier transform, we finally obtain Green’s function for propagation from a point source, eikr hr( ,ω) ∝> ,0 r (5.15) r 7 • This solution defines a (outgoing) spherical wave. z cos2πr 10 ;r = x2 + y2 + z2 r y x Figure 5-2. Spherical wave 8 5.3. Solution of the wave equation in cylindrical coordinates • The most common situation from an experimental point of view is light propagation in cylindrical coordinates with radial symmetry, the z coordinate defines the optical axes and the radial coordinate, r, defines the transverse coordinate. • The Fourier pair in this case can be written as ∞∞ ⋅ = ⋅⋅ikz z ⋅ ⋅ h( r,, z) ∫∫ h( k⊥ kzz) J0 ( k ⊥ r) e k⊥⊥ dk dk −∞ 0 ∞∞ (5.16) −⋅ = ⋅ ⋅ikz z ⋅⋅ h( k⊥⊥,, kz ) ∫∫ h( r z) J0 ( k r) e r drdz −∞ 0 • Due to the angular symmetry in both the spherical and cylindrical coordinates, the spatial frequency domain solution of the Helmholtz equation is identical in form with Eq. 13 and depends only on the modulus of the wave vector 9 1 hk( ,,ω) = 22 kk− 0 (5.17) 2 22 kk=⊥ + kz • Uk ( ,ω) was obtained from the procedure used in the spherically symmetric case, replacing the source term with (3)δ (rz, ) and using the property ∇22 →−k that holds for cylindrical coordinates with circular symmetry. • We now Fourier transform Eq. 13 back to the spatial domain, ∞∞ 1 ⋅ h r, z = ⋅J k ⋅⋅ r eikz z k dk dk (5.18) ( ) ∫∫ 2 220 ( ⊥ ) ⊥⊥z −∞ 0 kz −−( kk0 ⊥ ) • Eq. 18 can be interpreted as the Hankel transform (over k⊥ ) of a 1D Fourier transform (over kz ). In evaluating the integral over kz first, we recover the 1D case discussed in Section 5.1, the solution in Eq. 8, where we replace k0 with 22 kk0 − ⊥ and x with z. 10 • The remaining integral over k⊥ becomes ∞ eiqz = ⋅ ⋅⋅ h( r,, z) ∫ J0 ( k⊥ r) k ⊥⊥ dk 0 q (5.19) 22 qkk=0 − ⊥ • Equation 19a can be regarded as the exact Green’s function in cylindrical coordinates with circular symmetry. The integral can be evaluated analytically in the limit of low transverse frequencies, i.e. kk⊥ 0 . 22 • In this case we expand kk0 − ⊥ in Taylor series to obtain k 2 − ⊥ ik0 z1 2 eeiqz 2k0 . (5.20) qk0 • The amplitude term varies slowly with k⊥ , so the zeroth order term is sufficient, but the phase term is more sensitive, thus the second order term is necessary. 11 • Following this expansion, the Hankel transform remains to be calculated for a iz 2 −⋅k⊥ Gaussian phase function e 2k0 . Recall that Hankel transforms of Gaussian functions can be easily calculated and follows a similar transformation rule with ix2 22 − ib k that of 1D functions, e2b2 → be2 2 . • By combining Eqs. 20 and 19a, we obtain ik z ik e 0 0 ⋅r2 hrz( ,.) ∝⋅ e2z (5.21) z • This solution is characterized by a slow attenuation with distance, ∝ z , a plane wave-like phase delay vs. z, and a Gaussian transverse phase, of width that also grows as z . 12 5.4. Propagation of Gaussian beams • Often, experiments involve light beams. A light beam can be defined as a distribution of field that fulfills the approximation in Eq. 20, is characterized by a dominant wave vector component, kz kk xy, . A beam is the spatial equivalent of quasi-monochromatic light, where the field is characterized by a dominant (temporal) frequency component. • A Gaussian beam, such as that delivered by a single mode laser, has a field distribution described by a Gaussian function in the transverse coordinate. Green’s function in Eq. 21 provides a way to propagate this type of field. • Consider a Gaussian beam that at z=0 has the transverse distribution r2 − 2 1 w0 Ur0 ( ,0) = 2 e , (5.22) w0 • w0 defines the minimum beam waist. 13 • Upon propagating a distance z, the field becomes a convolution between U0 and h, U( rz,) = U0 ( r ,0) ∗ hrz( , ) (5.23) • We take the Fourier transform of Eq. 23, make use of the convolution theorem and use the approximation in Eq. 20 to obtain the frequency domain solution 2 22 izk⊥ wk ik0 z − − 0 ⊥ e 2k0 4 Uk ( ⊥ , z) =⋅⋅ e e (5.24) k0 • The spatial domain solution is obtained by taking the inverse Hankel transform of U , ∞ = ⋅ U( rz,,) ∫ U( k⊥ z) J0 ( krkdk ⊥) ⊥⊥ . (5.25) 0 14 • After propagation over distance z, the field remains Gaussian in the transverse direction and has the form r2 ik0 z − e cz2 U( rz, ) = e ( ) cz2 ( ) (5.26) 222z cz( ) = w0 + i k0 • The solution has a particularly simple form, thanks to the properties of Gaussian functions and their transforms. All the information regarding the beam change due to propagation is contained in function c(z). This function can be rearranged 22z cz( ) = w0 1 + i z 0 (5.27) kw22π w z =00 = 0 0 2 λ 15 • We can write explicitly the absolute value and phase of c2(z), express it in polar coordinates, as cz22( ) = wze( ) ⋅ izψ ( ) , (5.28) • where 2 22z wz( ) = w0 1 + 2 z0 (5.29) z ψ ( z) = arg z0 • The distance z0 is known as the Rayleigh range, over which the beam waist increases by 2 . We can re-write U(r, z) explicitly, zr2 r2 ikz+−ψ (z) − 0 2 2 1 z0 wz( ) wz U( rz, ) =⋅⋅ ee ( ) wz2 ( ) (5.30) =Arz( , ) ⋅ eiφ( rz, ) 16 • Equation 30 represents the fundamental solution of the Gaussian beam propagation. It can be seen that both the broadening in r and propagation in z requires lowering the amplitude, such that the energy is conserved, r2 − 2 2 w 1+z 1 0 z 2 Arz( ,.) = ⋅e 0 (5.31) 2 z2 w0 1+ 2 z0 z2 • The new beam waist is wz( ) = w0 1 + 2 . z0 • The phase dependence reveals more features 2 kr0 z φ (rz, ) =+− kz0 arg . (5.32) 2Rz( ) z0 17 • Besides the plane wave phase delay, kz0 , the wavefront radius of curvature R(z) induces additional phase shift, with R(z) defined as 2 z0 Rz( ) = z1 + 2 (5.33) z • The wavefront is centered a distance dz()= 2 z0 / z behind the z=0 plane. r θ w(z) w0 { z z R(z) d(z) Figure 5-3.
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