Gaussian Beams
Enrique J. Galvez Department of Physics and Astronomy Colgate University Copyright 2009

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Contents
1 Fundamental Gaussian Beams 1.1 Spherical Wavefront in the Paraxial region 1.2 Formal Solution of the Wave Equation . . 1.2.1 Beam Spot w(z) . . . . . . . . . . 1.2.2 Beam Amplitude . . . . . . . . . . 1.2.3 Wavefront . . . . . . . . . . . . . . 1.2.4 Gouy Phase . . . . . . . . . . . . . 1.3 Focusing a Gaussian Beam . . . . . . . . . 1.4 Problems . . . . . . . . . . . . . . . . . . . 1 1 3 6 8 8 9 10 12 15 15 17 20 21 25 25 26 26 27 29 30 31 31 33 35 35 36 39

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2 High-Order Gaussian Beams 2.1 High-Order Gaussian Beams in Rectangular Coordinates 2.2 High-Order Gaussian Beams in Cylindrical Coordinates . 2.3 Irradiance and Power . . . . . . . . . . . . . . . . . . . . 2.4 Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . 3 Wave-front interference 3.1 General Formalism . . . . . . . . . . . . . . . . 3.2 Interference of Zero-order Beams . . . . . . . . 3.2.1 Collinear Interference . . . . . . . . . . . 3.2.2 Noncollinear Interference . . . . . . . . . 3.2.3 Holographic Principle . . . . . . . . . . . 3.2.4 Interference of an Expanding Wave-front 3.3 Interference with High-Order Beams . . . . . . 3.3.1 Collinear Case . . . . . . . . . . . . . . . 3.3.2 Non-collinear Case . . . . . . . . . . . . 3.4 Generating Helical Beams . . . . . . . . . . . . 3.4.1 Spiral Phase Plate . . . . . . . . . . . . 3.4.2 Amplitude Diffraction Gratings . . . . . 3.5 Problems . . . . . . . . . . . . . . . . . . . . . . iii . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

iv 4 Energy and Momentum 4.1 Energy . . . . . . . . . . . . . . . . . 4.2 Linear Momentum . . . . . . . . . . 4.2.1 Optical Tweezers . . . . . . . 4.3 Angular Momentum . . . . . . . . . 4.3.1 Polarization and Spin Angular 4.3.2 Orbital Angular Momentum . 4.3.3 Rotation in Optical Tweezers 4.4 Problems . . . . . . . . . . . . . . . .

CONTENTS 41 41 43 46 49 49 51 52 53

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Chapter 1 Fundamental Gaussian Beams
In the last few decades the use of laser beams has become widespread. Lasers are used in both science and technology, from spectroscopical analysis to bar-code reading. It is therefore appropriate that we study the properties of laser beams in some detail. Laser beams of distinct types and colors have a few features in common: they are composed of a narrow band of wavelengths, to the point that we can call them “monochromatic” (i.e., of a single wavelength), and they are collimated, that is, the light energy is restricted in the direction transverse to the propagation direction to form a narrow beam, as shown in Fig. 1.1 In a plane that is transverse to the propagation direction

Figure 1.1: A laser beam from a HeNe laser seen due to scattering. the intensity of the beam decreases in a typical Gaussian shape. In this section we will discuss the basic properties of these (Gaussian) laser beams.

1.1

Spherical Wavefront in the Paraxial region

We will start by getting a rough idea of the mathematical representation of the light waves in Gaussian beams. Although we see that the beam is collimated, we also observe that the beam expands as it propagates. We will use the z axis as the propagation direction of the laser beam, leaving the x and y directions for describing the transverse extension of the beam. The simplest 1

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CHAPTER 1. FUNDAMENTAL GAUSSIAN BEAMS

types of 3-D waves are plane waves. A plane wave propagating along the z direction is represented by the wave function Ψ(x, y, z, t) = Aei(kz−ωt) . (1.1)

The amplitude A is constant. This equation cannot possibly describe a laser beam because the amplitude is infinitely extended in the transverse direction. A more appropriate amplitude would be one which decreases for points away from the z axis. We may “guess” something simple, like a Gaussian: A = A(x, y) = e−(x
2 +y2 )/w2

,

(1.2)

where w is proportional to the width of the beam, also known as the beam spot. It is defined as the distance from the beam axis where the amplitude has decreased by 1/e. Thus, the beam profile in Fig. 1.2a might be represented by the function in Fig. 1.2b. We note that the “intensity” of the beam shown in Fig. 1.2a is represented by the square of the amplitude, as we will see later in the course. The square of the √ Gaussian is another Gaussian with a width that is smaller by a factor of 2.

Figure 1.2: (a) False color image of the profile of a laser beam; (b) Profile of a Gaussian function in two dimensions. The expansion of the beam may be accounted by a beam spot that increases with z, although the exact dependence is yet to be determined. If the beam is expanding then its “wavefront” must have a spherical shape. This because a wave always propagates in a direction perpendicular to its wavefront (e.g., ripples on a pond). The wavefront is defined as the surface that contains all points of the wave that carry the same phase

1.2. FORMAL SOLUTION OF THE WAVE EQUATION

3

(e.g., the crest of the ripples). This spherical wavefront is not accounted fully by the plane-wave phase term eikz , so we must use a modified expression. The spatial part of a spherical wavefront has the form eikr , where r= x2 + y 2 + z 2 . (1.4) Let’s suppose that we are examining the wave far from the origin but close to the z-axis such that x ≪ z and y ≪ z. Then we can approximate Eq. 1.4 using the binomial approximation x2 + y 2 x2 + y 2 r=z +1≃z 1+ . z2 2z 2 (1.5) (1.3)

Thus a better approximation for the phase of the wave would include a phase similar to the one of a spherical wave. Our best guess for the expression of the wave function of the Gaussian beam may be Ψ(x, y, z) ∼ e−(x
2 +y2 )/w2

ei(kz−ωt) eik(x

2 +y2 )/2z

.

(1.6)

The first term contains the Gaussian profile, the second term has the unidirectional wave term, and the third term has the correction to the previous term that accounts for the curvature of the wavefront. Exercise 1 Suppose that we have a HeNe laser beam with a wavelength of 632.8 nm. Consider the beam stopping on a piece of paper 2 m away from the laser. A that point the beam has a beam spot of about 3 mm. The profile is smooth and decreasing with an apparent Gaussian shape. However, the phase variation across this profile is that of a spherical wavefront, so that in the transverse plane the phase also increases. √ Find the distance s = x2 + y 2 from the center of the beam at which the phase has increased by 2π with respect to the center.

1.2

Formal Solution of the Wave Equation

Obviously the expression for the light from a laser beam can’t be guessed. We must find the solution from solving the wave equation. The three-dimensional wave equation is given by 1 ∂ 2Ψ ∇2Ψ − 2 2 = 0, (1.7) v ∂t

4 where

CHAPTER 1. FUNDAMENTAL GAUSSIAN BEAMS

∇2 =

∂ ∂ ∂ + 2+ 2 ∂x2 ∂y ∂z

is the Laplacian. If we introduce a trial solution of the form Ψ(x, y, x, t) = E(x, y, z)e−iωt. (1.8)

Here E is the part of the wave function that depends only on the spatial coordinates. Replacing Eq 1.8 into Eq. 1.7 we get the Helmholtz equation: ∇2 E + k 2E = 0. (1.9)

Note that we could have also chosen to use e+iωt in Eq. 1.8. In that case the solution would represent a wave that moves toward the negative z direction. Thus, the solution is independent of the direction of propagation. Now we must simplify this equation given that we will restrict it to describe laser beams. We start by requiring that the solution must have the form E(x, y, z) = E0 (x, y, z)eikz . Replacing Eq. 1.10 into Eq. 1.9 and simplifying we get ∂ 2E0 ∂ 2E0 ∂ 2E0 ∂E0 + + + 2ik = 0. 2 2 2 ∂x ∂y ∂z ∂z Exercise 2 Verify that replacing Eq. 1.10 into Eq. 1.9 results in Eq. 1.11. The term eikz of Eq. 1.10 accounts for the wave oscillation along the propagation direction. The dependence of E0 with z is of a different nature. It likely accounts for the slow decrease in the amplitude of the wave as the wave propagates. Thus we can say that E0 varies slowly with z, and thus we can neglect the term ∂ 2E0 /∂z 2 in front of the other ones and drop it from Eq. 1.11. The resulting equation ∂E0 ∂ 2E0 ∂ 2E0 + + 2ik =0 ∂x2 ∂y 2 ∂z (1.12) (1.11) (1.10)

is called the paraxial wave equation. A “simple” solution to the wave equation is one where we insert the simplest possible form of the solution and find the exact form that obeys the wave equation. The more formal solution is one where we just solve the wave equation in its full generality.

1.2. FORMAL SOLUTION OF THE WAVE EQUATION We “guess” the simple trial solution to be of the form E0 (x, y, z) = Ae ik(x2 +y 2 ) 2q(z)

5

eip(z).

(1.13)

Note how we separated the x, y and z dependencies. Upon replacing Eq. 1.13 into Eq. 1.12 we get separate equations for q(z) and p(z). Here we are not going to cover the step-by-step derivations. We will just show the solutions. For q(z) we have that it is a complex function q(z) = z − izR, (1.14) where zR is a constant called the Rayleigh range. We will examine its significance later. Since q(z) appears in the denominator of a fraction in the exponential of Eq. 1.13, then a more appropriate way to express it is by 1 1 = 2 + q(z) z + zR z z2 zR

i . + zR

(1.15)

Inserting Eq. 1.15 into the first exponential term of Eq. 1.13 we get e ik(x2 +y 2 ) 2R(z)

e−

x2 +y 2 w(z)

.

(1.16)

These terms have the form that we guessed in the first section. R(z) is known as the radius of curvature of the wavefront. It is given by R(z) = z +
2 zR z

(1.17) For the second term of Eq. 1.16 we see that what we interpreted earlier as the beam spot of the beam has now a z-dependence w(z) = w0 1 + z2 2 zR

(1.18) where the constant w0 = is called the beam waist. zR λ π (1.19)

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CHAPTER 1. FUNDAMENTAL GAUSSIAN BEAMS Before we analyze what all these terms mean let’s look at the solution for p(z): w0 −iϕ(z) eip(z) = e , (1.20) w(z)

where ϕ(z) = arctan(z/zR) is known as the Gouy phase. Now let’s put it all together: Ψ(x, y, z, t) = w0 − A w(z) e x2 +y 2 w(z)2

(1.21)

ei(kz−ωt) e

ik(x2 +y 2 ) 2R(z)

e−iϕ(z) (1.22)

The first three terms specify the amplitude of the wave, and the next three terms define the phase of the wave embedded in the exponential terms. Now let’s analyze the meaning of all of these terms.

1.2.1

Beam Spot w(z)

Consider Eq. 1.18. The function w(z) represents the beam spot of the beam. Figure 1.3 shows a graph of w(z).

Figure 1.3: Graph of the beam spot of the beam. Note the two limits for z: • When z = 0 we have w(z) = w0 . Notice that this the smallest value that w(z) can have. Thus z = 0 is a special point in the propagation of the beam. For z > 0 the beam spot increases. In the negative z direction the beam spot also increases. Thus, Fig 1.3 describes the width of a beam when it is focused by a lens. When a light beam is focused there is always a minimum width of the beam at the focal point. This minimum beam spot can be understood in terms of diffraction.

1.2. FORMAL SOLUTION OF THE WAVE EQUATION

7

√ • When z = zR we have w(z) = 2w0. This is a turning point in the propagation of the light as the beam spot makes the transition from being nearly constant to increasing linearly (dashed line in Fig. 1.3). When z < zR the ray description of the propagation of the light breaks down. It is important to note that the Raleigh range is an indicator of the divergence of the beam: the smaller the value of zR the quicker it will turn to expand linearly. • When z ≫ zR the beam spot expands linearly as w(z) = w0 z/zR = zθ (i.e., approaching the dashed lines in Fig 1.3). At this limiting value θ is called the divergence angle of the beam θ= λ . πw0

(1.23) Note that the smaller the waist the larger the divergence angle. It is interesting that we get a similar result out of diffraction of light from a small aperture. If the aperture is a slit the equation for the first minimum of diffraction is b sin θ = λ, where b is the width of the slit. For small angles we can approximate this relation to λ . b For a circular aperture 2a, where a is the radius of the aperture, the angle at which the first diffraction minimum appears is given by θ= θ= 1.22λ . 2a

You can see that aside from a constant factor the expressions for diffraction angle are the same as that of the divergence of the beam. The reason is the same one. If we had an opaque piece of glass with a transparent circular region with a transparence that varies like a Gaussian, we will get a divergence angle that is given indeed by Eq. 1.23. In ordinary situations we have the following relation to hold λ ≪ w0 ≪ z R . (1.24)

Let’s do a numerical example. Suppose that we have a HeNe laser beam (λ =632.8 2 nm) that is focused to a spot w0 = 0.5 mm. The Raleigh range is zR = πw0 /λ = 1.2 m.

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CHAPTER 1. FUNDAMENTAL GAUSSIAN BEAMS

Exercise 3 Another important aspect related to the Gaussian waist is the amount of light that passes through a circular aperture. The intensity of the light at a given radius a is given by I = I0e−2a
2 /w2

.

1. Show that the fraction of the total power that gets transmitted through a circular aperture of radius a0 is 1 − exp(−2a2/w2 ). 2. Calculate the percentage of light that goes through an aperture of radius a0 = w

1.2.2

Beam Amplitude

The dependence of the amplitude with z is given by w0 /w(z). From the analysis of w(z) in the previous section we gather that for z < zR the amplitude is nearly constant, but for z ≫ zR it decreases as 1/z. The decrease of amplitude with z at large values of z leads us to conclude that in this limit the amplitude varies like the amplitude of a spherical wave (i.e., ∝ 1/r).

1.2.3

Wavefront

As seen in Eq. 1.22 there are three phase terms. The first and third phases, (kz − ωt) and φ depend only on z. The middle term contains the phase k(x2 + y 2)/2R(z). From the discussion in section 1.1 we interpreted the latter term as representing the curvature of the wavefront. The wavefront is spherical with a radius of curvature R(z). What is interesting is that the radius of curvature is not constant. Figure 1.4 shows sketchings of the wavefronts at different values of z. Notice first that when

Figure 1.4: Graph of the beam spot of the beam showing the shape of the wavefronts. we graph the beam spot of the beam into the negative values of z the beam spot

1.2. FORMAL SOLUTION OF THE WAVE EQUATION

9

increases in the same way that it does in the positive z direction. The thinner curves are sketches of the wavefronts with radii of curvature given by Eq. 1.17. Let’s consider the limits, but for this let’s rewrite R(z) as R(z) = z(1 +
2 zR ). z2

(1.25)

• When z = 0 we have R = ∞ (a vertical line in Fig. 1.4). It is the phase that we would expect of a plane wave. At this point all parts of the wave are moving in the same direction. • When z ≫ zR then R(z) ≃ z; the wavefront is a spherical surface traveling away from z = 0. This is the geometric optics limit. When we focus a beam of light we expect the light “rays” to go in straight lines to and from the focal point (z = 0). This is correct as long as |z| ≫ zR . For |z| < zR the wave aspect of the light manifests. Otherwise the ray-optics view would predict that the focal point is a singularity. • When z = zR then R(z) = 2z; the turning point between ray-optics and waveoptics.

1.2.4

Gouy Phase

There is an intriguing phase term in Eq. 1.22: the Gouy phase, given explicitly by Eq 1.21. This phase slightly shifts the phase of the wavefront of the wave as a whole. If we picture a wave being focused like that in Fig. 1.4 the Gouy phase varies as shown in Fig. 1.5.

Figure 1.5: Graph of the Gouy phase ϕ(z). As z → ±∞ the Gouy phase asymptotes to φ(z) → ±π/2. Thus the phase shift from z = −∞ to z = ∞ is only π. Notice that all of the action occurs between z = −zR and z = zR .

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CHAPTER 1. FUNDAMENTAL GAUSSIAN BEAMS

The Gouy phase is relevant only in situations where the wavefront is very complex. We will not consider it further here.

1.3

Focusing a Gaussian Beam

We must now consider the effect of a lens on a Gaussian wavefront. We have not discussed yet the physics behind lenses, but you may have seen some of it in a past course. In any of those contexts you may have seen a “lens equation” that describes a relation between the focal length of a lens and the locations of an object and the image that the lens forms of that object. Here we will treat lenses differently. We will treat them “locally.” That is, as a device that modifies the radius of curvature of a spherical wavefront, as shown in Fig. 1.6.

Figure 1.6: Change in the radius of cuvature of the wavefront by passage through a lens. If the beam that is incident on a lens of focal length f has a radius of curvature R1 then the radius of curvature of the outgoing wave R2 is given by 1 1 1 = − . R2 R1 f (1.26)

It is important to note that there is a sign convention: the radius is positive when the center curvature is to the left. In Fig. 1.6 R1 > 0 and R2 < 0. The typical cases of ray optics have similar correspondence with the wave optics form of Eq. 1.26 (the wave optics limit fully agrees with the ray optics in the limit that λ → 0). For example, if the input beam is fully collimated (R1 = ∞) then R2 = −f, or if R1 = f then R2 = ∞. Once we know the radius of curvature of the beam past the lens we then want to know the place where the minimum beam spot is going to be located and the size of

1.3. FOCUSING A GAUSSIAN BEAM the waist. The waist location will be given by1 z2 = R2 1+
R2 2 πw2 /λ 2,

11

(1.27)

where w2 is the beam spot of the beam just after the lens, which is the same as the beam spot just before the lens w1 (i.e., w2 = w1). Notice that under “ordinary” situations 2 (1.28) R2 ≪ πw2 /λ. For example. If a beam with a beam spot w1 = 1 mm is incident on a lens that 2 changes its radius to say R2 = 10 cm, then we see that πw2 /λ = 5 m, consistent with Eq. 1.28. When Eq 1.28 holds Eq. 1.27 becomes z2 ≃ R2 . This is the geometric optics limit. The waist will be given by w02 = 1+ w2
1/2 2 πw2 /λ 2 R2

.

(1.29)

Notice that for the “ordinary” situations that we mentioned (i.e., Eq. 1.28) the waist reduces to λR2 . (1.30) w02 ≃ πw2 In many situations we have reasons to focus the light to the smallest spot. For example, in laser welding you need the highest radiant energy in the smallest spot so that you can get enough energy to melt metal. Another reason to focus the light into the smallest spot stems from the need to send light to a small object but not to its neighbors. From Eq. 1.30 se see that to get a small waist one needs: 1. A small value R2 , which implies a lens with a small focal length f, 2. the smallest possible wavelength, and 3. a large beam spot w2 of the beam at the lens. Equation 1.30 can be put in a different form. If we use a lens with a focal length f and diameter d, then it has an f-number f# = f/d. If the incoming beam with R1 ≫ f fills the lens then R2 ∼ f and w2 ∼ d/2, so Eq. 1.30 reduces to w02 ∼ 2λf#/π. This implies that in order to focus the beam to a small spot one needs a lens with a small f#.
1

F.L. Pedrotti and L.S¿ Pedrotti in Introduction to Optics (Prentice Hall, 1993) p. 461

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CHAPTER 1. FUNDAMENTAL GAUSSIAN BEAMS

1.4

Problems

Problem 1 When a laser pointer is projected on a screen we see a red spot ( λ = 670 nm). The total power of the light reaching the screen is P = 3 mW. The beam spot of the beam reaching the screen is 2 mm. We then put a lens with a focal length f1 = 1 cm before the screen. 1. If the radius of curvature of the wavefront reaching the lens is 2 m, find the waist of the focused light.
2 2. The intensity of the light at the center of the beam is given by Imax = 2P/πw0 2 (in W/m ). Calculate the maximum intensity at the focal point.

3. If we place a second lens with f1 = 10 cm after the first one at a distance d = f1 + f2. (a) Calculate the beam spot of the beam at the second lens. (b) What application do you see for this system of lenses? √ Problem 2 Consider the plane wave traveling in free space where k = 2π(1, 1, 1)/(λ 3), where λ = 600 nm, and with an electric field amplitude of 10 V/m. The phase of the wave is 2π at the origin. 1. Find an expression for the wave using the complex notation. 2. Find the frequency f of the wave in THz. 3. Find a point other than the origin where the phase is 2π at t = 0.

Problem 3 A beam from a HeNe laser (λ = 632.8 nm) is traveling along the z-axis and has a waist w0 = 0.1 mm at z = 0. 1. What is the beam spot at z = 10 cm? 2. What is the radius of curvature of the wave front R at at z = 10 cm? 3. At what distance z1 is R = 2 m? 4. Make a graph of R/z vs. z from z = 0 to z = 1 m. 5. Explain the shape of the curve for part (4).

1.4. PROBLEMS 6. Make a graph of w vs. z from z = 0 to z = 0.5 m. 7. Comment of the shape of the previous graph and explain its significance.

13

8. Assuming that the phase of the wave is 2π at z = 0, find the phase of the wave (modulo 2π ) at z = 0.1 m. [If φ = 6.1π, φ(mod2) = 6.1π − 3(2π) = 0.1π.] 9. Find the phase of the wave (modulo 2π ) at z = zR , x = w(z = zR), y = 0.

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CHAPTER 1. FUNDAMENTAL GAUSSIAN BEAMS

Chapter 2 High-Order Gaussian Beams
2.1 High-Order Gaussian Beams in Rectangular Coordinates

If you recall, in Chapter 2 we solved the paraxial wave equation by introducing a “guess” solution. The solution was not the most general because it defined the x and y dependencies in Eq. 2.13. That dependence did not allow differences in the x and y solutions. It then led to the simplest solution, Eq. 2.22. A more general trial solution may be given by E0 (x, y, z) = Ag(x, z)h(y, z)eik(x
2 +y2 )/[2q(z)]

eip(z) .

(2.1)

By replacing this solution into the paraxial wave equation we get solution for g and h in terms of Hermite polynomials. The final solution is √ √ A 2x 2y −(x2+y2 )/w(z)2 ik(x2 +y2 )/[2R(z)] −iϕ(z) Hm ( )Hn ( )e e e . (2.2) Em,n (x, y, z) = w(z) w(z) w(z) The beam described by this solution is known as a Hermite-Gauss beam. The indices m and n of the Hermite polynomials provide a family of solutions. We define the order of the solution by N = n + m. Modes of the same order are degenerate in laser resonators. The lowest-order Hermite polynomials Hm (v) are H0 (v) = 1 H1 (v) = 2v H2 (v) = 4v 2 − 2 (2.3) (2.4) (2.5)

From Eq. 2.3 we can conclude that the solution of Gaussian beams discussed earlier (Eq. 2.22) is only the zero-order (N = 0) solution of Eq. 2.2. 15

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CHAPTER 2. HIGH-ORDER GAUSSIAN BEAMS

Figures 2.1 (a) and (b) show graphs of the amplitude of the HG10 and HG20 modes, respectively. To the right of the graphs are pictures of the corresponding laser-beam modes.

Figure 2.1: Graphs of the amplitude and pictures of Hermite-Gauss modes: (a)HG10, (b) HG20, (c) HG30.

Exercise 4 Consider the amplitude of the general solution of the wave equation in rectangular coordinates, Eq. 2.2, at z = z0. Make sketches of 1. E00 (x, 0, z0) vs. x and E00(0, y, z0) vs. y. 2. E01 (x, 0, z0) vs. x and E01(0, y, z0) vs. y. 3. E02 (x, 0, z0) vs. x and E02(0, y, z0) vs. y. 4. E11 (x, 0, z0) vs. x and E11(0, y, z0) vs. y. Draw a rough shape of the beam profiles for each of the cases above.

2.2. HIGH-ORDER GAUSSIAN BEAMS IN CYLINDRICAL COORDINATES 17 Aside from the Hermite polynomials, the other terms of Eq. 2.2 are the same as those in Eq. 2.22 except for the Gouy phase, which in Eq. 2.2 has a new factor: ϕ(z) = (N + 1) tan−1 (z/zR ). The constant term in Eq. 2.2 that normalizes the square of the amplitude is given by 1/2 21−N A= . (2.6) πn!m! This gives A= 2 π (2.7)

for N = 0. One can understand these modes with the analogy of a drum. The membrane of the drum can oscillate in its lowest-order mode: the whole membrane goes up an down. This mode has the lowest pitch. The next higher-order mode is one where the two halves of the membrane oscillate 180 degrees out of phase. This oscillation will have a higher pitch. There is a node along the middle of the membrane. This would be mode 01. A mode with a node orthogonal to it would be the mode 10. Notice that the placing of the node is arbitrary. Thus, modes 01 and 10 are “degenerate;” they oscillate at the same frequency. You can also imagine other higher-order solutions. All in analogy to the modes of the light.

2.2

High-Order Gaussian Beams in Cylindrical Coordinates

The physics of high-order modes gets much more interesting with the solutions of the paraxial equation in cylindrical coordinates. The relations between the cylindrical and Cartesian coordinates are: ρ = x2 + y 2
−1

(2.8) (2.9) (2.10)

θ = tan (y/x) z = z

We will not do any derivations because they are outside the scope of the discussion. Suffices to state the solution, given by √ ℓ 2ρ 2ρ2 A 2 2 2 Lℓ e−ρ /w(z) eikρ /[2R(z)]eiℓθ e−iϕ(z) . (2.11) Ep,ℓ (ρ, θ, z) = p 2 w(z) w(z) w(z) The solution now contains the associated Laguerre function Lℓ , and reason for calling p these modes Laguerre-Gauss. The subindices p and ℓ now account for the families of

18

CHAPTER 2. HIGH-ORDER GAUSSIAN BEAMS

solutions of order N = 2p + |ℓ|. The constant term in Eq. 2.11 that normalizes the solution is given by 1/2 2 A = p! . (2.12) πp!(|ℓ| + p)!) Notice that for N = 0 we aso get Eq. 2.7. One of the main characteristics of the solution is that the amplitude has a pure radial dependence, so that the intensity profile of the beams consists of one or more rings. When ℓ = 0 and p = 0 the beams have a characteristic single-ringed “doughnut,” with the radius of the doughnut proportional to ℓ1/2. Figure 2.2 shows a graph of the amplitude of a LG1 mode and a picture the same mode taken in the laboratory. 0

Figure 2.2: Graph of the amplitude and picture of the LG1 mode. 0

Exercise 5 The expression for the amplitude of an LG mode is Ep,ℓ (r, θ, z) ∝ √ 2r ω0
ℓ

Lℓ p

2r2 −r2 exp , ω(z0 )2 ω(z)2

(2.13)

ℓ where Lp (v) is the associated Laguerre polynomial of radial order p and azimuthal order ℓ. The first couple of polynomials are: Lℓ (v) = 1, and L0 (v) = 2 − v. Make 1 0 sketches of

1. E00 (r, θ, z0) vs. r 2. E02 (r, θ, z0) vs. r 3. E10 (r, θ, z0) vs. r 4. Draw a rough shape of the beam profiles for each of the cases above.

2.2. HIGH-ORDER GAUSSIAN BEAMS IN CYLINDRICAL COORDINATES 19 Returning to the full solution Eq. 2.11, the most interesting term is the one where the phase varies as ℓθ. Ignoring for the moment the correction due to the radius of curvature of the phase front (or equivalently assuming R(z) → ∞), we find that the wavefront has the form ψ(ρ, θ, z) = kz + ℓθ. (2.14)

Points of constant phase (i.e., the wave front) form a helix of pitch λ, as shown in Fig. 2.3 for the case ℓ = 1. For ℓ > 1 the wavefront consists of multiple helices. In a

Figure 2.3: Graph of the wavefront of a LG1 mode. 0 given transverse plane the phase depends on ℓθ, as shown in Fig. 2.4. At ρ = 0 the

Figure 2.4: Graph of phase of a LG1 mode across a transverse plane. 0 phase is undefined; it is said to be a phase singularity. For this reason the amplitude of the wave is zero at ρ = 0.

20

CHAPTER 2. HIGH-ORDER GAUSSIAN BEAMS

2.3

Irradiance and Power

In the section on electromagnetic waves we saw that the irradiance, or the time average flow of energy per unit time per unit area (W/m2 ), of a plane electromagnetic wave is given by cǫ0 2 I= E , (2.15) 2 0 where E0 is the amplitude of the electric field of the wave. For Gaussian beams the amplitude of the electric field of the wave is the value calculated earlier multiplied by the maximum amplitude of the electric field E0 E = ΨE0. For example, the amplitude of the field for the N = 0 beam is E0 = 2 E0 −ρ2 /w2 e , π w (2.17) (2.16)

where w is understood to be the width of the beam defined in Eq. 2.18, which varies with z. The irradiance then becomes cǫ0  2 E0  I = 2 π w I =
 2

(2.18) (2.19)

2 cǫ0 E0 −2ρ2 /w2 e . πw2

Normally we do not measure the electric field. Usually we have detectors that measure the total power of the beam P (in W). Thus we have to integrate the irradiance
2 cǫ0 E0 πw2 2 cǫ0 E0 . P = 2 ∞ 0

P =

e−2ρ

2 /w2

2πρdρ

(2.20) (2.21)

Replacing P from Eq. 2.21 into Eq. 2.19 we get I= 2P −2ρ2 /w2 e . πw2 2P . πw2 (2.22)

An important quantity in many fields is the maximum irradiance Imax = (2.23)

For example, by focusing a 5-mW HeNe laser beam to a waist of 25 µm with a 13-cm focal-length-lens we get that Imax = 509 W/cm2 .

2.4. PROBLEMS

21

2.4

Problems

Problem 4 Light waves with a transverse profile described by Hermite-Gauss functions (or modes) of indices (m, n) are normally labeled as HGmn . Make a sketch of the amplitude of the HG21. Problem 5 Light waves with a transverse profile described by Laguerre-Gauss functions (or modes) of indices (p, l) are normally labeled as LGℓ . p 1. Show that the radius of the LGl “doughnut” beam profile increases with ℓ1/2. 0 The Fig. 2.5 shows a diffraction pattern where the zero order is the LG0 and 0 the patterns on each side are the LGl modes, with p = 0 and increasing value 0 of ℓ. Note: Ll (v) = 1. 0

Figure 2.5: Picture of a diffraction pattern that shows the LG modes starting from ℓ = 0 in the middle and increasing ℓ by one on each direction. Modes at opposite sides have the same ℓ but with different sign. 2. Using the analytical expression for the LGℓ modes, show that the modes with p ℓ = 0 have an amplitude of zero at the center. Problem 6 c. If a Gaussian beam is in a LGℓ mode, determine the value of ℓ from 0 the interference pattern shown in Fig. 2.6. This pattern occurs at z = 0 when the Gaussian beam is superimposed with a plane wave of the same wavelength. The bright spots correspond to points where the two waves are in phase (also called constructive interference) and conversely the dark spots correspond to points where the waves are 180 degrees out of phase (also called destructive interference). Justify your answer.

Problem 7 A Gaussian beam from a HeNe laser is incident on a lens with a focal length of 10 cm. At the lens, before being refracted, the radius of curvature of the wave is infinite and its waist is 2 mm. The total power of the beam is 5 mW.

22

CHAPTER 2. HIGH-ORDER GAUSSIAN BEAMS

Figure 2.6: Simulation of the interference pattern between a LGℓ bean and a plane 0 wave. 1. What is the maximum intensity of the beam (in W/m2 ) at the focal point of the wave? 2. How much power will get transmitted through a 0.2-mm diameter aperture placed a distance of 1 cm from the smallest waist of the beam (focal point)? Problem 8 A Gaussian beam of wavelength 600 nm travels along the z-axis. It has a waist w0 = 0.5 mm located at z = 0. 1. At what value of z will the radius of curvature of the phase front be R = 2z? 2. What is the intensity at point z = zR , r = 2w(z) relative to the intensity at z = 0, r = 0 (r is the distance from the axis of the beam). 3. Calculate the Gouy phase at z = zR . 4. Calculate the phase difference between the point (z = zR , r = 0) and (z = zR, r = 2w(z)). 5. If the power of the beam is 5 mW, What is the maximum intensity at z = 5zR ? Problem 9 A “vector” beam is a beam in a Laguerre-Gauss beam with circular polarization. For example, consider a Laguerre-Gauss beam with angular index ℓ and circular polarization. The electric field of the wave is given by E(r, θ) = E0 (r) i cos(kz − ωt + ℓθ) ± j sin(kz − ωt + ℓθ)), ( (2.24)

2.4. PROBLEMS

23

where the plus is with right circularly polarized light and the minus is with left circularly polarized light If we look into the beam, the electric field in a given plane, say z = 0, and at a given time, say t = 0, points in a direction that depends on the angular position: E0 (r) E(r, θ) = i cos(ℓθ) ± j sin(ℓθ)). (2.25) ( If we pick right-circularly polarized light (use the “plus”) and ℓ = 1, the beam is called a “radially” polarized beam, as shown in the sketch of Fig. 2.7.

Figure 2.7: Direction of the electric field in a radially polarized vector beam. 1. Sketch the direction of the electric field when ℓ = −1 2. Sketch the direction of the electric field when ℓ = +2

24

CHAPTER 2. HIGH-ORDER GAUSSIAN BEAMS

Chapter 3 Wave-front interference
In this chapter we will discuss the formalism for understanding the wavefont of complex light. In the laboratory we study complex wavefronts via interference with known wavefronts. We can best do this with an amplitude dividing interferometer, such as a Mach-Zehnder interferometer in combination with wave-front reshaping optical elements in the arms of the interferometer.

3.1

General Formalism

Consider the wave-function of a beam of light to be given by the complex-notation expression Ψ = Aeiφ, (3.1) where A is the amplitude and φ is the phase. The amplitude can be constant, as it is for the case of a plane wave. It can also depend on the transverse coordinates (x and y, when the wave propagates along the z-axis). For the case of a Gaussian beam the amplitude decreases with a characteristic Gaussian shape given by exp[−(x2 + y 2)/w2 ], where w is the width of the beam or beam spot. The phase of the wave can take the simple form (kz − ωt), where k = 2π/λ is the wave number and ω is the angular frequency. This is the case for a plane wave traveling along the zdirection. The expression for the phase can be more elaborate and can have other terms. For example, an spherical wavefront expanding wave is accounted by the term k(x2 +y 2)/(2R), where R is the radius of curvature of the wavefront. There is another phase that Gaussian beams acquire due to propagation of the light. This phase is called the Gouy phase. It is given by ϕ = (N + 1) tan−1 (z/zR ), where zR is the Raleigh range and N is the order of the mode of the beam. Let us consider two light waves Ψ1 and Ψ2 . The interference pattern produced by 25

26

CHAPTER 3. WAVE-FRONT INTERFERENCE

the superposition of the two waves will have an intensity given by I = = = = = = |Ψ1 + Ψ2 |2 |A1eiφ1 + A2eiφ2 |2 (A1eiφ1 + A2eiφ2 )∗ (A1eiφ1 + A2eiφ2 ) (A1e−iφ1 + A2 e−iφ2 )(A1eiφ1 + A2 eiφ2 ) A2 + A2 + A1 A2(ei(φ1 −φ2 ) + e−i(φ1 −φ2 ) ) 1 2 A2 + A2 + 2A1 A2 cos(φ1 − φ2 ). 1 2 (3.2) (3.3) (3.4) (3.5) (3.6) (3.7)

For simplicity let’s consider A1 = A2 = A. In this case the intensity reduces to I = 2A2(1 + cos ∆φ), where ∆φ = φ1 − φ2 . Constructive interference fringes satisfy ∆φ = 2πn, where n is an integer. (3.10) (3.9) (3.8)

3.2
3.2.1

Interference of Zero-order Beams
Collinear Interference

A Mach-Zehnder interferometer is shown in Fig. 3.1. When the interferometer has

Figure 3.1: Mach-Zehnder interferometer. 50-50 beam splitters the first one divides the light into equal amplitudes traveling

3.2. INTERFERENCE OF ZERO-ORDER BEAMS

27

along the two arms. The second beam splitter recombines the light. The reflection √ √ and transmission coefficients of the beam splitters are r = i/ 2 and t = 1/ 2, respectively. Exercise 6 If the wave of Eq. 3.1 is incident on a Mach-Zehnder interferometer, show that the amplitudes at the two output ports of the interferometer are: i ΨC = (eiφ1 + eiφ2 ) 2 and 1 ΨD = (eiφ1 − eiφ2 ). 2 (3.11)

(3.12)

Exercise 7 Show that the intensity at the ports C and D are: IC = and ID = A2 (1 + cos ∆φ), 2 A2 (1 − cos ∆φ). 2 (3.13)

(3.14)

Let us consider projecting the output of port C of the interferometer onto a screen or CCD camera. If the arms of the interferometer differ by a length L, then the phase difference between the light coming from the two arms of the interferometer is ∆φ = 2π L. λ (3.15)

Ignoring the two turns that the light takes in going through the interferometer we can define the phase of the light going through the two arms of the interferometer as φ1 = kz1 and φ2 = kz2 . The beams travel distances z1 and z2 from the first beam splitter to the screen or CCD. The phase difference between the two beams is ∆φ = φ1 − φ2 = k(z1 − z2) = kL, which consistent with Eq. 3.15.

3.2.2

Noncollinear Interference

In the next case one of the beams, say beam 1, forms an angle β with the direction of propagation of beam 2, which is traveling along the z-axis. This is shown in Fig. 3.2. When a beam is incident on a screen at an oblique angle, its phase will not be constant along the plane of the screen. In the case of the beam 1 in Fig. 3.2 the phase will

28

CHAPTER 3. WAVE-FRONT INTERFERENCE

Figure 3.2: Noncollinear case of interference. Two beams with propagation directions that form an angle β meet at a screen. Dashed lines correspond constant-phase wave-fronts. vary along the x-direction. The greater the value of β the greater the phase variation with x. One can see this more formally by considering beam 1 to be a plane wave with wave-vector k = k(sin β, 0, cos β). The phase of the wave will be φ1 = k · r − ωt = kx sin β + kz cos β − ωt. The phase difference between the two beams is ∆φ = 2π 2πx sin β + z cos β − (z + L). λ λ (3.18) (3.16) (3.17)

The last two terms are constant at the screen. They would vary if we change the difference in length between the two arms L or the position of the screen. We can rewrite the phase difference as ∆φ = At maxima the phase difference is ∆φ = 2πn. The locus of points where there is constructive interference is given by x= λ n − x0 , sin β (3.21) (3.20) 2π sin β + φ(L). λ (3.19)

3.2. INTERFERENCE OF ZERO-ORDER BEAMS

29

Figure 3.3: Predicted fringes when one beam forms an angle β with the other one. where x0 = φ(L)λ/(2π sin β). The fringes constitute a set of vertical lines, as sketched in Fig. 3.4 for x0 = 0. The fringes are separated by a distance λ/ sin β. As β is increased the fringe density increases. If the path length difference L is changed then the fringes shift. We produce these fringes by misaligning the two beams, as shown in Fig. 3.4. In actuality, it is harder to align the two beams to be collinear, so the noncollinear case is the first situation that one encounters when putting together an interferometer.

Figure 3.4: Apparatus for observing vertical fringes shown next to a sample interference pattern.

3.2.3

Holographic Principle

If one puts a photographic film on the screen where the fringes are projected, and develop the film, the result is a diffraction grating with slits separated by a distance d = λ/ sin β. (3.22)

If we shine a zero-order beam to this photographic grating we get that the first-order diffracted beam appears at an angle given by the formula for diffraction: sin θ = λ/d. (3.23)

30

CHAPTER 3. WAVE-FRONT INTERFERENCE

Comparing Eqs. 3.22 and 3.23 we get that θ = β! This is the principle of holography. The pattern recorded on the film is the hologram of the beam coming at the angle β. When shining the “reference” beam (the one normal to the pattern) onto the grating we get the holographic reconstruction coming at the angle of the original beam: β.

3.2.4

Interference of an Expanding Wave-front

Suppose that beam 1 is traveling along the z-axis, but is expanding. The phase of beam 1 on an XY plane will not be constant. Because the expansion of the beam implies a spherical wavefront, the phase will vary with the distance to the axis of the beam: k(x2 + y 2) φ1 = , (3.24) 2R where R is the radius of curvature of the wavefront. Constructive interference fringes will satisfy the relation k(x2 + y 2) 2π + L. (3.25) 2πn = 2R λ This can be modified to look like the equation of a circle x2 + y 2 = r 2 . of radius r= √ 2Rλn − 2RL. (3.26) (3.27)

Thus the interference pattern consists of a set of concentric circles forming a “bull’s eye” pattern, as shown in Fig. 3.5. When L = 0 the radii of the circles increases as

Figure 3.5: Diagram of the expected locations of circular fringes between two beams with different radii of curvature. √ n. As L is changed the circles move inwards or outwards.

3.3. INTERFERENCE WITH HIGH-ORDER BEAMS

31

We create this situation in the laboratory by putting a diverging lens in one of the arms of the interferometer, as shown in Fig. 3.6. The expanding beam will interfere with the nonexpanding beam when the two beams meet at the screen.

Figure 3.6: Apparatus for observing circular fringes shown next to a sample of the bull’s-eye pattern.

3.3
3.3.1

Interference with High-Order Beams
Collinear Case

Now let’s suppose that one of the beams is in a Laguerre-Gauss mode that has a helical wave-front. The phase of the wave then has a term of the form φ1 = ℓθ. (3.28)

In this chapter we will limit ourselves to single-ringed Laguerre-Gauss beams. These have the shape of a doughnut. This is because at the center of the beam the waves with different phases interfere destructively. We can create an interference pattern for a Laguerre-Gauss beam by using a “forked” diffraction grating in one of the arms of the interferometer. As we will see later, the forked diffraction grating is a computer-generated hologram of the interference pattern of a Laguerre-Gauss beam with a zero-order beam. Let us first assume that the Laguerre-Gauss beam is collinear with the zero-order beam. The phase difference between the two beams is then ∆φ = ℓθ − 2π L. λ (3.29)

Constructive interference fringes will appear at θ= 2π n − θ0 , ℓ (3.30)

32

CHAPTER 3. WAVE-FRONT INTERFERENCE

where θ0 = 2πL/(λℓ). When ℓ = 1 this is just a line at a particular value of θ. For the case of a Gaussian beam, it will be a broad maximum–a “blob” at one side of the axis of the beam. This is shown in the simulation of Fig. 3.7. The value of θ0

Figure 3.7: Simulation of the interference of a first-order Laguerre-Gauss beam with a zero-order beam. increases with L. That is, the interference pattern will rotate if the path difference L changes. This is shown in the set of images of Fig. 3.8, taken in the laboratory, where the phase between the two beams was varied from one frame to the next one. Note

Figure 3.8: False color images of the interference of a first-order Laguerre-Gauss beam with a zero-order beam. The phase between the two beams was varied incrementally from frame to frame. that if ℓ = −1 then the interference will rotate in the other direction as L is changed. Exercise 8 Sketch the interference pattern between a zero-order beam with (a) ℓ = +2 Laguerre-Gauss beam, and (b) ℓ = −3 Laguerre-Gauss beam.

Exercise 9 Sketch the interference pattern of a Laguerre-Gauss beam with ℓ = 1 an an expanding beam (i.e., with a spherical wavefront.

3.3. INTERFERENCE WITH HIGH-ORDER BEAMS

33

3.3.2

Non-collinear Case

To understand the interference pattern for this case, let us first analyze the interference pattern produced by a Hermite-Gauss beam HG01 and a zero-order beam. The HG01 has two horizontal lobes. Within each lobe the phase is constant, but the two lobes are 180-degrees out of phase with respect to each other. The interference pattern between a HG01 beam and a zero-order beam incident at an angle consists of a set of vertical fringes. However, as shown in Fig. 3.9, the fringes of the two lobes are stagered. This stagger reflects the phase difference between the two lobes.

Figure 3.9: Image of the interference of a HG01 mode with a non-collinear zero-order mode. Now let us consider a single-ringed ℓ = 1 Laguerre-Gauss beam. The phase of the mode advances around the ring of the doughnut so that once we go one full turn around the center of the beam, the phase will have advanced by 2π. As a consequence, points at opposite sides from the axis of the beam are 180 degrees out of phase (e.g., directly above the center φ = π/2, and directly below φ = 3π/2). So points at opposite ends of the doughnut in the vertical direction should show fringes that are staggered, as seen for the case of HG01 mode. However, since the phase varies smoothly around the doughnut the fringes should become unstaggered as we move laterally away from the center, and connect smoothly along the sides of the doughnut (i.e., at θ ∼ 0 and π). Figure 3.10 shows the interference pattern of a ℓ = 1 mode with a non-collinear zero-order mode. It shows a peculiar forked interference pattern. The fork is a consequence of the phase dislocation. That is, as we go around the center in a loop we gain or lose a fringe, or a phase of 2π. Exercise 10 Using the previous argument predict the fork generated by the interference of a Laguerre-Gauss beam with ℓ = 3 with a non-collinear zero-order beam.

34

CHAPTER 3. WAVE-FRONT INTERFERENCE

Figure 3.10: Image of the interference of a LG1 mode with a non-collinear zero-order 0 mode.

Let us now proceed with a more quantitative approach. The phase difference between a Laguerre beam with “charge” (or vorticity) ℓ and a non-collinear zeroorder beam is 2π x sin β + ℓθ + φ(L) λ y 2π x sin β + ℓ tan−1 + φ(L). = λ x

∆φ =

(3.31) (3.32)

As discussed before, φ(L) = 2π (z sin β −z +L) is a constant at the screen that depends λ on the difference in length between the two arms of the interferometer. If we graph the points satisfying ∆φ = 2πn we get the patterns of Fig. 3.11. The theoretical description indeed explains the observations: the signature of a phase vortex is a fork. Notice also that the sign of the vortex is conveyed by the orientation of the fork; vortices that differ in sign produce oppositely oriented forks. The shape of the fork depends on the value of ℓ. Figure 3.12 shows how to recognize the charge ℓ of the beam that creates the fork. Each interference fringe is a change of 2π in the phase. If we start from a given point on a fringe and go in a path around the center of the fork counting fringes as we go along, we return to the initial spot having gained or lost fringes. Since the gain or loss is the difference between the fringes above the center and those below the center, the “charge” of the fork can be recognized as the number of tines of the fork minus one. This can be verified in the computer-generated fork for ℓ = 3, shown in Fig. 3.13

3.4. GENERATING HELICAL BEAMS

35

Figure 3.11: Graph of the locus of points that are in constructive interference when non collinear zero-order beam interferes with a Laguerre-Gauss beam with ℓ = +1 (left) and ℓ = −1 (right).

Figure 3.12: How to recognize the charge of the fork: number of tines minus one.

3.4

Generating Helical Beams

We conclude by discussing the generation of Laguerre-Gauss beams. There are various methods, and their use depends heavily on technology and cost.

3.4.1

Spiral Phase Plate

The phase ot the light in an ordinary laser beam is uniform and symmetric about the beam axis. The wave-fronts are either planar or spherical. One way to produce a helical beam is by brute force. That is, by introducing a phase lag as a function of the azimutal angle. This is done with the use of a spiral phase plate, shown on the left side of Fig. 3.14. The phase plate is made of a material with index of refraction n. It has different thicknesses at different angles. A beam of light going through it will experience phase

36

CHAPTER 3. WAVE-FRONT INTERFERENCE

Figure 3.13: Binary hologram for generating a Laguerre-Gauss beam with ℓ = 3 in the first diffracted order.

Figure 3.14: Spiral phase plate used for generating helical beams (left). It generates a helical beam by transmission through it (right). shifts that depend on the thickness, which depends on the angle: 2π(n − 1)x , (3.33) λ where x is the thickness increment of the phase plate at a particular angle from the minimum thickness. In order to get a helical beam of order ℓ the largest thickness increment h (see Fig. 3.14 must produce a phase shift ∆φ = 2πℓ, and thus be given by ℓλ h= . (3.34) n−1 There is one caveat: in practice the efficiency for making a given LG beam depends on the details [see M.W. Beijersbergen et al. Optics Communications 112, 321 (1994)]. For example, 78% of an LG0 gets converted into a LG1 , but a higher effeciency (90’s 0 0 %) is obtained when converting LGℓ to LGℓ+1 . 0 0 ∆φ =

3.4.2

Amplitude Diffraction Gratings

As we have discussed previously, a developed interference pattern serves as a hologram. Since we know the mathematical form of the interference pattern we can

3.4. GENERATING HELICAL BEAMS

37

calculate it, photo-reduce it, and use it as a grating. A binary grating for generating ℓ = 3 is shown in Fig. 3.13. The holographic reconstruction consists of simply sending a zero-order beam through a forked grating. As shown in Fig. 3.15 for a fork of charge 1, the reconstructed beams come in the first order diffraction. However, the binary grating

Figure 3.15: Sending a zero-order beam through a charge-1 binary forked grating. produces much more: in the higher diffracted orders we also get higher-order beams. That is, in the n-th diffracted order we get we get a beam with a charge n. At opposite sides of the zero order the beams have the opposite helicity. We can generalize the diffraction off a forked grating in a very interesting way. The separation ∆x between fringes in a forked grating satisfies 2π = 2π ∆x sin β + ℓ∆θ. λ (3.35)

The second term is responsible for the higher density of fringes above and below the center of the fork. The average fringe separation is given by ∆x0 = Rewriting Eq. 3.35 gives ℓ∆θ∆x0 . (3.37) 2π When a zero order beam is diffracted by the forked grating we get beams diffracted at angles α that satisfy nλ sin α = . (3.38) ∆x0 ∆x = ∆x0 − The diffracted light appears at angles specified by the average fringe separation. When analyzing diffracted light one usually considers the phase difference due to rays from λ . sin β (3.36)

38

CHAPTER 3. WAVE-FRONT INTERFERENCE

Figure 3.16: Schematic of two rays leaving adjacent slits of a diffraction grating at an angle α. adjacent slits (Fig. 3.16). This phase difference is given by ∆φ = k∆x sin α. Using Eqs. 3.37 and 3.38 we get that the phase between adjacent rays is given by ∆φ = 2nπ − nℓ∆θ. (3.39)

The equation implies that the helicity of the diffracted beam is given by nℓ. More interestingly, if the beam incident on the grating has a helicity ℓ′ then Eq. 3.39 becomes ∆φ = 2nπ − (ℓ′ + nℓ)∆θ. (3.40) Equation 3.40 implies that in the general case of an incoming beam of helicity ℓinc incident on a forked grating of charge ℓfork , the diffracted beam has a helicity ℓdiff = ℓinc + nℓfork . (3.41)

One could view this relation in a different way: as conservation of angular momentum. The fork imparts an angular momentum given by nℓfork . Thus the total angular momentum of the beam is the incident momentum plus the one given by the fork. It is interesting that one could use this to determine the helicity of the incoming light: the diffracted order that yields ℓdiff = 0 indicates that ℓinc = ℓfork/n. This relation has been demonstrated recently, as shown in Fig. 3.17. To diagnose the phase of a helical beam we place the fork in one of the arms of the interferometer and steer the beams such that the first diffracted order going through one arm is sent in a direction that is collinear with the zero-order coming from the other arm. Alternatively, we can place the grating before the interferometer and have the Mach-Zehnder components steer the first and zero orders so that they overlap at the screen. This is shown in Fig. 3.18

3.5. PROBLEMS

39

Figure 3.17: Diffraction pattern produced by ℓinc = 1, ℓf ork = 2 and n = 2. The pattern contains a fork with ℓdif f = 4, in agreement with Eq. 3.41 (Galvez and Baumann, to appear in Proc. SPIE, 2007).

Figure 3.18: Setup to investigate the phase properties of Laguerre-Gauss beams. One of the weaknesses of the binary diffraction grating is that since it spreads out the light into a large number of orders. If we wish to obtain one particular beam the efficiency is low, and is determined by the slit function of the grating. For example if the ratio of dark to light fringe thickness is 1, then the even diffraction orders are supressed. A more efficient way to produce beams in definite LG modes is by use of phase gratings.

3.5

Problems

Problem 10 Explain the experimental arrangement that gave rise to the following pattern. specify the possible angular momentum of the interfering beams.

40

CHAPTER 3. WAVE-FRONT INTERFERENCE

Figure 3.19: Interference pattern recorded by a CCD camera.

Chapter 4 Energy and Momentum
4.1 Energy
E = cB, (4.1)

Fundamentals of electromagnetic waves propagating in vacuum yield the relation

where E and B are the amplitudes of the electric and magnetic fields, respectively. The speed of light c is given by √ c = 1/ ǫ0 µ0 , (4.2)

where ǫ0 and µ0 are respectively the permitivity and permeability of vacuum. The electromagnetic energy density is given by the sum of the electric and magnetic energy densities: U = UE + UB 1 1 2 U = ǫ0 E 2 + B . 2 2µ0 (4.3) (4.4)

Exercise 11 Show that by using the relations 4.1 and 4.2 we get the energy density to become U = cǫ0EB. (4.5) The total energy crossing an area A in a time ∆t is ∆U = UAc∆t. 41 (4.6)

42

CHAPTER 4. ENERGY AND MOMENTUM

From the previous equation we get that the energy flowing per unit area per unit time is ∆U S= = cU = ǫ0c2 EB. (4.7) A∆t The flow of energy can be expressed vectorially by the Poynting vector S = ǫ0c2 E × B. (4.8)

The energy carried by the electromagnetic field can thus be transported through space and delivered at remote locations. Our daily lives are importantly affected by the energy carried by the electromagnetic waves, from the heating of our planet by the Sun, to radio communications via cellular phones. Gaussian beams restricted to narrow solid angles can be used to effectively channel energy. Gaussian beams produced by lasers can be used for interesting purposes. An interesting case is the one of adaptive-optics-based astronomy. In this case a laser beam is sent up to the atmosphere to a region withing the field of view of a telescope. The light excites a well defined area in the sodium layer of the upper atmosphere, and the fluorescence of the excited region can be seen by the telescope. While the image should be that of a point source, the atmospheric inhomogeneities distort the wavefront of the light, causing decreased resolution and blurring of the image. A deformable mirror in the ground telescope readjusts its shape via actuators to give a sharp image of the beacon, which we know is a point source. This effectively negates the wave-front distortion caused by the atmosphere, and along the way sharpens the other objects in the field of view of the telescope. Figure 4.1a shows the Keck telescope with its laser beacon (λ = 589 nm). Figures 4.1b and 4.1c show images of Neptune with and without adaptive optics.

Figure 4.1: Adaptive optics technology used in the Keck telescope (a), showing the image of Neptune with (b) and without adaptive-optics-correction. Courtesy W.M. Keck Observatory. The irradiance is the time averaged energy flow I =< S >= ǫ0 c 2 E , 2 0 (4.9)

4.2. LINEAR MOMENTUM

43

where E0 is the amplitude of the electric field. For example, the Sun produces an irradiance of 100 mW/cm2 on the Earth’s surface. A magnifying glass with an area of 25 cm2 concentrates 2.5 W of energy on the focal spot of the lens. This is enough energy to burn a piece of paper. The total power in the beam is the integral of the irradiance over the transverse profile of the beam P = I dA. (4.10) This expression has already been discussed in Sec. 2.3 for the case of a gaussian beam. From the quantal point of view, light is made of packets of energy hc . (4.11) λ It is convenient to use the result hc = 1240 eV-nm. It let us calculate the energy of photons. For example, the energy of a photon with λ = 500 nm is Eph = 2.48 eV. The number of photons of energy E traveling per unit time in a beam of power P is P . (4.12) N= Eph Eph = hf = A 1 mW beam of photons of wavelength 500 nm carries N = 2.5 × 1015 s−1 . Exercise 12 How many photons per second are emitted by a green laser pointer (5 mW, 532 nm)? CCD cameras and solar cells take the energy of individual photons to release electrons via the photoelectric effect. In the case of the CCD camera, the photoelectrons go to define an image, and in the case of a solar cell they go on to power an electrical circuit. When we make the image of a distant galaxy, we gather the energy of photons produced in those remote locations, and which traveled millions if not billions of light years to reach us. Figure 4.2 shows the image of quazar 3C454.3, which is about 14 billion light years away–almost the age of the universe! This quazar was imaged at Colgate when it had an outburst in 2005, displaying 12 magnitude. The photon flux that we get from it on Earth is 1 × 105 photons/s-m2 .

4.2

Linear Momentum

If all the energy of the electromagnetic field of a beam with power P is absorbed by an object, then the object experiences a force F = P , c (4.13)

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CHAPTER 4. ENERGY AND MOMENTUM

Figure 4.2: Image of quazar 3C454.3 taken at Foggy Bottom Observatory. with P being defined in Eq. 4.10. The field thus exerts a radiation pressure on the object. Since the energy per unit time per unit area is S, then the radiation pressure is given by S (4.14) Prad = . c Thus, the light can exert a force on the object The momentum associated to the electromagnetic field is u S p = = 2. (4.15) c c The momentum vector can then be defined as p = ǫ0 E × B = S . c2 (4.16)

The momentum associated to light is very small to make impact on macroscopic objects. However, microscopic objects and atoms are affected by radiation forces. It is convenient to define radiation dynamics in terms of photons. The momentum of a photon, defined by Einstein is given by pph = Eph h = . c λ (4.17)

The momentum of a single photon is a very small amount. For example, the momentum of a 500-nm photon is 1.3×10−27 kg m s−1 . A sodium atom traveling in a room-temperature vapor has a momentum pN a = 2×10−23 kg m s−1 . It would require a Na atom to absorb about 2×104 photons tuned to the D-line (λ = 589 nm) to slow it down. This is the basis of laser cooling of atoms. In the cooling process a beam of atoms is slowed down by absorption of photons traveling in the opposite direction in which they are going. Since re-emission goes in any direction, the Na atoms is

4.2. LINEAR MOMENTUM

45

effectively slowed down by consecutive absorptions followed by re-emission. This is the first step in a laboratory technique to slow and trap atoms to very low temperatures. Figure 4.3 shows a cloud of trapped Na atoms after being slowed down. The

Figure 4.3: Photo of a Na cloud trapped at a temperature of 1 mK inside a vacuum chamber. Credit: Kris Helmerson, NIST Atomic Physics Group. temperature of the cloud is 1 mK, which translates into atomic momenta of 4 × 10−26 kg m s−1 . If the trap is turned off it can easily be pushed away by a light beam. Let us consider now macroscopic objects. If the object is submerged on a medium of index of refraction nm then the radiation force of a beam of light carrying N photons per second is then FN = Nnm pph . (4.18) For example, a 1-mW beam of 500-nm light, such as that of a laser pointer, exerts a force of 3.3 pN on an opaque object. Notice, however, that this is larger than the weight of common microscopic objects. Consider a model object like a microscopic cube of side d = 50 µm and density ρ = 103 kg/m3 . Its mass is 125 pg and its weight is 1.2 pN. If the The light forces on this object can vary depending on the reflecting properties of the cube. This is shown pictorically in Fig. 4.4. If the cube is fully absorbent the force is FN . If it is fully reflective the force is 2FN . This is because the light gets a momentum change that is twice its incident momentum (i.e., pf − pi = −2pi ). If the cube is transparent then, only a fraction of the light gets reflected. For normal incidence the reflection coefficient is R = (nm − nc )2/(nm + nc )2, where nc is the index of refraction of the cube. For example, R = 0.04 for light going from air (nm = 1) to glass (nc = 1.5) and R = 0.004 for light going from water (nm = 1.33) to glass. Thus the force from the first reflection is 2FN R. There are more forces: the force from the second reflection involves a transmission into the cube (T = 1 − R), a reflection at the bottom surface and a transmission out of the cube at the top surface. The force will be 2FN T RT . The next term would involve now two

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CHAPTER 4. ENERGY AND MOMENTUM

Figure 4.4: Possible situations when light is incident on a cube: the cube absorbs all the light (a), the cube reflects all of the light (b), and the cube reflects a partial amount of the light (c). transmissions and three reflections, with a force 2FN T 2N 3 . The exact amount is an infinite series: FT = 2FN RT , (4.19) with RT = R + T 2R i=1 ∞

R2i .

(4.20)

The infinite series Therefore

i

x2 converges to 1/(1 − x) when x < 1. In our case x = R2 . RT = R 1 + T2 . 1 − R2 (4.21)

For a glass cube in air RT = 0.077, and for the cube in water it is RT = 0.0072.

4.2.1

Optical Tweezers

For example, a polystyrene microsphere with a diameter d = 5 µm has a mass of about 69 pg and a weight of about 0.67 pN. The effect of weight is further minimized if the sphere is submerged in water, where the buoyancy force is 0.64 pN. These forces scale as d3 . Consider photons incident on a polystyrene sphere, as shown in Fig. 4.5. A fraction of the photons get reflected at the first surface. Photons going in such a direction get a momentum kick by the sphere equal to the difference between the two momenta. By conservation of momentum the sphere gets an equal and opposite momentum kick. The same occurs with the photons exiting the sphere from the inside. If the sphere

4.2. LINEAR MOMENTUM

47

Figure 4.5: Trajectory of a light ray through a sphere. is uniformly illuminated it can be shown that the total force acting on the sphere is 0.4 FN . If instead of sending a light beam uniformly to a sphere we focus it we get a very interesting effect: optical forces work to create a position where the sum of forces on the sphere is zero, and when displaced from that position a restoring force appears. The ray-diagram view of this situation is shown in Fig. 4.6. The net effect is to trap

Figure 4.6: Trajectory of a light rays in an optical tweezer. the sphere. The trapping force is proportional to FN . The maximum trapping force

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1

on spherical object can be found to be given by Ftrap =

Qnm P , c

(4.22)

where Q is a constant that depends on the geometry. In a typical situation it ranges betwen 0.01 and 0.5. Figure 4.7 shows an example of transverse (top row) and longitudinal trapping. In the top row one can see the sphere labeled by an arrow in frame

Figure 4.7: Images of a sphere trapped on a microscopic slide in an aqueous solution. Frames (a)-(c) show transverse trapping and frames (d)-(f) show longitudinal traping. (a) at a fixed position relative to the background spheres. In this situation the microscopic slide was moved to the right, showing transverse trapping. These pictures were taken here at Colgate. In frames (d)-(f) of Fig. 4.7 the microscope was moved up and down, along the vertical direction of the trapping beam. As the microscope slide was moved vertically the spheres in the background got out of focus bu the trapped sphere stayed in focus. The trapping force in this longitudinal direction is much smaller. One could measure this force using the drag force due to viscosity. For spherical objects it is given by FD = 6πηav, (4.23) where η is the viscosity (ηwater = 10−3 Pl), a = d/2 is the radius of the sphere and v is the velocity of the sphere. For a sphere with d = 5 µm moving at a velocity v = 5µm/s the drag force is 0.24 pN. That is, it is also of the same order of magnitude as the other forces. We were able to move the sphere trapped for background sphere speeds of up to 60 µm/s. This translates into a drag force, and consequently maximum trapping force, of 3 pN.
A. Ashkin, “Forces of a Single Beam Gradient Laser Trap on a Dielectric Sphere in the Ray Optics Regime,” Biophysical Journal 61, 569-582 (1992)
1

4.3. ANGULAR MOMENTUM

49

Exercise 13 The maximum speed with which we could pull a 10-µ m sphere is 100 µm/s. What is the trapping force?

4.3

Angular Momentum

Light can also carry angular momentum. There are two distinct types of angular momentum. One is called “spin” angular momentum, and is due to the polarization of the light. The other type of angular momentum is due to dislocations in the wavefront of the light wave. Below we will discuss these two types of angular momentum. Angular momentum density j of a light wave is defined in the same way as for material objects j = r × p. (4.24) If we replace Eq. 4.16 for the momentum density of the field we have j = r × p = ǫ0 r × (E × B) or j= (4.25)

ǫ0 r × S. (4.26) c2 Notice something peculiar. If the light beam comes in the form of an infinite plane wave there is no angular momentum. This is because for a plane wave S is constant and parallel to the propagation direction at all points on the plane. Thus we need a beam of light for carrying angular momentum.

4.3.1

Polarization and Spin Angular Momentum

As seen before, light can be in a variety of polarization states. Circular polarization states are eigenstates of angular momentum. The electric field of a circularly polarized electromagnetic wave propagating along the z-axis can be represented as E0 E = √ (i cos(kz − ωt) ± j sin(kz − ωt)), 2 (4.27)

where the positive sign represents the state of right circular polarization, and the negative sign represents left circular polarization. In this state the direction of the electric field of the light changes as the wave propagates. For example, if we were to freeze in time a right circularly polarized wave, the electric field direction would rotate counter-clockwise along the direction of propagation. This is seen in Fig. 4.8a. Another way to view this polarization state is by looking at the electric field as a function of time at a fixed point in space. If

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CHAPTER 4. ENERGY AND MOMENTUM

Figure 4.8: Representations of the electric field of a right circularly polarized wave: (a) as a function of the propagation direction at a given instant, and (b) as a function of time on a fixed point in space. for example we set z = 0 in Eq. 4.27 we see that it describes an electric field that rotates clockwise as a function of time, as shown in Fig. 4.8. We are not going to derive an expression for the angular momentum of the light. The result is that the z component of the angular momentum of a beam of circularly polarized light is P Jz = σ , ω (4.28)

where P is, as before, the power of the beam. The quantity σ represents the helicity of the light, with σ = −1 for right circular polarization and σ = +1 for left circular polarization. For linearly polarized light σ = 0, while 0 < |σ| < 1 for states of elliptical polarization. If we replace P with N¯ ω, with N being the number of photons in the h beam we get that the angular momentum per photon is Jz = σ¯ . h N (4.29)

That is, each photon in a circularly polarized state carries h units of angular momen¯ tum. Birefringent materials can insert phases between the components of the electric field parallel and perpendicular to the symmetry axis of the material. A quarter wave plate inserts a phase of 90◦ while a half-wave plate inserts a phase of 180◦ . A quarterwave plate inserted in the path of a circularly polarized beam changes the polarization to linear. The converse is true if the quarter-wave plate is aligned properly. A halfwave plate changes the handedness of the circular polarization. Thus one can see waveplates as devices that exchange angular momentum with the light. In general, birefringent material in the path will exchange angular momentum with the light in

4.3. ANGULAR MOMENTUM

51

some degree that depends on specific details of the material and the polarization state of the light. One can observe this exchange of orbital angular momentum in an optical tweezer. By putting birefringent crystals in an aqueous sample on a microscope slide one can observe the rotation of the crystals by the angular momentum exchange with a circularly polarized trapping beam.

4.3.2

Orbital Angular Momentum

As discussed in a previous section, the complex amplitude of the field in Laguerre Gauss beams can have a phase term that depends on the transverse angular coordinate θ: E ∝ eiℓθ , (4.30) where ℓ is the phase winding number. As a consequence the wavefront consists of intertwined helices. In a wave that propagates in free space the direction of energy flow is perpendicular to the wavefront. In the case of a beam with a helical wavefront the linear momentum, which is perpendicular to the wavefront, has an axial component. If you imagine the circular ramp in a car garage, the normal to the pavement is not vertical: it is tilted toward points lower in the ramp. Thus, the linear momentum in a helical wavefront has a component along the propagation direction pz , a component along the radial direction if the beam is expanding pr , and a component along the azimuthal direction pφ due to the tilted helix. Thus, the linear momentum vector in cylindrical coordinates is p = (pr , pφ , pz ), as can be appreciated in Fig. 4.9. If we

Figure 4.9: Directions of the linear momentum of the light for a Laguerre-Gauss beam with ℓ = 1. calculate the angular momentum relative to the axis of the beam we get that the angular momentum has a non-zero z component: j = r × p = rpφ k. (4.31)

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CHAPTER 4. ENERGY AND MOMENTUM

The analytical evaluation of the z-component angular momentum density of a LaguerreGauss beam yields jz = ǫ0ωℓ|Epℓ (r, φ, z)|2. (4.32) The evaluation of this result involves getting the pφ in terms of the field amplitude of the beam. It is an evaluation that is beyond the scope of this discussion. As can be seen from Eq. 4.32, the angular momentum density depends on the field amplitude. That is, the angular momentum is in the field that surrounds the beam axis, which contains an optical vortex. The total angular momentum of the beam can be calculated by integration of the angular momentum density over the transverse profile of the beam. The result is surprisingly simple: P Jz = ℓ , (4.33) ω where P is the power of the beam. If we use P = N¯ ω, with N being the number of h photons in the beam. Thus the angular momentum of the beam per unit photon is: Jz = ℓ¯ . h (4.34) N If we compare this result with the one of Eq. 4.29 we arrive at a fundamental result: the angular momentum per photon in a beam in a Laguerre-Gauss beam that is circularly polarized is (ℓ + σ)¯ . We note the temptation to think that photons are h in different parts of the beam. That is incorrect. All photons in the beam are in the entire mode. If we restrict the beam by an aperture then we would have to recalculate the angular momentum of each photon going through the entire aperture.

4.3.3

Rotation in Optical Tweezers

If we send a beam carrying orbital angular momentum to an absorptive particle this particle will rotate. Figure 4.10 shows the rotation of a 10-µm particle that absorbs part of the light incident on it, which carries orbital angular momentum. The irregularities in the particle allow us to see its rotation. The optical torque on an absorptive particle is kℓP , (4.35) ω where k is the fraction of power that is absorbed. As can be seen the torque depends on ℓ. The rotational drag torque on a sphere of radius R due to a fluid of viscosity η is τD = 8πηR3Ω, (4.36) τ= where Ω is the frequency of rotation of the sphere. Thus one can use the orbital angular momentum of the light to do microrheology.

4.4. PROBLEMS

53

Figure 4.10: Images of a sphere trapped on a microscopic slide in an aqueous solution rotating by the action of orbital angular momentum.

4.4

Problems

Problem 11 Suppose that we have a d = 50-µm glass cube with a mass density ρ = 1.2 · 103 kg/m3 . 1. What is its weight? 2. If we send a beam of light upwards toward the cube. How much irradiance is needed to balance the cube against its weight? 3. How much irradiance would be needed to balance it in air if the cube is fully absorptive? 4. Now consider the cube to be a 50-50 beam splitter of the same size (d = 50 µm), such that when illuminated from below it transmits half of the light and reflects half of the light. (a) What irradiance is needed to keep it from falling down? (b) What is the total force acting on the cube? Make a diagram and explain your reasoning.

Problem 12 Consider Fig. 4.7. The diameter of the sphere is 5 µm. In frames (a), (b) and (c) it is being dragged at constant speed through the liquid. The time between frames is 1 s. 1. Estimate the speed of the sphere. 2. Find the force exerted by the light. 3. If this force is the maximum exerted by the trap, how much power is delivered onto the sphere?

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Problem 13 Consider Fig. 4.10. The diameter of the sphere is 10 µm. The average time between frames is 5 s. 1. Estimate the angular frequency of rotation of the sphere in rad/s. 2. What is the torque exerted by the light on the sphere? 3. If the total power incident on the sphere is 50 mW, what fraction of the light was absorbed by the sphere?