I am trying to intuitively understand the basics of the supplementary text of a recent publication from Eric Betzig's group on lattice light sheet microscopy (1). I am confused by the explanation of their non-diffracting beam (before the authors discuss Bessel beams at all). My naive understanding thus far is as follows: An annulus of infinitesimal size is placed at the rear pupil on a converging lens such that rays converge as a cone, and all wave vectors lie on the cone. If this were a paraxial optics problem, I would imagine a perfectly thin cone composed of many rays emerging from the peripheral lens. However, the annulus of infinitesimal size will cause diffraction, so I am picturing instead the sum of Huygens wavelets producing something like 2D plane waves stacked along the lens axis that fill the space of the cone, and propagate along the axis that the lens is on. I have a feeling my confusion is related to the difference between an ideal and pseudo non-diffracting beam. Here are some questions I have:

  1. If diffraction occurs as rays emerge from the annulus/lens, then does this mean that the non-diffracting beam will exist even immediately past the lens (ie at low values of the lens axis, before reaching focus where the cone converges), since diffraction from the annulus will send wavelets transversely toward the core of the cone.

  2. This, I sense, is something basic I should probably know...but how does amplitude vary away from the axis of a wave vector emerging from a slit/annulus? Amplitude is greatest along the wave vector (otherwise what would be the point of having a wave vector if uniform spherical waves were produced from the annulus) and thus the beam is strongest at the focus/tip of the cone, but there will be finite amplitude in other locations away from the focus and central wave vectors...how would this work in ideal and pseudo non-diffracting beam scenarios?

  3. In what sense is the beam non diffracting? I imagine a single beam centered on the lens axis corresponding to the maximum superposition of waves. But since the waves are curved (emerging from the annulus/slit), lobes that aren't in the center will be distorted at short distances near the lens, whereas further away from the lens where wave fronts more closely approximate plane waves, lobes will be more or less constant along the lens axis. Is the idea that a pseudo non-diffracting beam ignores the near-lens part of the beam and only considers near the cone tip? In that case, what does an ideal case look like? The wording of the paper implies that ideal involves an infinitesimally thin annulus, whereas pseudo involves a finite annulus size.

  4. What does a cross section of the cone look like, and does this change if you move up and down the lens axis? What is the meaning of the authors' statement "the electric field of the light beam propagates in the y direction without any change in its spatial distribution or amplitude in the xz plane"? Does this mean the magnitude of the electric field is uniform in the xz plane? Does the word "spatial distribution" refer to an interference pattern of some kind?

Feel free to answer only some of my questions or to explain in whatever way makes sense to you. Thanks!!

The following is an excerpt from the authors' supplemental materials:

" The electric field $\boldsymbol{e}(\boldsymbol{x}, t)$ of any continuous-wave, coherent light beam of free space wavelength $\lambda _o$ propagating in a homogeneous medium of real refractive index n...decomposed into a sum or integral over the electric fields $\boldsymbol{e_n}$ of a set of plane waves propagating in various directions defined by their wavewavectors $\boldsymbol{k_n}$:

$\boldsymbol{e}(\boldsymbol{x}, t) = \sum_{n=1}^{N} \boldsymbol{e_n}exp[i(\boldsymbol{k_n}\cdot\boldsymbol{x} - \omega t )]$

where $\omega = 2 \pi / \lambda, \lambda = \lambda_o/n$ is the wavelength in the medium, and c is the speed of light.

In the special case where all the wavevectors $\boldsymbol{k_n}$ lie on the surface of a single cone, $\boldsymbol{k_n}\cdot \boldsymbol{\hat{e}_y} = kcos \theta$ $\forall n$ where $k = 2 \pi / \lambda$ and the y axis is defined as the axis of the cone which has a half-angle \theta. Hence Eq 1 becomes:

$\boldsymbol{e}(\boldsymbol{x}, t) = exp[i(kycos \theta - \omega t)] \sum_{n=1}^{N} \boldsymbol{e_n}exp[i((k_x)_n x + (k_z)_n z)] = \boldsymbol{e}(x,z)exp[i(kycos \theta - \omega t)]$

In other words, the electric field of the light beam propagates in the y direction without any change in its spatial distribution or amplitude in the xz plane. Such a beam is termed non-diffracting. The constraint that all the wavevectors lie on a cone is equivalent to the constraint that the light entering the objective lens used to create the cone is confined to points on an infinitesimally thin ring in the rear pupil of the lens. "

  1. Chen, B. C., Legant, W. R., Wang, K., Shao, L., Milkie, D. E., Davidson, M. W., ... & Betzig, E. (2014). Lattice light-sheet microscopy: Imaging molecules to embryos at high spatiotemporal resolution. Science, 346(6208), 1257998.

1 Answer 1


Well done on your careful and correct thoughts, particularly about the diffraction from an infinitessimally thin annulus. If, the lens were perfect and we had a thin annulus (and by thin, we mean much less than the wavelength) then the output field would indeed be significantly different from the superposition of waves with wavevectors lying on a cone; certainly the output will be nothing like what the authors describe. So we need to kit you up with the knowledge to understand how the authors's theory breaks down in the face of diffractive effects. This is one of the most annoying things about modern research papers: I should think the authors certainly understand the following, but lack of space prevents them from explaining.

First of all, you've pretty much answered your own question 3. "In what sense is the beam non diffracting?" by your calculations at the end: for non-evanescent fields, solutions of Helmholtz's equation are superpositions

$$\int F(k_x,\,k_y,\,k_z)\,\exp(i\,k_x\,x)\,\exp(i\,k_y\,y)\,\exp(i\,k_z\,z)\, {\rm d}A\tag{1}$$

of plane waves $\exp(i(k_x,\,x+k_y\,y+k_z\,z))$ which each fulfill Helmholtz's equation so $k_x^2+k_y^2+k_z^2=k^2$. If all members of the superposition have the same $y$ component of the wavevector (equivalent to the wavevectors lying in the cone as you say), then we can pull the $\exp(i\,k_y\,y)$ as a common factor out of the superposition so that the latter is now in the form

$$\exp(i\,k_{y0}\,y)\,\int F(k_x,\,k_{y0},\,k_z)\,\exp(i\,(k_x,\,x+k_z\,z))\, {\rm d}A\tag{2}$$

and that's pretty much it. The superposition must have the functional form $\exp(i\,k_{y0}\,y)\,f(x,\,z)$, so the basic envelope $f(x,\,z)$ doesn't depend on $y$ and is simply scaled by the phase delay $\exp(i\,k_{y0}\,y)$ in moving from one $y$ to the other. The Cartesian components of all the field vectors fulfill the Helmholtz equation.

Now for what you are missing. This is the self-contradictory nature of a ray, which only describes an approximation to Maxwell's equations called the Eikonal equation, as I describe in my answer here. A true ray stands for the plane wave with wavefronts normal to it. It thus doesn't matter where the tail of the ray is in the wavefront: you can slide it in any direction normal to the ray and it means the same thing, the same plane wave! So a ray stands for a delocalised field. In the Eikonal approximation, we assume that waves are "locally plane", i.e. don't differ much from a plane wave over a region of at least several wavelengths. Thus the position of the ray's tail now takes on a meaning in the Eikonal approximation: it says "there is a little piece of EM field that is approximately plane in the neighbourhood of my tail". This means, for example, when you're calculating the contribution of bundle of converging rays to a focus, you can slide their tails normal to the rays (as long as you do this only over short distances, i.e. less than a wavelength or so) so that they all pass through the same point, work out what their phase is in the farfield and thus sum them all up as a Fourier integral. The self-contradiction is exactly the same as assuming a bandlimited and time-limited signal: it simply boils down to the fact that a function and its Fourier transform cannot both have compact support, and if you shrink one, you broaden the other.

Let's first look at the ideal situation is if you truly have only wavevectors on the cone $\vec{k}\cdot \hat{Y} = k_{k_y0}$, then the superposition of plane waves integrates a true ring in Fourier space and is thus, on the focal plane, proportional to:

$$\psi(r,\,\phi)\propto\int_{\theta=0}^{2\pi} \exp\left(i \,k\,\frac{a}{f}\,\left(r\,\cos\phi\,\cos\theta + r\, \sin\phi\,\sin\theta\right)\right){\rm d}\theta = \int_{\theta=0}^{2\pi} \exp\left(i \,k\,\frac{a\,r}{f}\,\cos(\theta-\phi)\right){\rm d}\theta= \int_{\theta=0}^{2\pi} \exp\left(i \,k\,\frac{a\,r}{f}\,\cos(\theta)\right){\rm d}\theta\propto J_0\left(\frac{k\,a\,r}{f}\right)\tag{3}$$

and, as you know, because this field comprises only plane waves with constant $k_y$ components (given by $k_{y0} = k\, \sqrt{1-\frac{a^2}{f^2}}$), it is a diffraction free field and varies only with $r$ and $y$ as:

$$\psi(r,\,y) \propto \exp\left(i\,\sqrt{k^2-k^2\,\frac{a^2}{f^2}}\right)\,J_0\left(\frac{k\,a\,r}{f}\right)\tag{4}$$

This is the ideal, diffraction free Bessel beam that the authors seek. If you care to substitute it, you find that it is indeed a solution of the Helmholtz equation in cylindrical polar co-ordinates (with axial co-ordinate $y$) with the propagation ($y$-dependence) as shown.

So now, how to visualise the diffraction? We can do this through linear superposition as follows. A plane wave propagating along the optical axis entering the imaging lens's pupil will lead to the following field (the so-called Airy disk) on the focal plane if the lens is free of aberration:

$$\psi(r) = \frac{a}{k\,f\,r}\,J_1\left(\frac{k\,a}{f}\,r\right)\tag{5}$$

where $a$ is the pupil radius, $f$ the lens's effective focal length (thus $a/f$ is roughly its numerical aperture), $k=2\pi/\lambda$ the field's wavenumber and $r$ the radial co-ordinate in the focal plane.

Now, if we subtract away a field given by (5) with a pupil radius of $a-\frac{w}{/2}$ from another field given by (5) with a pupil radius of $a+\frac{w}{/2}$, then we shall have the focal plane field when a thin annulus of mean radius $a$ and width $w$ is lit by an on-axis plane wave, i.e. the situation in your paper:

$$\psi(r) = \frac{(2\, a-w)\, J_1\left(\frac{k\, r\, (w-2\, a)}{2\, f}\right)+(2\, a+w)\, J_1\left(\frac{k\, r\, (2 \,a+w)}{2\, f}\right)}{2\, f\, k\, r}\tag{6}$$

Now, if we expand this field as a third order Taylor series in $w$ so as to model a thin annulus, we get:

$$\psi(r,\,0)\approx \frac{a\, w}{f^2}\,J_0\left(\frac{k\,a\, r}{f}\right) + \frac{w^3 \left(a\, k^2\, r^2\, J_2\left(\frac{k\,a\, r}{f}\right)-3\, f\, k\, r J_1\left(\frac{k\,a\, r}{f}\right)\right)}{24\, f^4}\tag{7}$$

As we know from (4), the first term is diffraction free and propagates according to:

$$\frac{a\, w}{f^2}\,J_0\left(\frac{k\,a\, r}{f}\right) \mapsto \exp\left(i\, \sqrt{k^2-k^2\,\frac{a^2}{f^2}}\right)\,\frac{a\, w}{f^2}\,J_0\left(\frac{k\,a\, r}{f}\right)\tag{8}$$

So it is the second term in (7) that represents the departure from the ideal cone field. It is not diffraction free (although it is too hard to calculate the closed form diffraction formula) and it is this term (and higher order Taylor terms) that you would numerically propagate to find out how your actual field deviates from the ideal Bessel Beam.

But wait! (8) says that the deviation from the ideal case is smaller and smaller as $w\to0$, whereas we have already noted that a very thin ($\ll \lambda$) ring of light should beget a field that is significantly different from a locally plane one. (8) is neglecting some diffractive effects that arise near the pupil. But there are two effects that quell these particular diffractive effects:

  1. A plane wave tilted through an angle $\theta$ but filling the lens pupil leads roughly to the field in (3) displaced laterally by a distance $\theta\,f$. But if his tilt is greater than the field of view of the lens then the field will be blocked by an aperture stop - it will be vignetted. So the non-paraxial part of the field from the diffractive effects of a very thin ring (i.e. plane wave components significantly skewed from the optical axis) will be blocked by vignetting.

  2. If the ring is much thinner than a wavelength, it will beget a significant evanescent field. I discuss evanescent fields more in my answer here. These cannot reach the focal plane; they are nonpropagating and represent stationary capacitive and inductive energy stores in the electromagnetic field. The vector description of the field shows that this part of the fkield's power is in fact reflected. So the output from a ring much less than a wavelength in width is the same as one with a wavelength width, only there is attenuation because the evanescent field is lost;

Therefore (8) should give you a reasonable picture of the system's deviation from the ideal Bessel beam output.

  • $\begingroup$ Thank you so much for your thoughtful answer. I will digest this for a while before getting back to you. $\endgroup$
    – Entangler
    Nov 12, 2014 at 8:01
  • $\begingroup$ @Entangler No problems - this is fairly subtle stuff and takes a while to digest $\endgroup$ Nov 12, 2014 at 9:05

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