I've spent some time reading wiki etc. What I get now is that apart from the normal light amplitude information, holograms also record the phase information of light. But this is so difficult for me to understand. In my understanding to display the 3D objects you have to reconstruct the 3D object field; the light source and exact light field at that moment, what does the phase mean here? why phase alone makes all these happen?

Apart from the principles, what is restraining holography display from being popular at the moment?


3 Answers 3


If I had to give a one sentence answer to this question, it would be as follows:

*That the phase and amplitude alone on one plane is enough to wholly define a three-dimensional light field arises from the various uniqueness theorems for Maxwell's equations within a connected volume given the solution on the volume's boundary; otherwise put: once you know a solution on a boundary, then the values within must follow from "reasonable" physical assumptions.

For simplicity let's sit with the scalar diffraction theory, so now we are essentially talking about uniqueness theorems for the Helmholtz equation $(\nabla^2 + k^2) \psi = 0$.

Uniqueness theorems when $k^2 > 0$ or when $k^2 \in \mathbb{C}-\mathbb{R}$ are much more complicated than when $k^2\leq0$. The latter case corresponds to static solutions of the Klein Gordon equation or to static solutions of the Maxwell equations with or without an assumption of a massive photon; see My answer here for more details. Such cases have very strong uniqueness theorems: once a solution's values are set on a compact volume's boundary, there is only one possible solution within the volume. This situation even extends to semi-infinite volumes. However the former situation includes $k^2>0$, the case for scalar diffraction in freespace or a lossless dielectric: uniqueness theorems need further strong assumptions about the field to make them work. Thankfully, some of these assumptions are reasonable physically.

We can restore simplicity to the solutions of the freespace Helmholtz equation (i.e. to the situation we have with a hologram) by making reasonable physical assumptions such as the Sommerfeld radiation condition or that the field is a tempered distribution; for more information on the latter condition, see my answers here or here.

Given these assumptions, together with the assumption that the field is propagating purely left-to-right, we can reconstruct a field from the hologram as follows. You begin with the Helmholtz equation in a homogeneous medium $(\nabla^2 + k^2)\psi = 0$. If the field comprises only plane waves in the positive $z$ direction then we can represent the diffraction of any scalar field on any transverse (of the form $z=c$) plane by:

$$\begin{array}{lcl}\psi(x,y,z) &=& \frac{1}{2\pi}\int_{\mathbb{R}^2} \left[\exp\left(i \left(k_x x + k_y y\right)\right) \exp\left(i \left(\sqrt{k^2 - k_x^2-k_y^2}-k\right) z\right)\,\Psi(k_x,\,k_y)\right]{\rm d} k_x {\rm d} k_y\\ \Psi(k_x,\,k_y)&=&\frac{1}{2\pi}\int_{\mathbb{R}^2} \exp\left(-i \left(k_x u + k_y v\right)\right)\,\psi(u,\,v,\,0)\,{\rm d} u\, {\rm d} v\end{array}$$

To understand this, let's put carefully into words the algorithmic steps encoded in these two equations:

  1. Take the Fourier transform of the scalar field over a transverse plane to express it as a superposition of scalar plane waves $\psi_{k_x,k_y}(x,y,0) = \exp\left(i \left(k_x x + k_y y\right)\right)$ with superposition weights $\Psi(k_x,k_y)$;
  2. Note that plane waves propagating in the $+z$ direction fulfilling the Helmholtz equation vary as $\psi_{k_x,k_y}(x,y,z) = \exp\left(i \left(k_x x + k_y y\right)\right) \exp\left(i \left(\sqrt{k^2 - k_x^2-k_y^2}-k\right) z\right)$;
  3. Propagate each such plane wave from the $z=0$ plane to the general $z$ plane using the plane wave solution noted in step 2;
  4. Inverse Fourier transform the propagated waves to reassemble the field at the general $z$ plane.

If you can understand these steps you should be other see how the solution to Helmholtz's equation, i.e. the full three-dimensional scalar light field, is reconstructed from its values on a plane. The latter of course is what a phase and intensity mask hologram encodes.

What hinders holography? I am not up with the latest hologram production techniques, but essentially making a hologram is a kind of interferometry and as such calls for low vibration and building of an interferogram between transmitted and reference light. One can't simply "snap" a hologram like one can with a digital camera (or even with an older style film camera). Moreover, the phase masking needed to make the equations above work is highly colour-dependent, so that any kind of colour holography is even more restrictive than the making of one-colour holograms. The Holography Wiki page gives you a good overview; the "rainbow" holographic technique is the nearest I know of to colour holography. Aside from this technique, most holograms need high coherence in the light source for reconstruction.

Another interesting technique is the manipulation of light by computer generated holography, where one computes by solving Maxwell's equation the phase and amplitude mask needed for e.g. nulling out the mean aberration from a lens before analysis by an interferometer.

  • $\begingroup$ Could you please link to a site that explains this part more in depth - "That the phase and amplitude alone on one plane is enough to wholly define a three-dimensional light field arises from..." $\endgroup$ Nov 2, 2014 at 5:31
  • $\begingroup$ @DevKanchen There are many different conditions that are taken as "reasonable" and which you must assume further to the values on the boundary to assure uniqueness. For a summary, see Theorem 3.17 of this document here. In particular, the Sommerfeld radiation condition (linked in my answer) assures uniqueness given boundary values. If you know both values AND normal derivatives, then uniqueness is simple and trivial by the Kirchoff Integral Theorem. Another "reasonable" assumption set is that (1) the field comprises only plane waves ... $\endgroup$ Nov 2, 2014 at 6:21
  • $\begingroup$ @DevKanchen .... or, equivalently, its value on one plane is a tempered distribution and thus is uniquely defined by its Fourier transfrom and (2) the plane waves have only positive components in the $z$ direction, i.e. the field propagates only left to right. Given these two assumptions, you can use the main equations in my answer to prove uniqueness. $\endgroup$ Nov 2, 2014 at 6:25
  • $\begingroup$ I can't say I understood all of that (the math is leagues out of my knowledge), but thanks for the explanation and link. I'll have to dig further. $\endgroup$ Nov 2, 2014 at 9:32
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    $\begingroup$ Is the term $\left(k-\sqrt{k^2 - k_x^2-k_y^2}\right)$ correct? Isn't it simply $\sqrt{k^2 - k_x^2-k_y^2}$? $\endgroup$ Jul 11, 2017 at 7:04

From an intuitive-divulgative point of view, you may think of an hologram as a plane or part of one that records the intensity and the direction distributions of light once passed through it. That's what the relative phase recording produces at the end.

When you look through a window you see part of whatever is in the other side, and you get 3D perception due to parallax: each eye gets different information (a 2D distribution of intensity), or if you move you get a different point of view. Then you say you see 3D objects. A possible model for that window may be not "glass+external world" but "plane whose points emit light in certain directions": if you trace a line on the glass in the points where you see some contour and you move, you see that contour through other points of the glass.

That directions distribution recording is achieved through, if you want, local diffraction gratings, that allows you to encode that 3D (2D$\otimes$directions) information in a plane (in surface holograms at least. In volume ones you have local Bragg gratings). The third parameter is the spatial frequency of the grating.

To reconstruct the scene you need to have part of the exact field it was recorded with, and that makes the "relative" in relative phase recover its meaning. That's why you usually chose a scene where there is an object (set of points "emitting" light) plus a collimated beam pointing to the plate. This allows to have an easy to reproduce part of the scene: the collimated beam. With it you can then reconstruct the complete scene, with the plate locally redirecting light to the exact directions, in a certain recording-reconstruction symmetry flavour. The hologram plate is the window through which you see the scene recorded.

I believe if you understand diffracting gratings theory you can make your own idea of holography by just thinking locally and with point/plane sources.

As for what @wetsavannaanimal-aka-rod-vance said about colour holography (I have no enough reputation to make comments) I can recomend to visit Yves Gentet site to at least see the videos and GIFs.


From the Experimentalist side:
each illuminated spot on the object will, via interference with the reference beam, create a Gabor Zoneplate on the hologram recording film. A Zoneplate is analogous to a glass lens, but rather than using a prism effect to bend light, it employs diffraction gratings. When the Zoneplate lens is later developed, illuminated, and viewed, the eye perceives a pointlike virtual image deep within the film; a spot of light at the same relative position as the original spot on the illuminated object. So basically a hologram is hundreds of thousands of inches-wide Zoneplate lenses stamped upon the film: one wide lens for each illuminated point on the original object.

And finally, if each Zoneplate lens happens to resemble a bulls-eye pattern, then it's the position of the rings in the bulls-eye which represent the recorded phase information.


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