# I polarize the slits (one H, the other V) of a Young's double-slit. If my source is H or V, do I see fringes? What about if my source is D or AD?

In my journey to understand light better, I build a "polarized Young's interferometer". Imagine the following polarizations: horizontal, vertical, diagonal, and anti-diagonal (H, V, D, and AD). I place an H sheet polarizer over one slit, and V over the other. I use a polarized laser source. Finally, I place a D sheet polarizer "mixer" between the polarzied Young's and the observation screen. I orient the source to explore the output of the four inputs above (H, V, D, and AD). What are the results of my journey? Do I see fringes for H?, for V?, for D?, for AD? If I see fringes, are they always the same phase? (Note that D input means colinear with the mixer, and AD input means perpendicular to the mixer. Also, the lines on the polarizers refer to the transmitted electric field axis of sheet polarizers and do not denote wire polarizers. Assume I conduct this experiment at a visible laser line with slits ~1.5 mm center-to-center, and each ~0.5 mm wide. I find it safe to assume that the bare slits have a negligible effect on polarization, and that the sheet polarizers are very efficient.)

• hint: For any polarization write the field $\mathbf E$ as the sum of two orthogonal pols: $\mathbf E = E_v \hat{\mathbf v} +E_h \hat {\mathbf h}$. Now pass this through the first polarizers. Then write the result (V or H) as the sum of two "diagonal" vectors and pass it through the final D polarizer. Commented Aug 14, 2021 at 23:13
• Every photon determines its own path, when you give the photon a choice of either slit, i.e. the path is unblocked or possible through either slit, then you will observe the pattern. Note that in dark areas of the "interference" pattern there are no photons, and bright areas have all the photons. The word "interference" is from Younge in 1801, but photons never cancel each other. Commented Aug 15, 2021 at 15:31
• Note that only D or AD photons have the option to pass either slit, also about 50% of A or AD will get thru the HV polarizers and as they emerge they become V or H (the opposite of how they went in). These newly polarized photons then have a 50% chance of getting thru the mixer and making the pattern. Commented Aug 15, 2021 at 15:48
• Thank you for your thoughtful comment. Please note that light just through the polarized Young's has no definite polarization (or, hint, is the polarization based on angle?) because I do not measure the polarization until the mixer, and I never measure the polarization as a function of choice of slit. Commented Aug 15, 2021 at 20:27
• A laser pointer is polarized and could be rotated from 0 (V) to 45 (D) or to 90 (H) degrees to carry out your experiment. The pattern will show for 45 D only! Commented Aug 16, 2021 at 3:46

I am going to assume a broad audience with some who have seen complex plane-wave analysis, and many who have not. Anyone who can do the ordinary Young's double-slit experiment should have no trouble doing the polarized-mixed version. Little or no math is needed, but there is a little art in making a polarized double-slit.

(See me showing the fringes going through all four outcomes just by rotating the mixer here.)

(A note about references: As much publicly-funded information is lately behind paywalls, I also include freely-available sources – some tertiary. I do not believe the paid sources are really needed to understand this answer, but they are included for completeness and lineage.)

The experiment really can be done in one's kitchen with no math needed. I make the slits by hand with roughly 1mm width, and spaced ~1.5mm apart. I use electrical tape to mask off the boundary between two abutted sheet polarizers whose transmission axes are perpendicular, and then cut and peal away the slit windows. I may include more detail on the art of making a polarized double-slit, if people request it. Laser pointers are usually polarized, but sometimes unstable. To be sure of the source's polarization, I use a polarizer right after the laser which I call a "prefilter". I find it easier to rotate the mixer between D and AD than to rotate the laser/prefilter assembly. The mixer can even be placed right against the observation screen and rotated (or removed) at that point. It is fun to watch the fringes shift phase and disappear altogether as I spin the mixer.

From the comments, I think people are catching on that diagonal and anti-diagonal (D and AD) inputs produce fringing output and that H and V input cannot make fringes. What takes maybe all of us by surprise (certainly me) is that AD makes a dark central (zeroeth) fringe. This means that when the mixer and input polarization angles are mutually perpendicular (the meaning of AD), the fringes look like a photo-negative of the familiar ordinary Young's fringes.

I start with cartoons of the qualitative output of the following setups: single-slit$$^{(7, 8)}$$, double-slit$$^{(9, 10)}$$, polarized double-slit, and polarized double-slit with mixer (Figures 0, 1). For a reminiscent layout, see the quantum eraser of Chiao, R., P. Kwiat and Aephraim M. Steinberg$$^{(11,12)}$$ (see Figs. 9b, 9c of v1 of the preprint, appearing at the last pages due to a LaTeX error).

I then show photographs of my experimental arrangement and real output (Figures 6-10). I finish with Table 0 summarizing the experimental results and a following complex plane-wave analysis (with accompanying Figure 5). The analysis harmonizes with experiment for ordinary, polarized, and polarized-mixed configurations. Figure 0 reviews the output characteristics of ordinary single- and double-slits, and builds to show the polarized double-slit, and mixed polarized double-slit. Figure 1 completes the results for the mixed polarized double-slit. Historically, I initially work out the plane-wave solution to make sure that I understand the problem. Having trouble with the answer I finally realize that the solution is correct, but the answer predicts a dark fringe for AD input. I then look back at my photographic data and realize the central fringe is dark. (Previously, I deliberately make the mixer AD to the input for some reason.) I then make the mixer D to the input and the bright central fringe returns.

Now for an overview of the math: I frame the problem using a complex plane wave allowing for insertion or removal of the polarizers (for example, removing the mixer is equivalent to using the identity matrix in its place which does not change the wavefunction at that point). I define the problem in Figure 5. I find some sense of confidence in getting a mathematical answer aligning with my experimental journey. The following complex plane-wave analysis gives the same results for fringe existence and phase I see for experiment.

Input Polarization Ordinary Double-Slit Polarized Double-Slit Polarized Double-Slit with Mixer
D (Aligned with mixer) yes(bright central fringe) no yes (bright central fringe)
AD (Orthogonal to mixer) yes(bright central fringe) no yes (dark central fringe)
H yes(bright central fringe) no no
V yes(bright central fringe) no no

TABLE 0 (All basic experimental cases are summarized here and proved in the complex analysis.)

I use the Jones Calculus$$^{(1, 2, 6)}$$. My approach is to make the general case include the ordinary Young's for confirmation that I am solving the problem correctly. I also simplify by keeping the specific geometry implicit and looking for a multiplicative sinusoidal term which indicates fringing. The equations still look a little scary, but they would be far worse if I were to try for a profile of the intensity versus diffraction angle. Anyway, I am only studying the existence and phase of the fringes for each input of H, V, D, and AD. The Jones calculus simply treats each optical element as a linear transformation of the complex plane wave, so the J.C. is a compact way of writing out the equations that one normally encounters when writing out the polarization components and projections as vector terms. Finally, the Born rule ("squaring" = the complex result dotted with its complex conjugate) tells us the observed intensity. The results match experiment for every input polarization for the naked double-slit, the polarized double-slit, and the polarized double-slit with mixer. I model each physical setup from the general equation, reduce algebraically, and then "square" separately, after which I apply the specific input polarization vectors. (See below the unique matrix arising for each case just before applying the Born rule.)

I write an incident plane wave$$^{(1,3)}$$, $$\Psi_0 = \vec E_0 e^{i(\vec k \cdot \vec x-\omega t)}\tag{0}\label{eq0}$$ where, $$\vec k$$ is the wavevector, $$\vec E_0= \left[\begin{matrix} E_{0y} \\ E_{0z} \\ \end{matrix} \right ]$$ (and $$E_{0y}, E_{0z} \in \Re$$), $$\vec k \parallel \vec x$$, and $$t$$ is averaged over many cycles at any given $$x$$, so that only relative phases are important. Therefore, I treat the exponential in $$\eqref{eq0}$$ as a complex constant of arbitrary phase and choose to set it to 1. I do not care about absolute intensity. I only care about the existence of output fringes, so I look for a surviving sinusoidal term, and use proportionalities liberally. The slits happen to the incident wave simultaneously, so the result is a sum. The slit illumination factors are the same while their relative complex phases differ by diffraction$$^{(7, 8)}$$ angle, $$\theta$$, slit separation$$^{(9, 10)}$$, $$d$$, and slit filter (polarizer). Removing an optical element is handled by the identity matrix$$^{(1, 2, 6)}$$. The final function, $$\Psi_3$$, is the product of the mixer$$^{(1, 2, 6)}$$ (if any) and the sum of the slit functions$$^{(1, 2, 6)}$$, $$\Psi_1$$ and $$\Psi_2$$. So, $$\Psi_3 \propto M_{(identity \lor mixer)}(\Psi_1+\Psi_2)$$, where $$M_{(identity \lor mixer)}$$ is the identity matrix (if no mixer is present) or the Jones matrix for a diagonal polarizer$$^{(1, 2, 6)}$$. The slit functions are $$\Psi_0$$ modulated by relative phase due to geometry ($$\theta$$ and $$d$$) and absolutely transformed by any polarizing filters. Expanding, $$\Psi_3 \propto M_{(identity \lor mixer)}(M_{(identity \lor H)}\Psi_1e^{i\phi_1}+M_{(identity \lor V)}\Psi_2e^{i\phi_2})\tag{1}\label{eq1}$$ where $$M_{H \lor V}$$ are the Jones matrices for horizontal and vertical polarizers, respectively, and $$\phi_1$$, $$\phi_2 \equiv \phi_1(\theta, +d), \phi_2(\theta, -d)$$. Any surviving sinusoidal dependence on $$\phi_{1,2}$$ indicates fringing, so $$\theta$$ and $$d$$ remain implicit from here on out. For slits only $$\eqref{eq1}$$ reduces to, $$\Psi_3 \propto I(I\vec E_0e^{i\phi_1}+I\vec E_0e^{i\phi_2})$$, where $$I$$ is the identity matrix. Therefore, $$\Psi_3 \propto \vec E_0(e^{i\phi_1}+ e^{i\phi_2})$$ and taking the complex vector dot product$$^{(4,5)}$$ (aka “squaring”) gives, $$\lvert\Psi_3\rvert^2 \propto \lvert E_0\rvert^2(e^{i\phi_1}+ e^{i\phi_2})(e^{-i\phi_1}+ e^{-i\phi_2})$$ which reduces to, $$\lvert\Psi_3\rvert^2 \propto \lvert E_0\rvert^2(1+\cos(\phi_2 - \phi_1))\tag{2}\label{eq2}$$ The sinusoidal term indicates fringing. Note that the basis components of $$\vec E_0$$ are unimportant, so any input polarization gives fringes and the whole “ordinary double-slit” first column of Table 0 is proven. Now, for polarized slits,

$$\Psi_3 \propto I \bigl ( \left(\begin{matrix} 1 & 0 \\ 0 & 0 \\ \end{matrix} \right ) \vec E_0 e^{i \phi_1} + \left(\begin{matrix} 0 & 0 \\ 0 & 1 \\ \end{matrix} \right ) \vec E_0 e^{i \phi_2} \bigr)$$

$$\Psi_3 \propto \left( \begin{matrix} e^{i \phi_1} & 0 \\ 0 & e^{i \phi_2} \\ \end{matrix} \right ) \left[\begin{matrix} E_{0y} \\ E_{0z} \\ \end{matrix} \right ]$$ $$\Psi_3 \propto \left [ \begin{matrix} E_{0y}e^{i\phi_1} \\ E_{0z}e^{i\phi_2} \\ \end{matrix} \right ]$$ $$\lvert\Psi_3\rvert^2 \propto \left [ \begin{matrix} E_{0y}e^{-i\phi_1} & E_{0z}e^{-i\phi_2} \\ \end{matrix} \right ] \left [ \begin{matrix} E_{0y}e^{i\phi_1} \\ E_{0z}e^{i\phi_2} \\ \end{matrix} \right ]$$ $$\lvert\Psi_3\rvert^2 \propto \lvert\vec E_{0y}\rvert^2 + \lvert\vec E_{0z}\rvert^2 = \lvert\vec E_{0}\rvert^2 \tag{3}\label{eq3}$$ So that any input does not give fringing, and the second “polarized Young’s” column of Table 0 is proven. Including the mixer, $$\eqref{eq1}$$ becomes $$\Psi_3 \propto \left(\begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right ) \bigl ( \left(\begin{matrix} 1 & 0 \\ 0 & 0 \\ \end{matrix} \right ) \vec E_0 e^{i \phi_1} + \left(\begin{matrix} 0 & 0 \\ 0 & 1 \\ \end{matrix} \right ) \vec E_0 e^{i \phi_2} \bigr)$$ $$\Psi_3 \propto \left(\begin{matrix} 1 & 1 \\ 1 & 1 \\ \end{matrix} \right ) \left ( \begin{matrix} e^{i \phi_1} & 0 \\ 0 & e^{i \phi_2} \\ \end{matrix} \right ) \vec E_0$$ $$\Psi_3 \propto \left (\begin{matrix} e^{i \phi_1} & e^{i \phi_2} \\ e^{i \phi_1} & e^{i \phi_2} \\ \end{matrix} \right ) \vec E_0$$ $$\Psi_3 \propto \left [ \begin{matrix} E_{0y}e^{i\phi_1}+E_{0z}e^{i\phi_2} \\ E_{0y}e^{i\phi_1}+E_{0z}e^{i\phi_2} \\ \end{matrix} \right ]$$ $$\lvert\Psi_3\rvert^2 \propto \left [ \begin{matrix} E_{0y}e^{-i\phi_1}+E_{0z}e^{-i\phi_2} & E_{0y}e^{-i\phi_1}+E_{0z}e^{-i\phi_2} \end{matrix} \\ \right ] \left [ \begin{matrix} E_{0y}e^{i\phi_1}+E_{0z}e^{i\phi_2} \\ E_{0y}e^{i\phi_1}+E_{0z}e^{i\phi_2} \\ \end{matrix} \right ]$$ $$\lvert\Psi_3\rvert^2 \propto \left [ \begin{matrix} E_{0y}e^{-i\phi_1}+E_{0z}e^{-i\phi_2} & E_{0y}e^{-i\phi_1}+E_{0z}e^{-i\phi_2} \end{matrix} \\ \right ] \left [ \begin{matrix} E_{0y}e^{i\phi_1}+E_{0z}e^{i\phi_2} \\ E_{0y}e^{i\phi_1}+E_{0z}e^{i\phi_2} \\ \end{matrix} \right ]$$ $$\lvert\Psi_3\rvert^2 \propto \lvert\vec E_0\rvert^2+2 E_{0y} E_{0z}\cos(\phi_2-\phi_1) \tag{4}\label{eq4}$$

Now, $$\lvert\vec E_0\rvert^2 = \lvert\vec E_{0y}\rvert^2 + \lvert\vec E_{0z}\rvert^2 = E_{0y}^2 + E_{0z}^2$$ , and for D, AD, H, and V input, $$\vec E_0= \left[\begin{matrix} \pm1 \\ \pm1 \\ \end{matrix} \right ]$$, $$\left[\begin{matrix} \pm1 \\ \mp1 \\ \end{matrix} \right ]$$, $$\left[\begin{matrix} 0 \\ \pm1 \\ \end{matrix} \right ]$$, $$\left[\begin{matrix} \pm1 \\ 0 \\ \end{matrix} \right ]$$, respectively. Plugging in to $$\eqref{eq4}$$ shows that the sinusoidal fringing disappears for H and V. For D and AD, the sinusoid remains, but is dark for AD wherever D is bright due to the sign inversion against the cosine term. So, the last column of Table 0 is proven.

So, I guess my theory skills are "correct" because I get the "right answer" for all cases, but that might be a long way from answering why this effect happens. In English, I would say that the complex phase of D is 180 degrees away from that for AD, so all else being equal (along the central fringe), the input amplitude has zero transmission through the mixer. That statement seems to make sense to me, if I imagine that the light vector components really are rotating in and out of imaginary space. For anyone not liking that image, I am not sure what I can do for you.

Acknowledgements
My gratitude to those who have supported and inspired my enthusiasm for science; there is not space here to mention them all. Zeroth, I thank Being One. I acknowledge myself for resolving to initiate this line of research at my personal expense which is not an immediate outgrowth of previous projects I have been blessed to work on. Of those not known to me in the flesh, Socrates who reminds me that I am allowed to start from zero, Huygens who proves more than he guesses, Tesla who trusts in the ether, B. Franklin who shows the way, J. C. Bose who sees life, Heaviside with his dogged quest for understanding, application, and proper credit, Newton and Kepler for sweeping away epicycles, Euler showing unity of cycle, Gauss empowering in surfaces and distributions,… Of those known personally: the late Robert Lin, Davin Larson, and Janet Luhmann who watched over my undergraduate scholarship at Space Sciences Lab, and my then other co-authors S. W. Kahler and N. U. Crooker; Jerry Edelstein, leader and mentor on SPEAR (Spectroscopy of Plasma Evolution from Astrophysical Radiation) and T-EDI (TripleSpec Exoplanet Discovery Instrument) projects, David Erskine, inventor and pioneer of EDI and X-EDI (Ordinary and Crossfading Externally Dispersed Interferometry); Julia Kregenow, Dae-Hee Lee, Mario Marckwordt, Matthew Ward Muterspaugh, Kaori Nishikida, Kwangsun Ryu, Michael Scholl, Kwang-Il Seon, Martin Sirk, the late C. H. Townes, Barry Welsh, and Ed Wishnow for many meaningful interactions. My late parents Esther and Bill Feuerstein, as well as being my teachers, always supported me in my sense of purpose.

1. Fowles, G. (1989). Introduction to Modern Optics (2nd ed.). Dover. p. 35.
2. Wikipedia contributors, "Jones calculus," Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/w/index.php?title=Jones_calculus&oldid=1020573977 (accessed August 22, 2021).
3. Wikipedia contributors, "Plane wave," Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/w/index.php?title=Plane_wave&oldid=1031803393 (accessed August 22, 2021).
4. Born, M. Zur Quantenmechanik der Stoßvorgänge. Z. Physik 37, 863–867 (1926). https://doi.org/10.1007/BF01397477
5. Wikipedia contributors, "Born rule," Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/w/index.php?title=Born_rule&oldid=1032140227 (accessed August 23, 2021).
6. H. Hurwitz and R. Jones, "A New Calculus for the Treatment of Optical SystemsII. Proof of Three General Equivalence Theorems," J. Opt. Soc. Am.  31, 493-499 (1941).
7. Chr. Huygens, Traité de la Lumière (drafted 1678; published in Leyden by Van der Aa, 1690), translated by Silvanus P. Thompson as Treatise on Light (London: Macmillan, 1912; Project Gutenberg edition, 2005), p.19.
8. Wikipedia contributors, "Huygens–Fresnel principle," Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/w/index.php?title=Huygens%E2%80%93Fresnel_principle&oldid=1040360110 (accessed August 24, 2021).
9. Wikipedia contributors, "Double-slit experiment," Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/w/index.php?title=Double-slit_experiment&oldid=1034409853 (accessed August 24, 2021).
10. Young, Thomas. "The Bakerian lecture. Experiments and calculation relative to physical optics". Retrieved 14 July 2021.
11. Chiao, R., P. Kwiat and Aephraim M. Steinberg. “Quantum non-locality in two-photon experiments at Berkeley.” Quantum and Semiclassical Optics: Journal of The European Optical Society Part B 7 (1995): 259-278.
12. arXiv:quant-ph/9501016v1


W. Michael Feuerstein is an independent researcher investigating orthogonal projection interferometry and interferometric basis discrimination. He has professional research experience in space physics (particularly the relationship between the interplanetary magnetic field and electron pitch-angle distributions), space-born extreme and far ultraviolet astronomy, and externally-dispersed interferometry for Doppler planet-finding.

• Pictures are not accessible to all users. Please type out the relevant parts of the picture. Also, it might help to explain things in your own words rather than just post text of another source. Alternatively you could link to the source and then state the relevant parts in your answer. Commented Aug 27, 2021 at 0:17
• I would type out the relevant parts of it. Copy and pasting might not be the best format here, but it is a step in the right direction. Commented Aug 27, 2021 at 0:32
• For equations and symbols, use mathjax Commented Aug 27, 2021 at 0:44
• Very good work ... and nicely presented. Does the very first polarizer (the one that conditions the light) have an effect ?.... if you were to rotate it 180, does then the central zero become the central max and the central max become the central zero. I'm trying to understand the symmetry of situation. (its possible that the first polarizer creates a purer form of light than the raw laser ... or maybe not? .. try just the raw laser) Commented Aug 28, 2021 at 16:11
• It might be interesting to run the experiment with no H and V polarizer on the slits, just the 2 diagonal ones ... or I mean the laser/prefilter and mixer. But maybe not .... I guess what you show is that D and AD give the "anti" pattern. P.S. I am not an expert of formatting items for SE .... but your paper above and its images are quite clear. Commented Aug 28, 2021 at 21:08

Since light must travel through both slits for an interference pattern to be seen, when the source light is H or V, no interference will occur since the incident light will only be able to pass through one slit.

When the incident light is D or AD, an interesting situation occurs. Depending on how narrow the slit widths are, the slit itself has a tendency to polarize the light (which we can define as the V orientation). These means that very little light will get through the slit with the H polarizer placed over it regardless of the polarization of the incident light.

Therefore, even in the case of D and AD essentially no interference pattern will be seen. You have basically designed an experiment that for all practical purposes prevents the incident light from going through both slits. And, this is the case for any and all polarizations of the incident light.

• How narrow does the slit have to be to act as a polarizer? My intuition is a few wavelengths, which would mean the effect is not a practical concern in a kitchen experiment.
– rob
Commented Aug 14, 2021 at 21:08
• @rob below 100GHz a standard technique to use wire polarizers in which the wires are separated by a fraction of a wavelength; for example, a 77GHz system (quasi-Cassegrain) has a wire mesh printed on the parabolic sub-reflector whose wires are about $\frac{\lambda}{3}$ apart. So when illuminated with a field whose $E$ is parallel with the wires they will act as a reflector whose focus is the transmit/receive port but it is transparent to a field whose $E$ is perpendicular to the wires. Commented Aug 14, 2021 at 21:57
• @hyportnex Thanks, that’s what I suspected. For visible light (half-micron wavelength) through an aperture made with a razor blade (perhaps 50–100 microns) I would expect zero impact on the polarization. That makes the final paragraph of this answer (v1) substantially incorrect.
– rob
Commented Aug 14, 2021 at 22:24
• Good comments. I actually answered this question before it was edited with the diagram and identified as a "kitchen" experiment. Therefore, I approached it more from the perspective of a thought experiment under ideal conditions (narrowest possible slits). Under these conditions the slits will indeed polarize most of the incident light. But, yes, for a kitchen experiment the slit widths will indeed be too wide to create any significant polarization.
– JRL
Commented Aug 15, 2021 at 14:44
• "Since light must travel through both slits for an interference pattern to be seen" ... I would restate this as ....Since light must -have an option to- travel through both slits for an interference pattern to be seen. Good answer, upvoted. Commented Aug 15, 2021 at 15:24

In your setup you cannot get two slit interference if the photon source is all vertical or all horizontal but keep in mind you will still have single slit interference. If the photon source is all diagonal or all anti-diagonal then your setup will get two slit interference.

• Thank you. And the fringes for AD are 180 degrees out-of-phase with the fringes for D. Commented Sep 6, 2021 at 22:50
• @feuerstein there are different ways to look at or derive the phasing. You should work your way up to the double slit by starting with a single edge. Check out my paper “Single Edge Certainty” at billalsept.com. Commented Sep 7, 2021 at 1:05