Experiment 3

Double slit +
Polarizers

Abstract

In this experiement, we will combined the two previous experiment. The double slit and the poalrizers. More precisly, we will cover both slit with different polarizer. We will discover that when those two polarizers are perpendicular, we will loose any kind of interference, only because the oscilation of the two resulting photons won't be in the same plan. But, if we put another polarizer after that, we could have a way to recombined the photon and regain interferance. By playing with the orientation of that third polarizer, we will be able to also simulate other results. Whenever we can predict in which slit the photon want, the double slit will act as a single slit, which mean no interference, just diffraction. We call that knowing the "which-way" data. We will use that term to tell you whenver weknow where the photo want and "no which-way" data whever we don't, just like in experiement 2.

As you may already have undertand, the reason we want to put polarizers in front of each slit is the be able to access the which-way data, without interfering with the course of the photon. We culd have placed a detector in front of on slit, or having someone watching the slit, but the problem there is that those two ideas completly absorb or block the photon. With polarizer, yes we might rotate the photon a little bit, and block some of them, but at least we still have a way to let it interact with another photon latter.

I don't want to spoil the next experiement, but in polarizers are not the only method to "not detect" a photon while still having a little interaction with it. We will discover a second method in experiement 4.

Try it yourself!

Here is an animation of the experiement 3. We assume the light coming out of the laser if diagonally polarized. You can rotate all the poalrizers using the pink dot and change the slit parameter in the control panel.  Thedark curve is the maximum diffraction you can get, the purple curve is the result of your setup.

Is that real?

The prove you that animation is not lying, here are the result we got from the experiement. The result will be explain in detail in the fallowing section.

Lab Experiment

Curious of what our setup looks like? Here is a few pictures of what we have:

Technical set-up and Equipment

Here is the representation of the setup you see in the picture. All of the equipement is shown, but the scaling is not representative.

The setup is the exact same as in experiement 2, but we added 3 polarizers. One in front of each slit, they are 90 degrees appart. The beam from both slit then passes through the same polarizer, a polarizer we can rotate.

Equipment list

Name
Description
Diode laser (405nm)
Two thin BBO crystals of type 1 placed at 90°
Double slit
Dimension (200-200-200 μm), made in-house
Mounted linear polarizers
Plano-convex lenses, L1 has f=50cm [LA-1908-B]
Adjustable slit (placed near 200μm)
Single photon detection module
Note: see more componant details and link in previous experiment

Result and analysis

Raw data

You can download all our data here:

To recreate a pattern, we moved an actuator along the beam. All those files has 4 columns. The first two are the angle of the polarizers, but only P1 is used. The third one is the distance the actuator has travelled in μm . The last column is the number of counts on D1, the detector. All acquisition are for a two seconds window.

Graphics

At 45°
At -45°
At 0°
At 90°

Let's take the time to analyse those graphics. First, we see that when we are at 0 degree are 90 degree, there's no interference. By placing the polarizer P1 in the same angle as one of the slit, we know for sure in which slit the photon went. We become in a single slit experiment. We see the pattern at 0 and 90 degree have a little shift. This shift represent the spacing between both slit. As each slit act as single slit, the product a diffraction pattern centered in front of them.

For the graphics at 45 degree and -45, we clearly see inteference, but they are both not the same. The 45 degree graphic has the same shape as the one we found in experiment 2, wit 3 big peak (we see the pattern is not perfect. There is actually a phase between our polarization which will be adjust only in the next experiment...). If you look closely, the pattern at -45 is the invert of the one at 45. All maximum are now minimum and vice-versa. And if you some both these plot, you will retreive a full diffraction envelop. (the same phenomena happens with the graphic at 0 and 90 degrees, if you add them up, you get the same diffraction pattern, but less mindblowing, because they were alredy diffraciton). So why the two plots are inverted? We have two reasons. First, if you think of it, with polarizers you are blocking part of the photons. Perpendicular polarizers will always block a different part of the photon. So went you add them up, you now have the some of all photons. The second explaination is because we have an initial beam at 45 degree. Whe P1 is at 45, it will restore the photons as they were, but wen at -45, it will reverse them of 180 degree which will instore a phase shift and they won't interact the same way.

Theory and equations

In the next sections, we will find the equation the solve this problem with P1's angle variable.

Mathematics

Full approach

1. Let's start with the initial state, we need to consider the poalrization state and the phase state.

\begin{align*}
\ket{\psi_0} &= \ket{\diag} \otimes \ket{S}\\
\end{align*}

2. We let the photons travel to reach either one of the slit. In both case, the distance it travels is l.

\begin{align*}
\ket{\psi_1} &= \ket{\diag} \otimes \Big(\frac{1}{\sqrt{2}}e^{ikl}(\ket{A}+ \ket{B})\\
\ket{\psi_1} &= \frac{1}{\sqrt{2}}e^{ikl}(\ket{\diag}\ket{A}+ \ket{\diag}\ket{B})\\
\end{align*}

3. We apply the two polarizors of 0 at position A and 90 at position B

\begin{align*}
\ket{\psi_2} &= \frac{1}{\sqrt{2}}e^{ikl}(P(0)\ket{\diag}\ket{A}+ P(90)\ket{\diag}\ket{B})\\
\ket{\psi_2} &= \frac{1}{\sqrt{2}}e^{ikl}(\ket{\updownarrow}\bra{\updownarrow}\ket{\diag}\ket{A}+ \ket{\leftrightarrow}\bra{\leftrightarrow}\ket{\diag}\ket{B})\\
\ket{\psi_2} &= \frac{1}{\sqrt{2}}e^{ikl}(cos(45-0)\ket{\updownarrow}\ket{A}+ cos(45-90)\ket{\leftrightarrow}\ket{B})\\
\ket{\psi_2} &= \frac{1}{\sqrt{2}}e^{ikl}(\frac{1}{\sqrt{2}}\ket{\updownarrow}\ket{A}+ \frac{1}{\sqrt{2}}\ket{\leftrightarrow}\ket{B})\\
\ket{\psi_2} &= \frac{1}{2}e^{ikl}(\ket{\updownarrow}\ket{A}+ \ket{\leftrightarrow}\ket{B})\\
\end{align*}

5. We add the distance needed to reach the point X, the detector. We need to consider that photons from both slit won't travel the same distance. Here, I cheat just a little bit, I'll add right away all the distance to reach the detector instead of adding it in two step, which would give the exact same result. We could say the polarizer is very close to the detector. The $\sqrt{2}$ arrivers because we are combining the 2 path.

\begin{align*}
\ket{\psi_3} &= \frac{1}{2}e^{ikl}\sqrt{2}(\ket{\updownarrow}e^{ika}\ket{X}+ \ket{\leftrightarrow}e^{ikb}\ket{X})\\
\end{align*}

6. Let's apply polarizor P1 at an angle $\delta$, pronounce delta. You will see the polarization state $\ket{\delta}$ popping up, it only means the photon is now in the same state as the polarization angle we choose.

\begin{align*}
\ket{\psi_4} &= \frac{1}{2\sqrt{2}}e^{ikl}(P(\delta)e^{ika}\ket{\leftrightarrow}+P(\delta)e^{ikb}\ket{\updownarrow}\\
\ket{\psi_4} &= \frac{1}{2\sqrt{2}}e^{ikl}(e^{ika}cos(0-\delta)\ket{\delta}+e^{ikb}cos(90-\delta)\ket{\delta}\\
\end{align*}

7. Apply an unknown angle might be trocky, because we need to

\begin{align*} \ket{\psi_6} &= \frac{1}{\sqrt{2}}\Big[\frac{1}{2}e^{ikl}\big(e^{ika}cos(-45)sin(\gamma)\ket{\diag}_1\ket{\gamma}_2 +e^{ikb}cos(45)cos(\gamma)\ket{\diag}_1\ket{\gamma}_2\big)\Big] \sqrt{2}\ket{C'}_1 \otimes \frac{1}{\sqrt{2}}\ket{D'}_2\\\end{align*}

8. Finaly, the photons will travel to the detectors.

\begin{align*} \ket{\psi_7}&= \frac{1}{\sqrt{2}}\Big[e^{ikl}\big( e^{ika}cos(-45)sin(\gamma)\ket{\diag}_1\ket{\gamma}_2 +e^{ikb}cos(45)cos(\gamma)\ket{\diag}_1\ket{\gamma}_2 \big)\Big] \otimes \frac{1}{\sqrt{2}} (\ket{X}_1 +\ket{W}_2)\\\end{align*}

9. The final step to solve get to pattern of the coincidence, we need to take the complex conjugate of this equation. To do a complex conjugate, you take the result of this last equation and you multiply it by itself, exept all |⟩ becomes ⟨| and vice-versa and all ei becomes e−i.

\begin{align*} I &= \frac{1}{8}\Big[sin^{2}(\gamma)\braket{\diag|\diag}_1 \braket{\gamma | \gamma}_2 + e^{ika}e^{-ikb}sin(\gamma)cos(\gamma)\braket{\diag|\diag}_1 \braket{\gamma | \gamma}_2 \\+ &cos^{2}(\gamma))\braket{\diag | \diag}_1 \braket{\gamma | \gamma}_2 + e^{-ika}e^{+ikb}sin(\gamma)cos(\gamma)\braket{\diag|\diag}_1 \braket{\gamma | \gamma}_2 \Big] \otimes (\braket{X | X}_1 +\braket{W | W}_2)\end{align*}\begin{align*} I &= \frac{1}{8} \Big( sin^{2}(\gamma) + cos^{2}(\gamma) + sin(\gamma)cos(\gamma) \big( e^{ik(a-b)} + e^{-ik(a-b)} \big) \Big)\end{align*}

10. This last equation is our interference pattern. We see that if γ is either 0 or 90, the term with the exponantial will be null. Which means no interference. Anything else will have interference. To have the final equation for the pattern, we need to multiply by the envelop of the diffration pattern. [ref]. Which give the final result:

\begin{equation*} \text{Pattern} = \frac{1}{8} \Big( sin^{2}(\gamma) + cos^{2}(\gamma) + sin(\gamma)cos(\gamma) \big( e^{ik(a-b)} + e^{-ik(a-b)} \big)\Big) \times \frac{sin(\alpha)}{\alpha}^2
\end{equation*}

Litterature and video

If you want to check out some papers that did the same experiment, here are the two best ressources we have found.

If you want a more interactive ressource, here are some Youtube video we found to be accurate and insightfull.

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AbstractLab ExperimentsResult and analysisTheory and equationsLitterature and video