Experiment 5

Putting it
all togheter

Abstract

In this section, we will work toward experiement 5: the delayed choice quantum eraser. We want to continue the built from experiement 4 until we get the delayed choice quantum eraser. We will take the previous setup and add the double slit and the polarizer on path 1. We wil also make the detector from path 1 move to recreate diffraction and interference patten.

This section will give you some challenges regarding entangle state. We will face one of the most confusing part of entanglement. Did you know that if we reproduce the exacte same experiemnt as experiement 3???, but with a pair of entangled photon, we will completly change the result of the experiement. It means that entanglement change the way a photon acts in a certain experiment. Even if you did know it was entangled in the first place.

In this section, we will add all the componant one at the time and see how it affect the final pattern. We will go strait up to the result, so for any questions on the setup, please refer to experiment 4.

We split the development in 3 experiement. In the first, 5.1,we only add a double slit. In 5.2, we will add to the double slit two polarizer right in front of each one. And in the final step, we will add a third polarizer to combine the wave lenght.
Try it first with this animation, and than check out the results!

Exp 5.1 : Adding the double slit

We start with the samesetup as in experiment 4, but we remove the two polarizers in front of the detectors. They were there to verify our polarization state. We also put the detector D1 on a linear actuator, because we want to recreate a full pattern.

Graphics

No polarizers

On both graphics, we can clearly see an interference pattern. The right-side graphic has way less counts than the left one because it is only the coincidence that are counted. When we say coincidence, we mean that D1 and D2 detected a photon at the same time. And when we say at the same time, we mean in an interval smaller than 10ns.

At this point, results are pretty obvious, a double slit creates interference.

Raw data

The data we acquired are all avalible here:

This file contains 6 columns, The first two are the angle on the polarizers, which is not applicable here. The third one is for the distance of the actuator. The forth is the sweep number. The last three columns are the counts on the detector. The first for D1, the second for D2 and the third is the coincidences between D1 and D2.
All the counts are taken in a 20 seconds window.

Exp 5.2 : Covering the slit with polarizers

We start with the samesetup as in experiment 4, but we remove the two polarizers in front of the detectors. They were there to verify our polarization state. We also put the detector D1 on a linear actuator, because we want to recreate a full pattern.

Graphics

The doouble slit is now polarised (Pa and PB)

We now see we lost both interference pattern. As we did in experiement ??? when we place perpendicular poalrizer on each slit, the photon cannot interact no more to create interference. This is why we only get diffraction. We also see a huge decrease in the number of counts because the polarizers blocks a big number of photon.

Raw data

You can download all our data here:

This file contains 6 columns, The first two are the angle on the polarizers, which is not applicable here. The third one is for the distance of the actuator. The forth is the sweep number. The last three columns are the counts on the detector. The first for D1, the second for D2 and the third is the coincidences between D1 and D2.
All the counts are taken in a 20 seconds window.

Exp 5.3 : Adding a third polarizer

We start with the samesetup as in experiment 4, but we remove the two polarizers in front of the detectors. They were there to verify our polarization state. We also put the detector D1 on a linear actuator, because we want to recreate a full pattern.

Graphics

Three polarizers PA + PB + P1

In this graph, we still see no interference... This is weird...

When we did the same experiemnt without entangled photons, in experiement 3, we did observe interference... and now we don't...

When we cover the double slit with polarizers, we get a way to know in which slit the photon want. We cannot acces this information yet, but it is still carried in the upper path photon. We will acces the information in the next experiement by placing a polarizer on the upper path. But still, for now, only the potential of knowing in which slit the photon went is enough to loose the interference.

It can actually make sens. Let's say we would see interference at this point, I could use a super still non-existant microscope to check the molecule interaction in the detector and figure out if the photon caught's polarization. Which mean we would change the result of something that happened in the past. It would break most of physics laws.

The result: even if we don't have any technologies yet to acces the information, the fact that the information exist somewhere is enough to loose interference. Let's remind that by saying "loosing interference" we mean that each slit act as a single slit and the pattern we see is the sum of the diffraction of each.

Raw data

You can download all our data here:

This file contains 6 columns, The first two are the angle on the polarizers, to ignore for P2. The third one is for the distance of the actuator. The forth is the sweep number. The last three columns are the counts on the detector. The first for D1, the second for D2 and the third is the coincidences between D1 and D2.
All the counts are taken in a 20 seconds window.

Mathematics and equations

Full approach

1. Let's start with the initial state

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

2. We transform it into the right Bell state depending on the crystal used.

\begin{align*} \ket{\psi_1} &= \frac{1}{\sqrt{2}}(\ket{\updownarrow}_1\ket{\updownarrow}_2+ \ket{\leftrightarrow}_1\ket{\leftrightarrow}_2) \otimes \frac{1}{\sqrt{2}}(\ket{S}_1+ \ket{S}_2)\\\end{align*}

3. We let the photons travel to reach either one of the slit. In both case, the distance it travels is l. We need to change the phase of the photon in both path of that distance.

\begin{align*} \ket{\psi_2} &= \frac{1}{\sqrt{2}}\Big(\ket{\updownarrow}_1\ket{\updownarrow}_2+ \ket{\leftrightarrow}_1\ket{\leftrightarrow}_2\Big) \otimes \frac{1}{\sqrt{2}}\Big(\frac{1}{\sqrt{2}}e^{ikl}(\ket{A}_1 + \ket{B}_1)+ \ket{S}_2\Big)\\ &= \frac{1}{\sqrt{2}}\Big[\frac{1}{2}e^{ikl}\big( \ket{\updownarrow}_1\ket{\updownarrow}_2\ket{A}_1+ \ket{\leftrightarrow}_1\ket{\leftrightarrow}_2\ket{A}_1 + \ket{\updownarrow}_1\ket{\updownarrow}_2\ket{B}_1+ \ket{\leftrightarrow}_1\ket{\leftrightarrow}_2\ket{B}_1 \big)\Big] \otimes \frac{1}{\sqrt{2}}\ket{S}_2\\\end{align*}

4. For experiment 5.1, the photons now go strait up to the detector. We need to add the phase shift in regard to their distances to the detector. Then, we take the complex conjugate

\begin{align*} \ket{\psi_{5.1}} =& \frac{1}{\sqrt{2}}\Big(\ket{\updownarrow}_1\ket{\updownarrow}_2+ \ket{\leftrightarrow}_1\ket{\leftrightarrow}_2\Big) \otimes \frac{1}{\sqrt{2}}\Big(\frac{\sqrt{2}}{\sqrt{2}}e^{ikl}(e^{ika}\ket{C}_1 + e^{ikb}\ket{C}_1)+ \ket{S}_2\Big)\\
I =& \frac{1}{2}\big(1+1 \big) \otimes \frac{1}{2} (e^{ikl}\times e^{-ikl})\Big( \big( e^{ika} + e^{ikb}\big)\big( e^{-ika} + e^{-ikb}\big)\braket{C} + \braket{S}\Big)\\
I =& 1 \otimes \frac{1}{2}\Big( (1 + e^{ik(b-a)} + e^{-ik(b-a)} + 1)\times 1 + 1\Big)\\
I =& \frac{1}{2}(3+2cos(k(b-a)))
\end{align*}

5. We continue on the second path by adding the 2 polarizer at 0 and 90 degrees.

\begin{align*} \ket{\psi_3} &= \frac{1}{\sqrt{2}}\Big[\frac{1}{2}e^{ikl}\big( sin(0)\ket{\leftrightarrow}_1\ket{\updownarrow}_2\ket{A'}_1+ cos(0)\ket{\leftrightarrow}_1\ket{\leftrightarrow}_2\ket{A'}_1 \\ +sin(90)&\ket{\updownarrow}_1\ket{\updownarrow}_2\ket{B'}_1+ cos(90)\ket{\updownarrow}_1\ket{\leftrightarrow}_2\ket{B'}_1 \big)\Big] \otimes \frac{1}{\sqrt{2}}\ket{S}_2\\\end{align*}

6. We add the distance needed to reach the point C. We need to consider that photons from both slit won't travel the same distance. Here, we cheat just a little bit, I'll add right away all the distance to reach the detector instead of adding it in two step. The reason is because it depends on the position of P1, but we could place it anywere and it won't change the result. So, let's say that the photons that went thourgh slit A travels a ditance a and the oneform slit B travel a distance b.

\begin{align*} \ket{\psi_4} &= \frac{1}{\sqrt{2}}\Big[\frac{1}{2}e^{ikl}\big( e^{ika}\ket{\leftrightarrow}_1\ket{\leftrightarrow}_2 +e^{ikb}\ket{\updownarrow}_1\ket{\updownarrow}_2 \big)\Big] \sqrt{2}\ket{C}_1 \otimes \frac{1}{\sqrt{2}}\ket{S}_2\\\end{align*}

7. Here we have everything for experiment 5.2. Let's take the conjugate complex. We find the intensity is a quarter, it still needs to be multiply by the diffraction, which says the diffraction is cut by 4 when adding the two polarizers.

\begin{align*} \ket{\psi_{5.2}} &= \Big[\frac{1}{2}e^{ikl}\big( e^{ika}\ket{\leftrightarrow}_1\ket{\leftrightarrow}_2 +e^{ikb}\ket{\updownarrow}_1\ket{\updownarrow}_2 \big)\Big] \sqrt{2}\ket{C}_1 \otimes \frac{1}{\sqrt{2}}\ket{S}_2\\
I =& \frac
{1}{4}(e^{ikl}e^{-ikl})\Big( e^{ika}e^{-ika} \braket{\leftrightarrow}_1 \braket{\leftrightarrow}_2+ e^{ika}e^{-ika} \braket{\leftrightarrow}_1 \braket{\leftrightarrow}_2 \Big)\braket{C}_1 \otimes \frac{1}{2}\braket{S}_2\\
I = \frac{1}{4}(1+1)1 \times \frac{1}{2}\\
I = \frac{1}{4}
\end{align*}

8. Let's go back and apply polarizor P1 of 45 degrees.

\begin{align*} \ket{\psi_{5}} &= \frac{1}{\sqrt{2}}\Big[\frac{1}{2}e^{ikl}\big(e^{ika}cos(0-45)\ket{\diag}_1\ket{\leftrightarrow}_2 +e^{ikb}cos(90-45)\ket{\diag}_1\ket{\updownarrow}_2\big)\Big] \sqrt{2}\ket{C'}_1 \otimes \frac{1}{\sqrt{2}}\ket{D}_2\\
\end{align*}

9. We now have finish experiment 5.3. Let's replace that and take the complex conjugate. We get as a final result the intensity is one eighth of the initial diffraction and that it is the half of the intensity with only two polarizers, which makes sens because to polarizer block half of the photon in that configuraton. This result proves we are not supposed to see interference with that setup.

\begin{align*} \ket{\psi_{5.5}} &= \Big[\frac{1}{2}e^{ikl}\big(e^{ika} \frac{1}{\sqrt(2)} \ket{\leftrightarrow}_2 +e^{ikb}\frac{1}{\sqrt(2)}\ket{\updownarrow}_2\big)\Big] \ket{\diag}_1\ket{C'}_1 \otimes \frac{1}{\sqrt{2}}\ket{D}_2\\
I =& \frac{1}{4} \Big( e^{ika}e^{-ika}\frac{1}{2}\braket{\leftrightarrow}_2 + e^{ikb}e^{-ikb}\frac{1}{2} \braket{\updownarrow}_2 \Big)\braket{\diag}_1\braket{C}_1 \otimes \frac{1}{2}\braket{D}_2\\
=& \frac{1}{4}(\frac{1}{2} + \frac{1}{2}) \otimes \frac{1}{2}\\
I = & \frac{1}{8}
\end{align*}

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