In this experiement, we will discover quantum entanglement and how it can be generated. Entanglement only means that one (or more) property of a particle are dependent of the other particle. Easy example, I have two marbles, one blue and one red. I give you one at random. Whenever you see the color pf your marble, you instantly know mine (if yours is blue, mine is red, and vice-versa). We could say their colors were entangled.
This experiment is interesting, because it seems to break a few laws in physics. It took many years before understanding entanglement, but we can assure you that entanglement doesn't make information go faster than the speed of light and cannot be use to change the past.
To create entanglement, we have two thin BBO crystals that entangles the photon's polarization. The two photons are emit at the same angle from the BBOs. We will add rotatable polarizers on each path and count the intensity of the photon we detect.
The BBO emits photons with parallel polarization. If one is vertical, so does the second. Have a try with the animation and try to understand why the intensity gets higher or lower with different combinasion of angles. You can change all polarizer's angle with the pink dots. You can also help yourself with the preset, they are the most intersting angles combination.
The prove you the animation is not lying, we built the setup in a lab and got those results. Those plots show all the angle's combination at the same time.
On the left side are the data we took in the lab. On the right side are theorical data. We can see a huge difference between both, but it is fine. When both polarizers are at the same angle, we should get a maximum of counts, because we are in a parallel Bell state. When they are at 90 degree, it is when we get no counts at all.
The fact we don't get a that great graphic is because our laser is probably not in the right state before creating the entangled pair. The Bell state will be out of balance. On the graph, we see we have way more pair varticaly polarized then horizontaly. We will see that imbalance following us for all the other experiment, but at least we no why.
Curious of what and entanglement optical setup looks like? Here is a few picture of what we have:
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Here is the representation of the setup you see in the picture. All of the equipement is shown, but the scaling is not representative.
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The setup starts on the top right corner with the laser emiting at 404 nm. Before reaching a beam splitter, the beam passes through a half-wave plate to make sure it is not one-sided polarized and tomodify the intensity of the beam. The beam splitter ensure the beam has a vertical polarization. The horizontal part is reject with the beam dump.
After, the beam passes in a set of a collimator (two lenses that reduce the size of the beam and make sure it is parallel). Thn, the set of two miror only place the beam were we want it to be. We did a U-turn only because we were out of space on the optic table. The third lens focalize the beam on the BBO, to ensure all photons are created at the same place.
Finally, we need to adjust the phase and polarization of the beam to be sure it is at 45 degrees with no phase. That is why we use a half-wave plate and a berek compensator.
The two BBOs creates pairs of photons in the state: $\ket{\psi_1} = \frac{1}{\sqrt{2}} (\ket{\uparrow}_1\ket{\uparrow}_2$.
Both path 1 and 2 only have a lens focalizing on the detector and a rotatable polarizer right in front of it. To rotate the polarizer, we are using a motor attached to the detector.
The last part is not shown in the drawing, but we need a way to recombined the 2 signals and count each time the detectors detected something. We call that a
The detectors we use are called single photon detector. Their mount are not shown in this setup, but the detector are linked to an optical fiber, which need a collider and a filter also attached to it. The detectors count the number of photon they see for a determine time. We move the detector D1 to count the number of photon at different position to recreate the full pattern. D2 doesn't need to move because we shoot a straight beam to it, no diffraction or interference.
You can download all our data here:
This file contains 5 columns, The first two are the angle on the polarizers. We go from -90 degrees to 90 degrees. 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 1 second window.
Let's take the time to analyse those graphics. As we said before, there is a huge gap between the theorical map and the experiment one, but it won't affect to much our result going forward. We still observe the same tendancy, tho. When the angles are parallel, we have a lot of counts and when we are perpendicular, we get something close to zero. The only part that is not following the theory is that we see way more counts above zero than bellow.
This plot reprensent the polarization state we are in. We are creating pairs in the Bell state $\ket{\psi_1} = \frac{1}{\sqrt{2}} (\ket{\updownarrow}_1\ket{\updownarrow}_2$. We could have choosen differently. We could represent the state by the polarization and phase of the beam. For example, $\ket{\psi_1} = cos(\theta)\ket{\updownarrow}_1\ket{\updownarrow}_2 + e^{i\phi}sin(\theta)(\ket{\leftrightarrow}_1\ket{\leftrightarrow}_2$, with $\theta$ being the polarization of the beam and $\phi$ the phase. The goal to be in the perfect state is to have $\theta = 0\pi $ and $\phi = \pi/4$. Instead, we have $\theta = 0.30\pi$ and $\phi = \pi/4.9$. With the Berek compensator, we should be able the get better result, but we couldn't somehow...
In the next sections, we will cover all the maths using the Dirac notation to find the final equation. (if you don't know the Dirac notation yet, all the notion needed as been covered in the two previous experiment).
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Full approach
1. Let's start with the initial state
2. We transform it into the right Bell state depending on the crystal used.
3. We apply the polarizers in respect to their path. In this scenario, we don't need to mind about the distance each photon is travelling, because it is the same for both path and no phase shift we take place.
4. Finaly, we want to find the intensity of the coincidence, which is done by taking the complexe conjugate of the last 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 (which doesn't apply here).
5. We can use some trigonometric indentities to simplify the last equation and we get our final answer. We can see easily that perpendicular angle give 0 and parallel angle will be maximum.
Note: the equations for intensity should always be normalized at 1. In this case, we see the highest value we can have is 1/2. The reason is because we have entangle photon. They can all be $\ket{\updownarrow}$ or $\ket{\leftrightarrow}$. If both polarizers are at 0 degrees, only half the photon will have the requirement to pass throught the polarizers, so the intensity is 1/2. If they are both at 45 degree, they all have 50% chance to pass, the intensity is again 1/2. There is no case were it is possible to have an intensity of 100% in this setup.
