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- Calculate the amount of light demonstrating interference in a 2 slit quantum eraser experiment

I'm trying to grasp a quantum eraser experiment as described in a book I read. It helps me to put numbers to things, so my goal is to calculate the amount (intensity) of the light that demonstrates interference.

The author describes a double slit quantum eraser experiment using polarized light:

Initial light is polarized to 45 degrees.

In front of each of the two slits is a polarizer, one slit has a vertical polarizer and the other slit a horizontal polarizer.

Interference is observed if a 45 degree polarizer is placed in front of the final target.

No interference is observed if a vertical or horizontal polarizer is placed in front of the final target.

To demonstrate interference, a photon has to be in superposition of slots (slot 1 and slot 2) and in superposition of polarity (horizontal and vertical). If I treat this as if the photon passes through both slits and both polarizers, I think it looks like the following:

1. Let the amount of light that could go through a single slit = I. This is determined by the source intensity and the area of the slit.

2. Since each slit has a polarizer at 45 degrees to the source, the amount of light that can go through a single slit is I*Cos^2(45) = .5I.

3. If I think of the .5 as a probability (50%) then the amount that goes through slit 1 AND slit 2, according to the addition law of probabilities, is .5*.5 = .25I.

4. The total amount of light, the light that goes through slit 1 OR slit 2, is .5+.5 = 1I.

5. So at this point, the total amount of light available to the final polarizer is 1I and the amount of light that can demonstrate interference is .25I.

6. After passing through the final polarizer, the total amount of light is 1I*Cos^2(45) = .5I and the amount of light demonstrating interference is .25I*Cos^2(45) = .125I.

So we started out with 2I (two slits), one quarter of that got through (.5I) and one quarter of that demonstrated interference (.125I).

Is this close? Or can I not treat superposition as having passed through both slots/polarizers?

Thanks in advance!

The author describes a double slit quantum eraser experiment using polarized light:

Initial light is polarized to 45 degrees.

In front of each of the two slits is a polarizer, one slit has a vertical polarizer and the other slit a horizontal polarizer.

Interference is observed if a 45 degree polarizer is placed in front of the final target.

No interference is observed if a vertical or horizontal polarizer is placed in front of the final target.

To demonstrate interference, a photon has to be in superposition of slots (slot 1 and slot 2) and in superposition of polarity (horizontal and vertical). If I treat this as if the photon passes through both slits and both polarizers, I think it looks like the following:

1. Let the amount of light that could go through a single slit = I. This is determined by the source intensity and the area of the slit.

2. Since each slit has a polarizer at 45 degrees to the source, the amount of light that can go through a single slit is I*Cos^2(45) = .5I.

3. If I think of the .5 as a probability (50%) then the amount that goes through slit 1 AND slit 2, according to the addition law of probabilities, is .5*.5 = .25I.

4. The total amount of light, the light that goes through slit 1 OR slit 2, is .5+.5 = 1I.

5. So at this point, the total amount of light available to the final polarizer is 1I and the amount of light that can demonstrate interference is .25I.

6. After passing through the final polarizer, the total amount of light is 1I*Cos^2(45) = .5I and the amount of light demonstrating interference is .25I*Cos^2(45) = .125I.

So we started out with 2I (two slits), one quarter of that got through (.5I) and one quarter of that demonstrated interference (.125I).

Is this close? Or can I not treat superposition as having passed through both slots/polarizers?

Thanks in advance!