Quantify Quantum Eraser Experiment

[glider77]

TL;DR Summary

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!

 

[Nugatory]

... in a book I read.

Which book? It is difficult to give a good answer if we don’t know what you’ve been reading.

 

[glider77]

Thanks for the reply. The book is "How to Teach Quantum Physics to Your Dog" by Chad Orzel. I've seen a few references to it on this forum.

 

[PeroK]

Summary:: 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!

You're using classical probabilities here. QM uses complex probability amplitudes, which are essentially the reason for quantum interference. Classical (real, positive) probabilities cannot cancel each other out, hence a classical particle cannot exhibit interference. Your whole line of reasoning, therefore, is based on a misconception.

I seriously doubt that any dog can learn QM.

 

[Cthugha]

Indeed, when it is fully sufficient to look at what classical waves do, when trying to consider how much light will arrive at the detector. 50% of the light at slit one will make it through the polarizer there and 50% of the light at slit two will make it through the polarizer there. Of all the light that makes it through the polarizers at the slits, another 50% will make it through the final polarizer. So when you start with a total of 2I, 0.5I will make it to the detector. All of this light has the same polarization, so you will ideally see full interference of all the light arriving at the detector.

I seriously doubt that any dog can learn QM.

Well, there was a famous PRL paper (Two-, Three-, and Four-Atom Exchange Effects in bcc 3 He) with the last author being a cat (F.D.C. Willard) and Nobel prize winner Andre Geim published a paper together with his hamster (Detection of Earth rotation with a diamagnetically levitating gyroscope). It might be a mistake to dismiss the potential of dogs so quickly. ;)

 

[glider77]

I figured I was barking up the wrong tree. My results, where some light that demonstrates interference mixes with light that doesn't demonstrate interference, doesn't match any of the images I've seen in books or on YouTube.

 

[glider77]

Not trying to beat a dead horse (or dog) here, just trying to figure out where the break occurs between my understanding of classical probabilities and QM. I agree that my reasoning is based on misconception, I just don't know how far back I need to go to relearn.

Let's look at the case where photons are sent one at a time, like particles, and see where I diverge.
For brevity's sake, let's call the polarizer in front of slit 1 p1 and the polarizer in front of slit 2 p2.
Assumptions:
1. Absent the polarizers, all photons (the ones we talk about anyway) will go through both slits and demonstrate interference.
2. To demonstrate interference, a photon must pass through both slits and, therefore, both p1 and p2.
3. Since p1 and p2 are physically different devices, a photon can pass through one without passing through the other. In other words, each polarizer 'discards' some photons, but not necessarily the same photons.
Classical particle approach:
1. Since each photon has a 50% chance of getting through each polarizer, it has a 25% change of getting through both polarizers. In other words, each photon has four possible outcomes:
not p1, not p2
p1, not p2
not p1, p2,
p1, p2
2. Only one of the four cases meets the requirement for interference.
3. Two of the other cases let a photon through, but don't meet the requirement for interference.

At this point, I think I've already broken from QM. If so, where?

 

[PeroK]

Try Feynman's lecture:



Or, Scott Aaronson:

https://www.scottaaronson.com/democritus/lec9.html

 

[PeroK]

Let's look at the case where photons are sent one at a time, like particles, and see where I diverge.
For brevity's sake, let's call the polarizer in front of slit 1 p1 and the polarizer in front of slit 2 p2.
Assumptions:
1. Absent the polarizers, all photons (the ones we talk about anyway) will go through both slits and demonstrate interference.

No.

2. To demonstrate interference, a photon must pass through both slits and, therefore, both p1 and p2.

No.

At this point, I think I've already broken from QM. If so, where?

One key point about QM is that:

a) The state of a quantum particle or system evolves. You need to understand the concept of a state first. In particular, a state can be a superposition of two other states.

b) We can only talk about particles definitely doing things if we measure them. If we don't measure them, then we cannot say (in fact, we cannot even ask) what the particles did.

This is the basis of QM thinking. You need to forget the classical concept that particles do definite things (in accordance with classical laws of motion) even when unmeasured.

 

[glider77]

Thanks for the Feynman lecture. I read a book that depicts that lecture to a high degree, including the distribution curves. And I kind of get it - now. But I don't understand the contribution of the front end polarizers in the quantum eraser experiment. Each of the polarizers (p1 and p2) block one slit half the time, sort of like Feynman describes when talking about electrons. With one slit blocked, alternating slits, N₁₂ = N₁ + N₂ i.e. not interference. I'm thinking that having both slits closed half the time is equivalent to both slits being simultaneously open only 25% of the time. And that would mean that some light would demonstrate interference and some wouldn't. Sort of like Feynman's light source behind the slits, if there are very few photons to show you which slit the electron went through, you get interference. As you turn up the light, you get less interference and more particle like distribution. In which case, P1 and P2 act like a light source tuned up to where you don't see 25% of the electrons.

 

[PeroK]

Let's try to describe this scenario using QM. There are three sections in the experiment:

1) Before the polarised slits: we have a certain intensity of incident light. Each incident photon will have an initial state with an unknown polarisation state:$$|\psi \rangle = |\psi_0 \rangle \otimes |P_0 \rangle$$

2) Between the polarised slits and the oblique polariser: we have a reduced intensity of light; or, a reduced number of photons (as some have been lost to the experiment). The state of a photon in this phase is a superposition of two states, representing each slit as a source and with a given polarisation. We can write this as: $$|\psi \rangle = \frac{1}{\sqrt 2} \big (|\psi_1\rangle \otimes |V \rangle + |\psi_2 \rangle \otimes |H \rangle \big )$$ Now the critical point is that this state does not give rise interference because interference requires states of the same polarisation.

Heuristically, we either get a vertical polarised photon or a horizontally polarised photon interacting with the screen.

If, however, we add a third oblique polariser:

3) After the oblique polariser: we have the same intensity of light, but the state can be represented as:
$$|\psi \rangle = \frac{1}{\sqrt 2} \big (|\psi_1\rangle \otimes |O \rangle + |\psi_2 \rangle \otimes |O \rangle \big )$$ And now we do get interference, as we have two states of the same polarisation.

Heuristically, we can interpret this as:

1) We have an intial incident photon of unknown polarisation.

2) If it gets through the slits, then the which-slit information is encoded in the polarisation: hence no interference.

3) The oblique polariser destroys the horizontal/vertical polarisation information and effectively we lose the which way information: hence we now have interference.

Note that any attempt to say precisely what each photon did in terms of the slit or slits it went through and the polarisation it had goes against the laws of QM in terms of state evolution. You must let the full state evolve throughout from source to screen and not assume that the photon must have taken one or other (or both) paths.

In short: in QM the state evolution replaces classical mechanics and classical trajectories.

 

[glider77]

Ok, this gives me an idea as to what I'm up against - new math. Thanks!

 

[PeroK]

Ok, this gives me an idea as to what I'm up against - new math. Thanks!

I don't know that there's a lot of maths in there - some fancy notation, perhaps - but it's more a new way of thinking about how an experimental setup produces a result. I can't overemphasise the importance of thinking about state evolution. It's the heart of QM.

 

[glider77]

I'm like a dog that won't let go of a bone. Let me try a different approach.

A two slit experiment using particles, say electrons.
In front of each slit is a device that randomly covers the slit half the time. We can't predict when a slit is open, but we know it's open 50% of the time.
Attached to each slit covering device is a recorder that tells us when the slit is open.
Lets call slit 1 S1 and slit 2 S2.
There are four cases for the slits:
Case 1 (C1): S1 closed, S2 closed
Case 2 (C2): S1 open, S2 closed
Case 3 (C3): S1 closed, S2 open
Case 4 (C4): S1 open, S2 open
The cases occur randomly, we can't predict them, but we know when they occur.
Case 1, an electron doesn't get through, we've lost 25% of our electrons.
Case 2 and Case 3, an electron gets through, but we know which slit it came through so the distribution is standard, N₁₂=N₁+N₂, and there's no interference. So 50% of our electrons don't demonstrate interference.
Case 4, an electron gets through and we don't know which slit it came through, so 25% of our electrons demonstrate interference.

At this point, 75% of the electrons got through (made it to the detector) and 25% demonstrated interference.

Questions:
1. Is the last statement correct?
2. What happens if we turn off the recorders on the slit covering devices and can no longer discern Cases and can't tell which slit an electron came through?
I'm thinking that because we could have known had we turned on the recorder, nothing changes.
3. Going back to the quantum eraser experiment that I originally referenced in this thread, how is having two polarizers, each of which absorbs 50% of the light presented, different from the two slit covering devices described in the above experiment?

 

[PeroK]

I'm like a dog that won't let go of a bone. Let me try a different approach.

A two slit experiment using particles, say electrons.
In front of each slit is a device that randomly covers the slit half the time. We can't predict when a slit is open, but we know it's open 50% of the time.

You're talking here about a classical device that follows classical probabilities. In this case you have a number of separate quantum experiments (depending on precisely the options for covering the slits), each of which must be analysed separately.

This is not likely to help understanding QM as this extra complexity is not QM in nature.

 

[glider77]

I hope to gain some understanding of QM. I've read a couple of entry level books and both stress the importance of understanding the quantum eraser experiment.

Feynman's lecture described a two slit experiment where both classical probabilities and probability amplitudes were demonstrated. A light source behind the slits is used to detect which slit an electron passes through. If the light source is very dim, few photons are emitted and no electrons are detected. In which case probability amplitudes describe the distribution. As the light source is increased, more electrons are detected and the detected electrons follow classical probabilities. If all the electrons are illuminated, the distribution is completely classical N₁₂=N₁+N₂. So it is possible to have a single experiment with both QM probabilities and classical probabilities. And I was thinking that the quantum eraser experiment should demonstrate both types of probabilities and that there should be a classical solution for the classical probabilities. In Feynman's experiment, if you detect half of the electrons then the distribution for those electrons is classical. The remaining distribution is QM. Adding the two distributions does not equal the total number of electrons.

I think I was looking at it backwards. My cases were based on photons passing through the polarizers. Because the polarizers are in front of the slits, I should look at the photons as being absorbed. Once a photon is absorbed, it's gone.
My new cases:
C1: S1 absorb, S2 n/a
C2: S1 n/a, S2 absorb
C3: S1 n/a, S2 n/a
In this scenario, 1/3 of the photons make through the front end polarizers and all of them will interfere if they pass through the final polarizer in front of the target. The final polarizer will absorb 50% of the photons that get to it, but that doesn't mean that 1/6 of the original photons will demonstrate interference because the number of photons entering the final polarizer is based on probability amplitudes, not 50% of the 1/3 that got through the front end polarizers.

Am I getting any closer?

 

[PeroK]

All that we are trying to understand here is that:

1) If we have different polarisers at the slits, then we do not get photon interference.

2) If we add a third polariser before the screen, then we get interference.

This is explained by the QM formalism. I don't really see the value in trying to look at it otherwise. I can't see what your explanation adds to the picture.

You can find essentially what I wrote in post #11 plus more detail here:

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Tutorials_(Rioux)/Quantum_Fundamentals/28:_The_Double‐Slit_Experiment_with_Polarized_Light

 

[glider77]

I'm not trying to add to the picture, only to correct my misunderstanding. After reading and rereading about the quantum eraser experiment I originally described, I didn't understand why there was no demonstration of classical distribution (non interference) after the final polarizer. So I tried to put numbers to it to see where I was going astray. I now believe it was because I was looking at a Feynman's detector behind the slits as being similar to polarizers in front of the slits. The polarizers limit what's available to the slits, but don't alter our knowledge of what gets through the slits so long as what gets through also goes through the final polarizer. That's all I was after. Hopefully my understanding is now less incorrect.

Thanks for the help!

 

[Nugatory]

1. Is the last statement correct?

Sort of, for some understandings of “demonstrate interference”. Every electron interacts with your double-slit-plus-shutters device, and this interaction is the preparation procedure for the quantum state of an electron after that interaction and before it interacts with the screen. We can use this state to calculate the probability of dots appearing at various points on the screen and some interference effects will appear in the resulting pattern. We cannot, however, attribute any particular dot to any particular state of the two randomly controlled shutters.

But note that this is a different preparation procedure than the traditional experiment. The state we’re preparing is a mixed state described by a density matrix, while the traditional double slit setup prepares a pure state that can described by a wavefunction that is a superposition of paths through the two slits.

What happens if we turn off the recorders on the slit covering devices and can no longer discern Cases and can't tell which slit an electron came through?
I'm thinking that because we could have known had we turned on the recorder, nothing changes.

Yes, it is irrelevant whether the recorders are on or off, but not for the reason you’re thinking. The absorption of a particle by a macroscopic shutter is a decoherent interaction; the result is not a quantum superposition of “went through” and “didn’t go through” but a classical “it did/did not happen”. Classical results are what they are, whether measured or not.

3. Going back to the quantum eraser experiment that I originally referenced in this thread, how is having two polarizers, each of which absorbs 50% of the light presented, different from the two slit covering devices described in the above experiment?

Because they are different preparation procedures. A photon that survives a 50% probability of being absorbed by a solid object is a different thing than a photon that has passed through a polarizing filter.

One of the mental barriers to understanding quantum mechanics is realizing that there really are no good classical equivalents of quantum systems. At some point we have to learn how the mathematical formalism works and start thinking in those terms.

 

[vanhees71]

All that we are trying to understand here is that:

1) If we have different polarisers at the slits, then we do not get photon interference.

2) If we add a third polariser before the screen, then we get interference.

This is explained by the QM formalism. I don't really see the value in trying to look at it otherwise. I can't see what your explanation adds to the picture.

You can find essentially what I wrote in post #11 plus more detail here:

https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Tutorials_(Rioux)/Quantum_Fundamentals/28:_The_Double‐Slit_Experiment_with_Polarized_Light

Yes, and you shouldn't start with such a pretty complicated example as the quantum eraser but simply with polarization experiments with single-photon states to start with quantum mechanics (though photons themselves are the most complicated example you can start with; maybe it's better to start with polarization/aka Stern-Gerlach experiments of some spin-1/2 particle).