I tried to recreate the Quantum Eraser experiment into a thought experiment with a few changes. It left me a little perplexed as to what outcomes I should expect. Any help would be appreciated.

Lets say you have 2 entangled photons so that when one is blue the other is always red. You shoot the 2 particles in different directions without measuring the color of the photon in either direction. Each particle, P1 and P2, hit a corresponding measuring device labeled D1 and D2. D1 only will measure the color of P1 when it hits it. D1 can also be turned on or off. When D1 measures the color of P1, P2 collapses into either red or blue state based on the measurement of P1. I assume prior to measurement from D1, P2 would be in both a red and blue state.

Now lets say that D2 has a single slit in front of it (like you would see in the double slit experiment, just with one slit) that has a yellow film over it that could alter the color of the photon passing through it to either orange or green based on if the photon was red or blue respectively . I imagine that if D1 was turned on you would gather data on D2 of either an orange or green mark based on D1's measurement collapsing P2. However when D1 was turned off would the results on the wall at D2 change to brown dots as the photons would pass through the yellow film in both a red and blue state simultaneously?

Any help with this is greatly appreciated.

  • 2
    $\begingroup$ A filter does not normally work by redistributing the light into different frequencies. Rather, it selectively removes light depending on its frequency. Each filter is characterized by an absorption spectrum. It is only your eyes that perceive this as a shift in colour --- in reality, colour is a property of perception and a "blue photon" is not a very precise concept. $\endgroup$ – Nanite Aug 20 '14 at 7:41
  • $\begingroup$ Photon in superposition doesn't mean a little from each flavor. When it hits the filter ( Cf the Nanite comment ) or anything acting like a detector ( ie a dust ) , it is 'measured' and its superposition collapses. $\endgroup$ – user46925 Jun 2 '15 at 1:47

I'm going to try to teach you the right way to think about this, but possibly that will be very difficult to visualize. So I wanted to give you a starter course on what you're getting wrong.

What you're getting wrong

Your colors are indeed orthogonal states that can be measured differently. On your second screen you'll see green and orange light hit the detector there independently, there will be no real "quantum" thing happening there. Your entire system is deeply noninteractive and you're not going to see quantum effects until you get quantum-coherent interaction. Entanglement isn't interaction, though it is often caused by interaction. Once you have two entangled photons going in different directions, you can do lots of interesting things with them: but intrinisically they are just correlated in a certain magical, nonclassical way when you compare measurements made at A versus measurements made at B. In general, with entanglement, everything that happens looks completely explicable to both A and B until they come back together and compare their measurements against each other to find that something really strange has happened.

The Eraser-like experiments in qubit form

So I like to think of the quantum eraser experiment in terms of qubits and quantum logic gates. A qubit is a bit that isn't purely either $|0\rangle$ or $|1\rangle$ but can be some quantum superposition $\alpha|0\rangle + \beta|1\rangle$ with $|\alpha|^2 + |\beta|^2 = 1$ for normalization.

Each vector $|\Psi\rangle = \alpha|0\rangle + \beta|1\rangle$ has a corresponding "complement" or "one-form" $\langle \Psi| = \langle 0| \alpha^* + \langle 1| \beta^*$, where $x^*$ is the complex conjugate of $x$. If you have never seen complex conjugates, don't worry: all of our numbers will be real numbers, which are the same as their complex conjugates, so for this post, $\alpha = \alpha^*$. The distinction is important when you start doing more sophisticated things with quantum mechanics. Putting a one-form next to a vector is called their "inner product"; $\langle 0|1\rangle = \langle 1|0\rangle = 0$ while $\langle 0|0\rangle = \langle 1 |1\rangle = 1$.

Quantum mechanics predicts averages of operators. An operator transforms one state into another state, like the operator $|0\rangle\langle 1| - |1\rangle\langle 0|$ transforms our state $|\Psi\rangle$ above into $\beta |0\rangle - \alpha |1\rangle$. Then the inner product of this with $\langle\Psi|$ is going to be $\alpha^*\beta - \beta^*\alpha$, which is 0 if both $\alpha$ and $\beta$ are real numbers, but might be nonzero when they start having imaginary components.

In general, quantum mechanics takes any "Hermitian" operator $H$ and assigns an average $\langle \Psi| H |\Psi\rangle$ to it, for state $|\Psi\rangle$.

When we have multiple qubits we write them "together"; the state $|01\rangle = |0\rangle\otimes|1\rangle$ for example, it means "the first qubit is definitely 0 and the second qubit is definitely 1."

Double slit

We start the "double slit experiment". We have a qubit in the state $|+\rangle = \sqrt{1/2}~|0\rangle + \sqrt{1/2}~|1\rangle$ and we evolve these to some $|0\rangle \rightarrow |\psi_0(z)\rangle$ on a screen with coordinate $z$, and $|1\rangle \rightarrow |\psi_1(z)\rangle$, and then we observe the pattern $\frac 12|\psi_0(z) + \psi_1(z)|^2$, which is an "interference pattern". (This is because they can have a complex-oscillating form $\psi_{0,1} \propto e^{i\omega z}$, which can interfere with each other.) Call the operator which does this operation $\text{screen}(z)$.

Measured slit

We now talk about the "interrupted" double slit experiment, where we try to detect the information about the qubit's path after it goes down path $0$ or $1$. In this case we have a second qubit which starts in state $|0\rangle$, and we perform the gate $\text{CNOT}_{1\rightarrow 2}$ operation, "perform a NOT on the qubit #2 if the qubit #1 is 1, otherwise let it be." This is a perfectly good, well-known quantum gate, and after it acts we are in the state $$|\Psi\rangle = \sqrt{\frac 12} |00\rangle + \sqrt{\frac 12} |11\rangle.$$Whether we measure the second qubit or not doesn't matter; when we observe $\psi(z)$ alone analogously to the previous experiment, we tensor-product it with the identity matrix, doing nothing to the second qubit. As a result we see:$$\langle\Psi|\text{screen}(z)|\Psi\rangle = \frac 12\left(\langle\psi_00| + \langle\psi_11|\right)\left(|\psi_00\rangle + |\psi_11\rangle\right)$$but the condition that $\langle 0 | 1 \rangle = \langle 1 | 0\rangle = 0$ on that second qubit gives just:$$\langle\Psi|\text{screen}(z)|\Psi\rangle = \frac 12|\psi_0(z)|^2 + \frac 12|\psi_1(z)|^2,$$and the interference pattern has "vanished": we see the two slits overlap without any interaction between them.

So here's the first bit of nuance: often we have to be very careful with what we mean by "measurement collapses the state." When you do this CNOT operation, the interference pattern disappears. It doesn't matter if you start trying to measure the second qubit.

Quantum eraser

Now we start trying to send an entangled pair through the two slits. This requires three qubits: the "which-way" qubit is going to be distinct from the information that each of these qubits has entangled. So we might start off with $\sqrt{\frac 12}|01\rangle + \sqrt{\frac 12}|10\rangle$ and add on a $|+\rangle$ to get$$|\Psi\rangle = \frac 12 |010\rangle + \frac 12 |011\rangle + \frac 12 |100\rangle + \frac 12 |101\rangle.$$When we measure $\langle\Psi|z|\Psi\rangle$ we again see $\frac 12|\psi_0(z) + \psi_1(z)|^2$; the entanglement doesn't "interact with" the "which-way" qubit so that is still going to behave like it's in state $|+\rangle$. And we see this because those first two bits repeat: we see $01$ twice and $10$ twice. That is actually super-important: if it doesn't look like the double-slit experiment "by default", what are we doing?

Now, as you may have guessed, we try to $\text{CNOT}_{3\rightarrow 2} |\Phi\rangle$, so that the entangled pair contains some of the "which way" information. This launches us into the state:$$|\Psi'\rangle = \frac 12 |010\rangle + \frac 12 |001\rangle + \frac 12 |100\rangle + \frac 12 |111\rangle$$Those first two bits now don't repeat, and if you carry out the calculation, massive orthogonality brings us back to $\frac 12|\psi_0(z)|^2 + \frac 12|\psi_1(z)|^2$. So this destroys the interference pattern, again, just like you'd expect: it looks like the "measured slit" above.

Here's where we get a little crazy with the quantum eraser. We map the first qubit -- just the first qubit! -- with the invertible transform $$\begin{array}{c} |0\rangle\rightarrow\sqrt{\frac 12}|0\rangle + \sqrt{\frac 12}|1\rangle, \\ |1\rangle\rightarrow\sqrt{\frac 12}|0\rangle - \sqrt{\frac 12}|1\rangle.\end{array}$$This is a valid operation in quantum mechanics known as a "Hadamard gate". Unlike CNOT, it has no classical equivalent. When we manipulate the first qubit this way we get$$\sqrt 8 |\Psi''\rangle = |010\rangle + |001\rangle + |000\rangle + |011\rangle + |110\rangle + |101\rangle - |100\rangle - |111\rangle,$$which we can write as$$\sqrt 4 |\Psi''\rangle = |01+\rangle + |00+\rangle + |11-\rangle - |10-\rangle.$$Now we still, on that detector, measure $\frac 12 |\psi_0(z)|^2 + \frac 12 |\psi_1(z)|^2$. However, this is actually the sum of two interference patterns: $\frac 14 |\psi_0(z) + \psi_1(z)|^2$ and $\frac 14 |\psi_0(z) - \psi_1(z)|^2$. The first pattern coincides with the first qubit being $|0\rangle$ and the second pattern coincides with the first qubit being $|1\rangle$.

In your typical experiment, the two detectors for the photons are connected by a "coincidence" circuit and a "measurement" is being made here in the form of a polarizer which drops all of the photons which go through the Hadamard and measure $|1\rangle$. Because of this, people thought (and continue to think) that it's really cool that this sort of "postselection" can restore an interference pattern where there wasn't one before. But there is not nearly as much mystery as some people attach to it: in truth, a lot of the "mystery" comes from the coincidence circuit discarding half of the photons.


Mathematically you have a superposition of two Eigenstates and you use it because you don't know all parameters of the source. And yes, you use a probabilistic source, otherwise you will not do any experiment with it.

Using your special source you get always the next result: Measuring one of the photons you know immediately the Eigenstates of the twisted photon. Why? Because your source is made so that it produces two photons with two different colors. Look closer. It is a composed quantum dot artfully selected by us to get this result.

Does this means that the two particles after they where created are in superposition? Mathematically yes because we do not know all parametres of the quantum dot and a mathematician has no way not to act so. But that does not mean that the photons are not "finished".

I'm free to say this because the result of my thought is the same like the thought about the "spukhafte Fernwirkung" (spooky long range effect). Imagine the same experiment with two color balls. Put them into a box with two exits and two tubes. Only when one of the balls came out from one tube you will be able to "predict" the color of the second ball. The same is the mathematical model of superposition of Eigenstates but obvious it have nothing to do with some kind of spukhafte Fernwirkung.

  • $\begingroup$ and what do you answer to experimental physicists that tell you the opposite, ie the 2 color balls analogy is not pertinent ? there is no other choice than giving credit to experiments $\endgroup$ – user46925 Jul 13 '15 at 22:56
  • $\begingroup$ @igael If it is not possible to see some mechanism due to the influence of the measurement instrument, when different explanations of the phenomenon could be true. Let us discuss the mistake in my explanation. $\endgroup$ – HolgerFiedler Jul 14 '15 at 5:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.