The Quantum Eraser

The key differences between quantum and classical mechanics is superposition. The granddaddy of superposition experiments is the double slit experiment with a single particle source. If we place polarizers in front of the slits such that at detection time we can measure the polarization and determine through which slit did the photon went, then the interference pattern is lost. The idea of the quantum eraser is to recover the interference pattern by erasing the "which way information". Today I want to talk in-depth about the quantum description of the experiment. We will see that in the actual mathematical description the things are quite straightforward.

The experimental setup is as follows:

A laser produces one photon at a time and this photon is passed through a Beta Barium Borate (BBO) crystal where by parametric down conversion it is converted into two photons entangled in a Bell state:

$$|\psi\rangle = \frac{1}{\sqrt{2}}(|x\rangle_s|y\rangle_i + |y\rangle_s|x\rangle_i)$$

where $$s$$ stands for signal, and $$i$$ stands for idler.

Here we need to introduce a few notations:
• $$|x\rangle$$ photon polarized horizontally,
• $$|y\rangle$$ photon polarized vertically,
• $$|+\rangle$$ photon polarized at 45 degrees
• $$|-\rangle$$ photon polarized at -45 degrees
• $$|L\rangle$$ photon left circular polarized
• $$|R\rangle$$ photon right circular polarized
and the relations between them:

• $$|x\rangle = \frac{1}{\sqrt{2}}(|+\rangle + |-\rangle)$$
• $$|y\rangle = \frac{1}{\sqrt{2}}(|+\rangle - |-\rangle)$$
• $$|R\rangle = \frac{1-i}{2}(|+\rangle + i|-\rangle)$$
• $$|L\rangle = \frac{1-i}{2}(i|+\rangle + |-\rangle)$$

After passing through the double slit, the wavefunction becomes:

$$|\Psi\rangle = \frac{1}{\sqrt{2}}(|\psi\rangle_1 + |\psi\rangle_2)$$

where

$$|\psi\rangle_1 = \frac{1}{\sqrt{2}}(|x\rangle_{s1}|y\rangle_i + |y\rangle_{s1}|x\rangle_i)$$
$$|\psi\rangle_2 = \frac{1}{\sqrt{2}}(|x\rangle_{s2}|y\rangle_i + |y\rangle_{s2}|x\rangle_i)$$

If we add quarter wave plates after the slits to convert the linear polarized photons into circularly polarized ones $$|\psi\rangle_1$$ and $$|\psi\rangle_2$$ become:

$$|\psi\rangle_1 = \frac{1}{\sqrt{2}}(|L\rangle_{s1}|y\rangle_i + i |R\rangle_{s1}|x\rangle_i)$$
$$|\psi\rangle_2 = \frac{1}{\sqrt{2}}(|R\rangle_{s2}|y\rangle_i - i|L\rangle_{s2}|x\rangle_i)$$

We see that $$|\psi\rangle_1$$ is orthogonal with $$|\psi\rangle_2$$ and that there is no interference possible. However, hidden in the no-interference photon distribution curve there are two interference patterns called "fringe" and "anti-fringe". Can we extract this pattern from it? Indeed we can and to see how we need to rewrite $$|\Psi\rangle$$ using a different basis:

$$|\Psi\rangle = \frac{1+i}{\sqrt{2}} \frac{1}{2}[(|+\rangle_{s1} - i|+\rangle_{s2})|+\rangle_{i} + i (|-\rangle_{s1} + i |-\rangle_{s2})|-\rangle_{i}]$$

You can try to do the simple but tedious algebaic manipulations to convince yourself that the equation above is indeed the same thing as the two equations before it.

Now if we place a  polarizer in the path of beam $$i$$ orientating it at +45 degrees to select $$|+\rangle_p$$ or at -45 degrees to select $$|-\rangle_p$$ then we get the fringe or the anti-fringe interference patterns. The experimental outcomes are not erased, only the interference pattern is extracted by using coincidence detection using the idler signal. What this means is that for each detector hit of the signal photon we know whether the idler photon has the 45 degree polarization or not based if it passed through the +45 degrees polarizer or not. Selecting only the ones for which the idler photon pass or did not pass recovers the fringes.

By  playing with the signal and idler path lengths one can make one photon to be detected before the other at will. If the idler is detected after the signal we have what is called "delayed erasure".

There are a lot of nonsense explanations about the meaning of the quantum eraser and there are a lot of baseless speculation about it like "erasing the past", but if you look at the math the explanation is straightforward. Granted, the name "quantum eraser" is a genius marketing ploy to get people excited about quantum mechanics.

On my end I wanted to present the math of this example because I will use it in the next post to look at it through the eyes of the equivalence relationship and see what it can teach us about the Grothendieck approach for solving the measurement problem. We will learn two lessons: contextuality and irreversibility. Please stay tuned.

2 comments:

1. Buna ziua,

Scuze, dar ma voi adresa in limba romana, datorita intrebarii mele:
Exista echivalent in limba romana al expresiei "quantum eraser"?
Mai precis, exista documentatie de specialitate (in romana) care sa trateze subiectul, si ce denumire este utilizata?
O intrebare similara si pentru cuvantul "entanglement".

Multumesc si o zi buna,
Emil Negrea

2. Buna ziua,

Din pacate nu cunosc echivalentul in limba Romana si cred ca cel ai bine termenii trebuie tratati ca neologisme, adica luati direct din engleza. Sa incerc totusi sa ii traduc:

Pentru entanglement poate cel mai potrivit ar fi: amestec, amestecare. Quantum eraser este si mai greu de tradus: stergerea (cuantica a) istoriei.

Numai bine,

Florin
PS: anul trecut am avut o prezentare de fizica la Academia Romana si am avut dificultati majore sa o prezint in limba Romana tocmai din cauza dificultatii traducerii termenilor tehnici.