## The contextuality of measurement

Now we have the machinery ready to put to the test the idea that measurement is a change of representation by distinguishing one Cartesian pair. In other words, the measurement process is a lifting of the representation degeneracy through equivalence breaking.

There is nothing special about being a measurement device, or an observer. All we need is to be able to establish the equivalence of the Cartesian pairs. Can we do this in the case of the quantum eraser?

After the parametric down conversion, the signal and the idler photons are entangled in a Bell state:

$$|\Psi\rangle = \frac{1}{\sqrt{2}} (|\phi_1(\mathbf{r})\rangle|M_1\rangle + |\phi_2(\mathbf{r})\rangle|M_2\rangle)$$

and measurement of the $$x$$ or $$y$$ polarization reduces $$|\Psi\rangle$$ to the appropriate state for passing through one or the other slit. In the quantum eraser scenario however, when we place the quarter wave plates, the point is to erase and recover the interference pattern and not to detect the which way'' information. But we can still borrow the idler photon designation as a measurement device and attempt to construct the Grothendieck Cartesian pairs and the corresponding equivalence relationships. We want to discuss the two cases: no-interference when we place a vertical or horizontal polarizer in front of the idler beam detector, and interference when we place the $$\pm45^0$$ polarizers.

In the no-interference case we can identify from above (and the prior post):

$$|\phi_1 \rangle = \frac{1}{\sqrt{2}} (|L\rangle_{s1} + |R\rangle_{s2})$$
$$|\phi_2 \rangle = \frac{1}{\sqrt{2}} (i|R\rangle_{s1} - i|L\rangle_{s2})$$
$$|M_1\rangle = |y\rangle_i$$
$$|M_2\rangle = |x\rangle_i$$

The equivalence is defined by the unitaries $$U_p$$, and $$U_n$$ such that:

$$U_p (|\phi_1 \rangle) |M_2\rangle = |\phi_2 \rangle U_n (|M_1 \rangle)$$

Such pairs of unitaries exist. For example they can be Pauli-Z gates.

Now onto the interference case. Here we make the following identifications:
$$|\phi_1 \rangle = \frac{1}{\sqrt{2}} (|+\rangle_{s1} + i|+\rangle_{s2})$$
$$|\phi_2 \rangle = \frac{1}{\sqrt{2}} (i|-\rangle_{s1} - |-\rangle_{s2})$$
$$|M_1\rangle = |+\rangle_i$$
$$|M_2\rangle = |-\rangle_i$$

In this case the equivalence is given by the unitaries $$V_p$$, and $$V_n$$ such that:

$$V_p (|\phi_1 \rangle) |M_2\rangle = |\phi_2 \rangle V_n (|M_1 \rangle)$$

Here $$V_p$$ could be the negative of Pauli-Y gate, and $$V_n$$ the Pauli-X gate

So far so good, but do we have an equivalence relationship across the two cases? This is the first test. If there is such an equivalence, then the Cartesian pair idea is too general to be of any value.

If we pick any index one wavefunction from one setup with any index two wavefunction from the other setup, then there are no $$U_p$$ unitaries such that the equivalence holds. For example there is no $$U_p$$ such that: $$U_p (|+\rangle_{s1} + i|+\rangle_{s2}) = |L\rangle_{s1} + |R\rangle_{s2}$$ because this relationship has to hold both for $$s_1 = s_2$$ and $$s_1 \ne s_2$$ and one precludes the other. As such the equivalence classes are disjoint in the two scenarios.

The result is general: it is not hard to show that any measurement configuration for which the statistics of the outcomes is different cannot share the Cartesian equivalence class. But this is nothing but contextuality of measurement. We get contextuality as a mathematical theorem.

Now above I stated that the measurement process is a lifting of the representation degeneracy through equivalence breaking. So can we undo the measurement by restoring the equivalence? Yes in theory, mostly no in practice. Let me explain.

If we are able to restore the equivalence and the information about the outcome did not propagate outside the experiment, then in theory the measurement outcome did not occur. However, the measurement outcome in quantum mechanics is inherently random due to operator non-commutativity. Then we have no means to control the process of restoring the equivalence relationship and the measurement outcome is hence irreversible. Measurement outcome irreversibility is a consequence of non-commutativity, not of information loss!

If we do not record the measurement outcomes we can restore the original state. This is a true quantum eraser. Here is how we can do it.

Suppose we have a source of vertically polarized electrons (say with spin up) and we pass this through a Stern-Gerlach device measuring the the y polarization. The beam splits into two and we get 50% electrons polarized on the y direction, and 50% with the opposite polarization.

Now suppose we pass each of the beams through a second S-G device measuring this time the vertical polarization. The original vertical polarization information seem to be lost as we get 50% up and 50% down on each beam. But now welcome to the marvel of superposition:

combine the two beams like in an interferometer  (effectively erasing the y-direction measurement information) and pass the combined beam through the vertical S-G device.

What do you get? Still 50%-50%? No, you get the original vertical polarization 100% of the time.

The explanation of what is going on in the Cartesian pairs formalism is trivial. When we recombine the beams (if there is no witness left in the environment) the distinguished Cartesian pair of the original vertical measurement is restored. Also the equivalence of the Cartesian pairs corresponding to the y measurement is restored as well because those Cartesian pairs acts as "which way" information markers and the "which way information" is no longer accessible.

The apparent information loss when we do not recombine the beams is a consequence of embedding different Hilbert spaces corresponding to different representations of different (but equivalent) Cartesian pairs into a single Hilbert space. More on this next time.

Did we solve the measurement problem? Not yet, but we are making good progress. Next time I'll talk about the von Neumann measurements of the first kind (also known as pre-measurements) and rigorously show they are mathematical nonsense. Please stay tuned.

## 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.

## Envariance and equivalence

I was having doubts about selecting the topic of this week. We had the exciting detection of gravitational waves and I was thinking to write about it, but then I decided not to break the logical sequence of the current series. I will do my best to make this post exciting...

So back to the measurement problem. When the wavefunction collapses, it projects to a subspace of the Hilbert space, and this means we need to use a basis. But the choice of a basis is arbitrary and we need a preferred basis. Do we have such a thing at our disposal? If we consider only the Hilbert space of the quantum system the eigenvectors come to mind, but the eigenvectors of which Hermitean operator? Since the measurement result does not exist before measurement, we need to consider the Hilbert space of the measurement device as well. And when we have two Hilbert spaces we have at our disposal the Schmidt decomposition which is unique when the coefficients are distinct. Now this decomposition depends on the original wavefunction, but if we remember last time, the original Cartesian pair contains a distinguished wavefunction as well: the wavefunction of the measurement device in the "ready" state, and so the Schmidt decomposition basis is indeed a preferred basis.

In the Grotendieck approach, we have pairs of wavefunctions which are physically indistinguishable: there is no physical device which can be constructed to distinguish one pair from another. (Why? Because quantum mechanics is probabilistic and not deterministic). What if I act with a unitary transformation on the physical system? Can I undo this operation by acting with a unitary transformation on the measurement device? This is not always possible but in the Schmidt basis there are such unitaries (which I will call Schmidt unitaries). If instead of the measurement device we consider the environment, this is Zurek's idea of envariance (environment assisted invariance):

“When a state $$|\psi_{SE} \rangle$$ of a pair system S, E can be transformed by $$U_S = u_S \otimes 1_E$$ acting soley on S, $$U_S |\psi_{SE}\rangle = (u_S ⊗ 1_E )|\psi_{SE} \rangle = |\eta_{SE}\rangle$$ but the effect of $$U_S$$ can be undone by acting solely on E with an appropriately chosen $$U_E = 1_S \otimes u_E : U_E |\eta_{SE} \rangle = (1_S \otimes u_E )|\eta_{SE} \rangle = |\psi_{SE}\rangle$$

I will not use directly Zurek's envariance, but this was a great inspiration for me on constructing the equivalence I need to solve the measurement problem.

Without ado here is my definition of the equivalence:

Suppose we have two wavefunctions $$|\psi_p \rangle, |\phi_p \rangle \in H_p$$ corresponding to a physical system, and another two wavefunctions $$|\psi_n \rangle , |\phi_n \rangle \in H_n$$ corresponding to another physical system, where $$H_p, H_n$$ are Hilbert spaces. Suppose also that $$|\psi_p \rangle \otimes |\psi_n \rangle$$ has the same biorthogonal decomposition base as $$|\phi_p \rangle \otimes |\phi_n \rangle$$and in this base m, n, p, q are the subspace dimensionality for $$|\psi_p \rangle, |\psi_n \rangle, |\phi_p \rangle, |\psi_n \rangle$$, respectively.

We call two pairs of a Cartesian product of wavefunctions equivalent:

$$(|\psi_p \rangle, |\psi_n \rangle ) \sim (|\phi_p \rangle, |\phi_n \rangle )$$

if the energy level of the composite system $$|\psi_p \rangle \otimes |\psi_n \rangle$$ is equal with the energy level of $$|\phi_p \rangle \otimes |\phi_n \rangle$$ and if there are unitary transformations $$U_p$$ acting on the left element $$(|\psi \rangle, \cdot )$$ and unitary transformations $$U_n$$ acting on the right element $$(\cdot, |\psi \rangle)$$ such that:

$$(U_p|\psi_p \rangle ) \otimes |\phi_n \rangle = |\phi_p \rangle \otimes (U_n |\psi_n \rangle )$$

subject to the constraint that: m + q + k = p + n + k

Under those conditions one can easily prove the reflexivity, symmetry, and transitivity properties needed to construct an equivalence relationship.

Before measurement, the Cartesian pairs (measurement outcomes, device states) corresponding to potential measurement outcomes respect this equivalence relationship. When an outcome is registered, the distinguished Cartesian pair corresponding to the outcome breaks the equivalence. For example, in a bubble chamber, an alpha particle interacts with an atom ionizing it. The wavefunction of the free electron prevents the existence of the unitaries which would transform the Cartesian pair corresponding to the outcome into another Cartesian pair. There is nothing non-unitary in the process. This is like we do integer arithmetic and the final answer is say 10. But which 10? Is it: (10,0) or (14,4), or perhaps (273, 263)? "Measurement" in this case would correspond to picking a particular representation say: (23, 13).  Similarly, in the quantum case, after the interaction of the quantum system with the measurement device we are forced to pick a single representation, because the others are no longer valid.

Now we can consider very interesting questions:
• Can we erase the measurement outcome by restoring the equivalence relationship?
• Can we change the outcome by changing the equivalence relationship?
• Is the measurement process irreversible?
I will try to work out those issue in detail in the next post when I will analyze the quantum eraser experiment in the Cartesian pair formalism. There is a lot to cover and perhaps I will need more than one post for this.

## What does quantum mechanics indeterminacy teach us?

Classical mechanics is deterministic, and quantum mechanics is probabilistic. Anyone who reads about quantum mechanics knows this. There are many approaches one can take explaining it. For example, one may argue that quantum mechanics is still like classical physics where one solves initial value problems, and the inderterministic behavior is due to some "hidden variables". If so, those hidden variables are not part of the theory, and quantum mechanics is incomplete. That was the position of Einstein in his debate with Bohr on the subject. Another take is mathematical: the observables in quantum mechanics do not commute. Say we measure the spin of an electron on the z axis, followed by a measurement on the x axis, and then again on the z axis. because of noncommutativity the second measurement on the z axis is unpredictable. Each quantum interpretation has a different explanation for indeterminism. My position is that we should embrace the inteterminism as a fundamental property and extract mathematical consequences out of it. Irreducible randomness is the key for solving the measurement problem.

Last time I talked about the Grothendieck construction and showed how integers arise out of a Cartesian pair of natural numbers. The key additional element was an equivalence relationship. What I am after is uncovering of an equivalence relationship out of quantum intederminism (there is a natural one suggested by Zurek's research). Then I can use the Grothendick construction to attempt solve the measurement problem.

Recall that I proposed the Cartesian pair to represent the quantum system and the observer (or the measurement device) like this:

(Quantum System, Measurement device) the same way in integers we have say:
3=(3,0)~(4,1)~(5,2)~... or
-5=(0,5)~(1,6)~(2,7)~...

So if there is an equivalence relationship then we have that:

(Quantum System before measurement, Measurement device before measurement) ~
(Quantum System after measurement, Measurement device after measurement)

Does this solve the measurement problem? No, this is only the first naive attempt.

What we have under this approach in the case of an electron spin is as follows:

(Quantum System before measurement, Measurement device is ready) ~
(Spin up wavefunction, Measurement device records up)~
(Spin down wavefunction, Measurement device records down)

and the one and only outcome comes as one of a distinguished Cartesian pairs. But what does distinguished pair means? It means that for itself the equivalence relationship is broken. The interaction between the quantum system and the measurement device breaks the equivalence and I will show how in mathematical details in subsequent posts (in one case this is actually similar with spontaneous symmetry breaking!).

All outcomes are possible, and in each Cartesian pairs we actually talk about different representations of the operator algebras by GNS construction. "Quantum System before measurement" does not evolve into "Spin up wavefunction": those are different wavefunctions residing in distinct Hilbert spaces. It is when we embed all those Hilbert spaces into a single one that measurement is described as collapse, but this is mathematical sloppy.

The electron does not have a definite spin before measurement, and the value of its spin is created by the interaction with the measurement device which selects one of the pairs as the only possible representation after interaction. Moreover, the equivalence breaking happens using only unitary evolution.

The equivalence breaking can be understood as information update, but there are no mystical ideas of consciousness involved here at all. If the right circumstances are present to construct and destroy the equivalence, then we have measurement outcomes. The universe did not have to wait until the first humans arrived on the scene for someone to reduce the wavefunctions.

The role of the observer is essential: there is no measurement outcome without an observer, but this is not a subjective approach: the observer changes as well!

Today I wanted to explain the paradigm, the conceptual roadmap to prevent the readers to get lost in dry mathematical abstractions. I will go in the mathematical details in subsequent posts and make all this mathematical rigorous.Then I will discuss concrete examples like Mott's problem and the quantum eraser and their implications for measurement irreversibility.

I only want to point out one more thing: this equivalence relationship arising out of indeterminism is not possible in classical mechanics, and this is a pure quantum effect. As such, in classical and quantum mechanics the role of the observer is fundamentally different.