## 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.