Are Einstein's Boxes an argument for nonlocality?

(an experimental proposal)

Today I want to discuss a topic from an excellent book by Jean Bricmont: Making Sense of Quantum Mechanics which presents the best arguments for the Bohmian interpretation. Although I do not agree with this approach I appreciate the clarity of the arguments and I want to present my counter argument.

On page 112 there is the following statement: "... the conclusion of his [Bell] argument, combined with the EPR argument is rather that there are nonlocal physical effects (and not just correlations between distant events) in Nature".

To simplify the argument to its bare essentials, a thought experiment is presented in section 4.2: Einstein's boxes. Here is how the argument goes: start with a box B and a particle in the box, then cut the box into two half-boxes B1 and B2. If the original state is $$|B\rangle$$, after cutting the state it becomes:

$$\frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)$$

Then the two halves are spatially separated and one box is opened. Of course the expected thing happens: the particle is always found in one of the half-boxes. Now suppose we find the particle in B2. Here is the dilemma: either there is action at a distance in nature (opening B1 changes the situation at B2), or the particle was in B2 all along and quantum mechanics is incomplete because $$\frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)$$ does not describe what is going on. My take on this is that the dilemma is incorrect. Splitting the box amounts to a measurement regardless if you look inside the boxes or not and the particle will be in either B1 or B2.

Here is an experimental proposal to prove that after cutting the box the state is not $$\frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)$$:

split the box and connect the two halves to two arms of a Mach-Zehnder interferometer (bypassing the first beam splitter). Do you get interference or not? I say you will not get any interference because by weighing the boxes before releasing the particle inside the interferometer gives you the which way information.

If we do not physically split the box, then indeed $$|B\rangle = \frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)$$, but if we do physically split it $$|B\rangle \neq \frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)$$. There is a hidden assumption in Einstein's boxes argument: realism which demands non-contextuality. Nature and quantum mechanics is contextual: when we do introduce the divider the experimental context changes.

Bohmian's supporters will argue that always $$|B\rangle = \frac{1}{\sqrt{2}}(|B_1\rangle+|B_2\rangle)$$. There is a simple way to convince me I am wrong: do the experiment above and show you can tune the M-Z interferometer in such a way that there is destructive interference preventing the particle to exit at one detector.

Gleason's Theorem

It feels good to be back to physics, and as a side note going forward I will do the weekly posts on Sunday. Today I want to talk about Gleason's theorem. But what is Gleason's theorem?

If you want to assign a non-negative real valued function $$p(v)$$ to every vector v of a Hilbert space H of dimension greater than two, then subject to some natural conditions the only possible choice is $$p(v) = {|\langle v|w \rangle |}^{2}$$ for all vectors v and an arbitrary but fixed vector w.

Therefore there is no alternative in quantum mechanics to compute the average value of an observable A the standard way by using:

$$\langle A \rangle = Tr (\rho A)$$

where $$\rho$$ is the density matrix which depends only on the preparation process.

Gleason's theorem is rather abstract and we need to unpack its physical intuition and the mathematical gist of the argument. Physically, Gleason's theorem comes from three axioms:

• Projectors are interpreted as quantum propositions
• Compatible experiments correspond to commuting projectors
• KEY REQUIREMENT: For any two orthogonal projectors P, Q, the sum of their expectation values is the expectation value of P+Q: $$\langle P \rangle + \langle Q\rangle = \langle P+Q\rangle$$
In an earlier post I showed how violating the last axiom (which is the nontrivial one), in the case of spin one particles, can be used to send signals faster than the speed of light and violate causality. But how does Gleason arrives at his result?

Let's return at the original problem: to obtain a real non-negative function p. Now add the key requirement and demand that for any complete orthonormal basis $$e_m$$ we have:

$$\sum_m p(e_m) = 1$$

For example in two dimensions on a unit circle we must have:

$$p (\theta) + p(\theta + \pi/2) = 1$$

which constrain the Fourier expansion of $$p (\theta)$$ such that only components 2, 6, 10, etc can be non zero. In three dimensions the constraints are much more severe and this involves rotations under SO(3) and spherical harmonics. I'll skip the tedious math, but it is not terribly difficult to show that the only allowed spherical harmonics must be of order 0 and 2 which yields: $$p(v) = {|\langle v|w \rangle |}^{2}$$.

The real math heavy lifting is on dimensions larger than three and to prove it Gleason first generalizes  $$\sum_m p(e_m) = 1$$ to $$\sum_m f(e_m) = k$$ where k is any positive value. He names this "f" a "frame function". Then he proceeds to show that dimensions larger than three do not add anything new.

If you are satisfied with the Hilbert spaces of dimension 3, the proof of the theorem is not above undergrad level, and I hope it is clear what the argument is. But what about Many Worlds Interpretation? Can we use Gleason's theorem there to prove Born rule? Nope. The very notion of probabilities is undefined in MWI, and I am yet to see a non-circular derivation of Born rule in MWI. I contend it can't be done because it is a mathematical impossibility and I blogged about it in the past.