Friday, January 29, 2016

The Grothendieck group construction

After "snowzilla" disrupted my priorities for about a week I am falling behind in my duties and today I only have time to present a small but essential topic. So let's talk about the Grothendick group construction which appears front and center in the categorical solution to the measurement problem. 

Alexander Grothendieck

There is no easier way to show this then to see it at work in the case of the integer numbers. Basically we introduce a Cartesian pairs of natural numbers and we call the left element positive integers, and the right element negative integers like this:

\(7\equiv (7,0)\)

\(-3\equiv (0,3)\)

Then we can add and subtract the numbers in the most natural way: \(7-3=7+(-3) = (7,0)+(0,3) = (7,3)\)

But this is not what we want: we want 7-3 to be 4 and not (7,3). How to do that? 4 is (4,0) and there is no problem if (7,3) is identical with (4,0). In other words, we need to introduce an equivalence relationship where distinct pairs represent the same thing.

How can we justify (7,3)=(4,0)? One way is to observe that if we subtract the right pair from the left one we get identical answers in both slots: (7-4, 3-0) = (3,3) but we are not allowed to use subtraction because we need to rely only on the operation available from the original commutative monoid. The answer is that we need to add the elements like this:

7+0 = 3+4

In other words, the equivalence relation we seek is as follows:

The pairs a-b and c-d are equivalent \((a,b)\sim (c,d)\)

if \(a+d = b+c\). Please notice the outer-inner pattern.

But is this an equivalence relation? To prove that we need to show three properties:

  • relexivity
  • symmetry
  • transitivity
Is this reflexive: \((a,b)\sim (a,b)\)? Indeed it is because: a+b (from outer) is the same as b+a from inner.

Is this symmetric: if \((a,b)\sim (c,d)\) is \((c,d)\sim (a,b)\)? Trivial.

Is this transitive: if \((a,b)\sim (c,d)\) and \((c,d)\sim (e,f)\) do we have: \((a,b)\sim (e,f)\)?

Let's see. We have: a+d=b+c and c+f=d+e. Can we prove a+f=b+e?

If we add the first two we get: a+d+c+f = b+c+d+e or a+f + k = b+e+k where k=d+c and we need to make a tiny generalization:

\((a,b)\sim (c,d)\) if and only if there is a k such that a+d+k = b+c+k.

So what does this have to do with quantum mechanics? It will turn out that those a,b,c,d numbers will be the dimensions of the Hilbert spaces involved in the measurement problem. Also if in the case of integers we have the Cartesian pair: (positive number, negative number) in the case of quantum mechanics we have the Cartesian pair: (Quantum system, Observer).

No more hokey pokey endless philosophical debates about the role of the observer in quantum mechanics which devolve into arguments about consciousness, but a sharp (and unique, and natural) mathematical construction which will allow us to bring about rigorous proofs. Please stay tuned.

Friday, January 22, 2016

What is the relationship between unitary evolution and collapse in quantum mechanics?

Last time we started discussing the measurement problem. In quantum mechanics textbooks, a quantum system is presented as undergoing two distinct time evolution: 
  • a unitary time evolution before measurement
  • a sudden change known as the collapse of the wavefunction
The epistemic explanation of it is very simple: the collapse corresponds to information update. QBism has self consistent explanations to all quantum puzzles, so why bother seek a different solution to the measurement problem? I mean besides trying to erase from your memory the creepy picture Chris Fuchs is using in his talks about qbism: 

For some reason this picture of an one-eye man with measurement dials instead of hands is like a bad song you cannot get out your brain.

The answer is that any non-unitary time evolution is fatal to quantum mechanics. Here is why:

Unitarity in the state space formulation of quantum mechanics, just like the Leibniz identity in the algebraic formalism are a consequence of the invariance of the laws of nature under time evolution and moreover each one can be derived from the otherI have showed in prior posts how to reconstruct quantum mechanics using Leibniz identity. What reconstruction of quantum mechanics shows is that breaking the Leibniz identity makes the whole quantum formalism inconsistent: no more Hilbert spaces or Hermitean operators before or after measurement. 

So for pure mathematical consistency arguments, can we describe the (non-unitary) update of information using only unitary time evolution? Yes we can, and the answer is that the non-unitary collapse is nothing but a change in the GNS representation, but to show it rigorously I need to build up the required machinery.

Last time I made a strong claim: the transformation

\((\lambda |\psi_A \rangle  + \mu |\psi_B \rangle )\otimes |M_0 \rangle\ \rightarrow \lambda |\psi_A \rangle \otimes |M_A \rangle + \mu |\psi_B \rangle \otimes |M_B \rangle\)

is not correct. In foundations of quantum mechanics the justification for the transformation above stems in part from the following argument: quantum mechanics is universal, and the measurement process should be described quantum mechanically in a Hilbert space. Asher Peres had a rebuttal to this but that was only handwaving inspired by Godel incompletness theorem. This rigorous  reason the argument is faulty is because it turns out that there are many Hilbert spaces involved. I will show that \(|\psi_A \rangle \otimes |M_0 \rangle\ \rightarrow |\psi_A \rangle \otimes |M_A \rangle \) and \(|\psi_B \rangle \otimes |M_0 \rangle\ \rightarrow |\psi_B \rangle \otimes |M_B \rangle \) should not be understood as unitary evolution, but as a change in representation. Since changes in representation do not happen in a Hilbert space, there is no superposition and they cannot be combined.

Interestingly enough, the relationship between unitary evolution and collapse is deeply related with the relationship between the addition and subtraction in the case of integer numbers. There is a category construction called the Grothedieck group construction which is at play in both cases. The Grothendieck group construction is an universal property, meaning it is both unique and natural.

For now I only want to point a problem in the easy integer numbers setting. When you learn arithmetic in elementary school , first you learn to add and subtract. Then you learn about multiplication and division. Fast forward to say your high school or early undergraduate years, you learn to formalize those operations in the concepts of groups, rings, and fields. The natural numbers for example are an abelian monoid. Going from monoids to groups one needs to add the inverse elements. Natural numbers N become the integers Z. And in the integers abelian group Z you have two operations: \(+\) and \(-\). But wait a minute, a group has only one operation! So what does the subtraction operation mean? Do we have two independent operations? In a field there are two independent operations, addition and multiplication, but not in a group. 

Starting from this trivial observation that subtraction must be not independent from addition, the problem is how to express one in terms of the other. The first try is to say: no problem at all, subtraction is addition by negative elements. I bet you already heard that in school. But this is mathematically sloppy and unacceptable answer because the definition of an operation should be decoupled from the nature of the elements it works with. It turns out that there is a unique way one can turn an abelian monoid into an abelian group, and this was figured out by Mr. Grothendieck.

So what does this have to do with quantum mechanics? Do you know a natural abelian monoid there? When we compose quantum systems we use the tensor product: \(\otimes\). This is associative: it does not matter the order in which we compose Hilbert spaces. It has a unit element: compose with nothing. It is also commutative. We will see that the inverse operation \({\otimes}^{-1}\) is deeply related with the collapse postulate, but it is not a straightforward relationship. Please stay tuned.

Thursday, January 14, 2016

What is the measurement problem?

After discussing the category theory approach of quantum mechanics reconstruction I want to start a series of posts discussing the so-called measurement problem. Is the wavefunction collapse real? The views on this topics varies wildly and touches a lot of raw nerves because it goes to the heart of quantum mechanics interpretation

Outside the foundations community, one prevalent attitude is that I know very well how to use quantum mechanics in my day to day computations which keep me very busy, and if I only have one free afternoon to think about it I will surely solve it. But this is actually a hard problem, and moreover I will attempt to show that all known solutions are incorrect/incomplete in one form or another. Because the category theory approach was successful in deriving quantum mechanics from physical principles, it is natural to expect that it also offers hints on solving the measurement problem. And the solution turns out to be completely unexpected, an entirely new paradigm. 

To set the stage, I can argue that the best solution so far to the measurement problem is offered by QBism. This is not without issues however (but not what people usually use against the epistemic interpretation), and I will attempt to make the epistemic interpretation mathematically rigorous. As an analogy consider the usage of \(ict\) in special relativity. Time is not an imaginary distance and the proper way to understand relativity is by using the metric tensor. Similarly in quantum mechanics if we are sloppy and ignore a mathematical construct which naturally appears in category theory, we can talk about collapse. 

The basic argument used to present the measurement problem is as follows:

Suppose I have a wavefunction \(|\psi \rangle\) and a measurement device \(|M\rangle\). To simplify the argument we can consider that there are  only two measurement outcomes: \(|\psi_A \rangle\) and \(|\psi_B \rangle\).  If the ready state of the measurement device is \(|M_0 \rangle\), the pointer state for the \(A\) outcome is \(|M_A \rangle\), and the pointer state for the \(B\) outcome is \(|M_B \rangle\), from the fact that a repeated experiment confirms the prior value we have:

\(|\psi_A \rangle \otimes |M_0 \rangle\ \rightarrow |\psi_A \rangle \otimes |M_A \rangle\)
\(|\psi_B \rangle \otimes |M_0 \rangle\ \rightarrow |\psi_B \rangle \otimes |M_B \rangle\)

Now by superposition:

\((\lambda |\psi_A \rangle  + \mu |\psi_B \rangle )\otimes |M_0 \rangle\ \rightarrow \lambda |\psi_A \rangle \otimes |M_A \rangle + \mu |\psi_B \rangle \otimes |M_B \rangle\)

This is interpreted differently by various quantum mechanics interpretations.

  • In many worlds, the world splits in two branches, and in each branch we have an outcome.
  • In Bohmian, we add a hidden variable, the position, and different initial conditions lead to the A or B outcomes.
  • In GRW the wavefunction spontaneously collapses to the A or B outcomes
  • In Copenhagen, the wavefunction only predicts potential outcomes and the collapse is only an information update. One criticism people level on this is the Wigner's friend problem. (QBism has a good answer to this criticism).
Sometimes the measurement problem is presented as a trilema: any 2 of the following 3 statements contradicts the other one:

S1: Quantum mechanics is complete
S2: Quantum mechanics predicts one outcome
S3: Quantum mechanics evolves linearly according to Schrodinger's equation

Bohmian violates S1 and respects S2 and S3.
MWI violates S2 and respects S1 and S3
GRW violates S3 and respects S1 and S2

What I will attempt to show in the next posts is that the trilema is false: quantum mechanics obeys all 3 properties: S1, S2, S3. The argument:

\((\lambda |\psi_A \rangle  + \mu |\psi_B \rangle )\otimes |M_0 \rangle\ \rightarrow \lambda |\psi_A \rangle \otimes |M_A \rangle + \mu |\psi_B \rangle \otimes |M_B \rangle\)

will turn out to be bogus. 

There is only one outcome from any experiment which occurs when the quantum system interacts with the measurement device, the Wigner's friend and quantum eraser has natural explanations, and we can talk about collapse when we sloppily ignore a mathematical structure. The epistemic information update will be rigurously described in a precise mathematical way. Please stay tuned.

Thursday, January 7, 2016

Musings over algebra and topology

One open problem in quantum mechanics reconstruction is the complete classifications of the realization of the algebraic properties. We know quantum mechanics can be formulated over the reals, complex, or quaternionic numbers, but is this all there can be? The problem is solved in the finite dimensional case, but is open in the infinite case. But why is this a hard problem? I think a recent unrelated cute recreational math video shows the heart of the mater. The video attempts to prove that:

1+2+4+8+16+... = -1

This is not the same as 1+2+3+...= -1/12 The rigorous treatment of the Riemann zeta function cannot be covered in only one post (you have to go past the usual Ramanujan tricks), but the current case is much simpler.

Formally, the algebraic manipulations are trivial:

if \(S = 1+q+q^2 + \cdots\) then \(Sq = q+q^2 + q^3 + \cdots = S-1\) and so \(S = 1/(1-q)\) which for q=2 results in -1.

But does it make sense to have this kind of algebraic manipulation? We learn in school that this is allowed only for convergent series which means that \(q\in (-1, 1)\) and 2 is outside the radius of convergence. Case closed, right? Wrong!

We have do dig deeper into what it means that an algebraic manipulation makes sense. The easiest thing to do is to consider the existence of a metric, which is a positive function which assign a number between any two points subject to the usual properties. The most important property of a metric is the triangle inequality and once we have it we can have the usual epsilon-delta arguments.

If you study functional analysis most of the concepts come from considering a metric. Take for example the notions of continuity or that of compactness. A function is continuous if when we approach a point from both ends the value of the function converges. Also a set is compact if it is bounded and closed. One difficulty in learning topology is in generalizing those common sense ideas and using only open sets. For example a function is continuous if and only if it returns open sets into open sets, and a set is compact if from any covering with open sets we can extract a finite covering.

But once we freed ourselves from the usual metric intuition we can see and appreciate things in a different light. For the problem above the key idea is to reorder the numbers to create a different topology and a different metric where the sum does converge. The trivial algebraic manipulation suggests how to do it and one arrives at the p-adic numbers.

So it looks that there is flexibility in messing with ordering and neighborhoods to satisfy algebraic identities. But how much flexibility is allowed by nature in the case of quantum mechanics? 

First, is there a p-adic quantum mechanics? Some publications claim there is, but they are all nonsense. p-adic numbers violate the so-called Archimedean property. While it is conceivable to mathematically imagine universes where probability predictions violate the Archimedean property, a non-Archimedean quantum mechanics must violate the Archimedean property for the Jordan algebra as well and this is where you get in trouble from the physical point of view.

p-adic quantum mechanics is not physical, but can we twist the order of the real numbers in a different way which respects the Archimedean property and yet we get a district topology? To me this looks highly unlikely but I do not have a proof for this impossibility. I did not even began to scratch the surface of topology and algebra in this post, but I hope I succeeded in highlighting the main issue: topology is not as rigid as naive metric epsilon-delta functional analysis proofs from college would made us believe. Categorical arguments nail the algebraic structure of quantum mechanics, but they have nothing to offer on the topological side.