Happy New Year!

Wishing everyone a Happy New Year! Let 2016 bring you hope, happiness, and prosperity.

No physics post today, only a brainteaser in honor of the new Star Wars: find the panda in the picture below.

Can quantum mechanics coexist with classical physics?

Continuing the discussion about quantum mechanics interpretations, today I want to look in depth on what it means to have a composite quantum-classical system. In standard Copenhagen interpretation, the measurement apparatus is considered to be described by classical physics. In physics there are only two theories of nature known and possible: quantum and classical mechanics and usually classical mechanics is described as the limit of quantum mechanics when $$\hbar \rightarrow 0$$. How can we introduce a fundamental theory of nature by using its limit case which is of a different character (obeys local realism)? This is one of the usual criticism of Copenhagen and a motivation for people to look for local realistic models of quantum mechanics. More important I think is to decide if there can be any consistent quantum-classical description of a physical system. But are there real world examples of such composite systems? I saw once this example given at a physics conference: a transistor. We do not see transistors in a superposition state and its inner workings are definite quantum mechanical.

So now that the stage is set, we can provide some answers. The framework is yet again the categorical approach to quantum mechanics, and part of what I will state today was discovered by one of Emile Grgin's colleague at Yeshiva University, Debendranath Sahoo: http://arxiv.org/pdf/quant-ph/0301044v3.pdf

So let's start at the beginning: is quantum mechanics defined by using one of its limits? The answer is no, but this is only a recent development with the complete derivation in the finite dimensional case of quantum mechanics from physical principles. Quantum mechanics stands on its own without classical mechanics help.

Is classical mechanics defined by the limit $$\hbar \rightarrow 0$$? Surprisingly, no again! This limit is mathematically sloppy, the proper limit is for $$\hbar$$ to become a nilpotent element: $$\hbar^2 = 0$$. If you remember the map $$J$$ between observables and generators, its dimension  is actually $$\hbar$$ and while in quantum mechanics $$J^2 = -1$$, in classical physics $$J^2 = 0$$, meaning $$\hbar^2 = 0$$ in classical physics. Another way to see this is by looking at deformation quantization approaches and convince yourself this is the proper limit. But if we are not sticklers for math and we adopt a physical point of view, $$\hbar \rightarrow 0$$ is good enough.

Can we combine consistently quantum and classical mechanics? No again because quantum and classical mechanics belong in disjoint composability classes. But what would happen if we try? This question was answered by Debendranath Sahoo in the paper above. Let's recall the fundamental composition relationships (in either quantum or classical mechanics):

$$\alpha_{12} = \alpha_1 \otimes \sigma_2 + \sigma_1 \otimes \alpha_2$$
$$\sigma_{12} = \sigma_1 \otimes \sigma_2 + J^2 \alpha_1 \otimes \alpha_2$$

If $$\alpha_1$$ is the commutator, $$\alpha_2$$ is the Poisson bracket, $$\sigma_1$$ is the Jordan product, and $$\sigma_2$$ is the regular function multiplication, what would $$\alpha_{12}$$ and $$\sigma_{12}$$ be? In quantum mechanics one can have superselection rules and there is nothing which prevents us to combine the 4 ingredients in a marriage of convenience. The penalty however is that we get something which lacks invariance under tensor composition! As such there cannot be any possible generalization of the commutator in the quantum-classical case. And people did try to invent such things but no such proposal withstood scrutiny. But this is not the only penalty!

What Sahoo found (working the problem Grgin-Petersen style) is that there is a lack of backreaction from the quantum to classical system! I double-checked the proof, it is correct, and because it involves a lot of Latex typing I will not repeat it here but you can read it in the paper. The result has two main implications:
• gravity has to be quantized: you cannot get a self-consistent theory of gravity in a mixed quantum-classical setting.
• The measurement devices should not be treated classically because they will not be able to measure anything: there is no information transfer from the quantum to the classical system.
If we were to solve the measurement problem we must do it solely in the quantum world. MWI is not the answer and it is no longer the only pure unitary quantum game in town. The categorical approach shows the way and provides a brand new solution. Despite its classical appearance, the transistor is still only a quantum object. The main problem is not interpreting quantum mechanics to appease our classical intuition but to explain the emergence of classical behavior. Decoherence only provides a partial answer. Please stay tuned.

PS: today is December 24

Can anyone defend the Many Worlds Interpretation?

Quantum mechanics has many interpretations or classes of interpretations with internal splits: Copenhagen, Bohmian, spontaneous collapse, many worlds, transactional, etc. Because my take on the matter falls within the neo-Copenhagen family, I do not follow very closely interpretations which fall outside my interest. But although I disagree with non-Copenhagen interpretations, I do understand the approaches they take with only one exception: the many worlds interpretation (MWI). Not for the lack of trying but as far as I dug into it, MWI did not make any sense whatsoever to me (except Zurek's approach which technically is not MWI). So here is my challenge: can anyone defend MWI in a way that will answer the issue I will raise below?

The ground rule of any interpretation is first and foremost to recover the standard quantum mechanics predictions, otherwise it cannot call itself a "quantum mechanics interpretation". Quantum mechanics has this novel feature called the Born rule. Let me digress for a bit and expand on why this does not occur in classical physics. If you recall from prior posts, in configuration space in classical mechanics one encounters the Hamilton-Jacobi equation, while in quantum mechanics one has the Schrodinger equation. In classical physics in phase space we need both the position and momenta of a particle to specify the trajectory, and therefore it should come as no surprise that in configuration space where we only have positions there can be crossing trajectories in the Hamilton-Jacobi case. Therefore the information content attached to a configuration point is ambiguous in classical physics: no Born rule in classical physics. However in the quantum case in configuration space we can attach an information interpretation to the Schrodinger wavefunction known as the Born rule. Born rule shows that quantum mechanics is probabilistic and initial conditions are not required. (In the Bohmian case you can add initial conditions only in a contextual (parochial) way respecting an additional constrain called quantum equilibrium otherwise you violate Born rule).

Is MWI compatible with Born rule?

But what is MWI, and why is it considered? MWI supposes to solve the measurement problem without resorting to the collapse of the wavefunction.

Suppose we have two outcomes, say spin up and down. Once spin is measured up, a quick subsequent measurement confirms the result, and the same for down. Since the wavefunction respects the superposition principle we can derive a superposition of up and down with the measurement device pointing up for the spin up, and pointing down for spin down. In other words, we arrive at the famous half dead half alive Schrodinger cat which does not occur in nature. Everett noticed that there is a correlation between how the measurement device points and the spin value and he proposed that the world splits between different outcomes: in each outcome the observer is only aware of his own unique measurement result. One proponent of this narrative was Sydney Coleman!!! (I have a big respect for the late Sidney Coleman, but in this instance I think he was shooting from the hip.) I grant that MWI is an appealing idea, but does it stand up to close scrutiny?

People naturally objected to the idea of split personality or "the I problem" to which supporters can fire back with "you do not take quantum mechanics seriously enough to trust what it shows". Also there is a "preferred basis problem" because the split can happen on an infinite number of basis. But to me the most important problem is the treatment of probabilities and agreement with Born rule. I think it is safe to say that anyone agrees that the original Everett argument of why MWI obeys Born rule is not satisfactory. If Everett's derivation were correct, then there will not be that many new "derivations" of Born rule in the MWI framework. However I found no satisfactory derivation to date of Born rule in MWI. Moreover, the only thing that makes sense to me is branch counting and this is definitely violating Born rule - also no disagreement here.

But why I am not convinced by the proposals of deriving Born rule? A common criticism is that those derivations are circular. I assert something stronger: when not circular, Born rule derivations in MWI are mathematically incorrect. Let me show why.

I have discovered long ago that the simplest problems are the hardest, and I will use this here. Instead of muddling the water with convoluted arguments and examples, let's streamline the basic system to the max. So consider a source of electrons which fire only one particle say once a minute. Pass this through a Stern Gerlach device and select only the spin up branch. In other words, we prepare a source of single electrons with a known vertical spin. Then we pass our electron through a second Stern Gerlach device and we measure spin on say the x axis. Half the time we will get the positive spin x and half the negative spin x. In MWI both outcomes occur and I am split into two "me" each observing one definite outcome. So far so good, but now rotate one of the two devices by some angle theta. The statistics changes!!! but what does MWI predict? The world still splits in two and in one world I detect up and in the other one down. In other words, no changes and this is the root cause of why MWI makes no sense.

Now supporters of MWI are well aware of this fact and attempt to derive Born's rule regardless starting from more or less natural assumptions. Let's dig deeper into their claims.

First we need to collect more data to make up a meaningful statistics. For the first electron we have two branches: one up and one down: u d. For the second electron we have 4 branches: uu ud du dd, for the 3rd electron we have 8: uuu ... Now let's count in those branches how many spins are up and how many are down regardless of the actual order of the events:

1st run:                                           1u    1d
2nd run:                                     1uu  2ud    1dd
3rd run:                          1uuu    3uud     3udd       1ddd
4th run:                1uuuu    4uuud   6uudd    4uddd      1dddd

We get Pascal's triangle and the binomial coefficients. This is nothing like Born rule, and the frequentist approach in statistics is rejected by the MWI supporters. Instead they adopt the Bayesian approach. For simple problems like this, the frequentist and Bayesian approaches predict the same things so something else must be thrown in the mix: "the rational observer".  A "rational observer" would have expectations of probabilities before the actual experimental outcome is obtained, and MWI supporters contend that to a rational observer making rational decisions, while branching is incompatible with Born rule, the sane way for such a person to behave is as Born rule appears to be true. Something like: the Earth moves around the Sun, but to us it appears that the Sun moves around the Earth. This line of reasoning was introduced by Deutsch and continued by Wallace.

Several natural sounding principles were proposed to justify this apparent emergence of Born rule in the MWI world. Now Born rule deals with the complex coefficients in front of the ket basis, and those coefficients are simply ignored by  branching because this is what it means to to have a relative correlation between the wavefunction and the measurement device. To derive Born rule you must deal with those coefficients and moreover you must do it in an indirect way. The only indirect way possible is for your "natural sounding principle" to say something nontrivial about a superposition. And in the best case scenario what you actually say about the superposition is nothing but the Born rule in disguise and you have a circular argument

But it gets worse if you claim you broke the circularity: you become mathematically inconsistent. Here is why:

1. To prove Born rule in MWI you need to reject branch counting.
Why? Because Born rule's prediction changes with changing complex coefficients, but branch counting does not.

2. Branch counting arises as a particular case of Born rule. When? In the particular case when the complex coefficients are equal.

So the very act of proving even an apparent Born rule inherently contains a contradiction. All mathematically consistent proposals of deriving Born rule in MWI I am aware of are circular arguments and all their "natural sounding principle" respects branch counting as well.

In summary, coming back to my physical example with the electron source and the two S-G devices, because branching happens the same way regardless of the orientation of the two devices, there is a one to (uncountable infinite) many degeneracy problem which MWI cannot hope to solve by relative arguments alone. In the frequentist approach it is impossible to derive Born rule which acts as a removal of this degeneracy, and MWI supporters pin their hopes on derivations of an apparent Born rule by using some "natural principles". However all the derivations I studied so far are circular, and I know one by Tippler which is mathematically incorrect-maybe I should write a rebuttal to that one, it was published last year. Moreover they cannot reject branch counting because this follows from Born rule when all the scalar coefficients are equal. If you claim you reject  branch counting you are killing your "apparent" Born rule too.

I am challenging MWI supporters to present a valid non-circular derivation of Born rule (either real or apparent). I don't have the time to closely follow MWI developments and maybe there is a recent proposal I missed which can stand up to scrutiny. However I contend it can't be done for the reasons outlined above.

The algebraic structure of Quantum Mechanics

Today we will conclude the mini-series on quantum mechanics reconstruction as we have all the required ingredients. We only need to perform one last computation to arrive at an associative product, the regular complex number multiplication in the Hilbert space representation.

But first, let me point out some strange notation I was using in all the prior posts about $$J$$ which obeys $$J^2 = -1$$. Why not simply call it $$i = \sqrt{-1}$$, the imaginary unit? The reason is that $$J$$ is more than the imaginary unit, and it is in fact a tensor of rank (1,1): $$J^{I}_{J}$$ and represents a map between the Jordan algebra to the Lie algebra. It also forms an "almost complex structure" which when multiplied (in the flat case) with the symplectic form gives rise to a metric tensor generating a Kahler manifold. So there is much more to it but I will not expand on this because we are led down the hard path of Poison manifold quantization. To put it in perspective, the proof of Poison manifold quantization got Maxim Kontsevich the Fields medal.

Only for the simplest 1-dimensional case we can consider $$J = i$$. In the finite dimensional case, by some algebraic magic (the Artin-Wedderburn theorem) one can show what the number system must be that of complex numbers in the case of transition probabilities (and I won't expand on this either, it is a large topic on itself too).

So why do we need associativity? When you run say an electron through s Stern Gerlach device you can pass the same electron through another Stern Gerlach device and concatenate two experiments: the output from one is the input to the next one. Now imagine three such devices (a,b,c) and ask yourself: what constitutes an experiment? The experiment separation occurs only in your mind, and to consistently reason about such a setup and define states you need associativity: $$(ab)c = a(bc)=abc$$.

But so far we only have two non-associative products: $$\alpha$$ and $$\sigma$$. Can they be combined to form an associative product? Up to $$\hbar /2$$ normalization factors we know that the $$\alpha$$ and $$\sigma$$ associators are equal:

$$[A, B, C {]}_{\sigma} + J^2 [A, B, C {]}_{\alpha} = 0$$

So how about a product beta:

$$\beta = \sigma \pm J \alpha$$

Its associator involves two beta products and it is not immediately obvious it is zero but it can be computed and it is not hard to show that:

$$[A, B, C {]}_{\beta} = [A, B, C {]}_{\sigma} + J^2 [A, B, C {]}_{\alpha} \pm$$
$$J ((A \sigma B) \alpha C + (A \alpha B) \sigma C − A \sigma (B \alpha C) − A \alpha (B \sigma C))$$

which is indeed zero after using Leibniz for the first two terms in the last line.

The sign of J does not matter, and traditionally the representation with plus is used by the usual Hilbert space representation, and the one with minus is used by the Moyal brackets in the phase space representation, but those are only historical accidents.

So now we can present the algebraic structure of quantum mechanics which is:

a real vector space equipped with two bilinear maps $$\sigma$$ and $$\alpha$$ such that the following conditions apply:

• $$\alpha$$ is a Lie algebra,
• $$\sigma$$ is a Jordan algebra,
• $$\alpha$$ is a derivation for $$\sigma$$ and $$\alpha$$,
• $$[A, B, C {]}_{\sigma} + J^2 / 4 \hbar^2 [A, B, C {]}_{\alpha} = 0$$
where J → (−J) is an involution, $$1 \alpha A = A \alpha 1 = 0$$, $$1 \sigma A = A\sigma 1 = A$$, and $$J^2 = −1, 0, +1$$.

Quantum mechanics corresponds to $$J^2 = -1$$, and classical mechanics corresponds to $$J^2 = 0$$.

From the associator identity it is immediately clear that in the classical case the product sigma is associative and in fact it is nothing but the usual function multiplication. It is not straightforward to prove, but the result it is available in books about Poisson manifolds, that alpha in this case is nothing but the usual Poisson bracket. And so in fact the two product algebra from above obtained by invariance under composition leads in a straightforward manner to Poisson manifolds and classical physics. To get to quantum mechanics we have three roads:

1. deform the associator identity changing J from $$J^2 = 0$$ to $$J^2 = -1$$ and use Kontsevich's result.
2. use positivity and GNS theorem to find a representation in a Hilbert space of the two product algebra
3. use hindsight and "guess" the usual representation as a commutator and the symmetrized product in the usual Hilbert space representation and verify they satisfy all the conditions of the two product algebra.

None of the three roads are satisfactory however, and more needs to be done. The third option gets you to the usual Hilbert space representation but does not say anything about uniqueness. The first option does not work in the case of spin, and the second option while better than the third, still has an open uniqueness problem in the infinite dimensional case.

Now we can look back and see what was obtained:

The categorical (algebraic) part of the reconstruction works because of the existence of an universal property which translates the tensor product physical principle of invariance under composition into algebraic consequences. The non-algebraic part is an open problem in the infinite dimensional case, and different quantization techniques corresponds to the second de Rham cohomology making the classification of representations a hard problem.

The classification of Jordan algebras is however well known and this restricts the possible number system representations over the reals, complex, and quanternions in the case of transition probability. There is no such thing as octonionic quantum mechanics because octonions are not associative, and the product beta is.

If the questions we ask nature do not result in a probability, but in a probability current then we actually get Dirac's equation by an unusual Hodge theory route. The quantum mechanics number system is SL(2,C) in this case and the Hilbert space generalizes to a Hilbert module. This shows again that the complete classification of the two-product algebra representations is not at all trivial.

The Jordan algebra of observables

In quantum mechanics the observables are represented by Hermitian operators: $$O^{\dagger} = O$$. In general Hermitian operators need not commute and this corresponds to incompatible observables like position and momenta.  Given two observables A and B, can we generate another observable out of it?

Let's try the simplest idea: AB. Is this self-adjoint? Let's see:

$${(AB)}^{\dagger} = B^{\dagger} A^{\dagger} = BA \ne AB$$

So the next idea to try is a symmetrized product:  $$1/2 (AB + BA)$$ .

This is called the Jordan product and the product $$\sigma$$: $$A \sigma B = 1/2 (AB + BA)$$ gives rise to the Jordan algebra of observables.

 Pascual Jordan

This product has two property:

1) symmetry: $$A \sigma B = B \sigma A$$
2) Jordan identity: $$(A \sigma B) \sigma (A \sigma A) = A \sigma (B \sigma (A \sigma A))$$

Can we derive those properties from categorical arguments like we derived the properties of α last time? Drum-roll....Yes we can!!!

Last time we have proven that α is antisymmetric: $$A \alpha B = - B \alpha A$$ and from the fundamental relationship:

$${\alpha}_{12} = \alpha \otimes \sigma + \sigma \otimes \alpha$$

it is trivial to show that σ must be symmetric thus respecting the first property of the Jordan algebra. Proving the second property is unfortunately much more involved project, about two orders of magnitude harder than what I left out last time as a simple exercise. Let me set the stage to where we want go. Both the products α and σ are not associative. To quantify the violation of associativity we introduce what is called the associator:

$$[A, B, C]_{o} = (A o B) o C - A o (B o C)$$

where "o" can be either $$\alpha$$ or $$\sigma$$. Then the Jordan and Lie algebras of quantum mechanics obey this identity:

$$[A, B, C]_{\sigma} + \frac{J^2 \hbar^2}{4} [A, B, C]_{\alpha} = 0$$

where $$J \hbar /2$$ is a one-to-one map between observables and generators obeying $$J^2 = -1$$the imaginary unit which when multiplying a Hermitian operators generates an anti-Hermitean operator.

You can convince yourself this is true using the usual realizations of the Jordan and Lie products:

$$A\sigma B = \frac{1}{2} (AB + BA)$$
$$A\alpha B = \frac{J}{\hbar} (AB - BA)$$

Now if this associator identity is true, proving the second Jordan identity is trivial because the identity can be rewritten as: $$[A, B, A \sigma A {]}_{\sigma} = 0$$

What is the alpha associator: $$[A, B, A \sigma A {]}_{\alpha}$$ ?  I can leave this as an exercise to show this is indeed zero by expanding the associator and using the Leibniz identity for α.

Now our goal is to derive $$[A, B, C {]}_{\sigma} + J^2 \hbar^2 / 4 [A, B, C {]}_{\alpha} = 0$$ from invariance under composability and time evolution.

But first where is the $$b_{11} = J^2 \hbar^2 / 4$$ coming from? If you recall two posts ago I derived the following fundamental composition relationship as a coproduct:

$$\Delta (\alpha ) = \alpha \otimes \sigma + \sigma \otimes \alpha$$
$$\Delta (\sigma) = \sigma \otimes \sigma + b_{11}\alpha \otimes \alpha$$

$$b_{11}$$ was a free parameter which can be normalized to -1 for quantum mechanics, 0 for classical mechanics and +1 for the unphysical "hyperbolic quantum mechanics". However I simply want to normalize it as:  $$b_{11} = J^2 \hbar^2 / 4$$. Why? To recover the usual definition of the commutator and the Jordan product. In other words, convenience.

If last time we used $$\Delta (\alpha ) = \alpha \otimes \sigma + \sigma \otimes \alpha$$, now it is time to employ:$$\Delta (\sigma) = \sigma \otimes \sigma + \frac{J^2 \hbar^2}{4}\alpha \otimes \alpha$$ as well.

The proof is rather long and was first obtained by Grgin and Petersen in 1976 and I won't present its gory bookkeeping detail, but I will only give the starting point:

$$(A_1 \otimes A_2 ) \alpha_{12} ((B_1 \otimes B_2)\sigma_{12} (C_1 \otimes C_2))$$

Use the bipartite Leibniz identity along with the two delta coproducts and the Jacobi identity and you'll reach the associator identity (from Jacobi). The first time I double checked Grgin's paper it took me one full week to work out the result given the intermediate steps in the paper, but now I can do it in about an hour of careful and boring bookkeeping.

We now have all the ingredients to put together the C* algebras of quantum mechanics and arrive at the usual Hilbert space formulation. Please stay tuned.

Historical notes:

I am very grateful to Emile Grgin for introducing me to his approach which I managed to expand into a full blown reconstruction of quantum mechanics. The root idea came from Bohr himself who passed his intuition to Aage Petersen, his personal assistant. Later on Petersen developed those ideas in collaboration to Emile Grgin at Yeshiva University. Unfortunately Grgin left academia and his ideas were forgotten. I got in contact with him after he retired and later on I realized the categorical origin of the approach. Completely independent from me, Anton Kapustin of Caltech noticed the same 1976 paper and we both came out with almost identical papers, mine written from the physics point of view, and his from the math point of view. I uploaded my paper on the archive 3 weeks before him and I did not notice his paper. John Preskill made me aware of Kapustin's paper when we met at a conference. He was the very first person who understood my research.