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Jordan algebras are neither natural nor essential for quantum mechanics

Pascual Jordan was one of the 3-5 true founders of quantum mechanics. His contributions have been largely erased from the history books and textbooks because of his Nazi faith and he hasn't even received a Nobel prize – which I consider completely unfair – but that doesn't mean that I am ready to worship everything that Jordan was obsessed with.

One of these things were Jordan algebras which he proposed as the modern correct formalization of the "algebra of observables" in quantum mechanics in 1933. If you're not familiar with those, you may look at the definition and the first strange thing that you quickly notice is that the Jordan algebras are non-associative. How can the multiplication of operators be non-associative?

In regular quantum mechanics as we're learning it, operators obey the associative law for multiplication,\[

(AB)C = A(BC).

\] So in these chains of factors, the parentheses don't affect the result. Only the ordering matters which is natural because these operators may be identified with some true "operations" acting on the physical system which may be denoted by the pure state \(\ket\psi\), for example. If you first act by \(C\), then by \(B\), and then by \(A\), it is the same thing as if you act by \(C\) and then by \(AB\) – which is first by \(B\) and then \(A\). And it is the same thing as if you act by \(BC\) – first by \(C\) and then by \(B\) – and then by \(A\).

A disadvantage of this simple associative multiplication is that the product isn't Hermitian whenever the factors are Hermitian. For example, \(xp\) isn't Hermitian because\[

(xp)^\dagger = p^\dagger x^\dagger = px = xp-i\hbar \neq xp

\] So \(xp\neq (xp)^\dagger\) which means that \(xp\) isn't Hermitian. You may fix this problem, as Jordan did, if you don't consider the simple multiplication but the symmetrization of it,\[

A\circ B = \frac 12 \zav{ AB + BA }.

\] Formally, you could call it the "anticommutator" but I personally reserve this special word only for the situation when \(A,B\) are Grassmann-odd or at least very analogous to them (tending to anticommute), for instance some combinations of gamma and similar matrices.

If you define the Jordan \(\circ\) operation as above, it will no longer be associative:\[

(A\circ B)\circ C \neq A\circ (B\circ C).

\] The left hand side contains the extra \(BAC/2\) term but the right hand side doesn't; the right hand side has the \(ACB/2\) term but the left hand side doesn't. But the Jordan operation is commutative, \(A\circ B = B\circ A\), because the product is symmetrized. It produces the Hermitian product of two Hermitian factors, as I said, and the associativity of the original product guarantees the so-called Jordan identity\[

(A\circ B) \circ (A\circ A) = A\circ (B\circ (A\circ A)).

\] Check it. Both sides are equal to\[


\] the symmetrization of \(A^3 B\) where \(B\) appears at all four possible places. This Jordan identity is clearly less natural than the simple identity for the associative operation we started with and it's probably the simplest identity among those that hold for the Jordan product and that are analogous to the associative law for the normal product.

Because of this complexity etc., the Jordan product wasn't really adopted and other pioneers of quantum mechanics – and researchers – haven't started to use it on a daily basis. It is not essential. The normal associative product is good enough. In fact, the normal product is better because it allows us to quickly access or write down a greater number of the operators that may be useful.

I would say that the Jordan product is a result of "misguided impatience". Jordan knew that Hermitian operators are "special" and "more interesting". When we measure an operator, it's ultimately a Hermitian one. But to use the Jordan product only – and not the general product – means to demand the Hermiticity for every single intermediate step of the calculation (composition of the operators). And that's simply not needed or justified. You may get lots of Hermitian operators if you simply work with an arbitrary functions based on the ordinary product, and you take the Hermitian part of the result at the very end!

So the whole motivation for replacing the normal product with the Jordan product seems fallacious to me and it's simply not a coincidence that Jordan's construction hasn't become popular or paramount. Researchers almost never use it. Textbooks don't teach it. They don't have to.

Nevertheless, the Jordan algebras appear at various places, usually "outside the things that are essential for doing physics". They are used by people who study Hermitian symmetric spaces and triple systems but so far, those things are esoteric mathematics or mathematical physics, not something that physics graduate students would have to learn.

My only personal pet among the algebraic structures that use the Jordan product is a representation of the exceptional Lie group \(F_4\). The 52-dimensional Lie group may be represented as the automorphism group of \(3\times 3\) traceless Hermitian octonionic matrices \(A,B\) equipped with the Jordan product \(A\circ B = (AB+BA)/2\). Note that Hermitian octonionic \(3\times 3\) matrices have 3 real numbers on the diagonal (or 2 if we demand the tracelessness – the scalar trace may be decoupled as another irreducible representation) plus 24 components of the 3 octonions above the diagonal (those below are determined by the Hermiticity).

In total, these traceless Hermitian octonionic matrices have \(24+2=26\) real parameters or components, exactly the right number for them to transform as the fundamental 26-dimensional representation of \(F_4\). But you know, this is not the only representation of \(F_4\) and not even the only way to visualize the fundamental representation and the role of \(F_4\) in physics is rather limited. It can't even appear as a grand unified group because it has no complex representations (because the 26-dimensional fundamental one we constructed is a real representation, and all others are therefore real as well).

So I am absolutely confident that if you want to get very far in any kind of modern quantum physics (condensed matter, optics, particle physics, string theory, foundations of quantum mechanics), you don't even need to learn what the Jordan algebras are – even though I would claim that you have effectively done so if you understood the blog post above.

This claim of mine dramatically contradicts comments by Florin Moldoveanu who argued that the Jordan algebras (plus their relationships to Lie algebras etc.) are fundamental on quantum mechanics. He wrote things like

The one-to-one map between the Lie algebra and the Jordan algebra is known in the literature as "dynamic correspondence". This has a deep relationship with Noether theorem...
But everyone who has the Internet may check what is the importance of these things and claims in the actual scientific literature. Open Google Scholar and search for
"dynamic correspondence" jordan lie (click)
You will see that the term "dynamic correspondence" appears in 4 papers only and they were written by the same Florin Moldoveanu. They have 2,0,1,0 citations, respectively, and only 2 of them are not self-citations. You may also replace the adjective "dynamic" by the more widespread word "dynamical" – Florin's English isn't quite "standard". You will find a few more papers but the top-cited two have 13 and 15 citations, respectively.

To argue that this term is some fundamental deep concept in the quantum mechanics literature is surely ludicrous according to a basic overview of the actual literature. But people like Moldoveanu are constantly deceiving their readers and sponsors. They want to intimidate you by references to the literature and consensus and all this stuff but if you carefully verify their bold statements, you will always find that there is basically nothing of value to back those statements up.

By the way, someone at Math Overflow has asked whether the Jordan algebras were really important in quantum mechanics. He apparently wanted some fair and penetrating appraisal such as the blog post above. He or she has gotten a few comments with examples where the Jordan algebras appear but none of the answers has really addressed the key question. Let me do so briefly again. The answer is No, you can do virtually any research, including the deepest and state-of-the-art research, in any discipline that relies on quantum mechanics without knowing what Jordan algebras are.

You may rewrite Jordan's \(A\circ B\) as \((AB+BA)/2\) everywhere and by doing so, you will get rid of all the dependence on Jordan's algebraic jargon, formalism, and identities, without losing any genuine mathematical or physics insight whose beef goes beyond conventions. To force yourself to express ideas using the Jordan anticommutator rather than the general normal product is similar to the task to talk about physics without using the letter/consonant "p". Can you say "ping-pong" while avoiding the contact between your upper and lower lips? Yes, you can insert the finger in between the lips and say "ping-pong". Or you can say "table tennis". (Too bad, it doesn't work in English because of the "b" in "table"; "stolní tenis" works in Czech, however.) Alternatively, you may also say that these restrictions are silly and ideas are more naturally formulated with the normal product and with the sound "p".

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