The Jordan Formulation of Quantum Mechanics

It is well-known that the Albert algebra, introduced in § 11.2, is the only exceptional realization of the Jordan formulation of quantum mechanics [Gürsey and Tze] ; this is in fact how it was first discovered. We summarize this formalism here.

Recall that the Cayley plane, introduced in § 12.5, consists of those Jordan matrices $\VVV$ which satisfy the restriction [Harvey] \begin{equation} \label{Cayley} \VVV \circ \VVV = \VVV; \qquad \tr\,\VVV = 1 \end{equation} Elements of the Cayley plane correspond to projection operators in the Jordan formulation of quantum mechanics. As shown in § 12.5, the conditions (\ref{Cayley}) force the components of $\VVV$ to lie in a quaternionic subalgebra of $\OO$ (which depends on $\VVV$). Basic (associative) linear algebra then shows that each element of the Cayley plane is a primitive idempotent, and can be written as \begin{equation} \label{ISq} \VVV = vv^\dagger \end{equation} where $v$ is a 3-component octonionic column vector, whose components lie in the quaternionic subalgebra determined by $\VVV$, and which is normalized by \begin{equation} \label{vnorm} v^\dagger v = \tr\,\VVV = 1 \end{equation} For given $\VVV$, the vector $v$ is unique up to a quaternionic phase. Furthermore, using (\ref{FreudSq}) and its trace (\ref{Sigma}), it is straightforward to show that, for any Jordan matrix $\BBB$, \begin{equation} \label{ISqF} \BBB*\BBB = 0 \Longleftrightarrow \BBB \circ \BBB = (\tr\,\BBB) \, \BBB \end{equation} which agrees with (\ref{Cayley}) up to normalization, and which is therefore the condition that that $\pm \BBB$ can be written in the form (\ref{ISq}) (without the restriction (\ref{vnorm})). Note further that for any Jordan matrix satisfying (\ref{ISqF}), the normalization $\tr\,\BBB$ can only be zero if $v$, and hence $\BBB$ itself, is zero, so that \begin{equation} \label{NormZero} \BBB*\BBB = 0 = \tr\,\BBB \Longleftrightarrow \BBB=0 \end{equation} since the converse is obvious.

We will need the following useful identities \begin{align} (\AAA*\AAA)*(\AAA*\AAA) &= (\det \AAA) \, \AAA \label{Springer}\\ (\tilde \AAA\circ \AAA)\circ(\AAA*\AAA) &= (\det \AAA) \, \tilde \AAA \end{align} for any Jordan matrix $\AAA$, where \begin{equation} \tilde\AAA = \AAA -\tr(\AAA)\,\III = -2 \, \III*\AAA \end{equation} which can be verified by direct computation. Finally, we also have the remarkable fact that \begin{equation} \label{OCross} \AAA*\AAA=0=\BBB*\BBB \Longrightarrow (\AAA*\BBB)*(\AAA*\BBB)=0 \end{equation} which follows by polarizing (\ref{Springer}) 1) and which ensures that the set of Jordan matrices satisfying (\ref{ISqF}), consisting of all real multiples of elements of the Cayley plane, is closed under the Freudenthal product.

In the Dirac formulation of quantum mechanics, a quantum mechanical state is represented by a complex vector $v$, often written as $|v\rangle$, which is usually normalized such that $v^\dagger v=1$. In the Jordan / formulation \cite{Jordan,JNW,GT}, / formulation [Gürsey and Tze] , the same state is instead represented by the Hermitian matrix $vv^\dagger$, also written as $|v\rangle\langle{v}|$, which squares to itself and has trace 1 (compare (\ref{Cayley})). The matrix $vv^\dagger$ is thus the projection operator for the state $v$, which can also be viewed as a pure state in the density matrix formulation of quantum mechanics. The phase freedom in $v$ is no longer present in $vv^\dagger$, which is uniquely determined by the state (and the normalization condition).

A fundamental object in the Dirac formalism is the probability amplitude $v^\dagger w$, or $\langle{v}|w\rangle$, which is not however measurable; it is the the squared norm $\left|\langle{v}|w\rangle\right|^2 = \langle{v}|w\rangle \langle{w}|v\rangle$ of the probability amplitude which yields the measurable transition probabilities. One of the basic observations which leads to the Jordan formalism is that these transition probabilities can be expressed entirely in terms of the Jordan product of projection operators, since \begin{equation} \label{TraceID} (v^\dagger w)(w^\dagger v) \equiv \tr(vv^\dagger \circ ww^\dagger) \end{equation} A similar but less obvious translation scheme also exits [Gürsey and Tze] for transition probabilities of the form $|\langle{v}|A|w\rangle|^2$, where $A$ is a Hermitian matrix, corresponding (in both formalisms) to an observable, so that all measurable quantities in the Dirac formalism can be expressed in the Jordan formalism.

So far, we have assumed that the state vector $v$ and the observable $A$ are complex. But the Jordan formulation of quantum mechanics uses only the Jordan identity \begin{equation} \label{JID} (A\circ B)\circ A^2 = A\circ \left(B\circ A^2\right) \end{equation} for 2 observables (Hermitian matrices) $A$ and $B$. The Jordan identity (\ref{JID}) is equivalent to power associativity, which ensures that arbitrary powers of Jordan matrices — and hence of quantum mechanical observables — are well-defined.

The Jordan identity (\ref{JID}) is the defining property of a Jordan algebra, and is clearly satisfied if the operator algebra is associative, which will be the case if the elements of the Hermitian matrices $A$, $B$ themselves lie in an associative algebra. Remarkably, the only further possibility is the Albert algebra of $3\times3$ octonionic Hermitian matrices. 2)

1) The necessary fact that $\det(\AAA+\BBB)=0$ follows from the definition (\ref{Det}) of the determinant in terms of the triple product, the cyclic properties of the trace of the triple product, and the assumptions on $\AAA$ and $\BBB$.
2) The $2\times2$ octonionic Hermitian matrices also form a Jordan algebra, but, even though the octonions are not associative, it is possible to find an associative algebra which leads to the same Jordan algebra [Jacobson] .

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