### Quaternionic Spin

In quantum mechanics, the eigenstates of a self-adjoint operator correspond to the physical states with particular values of the corresponding observable physical quantity. A fundamental principle of quantum mechanics further states that after making a measurement the system is “projected” into the corresponding eigenstate. It is therefore only eigenstates which are unaffected by the measurement process; they are projected into themselves. In particular, in order to simultaneously make two measurements, the system must be in an eigenstate of each; otherwise, the order in which the measurements are made will matter.

At the end of §8, we pointed out that the spin operators $\LL_a$ are imaginary multiples of the derivatives (“infinitesimal generators”) of the rotations $\RRR_a$ introduced in §5, Writing $\rrr_a$ for those derivatives, we have $$\rrr_a = {d\RRR_z\over d\alpha} \Bigg|_{\alpha=0}$$ where, as in §8, $\alpha=2\theta$ is the physical rotation angle.

Consider first the $z$-component of angular momentum. For reasons which will become apparent in later chapters, we henceforth write $\ell$, rather than $i$, for the complex unit. We therefore have \RRR_z = \begin{pmatrix} e^{-\ell\frac\alpha2}& 0\cr \noalign{\smallskip} 0& e^{\ell\frac\alpha2}\cr \end{pmatrix} and \rrr_z = \frac12 \begin{pmatrix} -\ell& 0\cr \noalign{\smallskip} 0& \ell\cr \end{pmatrix} This matrix is anti-Hermitian, that is \begin{align} \rrr_z^\dagger = -\rrr_z \end{align} In order to obtain real eigenvalues, one normally multiplies by $\ell\hbar$ to get a Hermitian matrix — which is just $\LL_z$.

As discussed in [ 25,28 ], however, even over $\HH$ care must be taken with this last step; we must put the factor of $\ell$ in the right place! The right place turns out to be on the right; we define the operator 1) $$\LLL_z[v] = (\rrr_z v) \, \ell\hbar \label{lop}$$ and consider the right eigenvalue problem $$\LLL_z[v] = v \lambda$$

What are the eigenvectors of $\LLL_z$? We content ourselves here with some intriguing examples. Unsurprisingly, we have $$\LLL_z \left[ \begin{pmatrix}1\cr 0\cr\end{pmatrix} \right] = \begin{pmatrix}1\cr 0\cr\end{pmatrix} {\hbar\over2}$$ (We could of course have written the eigenvalue on the left since it is real.) Somewhat surprisingly, this is not the only eigenvector with eigenvalue $\hbar\over2$. For instance, we have $$\LLL_z \left[ \begin{pmatrix}0\cr k\cr\end{pmatrix} \right] = \begin{pmatrix}0\cr k\cr\end{pmatrix} {\hbar\over2}$$ Note the crucial role played by the anticommutativity of the quaternions in this equation!

We can add these 2 eigenvectors, obtaining yet another eigenvector of $\LLL_z$ with the same eigenvalue, namely $$\eplus = \begin{pmatrix}1\cr k\cr\end{pmatrix} \qquad \label{eplus}$$ We will argue later that $\eplus$ represents a spin-$\frac12$ particle at rest. We have $$\LLL_z[\eplus] = \eplus \, {\hbar\over2}$$ This eigenvector has some remarkable properties.

Consider now the remaining spin operators, $\LLL_x$ and $\LLL_y$, constructed as in ($\ref{lop}$) from \rrr_x = \frac12 \begin{pmatrix} 0& -\ell\cr \noalign{\smallskip} -\ell& 0\cr \end{pmatrix} \qquad \rrr_y = \frac12 \begin{pmatrix} 0& -1\cr \noalign{\smallskip} 1& 0\cr \end{pmatrix} What happens if these operators act on $\eplus$? We have $$\LLL_x[\eplus] = {\hbar\over2} \begin{pmatrix}-k\cr 1\cr\end{pmatrix} = \eplus \left( -{k\hbar\over2} \right)$$ $$\LLL_y[\eplus] = {\hbar\over2} \begin{pmatrix}-k\ell\cr \ell\cr\end{pmatrix} = \eplus \left( -{k\ell\hbar\over2} \right)$$ This illustrates the fact that this quaternionic self-adjoint operator eigenvalue problem admits eigenvalues which are not real!. More importantly, as claimed in [ 25,28 ], it shows that $\eplus$ is a simultaneous eigenvector of the 3 self-adjoint spin operators $\LLL_x$, $\LLL_y$, $\LLL_z$!

This result could have significant implications for quantum mechanics. In this formulation, the inability to completely measure the spin state of a particle, because the spin operators fail to commute, is thus ultimately due to the fact that the eigenvalues don't commute. Explicitly, we have \begin{align} {4\over\hbar^2} \, \LLL_x \!\left[ \LLL_y [\eplus] \right] &= {2\over\hbar} \, \LLL_x [-\eplus \,k\ell] = -2r_x (\eplus \,k\ell) \,\ell = 2r_x (\eplus \,\ell) \,k\ell \nonumber\\ &= {2\over\hbar} \, \LLL_x [\eplus] \,k\ell = -\eplus \,k \, k\ell = + \eplus \,\ell \\ \noalign{\smallskip} {4\over\hbar^2} \, \LLL_y \!\left[\LLL_x [\eplus] \right] &= -\eplus \,k\ell \, k = - \eplus \,\ell \end{align} which yields the usual commutation relation in the form $$[\LLL_x , \LLL_y] [\eplus] = -\eplus \, [k,k\ell] \, {\hbar^2\over4} = \eplus \,{\ell\hbar^2\over2} = \LLL_z[\eplus] \,\ell\hbar$$

Furthermore, there is a phase freedom in ($\ref{eplus}$), since $$\LLL_z \left[ \eplus e^{\ell\theta} \right] = \left( \eplus e^{\ell\theta} \right) {\hbar\over2}$$ for any value of $\theta$. It is still true that $\eplus e^{\ell\theta}$ is a simultaneous eigenvector of all 3 spin operators, but the imaginary eigenvalues have changed. We have \begin{align} \LLL_x \left[ \eplus e^{\ell\theta} \right] &= \left( \eplus e^{\ell\theta} \right) \left( -{k \, e^{2\ell\theta}\hbar\over2} \right) \\ \LLL_y \left[ \eplus e^{\ell\theta} \right] &= \left( \eplus e^{\ell\theta} \right) \left( -{k\ell \, e^{2\ell\theta}\hbar\over2} \right) \end{align} so that the non-real eigenvalues depend on the phase. It is intriguing to speculate on whether the value of the non-real eigenvalues, which determine the phase, can be used to specify (but not measure) the actual direction of the spin, and whether this might shed some insight on basic properties of quantum mechanics such as Bell's inequality.

Finally, we point out that all eigenvectors of the complex operators $\LL_x$, $\LL_y$, $\LL_z$ turn out to be quaternionic; each eigenvector lies in some quaternionic subalgebra of $\OO$ which also contains $\ell$.

1) The operator $\LLL_z$ no longer corresponds to a Hermitian matrix, but it is self-adjoint with respect to a suitable inner product [ 25,28 ]. We will return to this issue later.