D.69 Awkward questions about spin

Now of course you ask: how do you know how the mathematical expressions for spin states change when the coordinate system is rotated around some axis? Darn.

If you did a basic course in linear algebra, they will have told you how the components of normal vectors change when the coordinate system is rotated, but not spin vectors, or spinors, which are two-di­men­sion­al vectors in three-di­men­sion­al space.

You need to go back to the fundamental meaning of angular momentum. The effect of rotations of the coordinate system around the $z$-​axis was discussed in addendum {A.19}. The expressions given there can be straightforwardly generalized to rotations around a line in the direction of an arbitrary unit vector $(n_x,n_y,n_z)$. Rotation by an angle $\varphi$ multiplies the $n$-​direction angular momentum eigenstates by $e^{{\rm i}{m}\varphi}$ if $m\hbar$ is the angular momentum in the $n$-​direction. For electron spin, the values for $m$ are $\pm\frac12$, so, using the Euler formula (2.5) for the exponential, the eigenstates change by a factor

\begin{displaymath}
\cos\left({\textstyle\frac{1}{2}}\varphi\right) \pm
{\rm i}\sin\left({\textstyle\frac{1}{2}}\varphi\right)
\end{displaymath}

For arbitrary combinations of the eigenstates, the first of the two terms above still represents multiplication by the number $\cos\left(\frac12\varphi\right)$.

The second term may be compared to the effect of the $n$-​direction angular momentum operator ${\widehat J}_n$, which multiplies the angular momentum eigenstates by $\pm\frac12\hbar$; it is seen to be $2{\rm i}\sin\left({\textstyle\frac{1}{2}}\varphi\right){\widehat J}_n$$\raisebox{.5pt}{$/$}$$\hbar$. So the operator that describes rotation of the coordinate system over an angle $\varphi$ around the $n$-​axis is

\begin{displaymath}
\fbox{$\displaystyle
{\cal R}_{n,\varphi} =
\cos\left(...
...\frac{1}{2}}\varphi\right) \frac{2}{\hbar}{\widehat J}_n
$}
\end{displaymath} (D.44)

Further, in terms of the $x$, $y$, and $z$ angular momentum operators, the angular momentum in the $n$-​direction is

\begin{displaymath}
{\widehat J}_n = n_x {\widehat J}_x + n_y {\widehat J}_y + n_z {\widehat J}_z
\end{displaymath}

If you put it in terms of the Pauli spin matrices, $\hbar$ drops out:

\begin{displaymath}
{\cal R}_{n,\varphi} =
\cos\left({\textstyle\frac{1}{2}}...
...ht)
\left(n_x \sigma_x + n_y \sigma_y + n_z \sigma_z\right)
\end{displaymath}

Using this operator, you can find out how the spin-up and spin-down states are described in terms of correspondingly defined basis states along the $x$-​ or $y$-​axis, and then deduce these correspondingly defined basis states in terms of the $z$ ones.

Note however that the very idea of defining the positive $x$ and $y$ angular momentum states from the $z$ ones by rotating the coordinate system over 90$\POW9,{\circ}$ is somewhat specious. If you rotate the coordinate system over 450$\POW9,{\circ}$ instead, you get a different answer! Off by a factor $\vphantom0\raisebox{1.5pt}{$-$}$1, to be precise. But that is as bad as the indeterminacy gets; whatever way you rotate the axis system to the new position, the basis vectors you get will either be the same or only a factor $\vphantom0\raisebox{1.5pt}{$-$}$1 different {D.70}.

More awkwardly, the negative momentum states obtained by rotation do not lead to real positive numerical factors for the corresponding ladder operators. Presumably, this reflects the fact that at the wave function level, nature does not have the rotational symmetry that it has for observable quantities. Anyway, if nature does not bother to obey such symmetry, then there seems no point in pretending it does. Especially since the nonpositive ladder factors would mess up various formulae. The negative spin states found by rotation go out of the window. Bye, bye.