Understanding Spin, (from Griffiths), a Paradox?

Question

In Griffiths', Introduction to Quantum Mechanics, 3rd ed, Sec. 4.4.1 Spin 1/2 the author represents the spin state by a spinor $$ \chi = \binom{a}{b} = a\chi_+ + b\chi_-, \qquad \chi_+ = \binom{1}{0},\quad \chi_- = \binom{0}{1},\quad |a|^2 + |b|^2 = 1 $$ where the latter two represent the spin up and the spin down along the $z$-axis, i.e., they are the eigenspinors of the $S_z$ for the eigenvalues $+\hbar/2$ and $-\hbar/2$, respectively. After he constructs the Pauli spin matrices he finds the eigenspinors of the $S_x$ to be $$ \chi_+^{(x)} = \binom{1/\sqrt{2}}{1/\sqrt{2}},\quad \chi_-^{(x)} = \binom{1/\sqrt{2}}{-1/\sqrt{2}} $$ for the eigenvalues $+\hbar/2$ and $-\hbar/2$, respectively. And therewith $$ \chi = \frac{a+b}{\sqrt{2}}\ \chi_+^{(x)} + \frac{a-b}{\sqrt{2}}\ \chi_-^{(x)} \tag{*} $$

Question: If I choose $a = \cos \theta, b = \sin \theta$ then $|a|^2 + |b|^2 = 1$ is fullfiled. But in the case $\theta = \frac{\pi}{4}$ (which means equal probabilities for $\chi_+$ and $\chi_-$) we get from ($*$) that $\chi = \chi_+^{(x)}$ which is a determinate (certain) state of $S_x$, and it means spin up along the $x$-axis, i.e., in the positive direction. But how does it know which is the positive direction ? why isn't it the other direction, since we could choose the opposite direction along $x$ as the positive ?

Stated another way: If the particle is in a state such that it is equally likely to get by measurement $+\hbar/2$ or $-\hbar/2$ along some axis ($z$-axis, say), How does measuring the spin at this state along some perpendicular line gives the positive direction along that line, though one may have not yet chosen which direction is the positive and which is the negative ?

Answer 1

I'm summing up, What @Cosmas Zachos said,

The first direction in space is unique but not it's representation. The one I'm calling positive $x$-axis, you can call $-x$-axis or vice-versa. We know that Spinors are generators of SU(2) group which correspond to rotation in $3D$ (naively) without changing right hand to left hand system.

For instance:

In our case : $$\chi=\frac{a+b}{\sqrt{2}}\chi^x_++\frac{a-b}{\sqrt{2}}\chi^x_-$$ Now apply a rotation of $\pi$ about $z$-axis so that

$$R(\pi\hat{z})=\begin{pmatrix} \cos(\pi) & \sin(\pi)\\ -\sin(\pi) & \cos(\pi) \end{pmatrix}=\begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix}$$ $$R(\pi\hat{z})\chi=-\frac{a+b}{\sqrt{2}}\chi^x_+-\frac{a-b}{\sqrt{2}}\chi^x_-$$ If you work out the whole algebra for the case $a=b$ lead you to $$\chi'=-\chi^x_-$$ That's the negative direction which is equivalent to other.

Answer 2

though one may have not yet chosen which direction is the positive and which is the negative ?

You did choose the direction when you chose the spinor basis. You have to prepare a system in the state $\binom{1/\sqrt{2}}{1/\sqrt{2}}$ before you do your measurement, and whatever method you use to do that will make use somehow of the fact that that state points in the +x direction in your basis. There's no way to create that state through measurements along the z axis, or a magnetic field directed along the z axis, or anything else that doesn't break the x/y symmetry.

The complex-vector representation of spin obscures the geometry. It's somewhat clearer to imagine a Pauli spinor as a quaternion representing a rotation from a canonical spinor orientation to the actual orientation. If you take the canonical spinor to be pointing in the +z direction, and $\binom10$ to represent the quaternion $1$, then $\binom01$ has to be a unit quaternion perpendicular to $1$, i.e. a pure imaginary quaternion. If you pick $j$ – i.e., a 180° rotation in the xz plane – then $\binom{1/\sqrt{2}}{1/\sqrt{2}}$ is $(1{+}j)/\sqrt2$, or a 90° rotation in the xz plane, and this leaves the spinor pointing in the +x direction (or -x; I wish there weren't so many sign conventions). The quaternionic representation makes it clearer that $\binom10$ and $\binom01$ aren't (can't be) just "+z" and "-z"; they have another spatial direction hidden in them.