In a 2D complex vector space, are there more than 3 orthonormal bases that are all "45 degrees" from each other? If I understand correctly, if a quantum spin is measured/prepared in the $+x$ direction, then there is a .5 probability of subsequently measuring it in the $+y$ direction and .5 probability of measuring it in the $-y$ direction. This is because
$$| \langle y+|x+ \rangle| = | \langle y-|x+ \rangle| = 1/\sqrt{2},$$
i.e. the $y$ eigenstates are each at a "45 degree" angle to each of the $x$ eigenstates. By symmetry this is true for the $z$ vs $x$ case and the $z$ vs $y$ case also.
One can see this by choosing a basis where
$$|x+\rangle = [1, 0]^*, |x-\rangle = [0, 1]^*$$
$$|y+\rangle = [1/\sqrt{2}, 1/\sqrt{2}]^*, |y-\rangle = [1/\sqrt{2}, -1/\sqrt{2}]^*$$
$$|z+\rangle = [1/\sqrt{2}, -i/\sqrt{2}]^*, |z-\rangle = [1/\sqrt{2}, i/\sqrt{2}]^*$$
and computing the inner products to show each basis is orthonormal and is "rotated 45 degrees" from each of the others.
My question is: can one find a fourth orthonormal basis $\{|w+ \rangle, |w- \rangle\}$ that is itself 45 degrees from each of the other three? Intuitively it seems like there are too many constraints for this to be possible, but I'm having a hard time visualizing or calculating exactly why.
If this is indeed impossible, is there a deeper relationship between the 2D complex vector space in which spin is introduced in QM and the 3 dimensions of ordinary space? Is the 2D nature of QM spin states somehow derivable from the symmetry of space?
If we lived in $k$-dimensional ordinary physical space, what dimension complex vector space would QM spin states live in?
 A: It is not possible to find a fourth set.  Your question is related to the existence of mutually unbiased bases.  In your case, this problem is one of finding sets of pairs of states like $\vert x,\pm\rangle$ etc so that any overlap between elements in different sets satisfies
$$
\vert \langle y;\pm\vert x\pm\rangle\vert^2=\frac{1}{2} \qquad \hbox{etc.}
$$
In dimension 2 it is know there are only 3 sets and they are the ones you listed.  They are eigenvectors of $\sigma_x,\sigma_y$ and $\sigma_z$ respectively.
In dimension $k$, where $k$ is a prime or power of a prime, it is known that one can construct $k+1$ sets, where in each sets there are $k-1$ commuting operators, and the common eigenvectors of these $k-1$ commuting operators will be orthogonal and satisfy $\vert \langle \alpha,j\vert \beta,m\rangle \vert^2=1/k$, where $\alpha$ labels a set and $j$ an eigenvector in this set.  For $k=2$, you have three sets each containing one operator:
$\{\sigma_x\}, \{\sigma_y\},\{\sigma_z\}$.
If the dimension $k=p^n$ is a power of a prime, one must use the algebraic field extension of $\mathbb{Z}_p$, but it is possible to find $k+1$ sets each containing $k-1$ vectors to satisfy the condition that the modulus squared of the overlap is $1/k$.
If $k$ is a composite number (like $6=2\times 3$), it is not known if one can find the complete collection of $k+1$ sets of $k-1$ vectors.  In fact it is conjectured that one cannot, and extensive numerical work has been done in dimension $6$ (by Brierley and Weigert, see this arXiv post), since this is the lowest composite number.
MUB have applications in foundational physics, in quantum tomoagraphy, and in state reconstruction.  They are also linked to Latin squares (see for instance this arXiv post or this arXiv post; a GoogleScholar search returns more hits), which are themselves linked to algebraic field theory (not the fields as in E&M field, but fields in the in the mathematical sense).
