Are all spin states orthogonal? For a spin 1/2 particle you have two spin states, either up or down which are orthogonal. But what about a spin 1 particle which has 3 spin states, either up, down, not up/not down?
 A: It is a common misconception that a spin-½ particle can only be in spin "up" or "down" states. For one, it is well known that such a quantum particle can also be in a superposition of such states. Additionally, the up and down states are related to a given spatial direction, but the particles are spherically symmetric, so what about the other directions?
It turns out that both those concerns are the same. If a particle has spin states $|\!\uparrow\rangle$, "up", and $|\!\downarrow\rangle$, "down", which represent positive and negative projections of the angular momentum along the $z$ axis, then there are also "right" and "left" states,
$$
|\!\rightarrow\rangle=\frac{|\!\uparrow\rangle+|\!\downarrow\rangle}{\sqrt{2}}
\text{ and }
|\!\leftarrow\rangle=\frac{|\!\uparrow\rangle-|\!\downarrow\rangle}{\sqrt{2}},
$$
which have positive and negative projections of the angular momentum along the $x$ axis. The most general state a spin-½ particle can have can be written in the form
$$|\psi\rangle=\cos\frac\theta2|\uparrow\rangle+e^{i\phi}\sin\frac\theta2|\!\downarrow\rangle,$$
where $\theta\in[0,\pi]$ and $\phi\in[0,2\pi)$ are the spherical coordinates of the unit vector of the direction in which $|\psi\rangle$ has a definite angular momentum projection.
These states are not all orthogonal. In fact, for every state $|\psi\rangle$ there exists a single other state $|\bar\psi\rangle$ which is orthogonal to it (and whose unit vector, of course, points in the antipodal direction).

A similar thing happens with a spin-1 particle. If you fix a spatial direction as your $z$ axis, you will have a state $|\!\uparrow\rangle$ which has positive $L_z$, a state $|\!\downarrow\rangle$ which has $L_z<0$, plus a third state $|-\rangle$ with $L_z=0$. These three states are all orthogonal to each other. However, there are a multitude of other states, which can be written in general as
$$
|\psi\rangle=\alpha|\!\uparrow\rangle+\beta|-\rangle+\gamma|\!\downarrow\rangle,
$$
where $\alpha,\beta$ and $\gamma$ are complex numbers, unique up to a global phase, such that $|\alpha|^2+|\beta|^2+|\gamma|^2=1$. In general, two given states $|\psi\rangle$ and $|\psi'\rangle$ will not be orthogonal. Given a state $|\psi\rangle$, you can choose a maximum of two other states which are orthogonal to $|\psi\rangle$ and to each other. (Additionally, linear combinations of those two states will also be orthogonal to $|\psi\rangle$.)

The overall reason for these facts is that the state space of a particle of spin $s$ is a (complex) vector space of dimension $d=2s+1$. In such a vector space, one can only ever find up to $d$ vectors which are orthogonal to each other; two general vectors are, of course, not orthogonal.
