Are all spin states orthogonal?

Question

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?

Answer 1

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).

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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 $|\psi\rangle$.)

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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.