Components of electron spin towards three dimensional coordinate axes

Question

The electron spin can be described using two orthonormal state vectors in two dimensional Hilbert space. If we assume the two orthonormal vectors as $|+\rangle$ and $|-\rangle$, we can represent the general spin state as $$|S\rangle = c_1 |+\rangle + c_2 |-\rangle$$ with condition that $|c_1|^2 + |c_2|^2 = 1$.

We now determine components $S_x$, $S_y$ and $S_z$ of spin state $|S\rangle$ along the three axes of the Cartesian coordinate system. This is done by using Pauli matrices which serve as the operators for observables $S_x$, $S_y$ and $S_z$.

Now my question is this: While studying Pauli matrices, it appears to me that state vector $|+\rangle$ lies along $+z$ axis while $|-\rangle$ lies along $-z$ axis. Is this by assumption or is there any mathematical reason why the orthogonal vectors $|+\rangle$ and $|-\rangle$ model directions that are at 180 degree with each other in three dimensional space?

In other words, although $|+\rangle$ and $|-\rangle$ are at 90 degree (orthogonal) in Hilbert space, they model $+z$ and $-z$ directions that are at 180 degree in three dimensional space. Is this by assumption or is there any mathematical reason for it?

If the answer requires mathematics, please keep it as simple as possible.

ABOUT THE EDIT

Edited the question a little bit to reflect that I am not asking why orthogonal spin states 'point towards' directions that are at 180 degree but, instead, why orthogonal spin states are chosen to 'model' directions that are at 180 degree. Is it by assumption or is there any deeper meaning?

Answer 1

Because quantum mechanical states do not live in ordinary space, the notion of orthogonality of quantum mechanical states cannot be understood as orthogonality in 3d space.

In quantum mechanics, states are orthogonal when they correspond to distinct outcomes of an observable. This orthogonality is a mathematical property of eigenstates of Hermitian operators (which represent observables). In particular, the dimension of the space is the number of possible distinct outcomes of the observable.

There are two possible outcomes of measuring spin along any direction so the space will be 2-dimensional for any direction of spin, and the states corresponding to spin-up and spin-down will be orthogonal in this 2-dimensional space.

In the same way, there are three angular momentum states with $L=1$, and they have projections $m=+1,0,-1$. The states with outcomes $m=+1$ and $m=-1$ are orthogonal in the space of states even if they can be at $180^\circ$ in 3d space. There are 5 orthogonal angular momentum states with $L=2$, even if these "orbitals" live in 3d space. The solutions to the particle in the box problems are orthogonal in the (infinite dimensional) space of states even if all the particle "lives" in 1d. Ditto for the harmonic oscillator.

Answer 2

The state vectors ($|+\rangle$, $|-\rangle$, and their superpositions $|c_1 |+\rangle + c_2 |-\rangle$) are members of the Hilbert space $\mathbb{C}^2$.

The two states $|+\rangle$ and $|-\rangle$ are said to be orthogonal to each other just because their scalar product is zero, $\langle +|-\rangle=0$. Only in this sense the "angle" between them is $90°$. This may seem counter-intuitive. But this is only because human intuition isn't strong in the $\mathbb{C}^2$ space. Our intuition is much stronger in our every-day space $\mathbb{R}^3$ (containing vectors with 3 real components in $x$, $y$ and $z$ direction).

There is a mathematical procedure using the Pauli matrices to get a vector $\vec{n}$ in $\mathbb{R}^3$ from a state vector $|\chi\rangle$ in $\mathbb{C}^2$. The physical meaning of this is, that for a state $|\chi\rangle$ the vector $\vec{n}$ points along the spinning axis of this state. $$\vec{n} = \langle\chi|\vec{\sigma}|\chi\rangle$$ or written in components $$\begin{pmatrix}n_x\\n_y\\n_z\end{pmatrix}= \begin{pmatrix} \langle\chi|\sigma_x|\chi\rangle \\ \langle\chi|\sigma_y|\chi\rangle \\ \langle\chi|\sigma_z|\chi\rangle \end{pmatrix}$$

Note, there is a correspondence between $|\chi\rangle$ and $\vec{n}$, but they are not the same thing.

For your two special states you get $$|+\rangle=\begin{pmatrix}1\\0\end{pmatrix} \Rightarrow \vec{n}_+ =\langle +|\vec{\sigma}|+\rangle =\begin{pmatrix}0\\0\\1\end{pmatrix}$$ $$|-\rangle=\begin{pmatrix}0\\1\end{pmatrix} \Rightarrow \vec{n}_- =\langle -|\vec{\sigma}|-\rangle =\begin{pmatrix}0\\0\\-1\end{pmatrix}$$

Here you see: The angle between $\vec{n}_+$ and $\vec{n}_-$ (two vectors in $\mathbb{R}^3$) is $180°$.

Do not confuse the $\mathbb{C}^2$-vectors $|+\rangle$ and $|-\rangle$ with the $\mathbb{R}^3$-vectors $\vec{n}_+$ and $\vec{n}_-$. They live in two very different vector spaces. And therefore it should not come as a surprise that the angles between these two pairs of vectors are different.