Intuition for Spin operator in arbitrary direction

Question

I understand why the Spin operators in $x$, $y$ and $z$ direction are given by : $$ S_x = \begin{pmatrix} 0 &\hbar/2\\ \hbar/2 & 0 \end{pmatrix} \qquad S_y = \begin{pmatrix} 0 & -i\hbar/2\\ i\hbar/2 & 0 \end{pmatrix} \qquad S_z = \begin{pmatrix} \hbar/2 & 0\\ 0 & -\hbar/2 \end{pmatrix} $$

But why is the spin operator along an arbitrary direction $\vec{n}$ given by : $$S_{\vec{n}} = n_x \cdot{S_x} + n_y \cdot{S_y} +n_z \cdot{S_z} \qquad ? $$

I can see that it works along the $x$, $y$ and $z$ axis, and that is look like a scalar product between $\vec{n}$ and $ \textbf{S} = (S_x,S_y,S_z)$. I don't need a rigorous proof, a more physical explanation would be ok. I saw this post related, but no satisfactory answer.

EDIT:

Little precision, what is not clear for me is why I can do stuff with $\textbf{S}$ like if it was a vector. Also I would not be satisfied if you just say "it transforms like a vector". It is also not really clear what it would mean to take a scalar product with $\textbf{S}$.

Answer 1

I'm not sure what reasoning took you to accept that,

$$S_{x}=\boldsymbol{e}_{x}\cdot\boldsymbol{S}$$

is OK, but let's take it from there. There's nothing special about “$x$”, as opposed to “$y$”, or “$z$”. You could have said $\boldsymbol{n}$, instead of $\boldsymbol{e}_{x}$, and you would have,

$$S_{\boldsymbol{n}}=\boldsymbol{n}\cdot\boldsymbol{S}$$

IOW, $x$ is a dummy parameter there.

As a simpler example that hopefully will clarify the question (Pauli matrices carry "internal space" representations), consider this: Suppose you have a physical law that tells you that a certain scalar $\sigma$ that depends on a given direction, x –that's relevant to your problem– is:

$$\sigma=x$$

This is unsatisfactory as a physical law, because it does not have any definite transformation law under rotations. $\sigma$ is a scalar, but $x$ is not. If you want to make it right, you have to upgrade it to a physical law but referring it to an arbitrary direction and make the transformation law transparent:

$$\sigma\left({\boldsymbol{n}}\right)=\boldsymbol{x}\cdot\boldsymbol{n}$$

Answer 2

Formally speaking, the spin-half representation is a linear map: $$\rho_{1/2} : \mathfrak{su}(2) \to M_{2 \times 2}(\mathbb{C})$$ You should also keep in mind that there is a linear isomorphism $$\rho_{\mathrm{iso}} : \mathfrak{so}(3) \to \mathfrak{su}(2)$$ Note that elements in the group $SO(3)$ are determined by an axis of rotation, and a degree $\theta$ describing how far to rotate around the axis of rotation. There are normalised elements $v_x, v_y, v_z \in \mathfrak{so}(3)$ corresponding to rotations about the $x,y,z$ axes respectively.

The actual thing to check here is that exponentiating $\vec{n} \cdot \{v_x,v_y,v_z\} \in \mathfrak{so}(3)$ corresponds to rotations about the axis $\vec{n}$. A simple way to check is just to notice that it leaves $\vec{n}$ invariant.

Composing the maps from above, $\rho_{1/2}(\rho_\mathrm{iso}(v_x)) = \sigma_x$, similarly for $\sigma_y,\sigma_z$. So, the spin along the $\vec{n}$ axis should be interpreted (by linearity of the $\rho$ maps) as $\vec{n} \cdot \vec{\sigma}$.