Intuition for Spin operator in arbitrary direction

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

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

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 $$\textbf{S}$$.

• ? S transforms like a vector (with matrix components). The component of a vector $\vec V$ in a direction $\hat n$ is $\hat n \cdot \vec V$. Nov 12 '21 at 21:23
• @CosmasZachos surely what's being asked for is an argument that $\vec{S}$ so defined does actuall transform like a vector. You could find $S_z,S_y,S_z$ just by guessing matrices to satisfy the $SU(2)$ relations but it's not immediately obvious how they transform under rotations. Nov 12 '21 at 21:57
• @jacob1729 you guessed right as per his subsequent edit. XY problem… Nov 13 '21 at 12:12
• Is the OP comfortable with the rotation group in, e.g., 't Hooft's notes, or any of the nice texts of this? Nov 13 '21 at 14:24

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:

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