# Intuition for Spin operator in arbitrary direction

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 $$\textbf{S}$$.

• ? S transforms like a vector (with matrix components). The component of a vector $\vec V$ in a direction $\hat n$ is $\hat n \cdot \vec V$. Commented Nov 12, 2021 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. Commented Nov 12, 2021 at 21:57
• @jacob1729 you guessed right as per his subsequent edit. XY problem… Commented Nov 13, 2021 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? Commented Nov 13, 2021 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:

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

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 $$\vec{n} \cdot \vec{\sigma}$$.