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.


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 $\textbf{S}$.

  • $\begingroup$ ? S transforms like a vector (with matrix components). The component of a vector $\vec V$ in a direction $\hat n$ is $\hat n \cdot \vec V$. $\endgroup$ Nov 12, 2021 at 21:23
  • $\begingroup$ @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. $\endgroup$
    – jacob1729
    Nov 12, 2021 at 21:57
  • $\begingroup$ @jacob1729 you guessed right as per his subsequent edit. XY problem… $\endgroup$ Nov 13, 2021 at 12:12
  • $\begingroup$ Is the OP comfortable with the rotation group in, e.g., 't Hooft's notes, or any of the nice texts of this? $\endgroup$ Nov 13, 2021 at 14:24

2 Answers 2


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


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,


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:


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:



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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.