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.


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 '21 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 '21 at 21:57
  • $\begingroup$ @jacob1729 you guessed right as per his subsequent edit. XY problem… $\endgroup$ Nov 13 '21 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 '21 at 14:24

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:



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