# Representation of $SU(2)$, i.e., spin

Let $$$$X= \begin{bmatrix} 0 & 1\\ 0 & 0\\ \end{bmatrix}, \qquad Y= \begin{bmatrix} 0 & 0\\ 1 & 0\\ \end{bmatrix}, \qquad H= \begin{bmatrix} 1 & 0\\ 0 & -1\\ \end{bmatrix}\tag{1}$$$$ If $$V_m$$ is the $$(m+1)$$-dimensional complex representation of $$\text{sl}(2,\mathbb{C})$$, then we know that there exists a basis $$u_m,u_{m-2},...,u_{-m}$$ such that $$u_k$$ is the eigenvector of $$H$$ with eigenvalue $$k$$ and $$Y u_k = u_{k-2}$$.

In physics, on the other hand, we write $$S_{\pm} =X,Y$$ and $$S_z = H/2$$ and use the basis $$|s,m_s \rangle$$ such that \begin{align} S_z |s,m_s \rangle &= m_s |s,m_s \rangle \\ S_{\pm} |s,m_s \rangle &= \sqrt{s(s+1)-m_s (m_s \pm 1)} |s,m_s \pm 1\rangle \end{align}\tag{2} and that $$|s,m_s \rangle$$ are orthogonal.

Is there a deeper reason why we put the coefficients in front of $$|s,m_s \rangle$$, when $$S_{\pm}$$ acts on it and why $$|s,m_s \rangle$$ are orthogonal?

In the specific case of the 2-dimensional representation, the coefficients are 1 so it doesn't matter much. On the other hand, for the higher-dimensional reps of $$SU(2)$$, the coefficients in front are not trivial, v.g. your raising operator $$X\to \sqrt{2}\left(\begin{array}{ccc}0&1&0\\0&0&1\\ 0&0&0\end{array}\right)$$ and for even larger representations the coefficients are not all the same, $$v.g.$$ $$X\to \left(\begin{array}{ccc}0&2&0&0&0\\0&0&\sqrt{6}&0&0\\ 0&0&0&\sqrt{6}&0\\ 0&0&0&0&2\end{array}\right)\, .$$
If you don't have the correct coefficients your matrices will not be a hermitian representation and thus will not exponentiate to a unitary rep. Alternatively, if your basis states are not properly normalized you will not get a unitary rep. either. We want unitary because it preserves the (complex) inner product $$\langle \phi\vert\psi\rangle$$ and (for instance) probabilities of outcomes depend on such overlaps.
The difference between an abstract (finite-dimensional) $$su(2)$$ Lie algebra representation $$\rho:su(2)\to V$$, and applications in quantum physics, is that the vector space $$V$$ in quantum physics is typically a complex Hilbert space. In other words, in quantum physics $$V$$ additionally comes equipped with a sesquilinear form, and the image $$\rho(su(2))$$ should consist of (anti)Hermitian$$^1$$ operators, i.e. the representation should be unitary. This in turn guarantees that operators are diagonalizable (although not simultaneously), and eigenvectors are orthogonal. The eigenvalues in OP's eq. (2) are (to a large extent) dictated by the unitary representation.
$$^1$$ Hermitian or anti-Hermitian depending on conventions for factors of $$i$$.