Derivation of Raising operator of $\rm SU(2)$ I'm reading a paper called: "A Simple Introduction to Particle Physics Part I - Foundations and the Standard Model" and i have some questions regarding the derivation of the raising and the lowering operators of $SU(2)$.
in the adjoint representation for $j=1$ the only Cantan generator is the the $J^3$ matrix, so the root vectors are the following:
$$t_1=1, t_2=0, t_3=-1 \ \ \text{Equation(s) (1)}$$
Since we want $[J^3,E^{\pm}]=\alpha E^{\pm}$, the raising and the lowering operator needs to be in this form:
$$E^{\pm}=\alpha(J^1\pm iJ^2)$$
Next we evaluate:
$$[J^3,E^{\pm}]=2\alpha^2J^3$$
Now the paper I'm reading states the following:
From equations $(1)$ and the definitions of $E^{\pm}$ we see that $\pm(t_1-t_2)=\pm 1$, So we therefore set $\alpha^2=1 \Rightarrow \alpha=\frac{1}{\sqrt{2}}$ and we find the approriate non carter generators:
$$E^{\pm}=\frac{1}{\sqrt{2}}(J^1\pm iJ^2)$$
Question 1: During the derivation of the relations that the generators should obey the authors state that:
$$[H^a,E^{e_b}]=e^a_bE^{e_b}$$ where $H^a$ a Cartan and $E$ a non-Cartan generator. Why can the raising operator and the lowering operator be the combination of two Non-Cartan generators $J^1$ and $J^2$?
Question 2: What's the relation between the final result and equations $(1)$? While i understand what these equations mean, i don't understand how one result leads to the final result. Why do we only care about $\pm(t_1-t_2)$ and not $\pm(t_1-t_3)$ for example?
 A: I think you mean "Cartan" (after Elie Cartan, the French mathematician) rather than "Cartar"? A semi simple Lie algebra  is the direct sum of the Cartan algebra composed of a mutually commuting generators and the rest of the generators, whose skew adjointness with respect to  the Killing form    shows that they  can be gathered into pairs of ladder operators.
I have no idea, though, what you mean by $t_1$ and $t_2$. I do not have  time to   try and  find the paper you cite so as to figure out what they are. If you want help here, you need to explain what is puzzling.
If the Cartan algebra generators are $h_i$  (in  $\mathfrak {su}(2)$ there is only one, which is $J_3$) the   ladder operators  ${\bf e}_{\boldsymbol \alpha}$ are the simultaneous eigenvectors of the adjoint action
$$
{\rm ad}(h_i) {\bf e}_{\boldsymbol \alpha} \equiv[h_i, {\bf e}_{\boldsymbol \alpha}]=  \alpha_i {\bf e}_{\boldsymbol \alpha}.
$$
of the maximally commuting set $h_i$.
This means that ${\bf e}_{\boldsymbol \alpha}$ changes the eigenvalues of $h_i$ by $\alpha_i$, the $i$'th components of the root vector ${\boldsymbol \alpha}$.  For $\mathfrak {su}(2)$ the root  vector ${\boldsymbol  \alpha}$ has only one component "1" and so $J_3$  has its eigenvalue increased by unity by ${\bf e}_{\boldsymbol \alpha}=J_1+iJ_2$.  For each   ${\boldsymbol  \alpha}$ there is a second ladder operator ${\bf e}_{-{\boldsymbol \alpha}}$, in this case it is $J_1-iJ_2$ that reduces the eigenvalue of $J_3$ by unity. The set $h_i$ and the ${\bf e}_{\pm{\boldsymbol \alpha}}$ form a basis for the whole (complexified) Lie algebra.
I do not understand why you have three root vectors $t_{1,2,3}$ in your case.
