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The generators for the isospin symmetry are given by

$$ T_{+}=\left|\uparrow\rangle\langle\downarrow\right|, \qquad T_{-}=\left|\downarrow\rangle\langle\uparrow\right|, \qquad T_{3}=\frac{1}{2}\left(\left|\uparrow\rangle\langle\uparrow\right|-\left|\downarrow\rangle\langle\downarrow\right|\right),$$

where $\left|\uparrow\right\rangle$ and $\left|\downarrow\right\rangle$ form a $2$-dimensional basis of states.

In the $2$-dimensional basis of states $\left|\uparrow\right\rangle$ and $\left|\downarrow\right\rangle$, the generators $T_+$, $T_-$ and $T_3$ can be written as

$T_{+}=\begin{bmatrix} 0 & 1\\ 0 & 0\\ \end{bmatrix},\qquad T_{+}=\begin{bmatrix} 0 & 0\\ 1 & 0\\ \end{bmatrix},\qquad T_{3}=\begin{bmatrix} 1/2 & 0\\ 0 & -1/2\\ \end{bmatrix}.$

  1. The generators $T_+$, $T_-$ and $T_3$ obey the $SU(2)$ algebra $[T_{i},T_{j}]=i\epsilon_{ijk}T_{k}$ (with $\epsilon_{+-3}=1$). However, I do not get the correct commutation relations using the matrix representations of the generators.
  2. Is it possible to rewrite the generators $T_+$, $T_-$ and $T_3$ in terms of an arbitrary number of states?
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  • $\begingroup$ Are you familiar with the representation matrices of the rotation algebra? $\endgroup$
    – Cosmas Zachos
    Commented Nov 17, 2016 at 23:49
  • $\begingroup$ Yes, I am! I forgot that $T_+$ and $T_-$ are ladder operators. I need to write them in terms of $T_1$ and $T_2$ and then show that the commutation relations are satisfied. $\endgroup$
    – nightmarish
    Commented Nov 18, 2016 at 10:41
  • $\begingroup$ Do you have any answer to my second question? $\endgroup$
    – nightmarish
    Commented Nov 18, 2016 at 10:41
  • $\begingroup$ You Kronecker-multiply your states (add their spins) and Clebsch decompose into a series of arbitrary spins, as your angular momentum text should tell you. You then rotate those as per the above link. To add N spin 1/2s, see, e.g. The Catalan ladder in 249635. $\endgroup$
    – Cosmas Zachos
    Commented Nov 18, 2016 at 13:08

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