# A Isospin symmetry as an $SU(2)$ symmetry

1. Nov 17, 2016

### spaghetti3451

The generators for the isospin symmetry are given by

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

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

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

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

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 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?

2. Nov 17, 2016

### eys_physics

For your first question, the SU(2) algebra your are referring to is wrong. It should be $[T_{i},T_{j}]= i\epsilon_{ijk}T_{k}$ where $i,j,k=1, 2, 3$. Here $T_\pm=T_1\pm i T_2$. From your representation of $T_\pm$ you can get $T_{1,2}$ and the matrices will satisfy the commutator relation.our

I don't really understand your second question. The above generators correspond to isospin $1/2$ and you can of course consider other values of $T$, as you do for angular momentum. Is this what you mean?

3. Nov 18, 2016

### ChrisVer

No they don't.... $[T_3,T_\pm] = cT_\pm$. That's the reason why the T+/- are the ladder operators.
For using commutators with matrices, just do some matrix multiplication.

4. Nov 18, 2016

### spaghetti3451

Yes. That is exactly what I meant.

For spin-$1/2$, we have a $2$-dimensional matrix representation for each of the generators, and therefore two basis states. These basis states can be denoted by $|\uparrow\rangle \equiv \begin{pmatrix} 1\\ 0\\ \end{pmatrix}$ and $|\downarrow\rangle \equiv \begin{pmatrix} 0\\ 1\\ \end{pmatrix}$.

For spin $1$ (or higher spins), we have a $3$(or higher)-dimensional matrix representation for each of the generators, and therefore three (or more) basis states.

Am I right?

5. Nov 18, 2016

### eys_physics

Yes. You are correct. One example with $T=1$ is the description of the pions ($\pi^\pm$ and $\pi^0$) , as one particle having three states (projections of the isospin). Notice, also that same coupling rules apply for isospin as for spin.

6. Nov 18, 2016

### spaghetti3451

Say that you now have the states $$|\uparrow,\uparrow\rangle = |\uparrow\rangle\otimes|\uparrow\rangle,\qquad|\uparrow,\downarrow\rangle = |\uparrow\rangle\otimes|\downarrow\rangle,\qquad |\downarrow,\uparrow\rangle = |\downarrow\rangle\otimes|\uparrow\rangle\qquad |\downarrow,\downarrow\rangle = |\downarrow\rangle\otimes|\downarrow\rangle.$$

and you want to know how the generators $T_+$, $T_-$ and $T_3$ act on this set of four basis states.

Is it legit to define the generators $T_{++}$, $T_{+-}$, $T_{+3}$, $T_{-+}$, $T_{--}$, $T_{-3}$, $T_{3+}$, $T_{3-}$ and $T_{33}$ so that, for example,

$$T_{++}=\begin{pmatrix}T_{+} & 0\\ 0 & T_{+}\end{pmatrix}$$

and

$$T_{++}|\uparrow,\uparrow\rangle = T_{+}|\uparrow\rangle\otimes T_{+}|\uparrow\rangle$$?