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

  1. Nov 17, 2016 #1
    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. jcsd
  3. Nov 17, 2016 #2
    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?
     
  4. Nov 18, 2016 #3

    ChrisVer

    User Avatar
    Gold Member

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

    can you please clarify?
     
  5. Nov 18, 2016 #4
    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?
     
  6. Nov 18, 2016 #5
    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.
     
  7. Nov 18, 2016 #6
    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

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