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I was reading about the pion ($\pi$) SU(2) representation and stumbled upon an expression for the isospin operator, $$ I_i=\epsilon_{ijk}\int d^3x\phi_j \dot\phi_k=-i\int d^3x \dot{\phi}^T t_i \phi ,$$ where $\phi$ stands for the isospin-vector field that can be decomposed into three real components $\phi_1, \phi_2,\phi_3$, or one complex field $\Phi=\frac{1}{\sqrt{2}}(\phi_1-i\phi_2)$ plus one real field $\phi_3$.

And then it's said that the $t_i$ for which $$ [t_i,t_j]=i\epsilon_{ijk} t_k \qquad \hbox{and} \qquad (t_i)_{jk}=-i\epsilon_{ijk}$$ form the adjoint representation of the isospin group.

And here I am a bit confused: if the $t_i$ form the adjoint representation of the isospin group, what are the $I_i$?

I also read in one book that there exist two representations of the isospin group the canonical one (written in the $\pi^+,\pi^0,\pi^- $ basis):

$T_1=\sqrt{2}\begin{bmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{bmatrix} $ $, \quad T_2=\sqrt{2}\begin{bmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end{bmatrix} $

$T_3=\begin{bmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{bmatrix} $

and the above-mentioned adjoint one. Are the $I_i$ maybe the canonical representation?

If the $t_i$ are written as: $(t_i)_{jk}=-i\epsilon_{ijk}$, and I calculate a matrix for the $t_i$ operator, which basis is meant? the $\pi^+,\pi^0,\pi^-$ basis or the $\pi_1,\pi_2,\pi_3$ basis?

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  • $\begingroup$ Review your spherical basis. The ts and Ts are matrix representations, while the Is are pion quantum field operator representation charges. Choose a book you understand. $\endgroup$ – Cosmas Zachos Sep 26 '19 at 21:40
  • $\begingroup$ Note footnote 3 here giving the explicit similarity transformation changing basis from your t to your T matrices: they are the same adjoint representation in different bases. Your I are the Jordan realization of the t basis matrices. $\endgroup$ – Cosmas Zachos Sep 27 '19 at 11:08
  • $\begingroup$ Jordan realization. $\endgroup$ – Cosmas Zachos Sep 27 '19 at 11:20
  • $\begingroup$ @CosmasZachos, I think, I understand things now. The $t_i$ form a (matrix) representation for the isospin $SU(2)$ group in the $\{\pi^1,\pi^2,\pi^3\}$ basis, whereas the $T_i$ are just the $t_i$ transformed to the $\{\pi^+,\pi^-,\pi^0\}$ basis and the transformation is given in terms of an orthogonal operator. The $I_i$ on the other hand represent the isospin operators $T_i, t_i$ analogy in QFT, for example: $I_3$ gives $1,0,-1$ if the field $\phi$ is in the $\pi^+,\pi^0,\pi^-$ state respectively. $\endgroup$ – Luka8281 Sep 27 '19 at 12:46
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    $\begingroup$ Perhaps you might answer your own question concisely and explicitly for future students in the same straits... "Canonical" for the spherical basis refers to the fact it works for all spins, not just spin one, and amounts to the ladder operators used in the QM $|lm\rangle$ rep. $\endgroup$ – Cosmas Zachos Sep 27 '19 at 13:17
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I think I've managed to clear up my confusion and would like to summarize the above comments in an answer. Pions are described in QFT using three different fields, all satisfying the K-G equation. Due to their (almost) equal mass the three fields correspond to the same isospin number $T=1$ and form a basis $\{\pi^+,\pi^0,\pi^-\}$ for the $SU(2)$ representation, when stacked in matrix form:

$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\phi=$$ \begin{pmatrix} \pi^+ \\ \pi^0 \\ \pi^- \\ \end{pmatrix} $.

The isospin generators of $SU(2)$ in the 3-dimensional isospin space spanned by $\{\pi^+,\pi^0,\pi^-\}$, then take the "canonical" form:

$\qquad\qquad\qquad T_1=\sqrt{2}\begin{pmatrix} 0 & 1 & 0\\ 1 & 0 & 1\\ 0 & 1 & 0 \end{pmatrix} $ $, \quad T_2=\sqrt{2}\begin{pmatrix} 0 & -i & 0\\ i & 0 & -i\\ 0 & i & 0 \end{pmatrix}, $ $\quad T_3=\begin{pmatrix} 1 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & -1 \end{pmatrix} $.

We can also choose a different basis, for example the "adjoint" basis $\{\pi^1,\pi^2,\pi^3\}$. Now the generators take the "adjoint" form:

$\qquad\qquad\qquad t_1=i\begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & -1\\ 0 & 1 & 0 \end{pmatrix} $ $, \quad t_2=i\begin{pmatrix} 0 & 0 & 1\\ 0 & 0 & 0\\ -1 & 0 & 0 \end{pmatrix}, $ $\quad t_3=i\begin{pmatrix} 0 & -1 & 0\\ 1 & 0 & 0\\ 0 & 0 & 0 \end{pmatrix} $,

the transformation $\{\pi^+,\pi^0,\pi^-\}\rightarrow \{\pi^1,\pi^2,\pi^3\}$ is given in terms of an unitary operator $t_i=U^{\dagger}T_iU$, as can be read here (see answers). The $t_i$ form the adjoint represntation, so that their elements are given as structure constants of the $SU(2)$ group, meaning $(t_i)_{jk}=-i\epsilon_{ijk}$, where $\epsilon_{ijk}$ are the structure constants: $[t_i,t_j]=i\epsilon_{ijk} t_k$.

The $ I_i=-i\int d^3x \dot{\phi}^T t_i \phi$ on the other hand represent the QFT analogy of $T_i$ and $t_i$ and are the isospin operators in QFT, for example the $I_3$ gives $+1,0,-1$ if the field $\phi$ is in the $\pi^+,\pi^0,\pi^-$ state respectively. This is realized in terms of the Jordan map, but in a QFT sense, where each field is treated as an infinite sum of oscillators.

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