In Electroweak Interactions by Renton (page 407) there is the calculation of matrix elements for electromagnetic interactions/weak interactions involving nuclei:

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A point of confusion for me is the reference to 8.31 is an isovector part of the matrix element. $\tau_3$ is a 2x2 Pauli matrix, which forms part of the spin 1/2 representation of $SU(2)$.

I thought isovector would refer to matrix elements of the spin 1 representation, involving 3x3 matrices. Have I misunderstood the language or is there something else going on here?

  • $\begingroup$ Both $u$ and $\bar u$ belong to the spin-1/2 representation of SU(2), so their product belongs to a direct product of two spin-1/2 representations. Inserting the Pauli matrices simply projects out the spin-1 component of the direct product. Inserting the unit matrix, on the other hand, projects out the singlet (spin-0) component of the direct product. $\endgroup$ – Tomáš Brauner Oct 25 '19 at 15:26
  • $\begingroup$ Thanks for the comment Tomas. I'm not familiar with a direct product of representations, is this the same as a tensor product? ie you have an SU(2) rep that acts on $u$ and one that acts on $\bar{u}$ and the matrix element then transforms as the tensor product of the two, $2\otimes 2 = 3 \oplus 1$. $\endgroup$ – CT1234 Oct 25 '19 at 15:50
  • $\begingroup$ Why does inserting a Pauli matrix project out the spin 1 component specifically? Consider the situation where you had two spin 1 representations in the matrix element, which would decompose as $3\otimes 3 = 5 \oplus 3 \oplus 1$, what are the projection operators for the quintet and triplet parts then? $\endgroup$ – CT1234 Oct 25 '19 at 16:00
  • $\begingroup$ The 3 is the antisymmetric piece (the cross product!) and the remaining 5 (traceless symmetric tensor) and singlet (dot product) are the symmetric pieces. $\endgroup$ – Cosmas Zachos Oct 25 '19 at 16:12

Leaving out irrelevant parts, $\bar u ' \gamma^\mu\tau_3 u$ is a "piece" (the third component) of an isovector $\bar u ' \gamma^\mu {\boldsymbol \tau} u$.

You have built this isovector (triplet rep) out of two isospinors (doublet reps) u, which transform under su(2) via 2×2 generator matrices, like the ${\boldsymbol \tau}/2$. In turn, the isotriplet you have transforms under su(2) under 3×3 matrix generators ${\boldsymbol T}$, (thus obeying the same su(2) algebra!) essentially rotating a 3-vector infinitesimally. To see how rotating two isospinors by some angle amounts to precisely rotating an isovector by the very same angle explicitly see this answer: acting on both sides of Pauli matrices amounts to rotating their adjoint (triplet) indices by the Rodrigues formula--the sweet magic of group theory$^\dagger$.

Physicists summarize this composition of irreps by 22 = 31 , formally identical to adding two spin 1/2s to get a spin 1 and a spin 0. The triplet is the symmetric part of the product, enforced by the symmetry structure of the isospinors, whereas the singlet is the antisymmetric part, (e.g. $\bar u' u$. Here, you need consider the reversal of up and down components of the spinor with the telltale - sign, a feature of the conjugate rep of su(2)).

$^\dagger$ Compute $$ \exp{(i{\boldsymbol \theta}\cdot {\boldsymbol \tau})}~~{\boldsymbol \tau}~~ \exp{(-i{\boldsymbol \theta}\cdot {\boldsymbol \tau})} = \bigl ( \boldsymbol{\tau} ~ \cos (2\theta)+ \boldsymbol{ \hat{\theta} }\times \boldsymbol{\tau} ~\sin (2\theta)+ \boldsymbol{\hat{\theta}} ~ \boldsymbol{\hat{\theta}} \cdot \boldsymbol{\tau} ~ (1-\cos (2\theta))\bigr ) ~, $$ which amounts to the Rodrigues rotation on triplet indices by $\exp (i2\boldsymbol { \theta \cdot T})$, since the Pauli matrices are normalized by 1/2, so $\theta$ is the half angle.

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