0
$\begingroup$

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

enter image description here

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?

$\endgroup$
  • $\begingroup$ 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 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 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 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 at 16:12
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.