# Georgi-Glashow model of W bosons/photons

• I
Gold Member

## Main Question or Discussion Point

The following is from my notes:

In 1972, a model was proposed by Georgi and Glashow as a candidate theory describing W bosons and photons with Lagrangian $$\mathcal L = -\frac{1}{2} \text{Tr} F^{\mu \nu}F_{\mu \nu} + (D_{\mu} \phi)^T (D^{\mu} \phi) - \mu^2 \phi^T \phi - \lambda(\phi^T \phi)^2$$ with $F_{\mu \nu}$ the field strength tensor and $A_{\mu}^a$ the gauge fields of the gauge group SO(3), $D_{\mu} = \partial_{\mu} + ig A_{\mu}^a \tau^a$, and $\phi$ is a 3 component real scalar field.

It's clear that the generator basis is $i(\tau)_{jk} = \frac{1}{2} (\delta_{jk} - \delta_{kj})$ for $1 \leq j < k \leq 3$ which may be applied to the vacuum expectation value $$\phi_{min} = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ 0 \\ v \end{pmatrix}$$ to deduce the number of Goldstone bosons, in accordance with Goldstone's theorem....

My question is simply, why is that a generator basis? First of all there are three generators in SO(3) so I expected to see another index on the $\tau$ to label each generator. Secondly, for any $1 \leq j < k \leq 3$ the components $\tau_{jk}$ are all identically zero (!). So clearly I am misunderstanding something here. Can anyone help?

Thanks!

Related High Energy, Nuclear, Particle Physics News on Phys.org
nrqed
Homework Helper
Gold Member
The following is from my notes:

In 1972, a model was proposed by Georgi and Glashow as a candidate theory describing W bosons and photons with Lagrangian $$\mathcal L = -\frac{1}{2} \text{Tr} F^{\mu \nu}F_{\mu \nu} + (D_{\mu} \phi)^T (D^{\mu} \phi) - \mu^2 \phi^T \phi - \lambda(\phi^T \phi)^2$$ with $F_{\mu \nu}$ the field strength tensor and $A_{\mu}^a$ the gauge fields of the gauge group SO(3), $D_{\mu} = \partial_{\mu} + ig A_{\mu}^a \tau^a$, and $\phi$ is a 3 component real scalar field.

It's clear that the generator basis is $i(\tau)_{jk} = \frac{1}{2} (\delta_{jk} - \delta_{kj})$ for $1 \leq j < k \leq 3$ which may be applied to the vacuum expectation value $$\phi_{min} = \frac{1}{\sqrt{2}} \begin{pmatrix} 0 \\ 0 \\ v \end{pmatrix}$$ to deduce the number of Goldstone bosons, in accordance with Goldstone's theorem....

My question is simply, why is that a generator basis? First of all there are three generators in SO(3) so I expected to see another index on the $\tau$ to label each generator. Secondly, for any $1 \leq j < k \leq 3$ the components $\tau_{jk}$ are all identically zero (!). So clearly I am misunderstanding something here. Can anyone help?

Thanks!
Your understanding is correct and the notes are clearly wrong. It is better to simply take the three Pauli matrices and forget about this definition in terms of Kronecker delta (which does not make any sense as it is identically zero since the delay is symmetric as you know, and you are correct that one index is missing).

Gold Member
Hi nrqed,
Your understanding is correct and the notes are clearly wrong. It is better to simply take the three Pauli matrices and forget about this definition in terms of Kronecker delta (which does not make any sense as it is identically zero since the delay is symmetric as you know, and you are correct that one index is missing).
Ok thanks. The triplet field vector would transform under the fundamental representation of SO(3) or adjoint representation of SU(2), right? But the Pauli matrices are used for the fundamental representation of SU(2) so is it a case I should use the linearised versions of the matrix representations for the generic rotations around the x,y,z axes for example to constitute the generators of SO(3)?