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## Hot answers tagged group-theory

4

The confusion here arises because we are not fully analogous to non-relativistic QM here. Given a (quantum or classical) field $\phi$, we usually specify whether it is a "scalar", "spinor", "tensor", whatever field. This refers to a finite-dimensional representation $\rho_\text{fin}$ of the Lorentz group the field transforms in as an element: $$\phi ... 3 From the way it is defined \left| \Psi \right\rangle is not a vector on the sphere, but rather a vector along the z-axis between -\hat{z} and \hat{z}, because it is a linear combination of \left|0\right\rangle and \left|1\right\rangle which are both vectors along the z-axis. Now we want \left|\Psi(\theta = 0 , \phi =0)\right\rangle = ... 2 I) The main point is that the half-angle \frac{\theta}{2} doubles when we go from the ket$$\tag{1} |\psi\rangle~=~\begin{bmatrix}\cos\frac{\theta}{2} \cr e^{i\phi}\sin\frac{\theta}{2}\end{bmatrix}, \qquad ||\psi||~=~1, $$to the density matrix/operator$$\tag{2}\rho~=~| \psi\rangle \langle\psi | ~=~\frac{1}{2}\left({\bf 1}_{2\times 2}+ \vec{r}\cdot ...

2

$SU(N)$ is the $N$-fold cover of $PSU(N)$. They share the same Lie algebra, so the Yang-Mills action would look identical locally. The center of $SU(N)$ is just $Z_N$. At the level of representations, the fundamental representation of $SU(N)$ is a projective representation of $PU(N)$, and only the adjoint ones are linear representations of $PU(N)$. If the ...

2

The definition suggested by joshphysics and clarified by Qmechanic already exists in the literature under then name of representation operator. This is discussed in, e.g., Sternberg's Group Theory and Physics, as well as the somewhat more elementary text An Introduction to Tensors and Group Theory for Physicists by Jeevanjee.

1

Baez actually has another paper (with Huerta) that goes into more detail about this. In particular, Sec. 3.1 is where it's explained, along with some nice examples. The upshot is that the hypercharges of known particles work out just right so that the action of that generator is trivial. Specifically, we have Left-handed quark Y = even integer + 1/3 ...

1

How can we see that the group $N$ generated by $$g = (e^{2\pi i/3} I, -I, e^{i\pi /3}) \in SU(3)\times SU(2)\times U(1)$$ acts trivially on all fields in the Standard Model? First of all, note that $g$ is in the center of $SU(3)\times SU(2)\times U(1)$. Therefore its representative in the adjoint representation is the identity. Since gauge bosons ...

1

Tricritical Ising model belongs to the family of minimal models ($M(5,4)$). There are several different coset constructions that represent them, one of them is the following: $M(m+1,m)=SU(2)_{m-2} \times SU(2)_1/SU(2)_{m-1}$

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$g$ denotes the metric. For Euclidean space the metric is just the unit matrix $I$. For Minkowksi space, which is of interest when talking about the Lorentz group it's the Minkowski metric $\eta_{\mu \nu}$. The lower right matrix inside the Minkowski metric is the 3-dimensional unit matrix and therefore for the space-like components of the Minkowski metric ...

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I may add a few recent works that use coadjoint orbits to better understand the space of solutions of 2+1 gravity. With a cosmological constant you get Virasoro group coadjoint orbits, and without a cosmological constant you get BMS$_3$ coadjoint orbits. These are the symmetries of the spaces of solutions of the corresponding gravitational theories. The ...

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