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2

The $SU(2)$ triplet results from the Adjoint Representation $\mathrm{Ad}: SU(2)\to SO(3)$ of $SU(2)$, whereby $SU(2)$ acts on its own Lie algebra. As a $2\times2$ matrix, an element of the Lie algebra $\mathfrak{su}(2)$ can be written: $$X=\left(\begin{array}{cc}i\,z&i\,x - y\\i\,x + y&-i\,z\end{array}\right)=i\,(x\,\sigma_x+y\,\sigma_y + z\,\...


2

In quantum mechanics, operators $\{J_x,J_y,J_z\}$ measuring the angular momentum of a state are required to obey the commutation relations \begin{equation} [J_i,J_j]=i \sum_k \epsilon_{ijk} J_k. \end{equation} If we only care about the spin of a particle, which does not know about the wavefunction, the state of a particle becomes a length $n$ vector (we do ...


3

The group elements are in principle abstract objects defined by the way they act on some structure. For example, the rotation group in three dimensions is formed by elements that rotate coordinate systems in some appropriate way. In order to make things easier to understand and visualize we assign linear representations, i.e. matrices to the elements of ...


6

This is what happens when physicists try to do group theory but don't bother introducing the proper mathematical notions. There is no isomorphism between $\mathrm{SO}(1,3)$ and $\mathrm{SU}(2)\times\mathrm{SU}(2)$, the former is non-compact, the latter is compact. More around this apparently confusing topic can be found in this answer. Furthermore, using ...


3

When the gauge theory is quantized in the proper way, you cannot even meaningfully talk about the action of an $\mathrm{SU}(3)$ transformation on the space of states because the quantization of a gauge theory essentially requires you to quotient out the gauge transformations (all of them, including the global ones), so that they are "do nothing" ...


1

A 2-dimensional vector space requires 2 basis vectors $v_1$ and $v_2$ to span it. These vectors should not be thought of as either real or complex per se. Instead, for a real vector space arbitrary combinations $a v_1 + b v_2$, $a$ and $b$ real, are also in the space, while for a complex vector space $a$ and $b$ can be complex. An illuminating example is ...


1

A rotation is of the form $$\begin{bmatrix} \cos(\theta) & - \sin(\theta) \\ \sin(\theta) & \cos(\theta) \end{bmatrix}$$ A reflection is of the form $$\begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ \sin(2\theta) & - \cos(2\theta) \end{bmatrix}$$ If we want to find a $2-$dimensional representation of a $3-$dimensional rotation then we can ...


0

$j=\frac{1}{2}$ representation is the fundamental representation of the group $SU(2)$. $SU(2)$ is the group of $2\times 2$ Unitary matrices with determinant $+1$. The group $SU(2)$ does act on a two dimensional complex space that you have described. $SU(2)$ has an algebra $su(2)$ whose representations may act on vector spaces of different dimensions. ...


0

One place you could look for a rather neat derivation (that I haven't really found elsewhere) is Lecture 38 and 39 from the series that Sidney Coleman gave at Harvard in 1976. The series is available online at the Harvard physics website. He says he learnt that method himself from Smorodinsky (Russian mathematician) at the Dubna conference (probably in the ...


1

When we say scalar, spinor, vector, and so on, field, we mean which representation of the frame bundle the field belongs to. Or in index notation, which spacetime indices the field has: none, spinor, vector, and so on. We can combine this with internal symmetries which are $G$-bundles for some gauge group $G$, for example $SU(2)$. In indices this is some ...


3

Let us consider an example and take the Weinberg-Salam Lagrangian: $$ \mathscr{L} = i\bar{\psi}\gamma\cdot\partial\psi - m\bar{\psi}\psi $$ and let us adapt it to the case describing electrons and neutrinos as $$ \mathscr{L} = i\bar{\textrm{e}}_R\gamma\cdot\partial\textrm{e}_R + i\bar{\textrm{e}}_L\gamma\cdot\partial\textrm{e}_L + i\bar{\nu}_L\gamma\cdot\...


6

I can't give an answer using fiber bundles, but I don't think it is important as the confusion is at a much simpler level. A field can be in different representations for different symmetry group. The Higgs field is in the trivial representation of the Poincarre group, that is, under Lorentz transformations, $\phi(x)\to \phi(\Lambda x)$, but in non-trivial ...


1

To answer first to your last question, it is an experimental fact that only left-handed fermions are affected by weak interactions. The couplings of the fermions to the $W$ bosons is proportional to their weak isospin $T_3$. Therefore, right-handed fermions must have $T_3=0$. Left-handed fermions are observed to have the same isospin (charged currents ...


2

The labelling of finite-dimensional irreducible representations of (the universal cover of the connected component of) the Lorentz group $\mathrm{SO}(1,3)$ by two half-integers $(s_1,s_2)$ arises as follows: The complexified Lorentz algebra $\mathfrak{so}(1,3)_\mathbb{C}$ has $\mathfrak{su}(2)\times\mathfrak{su}(2)$ as a compact real form, and the finite-...


0

Before going into the details, let me describe pictorially how the Hamiltonian, the Symmetry group, and the Dynamical group look in a basis in which the Hamiltonian is diagonal. Hamiltonian $$ H = \begin{bmatrix} \begin{bmatrix} \lambda_1 \mathbf{1} \end{bmatrix} & & & \\ & \begin{bmatrix} \lambda_2 \mathbf{1} \end{...


1

It seems like you're fixing the representation of $\mathrm{SU}(2)$ to be $T^i=\frac{1}{2}\sigma^{i}$ (i.e., the fundamental representation). This makes sense if you're talking about the Higgs mechanism. Now you want to find the generator $Y$ for the $\mathrm{U}(1)$ part of $\mathrm{SU}(2)\times\mathrm{U}(1)$. Let's call that generator $Y$. Now, any ...


4

You already got your answer, all right, several times over, but I will emphasize the central puzzle of your question which you only got indirect answers for, connected to the peculiar special structure of SO(4). Any self-respecting text introducing the standard model more or less has it. I'll skip all superfluous issues like lagrangian terms, the U(1)s, etc....


7

The boost generators are First of all, note that You've chosen specific representation of Lie algebra generators of Poincare group, which is vector-like matrix representation. There are many representations in general (below I'll write about them). In Your question, You've chosen the matrix representation of Poincare group algebra generators in pseudo-...


1

Answer of this question is quite subtle. First let us consider the most general Higgs potential which is renormalizable and invariant under $SU(2)_{L}\otimes U(1)_{Y}$ gauge transformations, which has the form \begin{equation} V = \lambda(\phi^{\dagger}\phi-\mu^{2})^{2} \end{equation} Where \begin{equation} \phi = \frac{1}{\sqrt{2}}\begin{pmatrix} \phi_{1}+...


0

After a bit of discussion I believe there is actually a $SU(2)\times SU(2)$ symmetry in a sense. So in principle there is a $U(2)$ symmetry if $\phi=(\phi_1,\phi_2)^T$, $\phi^\dagger=(\phi_1^*,\phi_2^*)$ and the lagrangian $$\mathscr{L}=\partial_\mu \phi^\dagger\partial^\mu \phi-m\phi^\dagger\phi-\lambda(\phi^\dagger\phi)^2,$$ simply sent $\phi\to U\phi$, ...


-1

If the field is a simple complex scalar field, than the symmetry is just $U(1)$. For a higher symmetry, $\phi$ need to be higher dimensional too, for instance you can add a vector index $\phi_i$ with $i=1,2$ for simplicity, which means that you add an additional complex field. If these two fields interact, you can have two cases now: Each field has a $U(1)$ ...



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