New answers tagged

1

Your feeling looks very misguided. Whatever you do, stay away from SU(3) for rotations. The rotation group and its Lie algebra are always linked to SO(3) ~ SU(2), to avoid formal forays into double covers and half angles. Read up on the spin matrices for any representation of the very same group (any spin). There are, in fact, simple systematic ...


1

Well, you might have spared yourself confusion and grief by checking your peculiar language in SO(3), which any undergraduate is familiar with. Let me illustrate this for SO(3), before moving on to the much messier SU(3). For SO(3), Kronecker-composing two vectors (spin 1, so 3 s) yields a spin 2 quintet (call it φ, so 5), a triplet (π) and a singlet (s), ...


2

Let $g\in G$ (more strictly let $g$ be an element of a unitary representation of $G$ on our Hilbert space $\mathbb{H}$) then $g$ leaves the SE invariant: $${\mathbf H} g\psi = Eg\psi$$ Since $E$ is just a number we can commute $E$ and $g$ freely so: $${\mathbf H} g\psi = gE\psi$$ Left multiplying the original symmetry equation by $g$ gives: $$g{\mathbf ...


2

The spin $s$ of a particle characterizes how the rotation generators act on it. In $D$ dimensions, you represent the little group $SO(D-1)$ for massive particles and $SO(D-2)$ for massless ones. In fact, you really need to consider its universal cover $\textrm{Spin}(n)$ which happen to be just its double cover. Now, you can define the spin to be the largest ...


2

Well, you might not really want to overthink it. It all hinges on the Pauli matrix $\sigma_1$ which, yes, is the permutation group with two components, with eigenvalues $\pm 1$ for $v^T=(1,1)$, symmetric, and $w^T=(1,-1)$, antisymmetric, respectively. Equivalently to your decomposition, you may write the hamiltonian as $$ H= \epsilon_0 I_4 + ...


1

To a given Lie algebra $\mathfrak g$ there is a unique group $\tilde G$, called the universal covering group, with the property of being simply connected. For example, the covering group of the algebra $\mathfrak{su}(2)$ is $SU(2)$. The other groups, $\{G\}$, associated to the same algebra can be obtained from the covering group in the following way ...


4

The vector $(0,0,0,v)$ is left invariant by the set of matrices of the form \begin{align*} M=\begin{bmatrix} R & \vec 0 \\ \vec 0^T & 1\end{bmatrix} \end{align*} where $\det(M)=\det(R)=1$ and $M^{-1}=M^T$ implies $R^{-1}=R^T$. By definition, $SO(3)$ is the group of 3 by 3 orthogonal matrices with determinant 1. In general, you need to know the Lie ...


0

(I) Assuming there are $N$ distinct fermions in your Lagrangian (e.g. quarks in the standard model), the kinetic term will have the $U(N)\times U(N)$ symmetry. This symmetry is broken by the mass term, however, which couples fermions with different handedness. (II) If $V$ is the fundamental representation of a Lie group (with dual/conjugate representation ...


1

Yes, the first part of your question is appreciated and answered soundly. The fermion kinetic term splits into two independent parts involving left and right Weyl spinors respectively, so each is independent under a separate U(N) as your wrote down. The second question is a matter of language. A generator is a matrix with one adjoint index, a here, ranging ...


0

I will talk about $SU(3) \times SU(2)$. First, a matrix $T_3 \in SU(3)$ acts in the fundamental representation on $\mathbb C^3$ in the following way: A vector $\vec v \in \mathbb C^3$ with components $v_i$ is mapped to $v'_i = (T_3)_{ij} v_j$. Similarly, a $T_2 \in SU(2)$ acts on $\vec w \in \mathbb C^2$ as $w'k = (T_2)_{kl} w_l$. The bifundamental ...


2

Let $\phi : \mathbb{R}^4\to V$ be a field with (complex) target vector space $V$, transforming in a finite-dimensional projective representation $\rho_\text{fin} : \mathrm{SO}(1,3)\to\mathrm{U}(V)$. As it is a field, the representation of the translations $\mathbb{R}^4$ on $V$ is the trivial one, since the field transforms as $\phi(x)\overset{x\mapsto ...


0

Yes In 5 dimensions, the R-Symmetry group is $SU(2)$. Any spinor, therefore, has to transform under $SU(2)$ as $\delta_{SU(2)} (\Lambda^{ij}) \lambda^i = - \Lambda^{i}{}_{j} \lambda^j$. You cannot consider a spinor $\psi$ without the $SU(2)$ index, i.e. $\delta_{SU(2)} (\Lambda^{ij}) \psi =0$. The algebra would not close on such a field, thus $\psi$, if not ...


3

There most definitely is, and your text should have used it in defining the unitary gauge more conventionally: the SU(2) group element parameterization of physics, that is the rotation matrix for spinors R. Absorb v into the definition of σ, where it belongs and from where it can re-emerge at will. $$R=\exp (i\theta ~\hat{n}\cdot\vec{\sigma})=I \cos \theta ...


3

Quantum Field theory is not a study of groups, but a study of physical states and observables. The groups are interesting only because they act on the objects of actual interest. The representation theory is just the study of such actions; if there is no representation of a group on interesting objects, the group itself is not interesting for QFT. Thus the ...


3

Representations represent group (algebra) elements as linear operators on a vector space, in physics the vector spaces are also usually Hilbert and the operators are often orthogonal or unitary (self-adjoint), so they do indeed introduce quite a bit of extra structure. The simplest piece to see is that the representation does not have to be faithful, i.e. ...


1

Let $\mathfrak{g,h}$ denote the respective Lie algebras of $G,H$. Note that $\mathfrak{g} = \mathfrak{h}\oplus\mathfrak{g}/\mathfrak{h}$ as vector spaces. Obviously, there is the adjoint representation of $\mathfrak{h}$ on the whole of $\mathfrak{g}$. Also, the adjoint action of $\mathfrak{h}$ on itself is a subrepresentation, since the algebras of Lie ...



Top 50 recent answers are included