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0

I'm not an expert but I think there's some misunderstanding since $SU(4)$ has $4^2-1 = 14$ generators, and not $6$. $6$ is the dimension of the chosen representation. Moreover $SU(2)$ has dimension $2^2-1=3$ and so requires $3$ generators which in the fundamental representation, the $2$ you mention, are given by the Pauli matrices. Moreover, beside the ...


0

The confusion here is that conjugating a Young diagram does conjugate the representation. This is compounded by the regrettable use of the dimension to label the representation. In the specific case of $\mathfrak{su}(3)$ (and more generally for $\mathfrak{su}(n)$), a better solution is to use the Dynkin labels $(a,b)$. For $\mathfrak{su}(3)$, an irrep ...


1

I'm taking the Clifford algebra route as pointed out by non-user38741 and Giorgio Comitini, but I'll try to justify intuitively how to end up there and how the spinor transformation law appears inevitable. So I start with geometric algebra, which is simply another name for Clifford algebra when used in a physics context, and the vectors are taken to be ...


13

I think you're asking for intuition in the wrong direction here. Suppose that somebody is already familiar with vectors, and wants to understand tensors. That's possible, because tensor representations are built out of vectors, i.e. the rank $2$ tensor representation is just the product of two vector representations, or equivalently a rank $2$ tensor is a ...


21

The proper analogous formalization of spinors is not to view them as some sort of different functions from tensors on the same underlying vector space $V$, but instead to expand our idea of the underlying geometry: Where tensors are multilinear functions on vector spaces, tensors with "spinor" and "vector" parts are multilinear functions on super vector ...


9

Yes. Spinors are elements of representation spaces of objects known as a Clifford algebras. A Clifford algebra is basically a vector space turned into an algebra via the product rule $$ v\cdot w=2g(v,w)\Bbb{1} $$ where $g$ is some metric on the vector space itself. The most famous Clifford algebra is the Dirac algebra, i.e. the algebra of the Dirac ...


9

Well, you should look at (irreducible) representations of the Lorentz group. Basically you want all of your ingredients to have correct and consistent transformations under the Lorentz group. The Weyl and Dirac spinors are the most basic objects satisfying that requirement. Starting from those you can build vectors as a (multiplicative) combos of two ...


1

In general, any $n\times n$ complex matrix will have $2n^2$ independent elements for real and imaginary parts. Gamma matrices must satisfy $\{\gamma^\mu,\gamma^\nu\}=2g^{\mu\nu}$. If we are considering $n$ dimensional $\gamma$ matrices, this yields $2n^2$ equations of $4n^2$ elements of $\gamma^\mu,\gamma^\nu$. So this characteristic property reduces half of ...


2

A bilinear, by definition, is an expression of the form $\bar{\psi}_i A_{ij} \psi_j$, with $A_{ij}$ some matrix. For example, the scalar has $A_{ij} = \delta_{ij}$, while the four vector bilinears are when $A$ is equal to one of the gamma matrices. And since $A$ is a $4\times 4$ matrix, the (complex) vector space of all such matrices has dimension $4\times 4 ...


0

I'm probably misunderstanding your question screened behind a wall of formalism. Looks like you are really asking a question about the spherical basis. Rather than a thrust-and-parry avalanche of comments, I'm reviewing the obvious. So, let's illustrate it out for elementary QM, spin 1/2, so J=σ/2, hence, for $\hat n= (n_1,n_2,n_3)$, $n_1^2+n_2^2+n_3^2=1$,...


1

Two reasons that come to my mind: Physicists use Hilbert spaces to describe state spaces. So a model of a particle is set to a (inner product) linear space naturally in this stream. The theory has (group) symmetry. Encoding it gives us a group linear representation on Hilbert spaces. You might worry that this could be an over-simplification. But by ...


1

You are giving very little information on what you know, your conventions, etc... With the conventions of T P Cheng & L F Li, $$ \lambda_0=\frac{1}{\sqrt{15}}\operatorname{diag}(2,2,2,-3,-3)=-\frac{6}{\sqrt{15}} \frac{Y}{2},\\ \lambda_3= \operatorname{diag}(0,0,0,1,-1)=2T_3,\implies\\ Q= \operatorname{diag}(-1/3,-1/3,-1/3,1,0), $$ by Gell-Mann–...


3

They do. Obviously they are not generated by the action of $J_\pm$ since this action cannot change the $J$ quantum number but if my typesetting is right two of them are \begin{align} &\sqrt{\frac{2c(-a+b+c)(a+b+c+1)}{2c+1}}C^{c\gamma}_{a\alpha b\beta}\\ &=\sqrt{(b-\beta)(c-\gamma)}C^{c-1/2,\gamma-1/2}_{a\alpha,b\beta-1/2} +\sqrt{(b+\beta)(c+\gamma)}...


3

This is a Lie group, and its connected component around the identity results by exponentiating the Lie algebra, whose most general element you already wrote down. Expanding the generic group element around the identity, you find, as is standard in Lie theory, $$G(\epsilon, \omega)=e^{i\epsilon_{\mu}P^{\mu}-i\omega_{\mu \nu}M^{\mu\nu}/2}= 1\!\!1 + i\...


2

OP explicitly considers the trivial representation $SU(2)\to GL(3,\mathbb{R})$. However, modulo equivalence of representations, the sought-for irreducible representation $\rho:SU(2)\to GL(3,\mathbb{R})$ is undoubtedly $\rho=R\circ\pi$, where $\pi:SU(2)\to SO(3)\cong SU(2)/ \mathbb{Z}_2$ is the covering homomorphism, and $R:SO(3)\to GL(3,\mathbb{R})$ is the ...


3

Indecomposable representations (which are non-unitary and finite dimensional) of Poincaré appear in theories of unstable particles. It's not an easy topic but was explored in this paper: Raczka, R. "A theory of relativistic unstable particles." Annales de l'IHP Physique théorique. Vol. 19. No. 4. 1973.


0

I'm at around the same point as you, so I'm not sure how well I'll be able to answer, but I'll try. I don't know if you've come across the Einstein-Cartan formulism for GR yet, but it's basically treating gravity as a gauge theory, in an analogous way to Yang-Mills. Gravity is the gauge theory of the Poincare group so spacetime does have a Poincare symmetry,...


0

In classical physics you need to describe mathematically physical quantities whose meaning in related to some directions. Some of this quantities are described with vectors, their rotation can be described with the rotation matrices of $SO(3)$. However there are some quantities (like spin in quantum mechanics) which are described with mathematical objects ...


2

To a physicist, a "group representation" is a set of matrices (i.e elements of ${\rm GL}[{\mathbb F}]$ where $\mathbb F$ is a field, often ${\mathbb F}={\mathbb C}$) having the same multiplication table as the abstract group. The representation is faithful if different group elements correpond to different matrices. To a mathematician the representation is ...


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