# Tag Info

1

Left handed spinors $\psi_{\text{L}}$ transform as $$\psi_{\text{L}} \mapsto \exp\left((-\mathrm{i}\boldsymbol{\theta} +\boldsymbol{\beta})\cdot\frac{\boldsymbol{\sigma}}{2} \right)\psi_{\text{L}}\tag{1}$$ and right handed spinors $\psi_{\text{R}}$ transform as $$\psi_{\text{R}} \mapsto \exp\left((-\mathrm{i}\boldsymbol{\theta} -\boldsymbol{\beta})\cdot\... 0 The projectors P_R, P_L project \psi \in \mathcal{H}\cong \mathbb{R}^4 onto the right- and left-handed sectors of the representation of the Lorentz algebra, which are each a two-dimensional vector space, hence (locally) isomorphic to \mathbb{R}^2. What "sum" is meant here is the direct sum \oplus:$$\begin{pmatrix} \psi_1 \\ \psi_2 \end{...

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This is just a consequence of the fact that the set $\{\sigma_i\otimes \sigma_j\}_{i,j=0,\ldots,3}$ is a basis of the real vector space of hermitian $4\times4$ matrices. This vector space is equipped with an inner product such that for two elements $\alpha$ and $\beta$ we have that $$(\alpha,\beta) \equiv \frac{1}{4} \mathrm{Tr}(\alpha\,\beta) \quad.$$ Any ...

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This depends on which one is bigger. If $\ell$ is bigger than $s$, then $|\ell-s|=\ell-s$, so $j$ goes from $\ell-s$ to $\ell+s$ in unit steps, which means $2s+1$ different values. If $\ell$ is smaller than $s$, you get the same story with swapped symbols, so you get $2\ell+1$ different values. If they're equal, then both options apply.

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It sounds like Georgi's notion of 'complete reducibility' is synonymous with the notion of 'decomposability' that is more commonly used in mathematical settings. In that case, a representation is said to be indecomposable if it cannot be expressed as the direct sum of two or more proper subrepresentations. Note that this is a weaker condition than ...

4

Here are some group theory references with a focus on particle physics, basically transcribed from the introduction of my group theory notes. Nick Dorey's Symmetries, Fields, and Particles lectures as transcribed by Josh Kirklin. Introduces Lie algebras and Lie groups and outlines the Cartan classification. Rather brief, but covers the essentials for ...

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I don't think this is true, $\pi_5(G)$ has little to do with anomalies, at least not in any direct way. The general statement is: triangle anomalies for a given symmetry $G$, in $d$ dimensions, are classified by the free part of $\Omega_{d+2}(G)$. Here $\Omega_n(G)$ denotes the cobordism group of $G$, namely the collection of $n$ dimensional manifolds $M_n$ ...

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Any $4$-vector forms an irreducible representation of the Lorentz group, since any Lorentz transformation mixes all four components. But from the point of the $SO(3)$ subgroup it is reducible since spatial rotations do not mix the temporal component $V^0$ with the spatial components $V^i$ of the $4$-vector. Clearly the temporal component is invariant under ...

1

Hint: Using Young diagrams the tensor decomposition ${\bf 15}\otimes{\bf 4}\cong{\bf 36}\oplus{\bf 20}\oplus{\bf 4}$ amounts to \begin{array}{rl} [~~]&[~~]\cr [~~]\cr [~~] \end{array} \quad\otimes\quad[a] \quad\cong\quad \begin{array}{rl} [~~]&[~~]&[a]\cr [~~]\cr [~~] \end{array} \quad\oplus\quad \begin{array}{rl} [~~]&[~~]\cr [~~]&[a]\... 2 Yes, the color charge of a quark can be represnted by a vector in \mathbb{C}^3. What's more, if you were to "hold" the quark (which you cannot actually do, but let's put that aside) and move it around a background gluon field, that \mathbb{C}^3 vector would be be "rotated" by an SU(3) matrix. This is a lot like how if you parallel ... 2 Maybe this link will help: Color charge is a property of quarks and gluons that is related to the particles' strong interactions in the theory of quantum chromodynamics (QCD). The "color charge" of quarks and gluons is completely unrelated to the everyday meaning of color. The term color and the labels red, green, and blue became popular simply ... 1 A choice be made for all eigenvalues to be either even or odd ? The analysis in group theory can classify how many different sysmmetry styles will be in the eigen functions according to the symmeytry group of the Hamiltonian. It can tell from the given basis functions to determine what irreducible representations (symmetric styles) will show up in the ... 1 The gamma matrices are a representation of a Clifford algebra, whilst the 'generators' is a representation of a Lie algebra. These are different things. It's best not to mix up these representations of different objects. 1 We should distinguish between two things: A global symmetry transform (e.g., global rotation) clearly has no observable physical effect. If we rotate the whole universe, we haven't changed anything observable. In contrast, the existence of a global symmetry (I mean the mathematical exisitence, in the model) typically does have observable consequences. That'... 1 Your question is founded on a false premise. Spacetime symmetries are not redundancies. Picking up my tea cup and moving it across the room is a genuine transformation, not a gauge transformation of some kind, and claiming that the laws of physics are invariant under such transformations is making a genuine statement about the universe. One could imagine a ... 4 Now by the last statement we should have a map from \mathrm{SO}^+(1,3) to \mathrm{SL}(2,\mathbb C). How can we define this map? That's not quite right. What that statement says is that we should have a map from paths^\ddagger on \mathrm{SO}^+(1,3) to paths on \mathrm{SL}(2,\mathbb C). In other words, if we start at some point p\in \mathrm{SO}^+(... 1 The answer by Jeanbaptiste Roux contains the exact form of the mapping (Lie group homomorphism) from \mathrm{SL}(2,\mathbb C) to \mathrm{SO}^{+}(1,3;\mathbb{R}) (the latter is also denoted in the literature as \mathfrak{L}^{+}_{\uparrow} or \mathfrak{Lor}(1,3)). As a mapping, it is not bijective, so it does not an inverse in the proper mathematical ... 0 Since passive rotation and active rotation are equivalent, if we take the active point of view things should be more clear. What I mean is this : instead of taking the approach where it is the apparatus that is rotating we take the approach that is the spinor which is rotating. 0 First there is a bijection between Minkowski space, M and Hermetian 2x2 matrices, Hm[2], this is given by X: M \rightarrow Hm[2] where x=(x^\mu) \rightarrow x^\mu.\sigma_\mu. We then see that det[X(x)] = |x|^2, where the latter is the square of the Minkowski norm. Next, we define an action of SL(2,C) on Hm[2] by S.A:=SAS^\dagger where S \in ... 0 Let's define \tilde{\sigma}^\mu \equiv \tilde{\sigma}_\mu the Pauli matrices, with \sigma^0=\text{Id} that does not change when we raise or lower the index. There exist an M isomorphism between the set of 2\times 2 hermitian matrices and \mathbb{R}^{1,3}: \begin{align*} M : \mathbb{R}^{1,3} & \longrightarrow MH(2,\mathbb{C}) \\ x^\mu &\... 1 OP particular question boils down to the use of the polarization identity, cf. e.g. this related Phys.SE post. More generally for the group homomorphism, see e.g. this related Phys.SE post. 2 There is no relation at all between the particle fields \phi that are sections of associated bundles and the gauge field A. A is the field of the gauge bosons, \phi is a field of a particle charged under the interaction with that boson. The Standard Model does not contain particles charged in the adjoint of some gauge group, so there is no analogue ... 1 It seems you are essentially asking two questions which aren't really related. One of the two can be phrased as: given a Lie algebra-valued 1-form A (whether it's a connection or not is irrelevant to this question), does there exist a spacetime vector field v such that on some patch U of spacetime we may obtain A(v)=\phi for an arbitrary Lie algebra-... 0 First of all there are covariant and contravariant spinors which transform according to representations of SL(2,\mathbb{C}) like: \psi'_A = M^{\,B}_A \psi_B \quad\text{and}\quad \psi'^C = M_{\,D}^C \psi^D \quad\tag{1} $$. The matrix M^{C}_{\,D} = (M^{-1T})_{C}^{\,D} is the contragredient to M. So it is a different matrix, but actually it is ... 2 I think this is a matter of convention because (m+n) will be the highest spin irrep. present. What is certainly true, however, is that the sum is insufficient to specify all the information about how the particle transforms. For example the (1,0) and (1/2,1/2) representations are certainly not the same, though the sum matches. It's still interesting, ... 5 Well, do they? In your ("modern", half-scale) conventions, Q=T_3+Y_W, you observe the lepton isodoublets with Y_W=-1/2,$$ l_L = \begin{pmatrix} \nu\\ e^-_L\end{pmatrix} $$the isosinglets with Y_W=-1$$ e^-_R,  (and, arguably, $\nu_R$ with $Y_W=0$). They all have lepton number 1, which is thus lacking from the Higgs doublet, with \$Y_W=1/...

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