# Tag Info

7

This would be the number of parameters needed to specify an element in the group $SO(n)$, and this number is $n(n-1)/2$. Thus: $n=2$ requires one parameter $n=3$ requires $3$ real parameters, $n=4$ requires $6$, $n=5$ requires $10$ etc.

5

Consider any Lorentz-symmetric quantum field theory that has a single-particle sector. As a specific example, consider the Proca model. This is the QFT for a non-interacting quantum relativistic massive vector field. It is arguably the most straightforward example of a QFT after the free scalar model. It has a single-particle sector which is self-contained ...

5

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.

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 ...

4

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 $M_n$ ...

4

As the three-form $(g^{-1} dg)^3$ is proportional to the Haar volume form on ${\rm SU}(2)$, and there is a tacit pull-back of this to $S^3$, the quantity $Q/24 \pi^2$ will be the integer winding number (Brouwer degree) of the map $g: S^3 \to {\rm SU}(2)\equiv S^3$.

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 $p\in \mathrm{SO}^+(... 3 OP, you are exactly right. This statement in Schwartz's book is completely wrong, and for exactly the reason you mention. I am now going to do something naughty and paste my answer from a previous question because I think it applies here. Here I go. ... It seems to me that you got derailed by a rather unhelpful discussion in Schwartz. There's a lot going on ... 3 As ZeroTheHero says,$n(n-1)/2. Take an n dimensional cube in n dimensional space oriented aligned with an X, Y, Z, ... coordinate system. How do we give it an arbitrary new orientation? Give the X edge an arbitrary orientation. There are n degrees of freedom. You can specify its new direction as a vector by specifying n numbers. Give the Y edge an ... 2 You have the group of unitary matrices\mathrm U(2)$, and you have the generators of that group, which constitute an algebra. It's important not to get these things confused.$\mathrm U(2)$consists of all$2\times 2$complex matrices$ U$such that$ U^\dagger U = U U^\dagger = \mathbb I$. The set of such matrices forms a (non-abelian) group with the ... 2 One important point to add to the obvious answer "yes, that is right" is that you should be careful about your signature. Specifically, Euclidean conformal group$SO(1,d+1)$describes symmetries of Euclidean correlation functions and has nothing to do with unitary operators on the Hilbert space of the CFT. On the other hand, the Lorentz group$SO(1,...

2

To see why it's true, consider the set of all lightlike rays through the origin in Minkowski spacetime. This set has the topology of a two-dimensional sphere $S^2$, with each point on the sphere corresponding to one lightlike ray through the origin in Minkowski spacetime. What do Lorentz transformations in 4d Minkowski spacetime do to the points on that ...

2

There are a number of reasons why the single particle theory of QM can't be extended into a quantum theory of fields and this is another one. Generally speaking we have to move into a picture where particles can be created and annihilated. This is the multi-particle picture. Freeman Dyson discusses some of the reasons in his Advanced QM. The reason why there ...

1

The first notation hides the Lie algebra generators by writing $A_\mu = A_m^a {\boldsymbol \lambda_a}$ where the implied generators obey $$[{\boldsymbol \lambda}_a, {\boldsymbol \lambda}_b]= i {f_{ab}}^c {\boldsymbol \lambda}_c.$$ The second notation makes the generators explicit and writes $${\bf A}_\mu\cdot {\boldsymbol \lambda}.$$ For SU(2), ...

1

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 ...

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Hint: Using Young diagrams the tensor decomposition ${\bf 15}\otimes{\bf 4}\cong{\bf 36}\oplus{\bf 20}\oplus{\bf 4}$ amounts to $$\begin{array}{rl} [~~]&[~~]\cr [~~]\cr [~~] \end{array} \quad\otimes\quad[a] \quad\cong\quad \begin{array}{rl} [~~]&[~~]&[a]\cr [~~]\cr [~~] \end{array} \quad\oplus\quad \begin{array}{rl} [~~]&[~~]\cr [~~]&[a]\... 1 A choice be made for all eigenvalues to be either even or odd ? The analysis in group theory can classify how many different sysmmetry styles will be in the eigen functions according to the symmeytry group of the Hamiltonian. It can tell from the given basis functions to determine what irreducible representations (symmetric styles) will show up in the ... 1 Symmetries form groups because (a) they have an identity ("do nothing"), and (b) each operation is invertible. Non-invertible transformations would form a monoid, but such transformations do not describe a "symmetry" because some property of the system is destroyed by the transformation (you can't get back to the original state -- if you ... 1 Terms like$$\partial_\mu\pi^i\epsilon^{ijk}(\partial^\mu\pi^j)\alpha^k(x)\\ \epsilon^{ijk}(\partial_\mu\pi^j)\alpha^k(x)(\partial^\mu\pi^i)$$vanish since \epsilon^{ijk} is totally antisymmetric and \partial_\mu\pi^i\partial^\mu\pi^j is symmetric in the two indices. Terms depending on \alpha^2 should be cancelled since the expansion should be done to ... 1 The book coddles you and only bothers to give you the infinitesimal expression for this adjoint action, which is the only thing it needs for that purpose. It is but a bland plug-in:$$ R^{ij}(\boldsymbol{ \alpha})=\frac{1}{2}\operatorname{Tr}(\tau^i \exp(-i\boldsymbol{\tau\cdot\alpha}/2)\tau^j \exp(i\boldsymbol{\tau\cdot\alpha}/2)) ~~~\leadsto \\ =\frac{1}{...

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 ...

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.

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