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1

You have conservation of angular momentum and parity in electromagnetic interactions. The vector boson must be massive to be able to decay into two photons due to energy-momentum conservation. So we can consider its decay in its rest frame. So if it decays in two photons their momenta must be back-to-back. We assume those vectors to be $\pm k \vec{e}_z$. The ...


1

Isotropy isn't a property of a particle; it's a property of spacetime. A coupling between the direction of the vector particle's spin and the direction of the linear momentum of a decay photon would be a parity-odd observable; all such observables are forbidden in purely electromagnetic decays. (Parity-odd observables are allowed in weak decays, but you ...


0

I believe that the simplest explanation, in the plainest language, is that the Wigner-Eckhart theorem is a quantum-mechanical expression of conservation of angular momentum. This may not be self evident, but it's hard to get simpler.


1

A rather recent book is An Introduction to Tensors and Group Theory for Physicists. It also speaks of vectors and tensors at a good level. In my opinion it clears up the confusion physicists tend to make when speaking of these topics. Moreover the book is disseminated with examples and applications from mechanics, EM and QM, so is a great introduction to ...


3

Stack all the $\phi^i$s into a column "vector" $\vec\phi$. The mass term $m^2\vec\phi\cdot\vec\phi$ is obviously invariant by $R^{-1}=R^T$. The same with the kinetic term $(\partial_\mu\vec\phi)\cdot(\partial^\mu\vec\phi)$ because $\partial_\mu R=0$. It is $SO(n)$ invariant because I take it $i$ runs over $n$ values. Thus your $r^i_j$ generates $SO(n)$. ...


0

The algebraic form of the quadratic casimir $T^2$ depends only on the structure constants, and are therefore the same in any representation. As in $SU(2)$, of course, it's matrix form is representation dependent. Edit: To see why this is the case, suppose you constructed $T^2$ out of other elements of the algebra, and you did this for say, the fundamental ...


4

Your prof is using slightly wrong words: the group is a Lie group of dimension 10, not order 10. A group's order is the number of its elements, which is here uncountably infinite. A group of order 10 is a finite group (and indeed there are only two possible groups with 10 elements). I'm not sure how much continuous group theory (Lie group) theory you have ...


0

I) Representation theory of the Lorentz$^1$ group is a fairly broad subject covered in many textbooks, see e.g. Ref. 1 for further information. II) The irreducible representation $$\tag{1} (j_L,j_R)~=~j_L\otimes_{\mathbb{C}} j_R, \qquad j_L, j_R~\in~ \frac{1}{2}\mathbb{N}_0,$$ is a tensor product of $V=V_L\otimes_{\mathbb{C}} V_R$ of two complex vector ...


0

The ambiguity is resolved by choosing the coordinates first. So: 1)Set up a cartesian system with x,y,z coordinates. 2)Pick the group you would like to study. Let's say $P4_32_12$ from page 1151 of this document: http://mcl1.ncifcrf.gov/dauter_pubs/284.pdf that DavePhD recommended. 3)You can then see that the asymmetric unit is given by $$0\leq ...


0

In the context of e.g. a pseudo-orthogonal Lie group $$\tag{1} O(p,q)~:=~ \{\Lambda\in {\rm Mat}_{n\times n}(\mathbb{R}) ~|~\Lambda^T\eta\Lambda= \eta \} $$ of pseudo-orthogonal matrices $\Lambda$ for the metric $$\tag{2} \eta_{\mu\nu}~=~{\rm diag} (\underbrace{1,\ldots,1}_{p~\text{times}},\underbrace{-1,\ldots -1}_{q~\text{times}}), \qquad n~=~p+q,$$ ...


3

The equation you gave is indeed the definition of matrix multiplication, applied to a $d\times d$ matrix and a $d\times 1$ matrix. But the underlying concept is something more. The thing about vectors is that they exist, in some sense, independent of the numbers used to represent them. For example, an ordinary 3D displacement vector represents a physical ...


0

Yes, he defined the vector as behaving that way (a vector rotation is equivalent to a change of basis), otherwise it would not be a vector. A tensor is a different kind of object, it has at least two indexes and behaves different that a vector under transformation of coordinates (as defined in your book). You might probably read more about linear algebra up ...


4

We interprete OP's question (v3) as essentially asking Is $SU(3) \times SU(2)\times U(1)$ a normal subgroup of $SU(5)$? Or in terms of the corresponding Lie algebras, Is $su(3) \oplus su(2)\oplus u(1)$ an ideal of $su(5)$? Here we identify $su(5)$ with antihermitian $5\times 5$ matrices; $su(3)$ with antihermitian $3\times 3$ block matrices in ...


0

They do not lie in $\mathfrak{so}(3)$ but they lie in its complexification, which would be $A_1$ in the usual mathematical classification. Much of Lie representation theory is set up this way: you work at the level of the complexification then go back to the real form. For compact groups it's not a big deal; for non-compact groups extra care is needed. So ...


1

How to understand non-associative composition of velocities in STR? Special relativity introduces a weirdness about how your axes can be related to other observers' axes: if your axes are aligned with observer A's axes and theirs are aligned with observer B then special relativity (i.e. the Lorentz transformations) say that B's axes will be rotated with ...


3

Are Lorentz transformations more adequate representations of motion, than the more intuitive velocities? Yes. The non-associativity that bothers you simply arises because there is no group of three dimensional boosts. Confined to one dimension, boosts form a rather lovely one parameter subgroup of the Lorentz group $SO^+(1,3)$. So everything "works", ...


3

why it is such a prevalent idea. In elementary particle physics and nuclear physics groups and their representations have played a very crucial role in developing the standard models. The elementary particles in the table in the link above have a lot of quantum numbers. These quantum numbers have lead to observed symmetries, that can be described by ...


9

It's an enormous subject, but briefly: Lie groups are smooth groups. Technically, Lie groups are sets that are both a smooth manifold, like a sphere for instance, and also have a group structure (multiplication operator, inverses, and an identity). The group multiplication and inverse must be smooth (differentiable) functions on the manifold. As you ...


2

I guess what you are missing is the following: given a representation $\rho(g)$ of $g\in$SU(2) acting on some vector space $V$. We define the representation $\rho_\otimes$ of SU(2) (not of SU(2)$\times$SU(2)) on $V\otimes V$ as $$\rho_\otimes(g) (v_1 \otimes v_2) = \rho(g) v_1 \otimes \rho(g) v_2.$$ So in fact we are defining the tensor product of two ...



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