New answers tagged group-representations
0
In the inhomogenous Lorentz group $ISO(1,3)$, you have the space-time translation group, and the Lorentz group $SO(1,3)$.
You begin to find a representation of the space-time translation group, by choosing a momentum $p$. So your representation must have a $p$ indice :
$$\psi_p$$
After this, you will have to get the full representation, by finding a ...
2
To be honest, I have a hard time interpreting geometrically what's going on here when using quaternions or biquaternions or anything else. All the algebra of rotations in 4d is adequately handled by a geometric algebra, with the elements of that algebra having clear geometric interpretations. The mathematics is similar to quaternions, but differs in some ...
2
The problem is that your coordinates aren't well defined at $\theta=0$ and $\phi=\pi/2$. Note in particular that
$$
U|_{(0,\frac{\pi}{2},\gamma)} = \begin{pmatrix}1&0\\0&1\end{pmatrix}
$$
for any value of $\gamma$. A simpler choice is
$$
\tilde{U} =
\begin{pmatrix}
x+iy & z+iw \\
-z+iw & x-iy
\end{pmatrix},
$$
with
$$
x = \sqrt{1 - y^2 - z^2 ...
-1
I don't see what you're missing.
$$\frac{\partial e^{ \pm i \gamma}}{\partial \gamma} = \pm i$$ which gives you the third generator, doesn't it?
3
User twistor59 has addressed the part regarding the "generator" terminology, but let me give a bit more detail on the second part of the question. I'm going to restrict the discussion to matrix Lie groups for simplicity.
Some background.
Given a Lie group $G$ with Lie algebra $\mathfrak g$, there exist two mappings $\mathrm{Ad}$ and $\mathrm{ad}$, both ...
3
If you have a basis for the Lie algebra, you can talk of these basis vectors as being "generators for the Lie group". This is true in the sense that, by using the exponential map on linear combinations of them, you generate (at least locally) a copy of the Lie group. So they're sort of "primitive infinitesimal elements" that you can use to build the local ...
4
Here we will only discuss the case of finite-dimensional irreducible representations (irreps) of a complex semisimple Lie algebra $L$.
Recall that the set $Z$ of Casimir invariants is the center $Z(U(L))$ of the universal enveloping algebra $U(L)$, cf. e.g. this Phys.SE post.
OP's question is answered without proof on p. 253 in Ref. 1:
Theorem 2. For ...
2
It can be proven (Racah's theorem) that the number of Casimir operators is the same as the rank of the algebra (number of simultaneosly conmuting generators). This is at least true for semi simple algebras.
3
For Poincare algebra there are (as far as I know) two different approaches to find its representations. In first approach one begins from a finite dimensional representation of (complexified) Lorentz algebra, and using it one constructs a representation on space of some fields on Minkoski space. Representation so obtained is usually not irreducible and an ...
4
With respect to the discussion of momentum-eigenstates and the following derivation in Weinberg's book, $\sigma$ is just a label that denotes any degree of freedom that is not momentum. Even though it can be identified with spin, its nature is not relevant for the discussion at hand.
1
You can write an infinitesimal transformation, with generator $J$, as
$$
R(\delta\theta) = 1 + iJ\delta\theta
$$
A finite transformation is a succession of $N\to\infty$ infinitesimal transformations,
$$
R(\theta) = (1 + iJ\theta/N)^N = e^{iJ\theta}
$$
The rotations $O(3)$ are isomorphic to $SU(2)$, with generators $J = \sigma/2$. The Lorentz ...
3
OP has already answered his own question with help from other answers, especially Peter Kravchuk's answer. Here we make some comments on how the fusion rule mentioned by Peter Kravchuk should be concretely realized.
The first point is that the adjoint representation $Ad_{SU(N)}$ of $SU(N)$ is the real vector space of Hermitian traceless $N\times N$ ...
6
Noncompact internal symmetries – and R-symmetry is an internal symmetry (it doesn't transform positions in the spacetime) – are unacceptable in a physical theory because they would lead to negative-norm states.
Consider the $i$-th superpartner of a bosonic particle state, $|i\rangle$, where $i=1,2,\dots,N$. The inner product $\langle i|j\rangle$ of such ...
3
With the help of Peter Kravchuk and joshphysics, I have completed a proof of the trace identity. I will post it here as a reference. By the method of Kravchuk, we find
$$\begin{split}
\mathrm{tr}\big(t^a_Gt^b_Gt^c_Gt^d_G\big)&=2\big[
\mathrm{tr}\big(t^a_Nt^b_N\big)\mathrm{tr}\big(t^d_Nt^c_N\big)+
...
3
Wow. I just had an amazing experience of discovering the following fact:
It is known that for an element $U$ of the group, in matrix sence:
$$
Ad_Ux=UxU^{-1}.\,\,(1)
$$
Now, we note that the target space of the adjoint rep is spanned by $N^2-1$ traceless matrices $t_a$. So, if we add the unity matrix, we get a full basis in $\mathrm{Mat}_N(\mathbb{C})$. We ...
2
Here is my recommendation on how to proceed. Notice that you are given the trace of the product of any two generators. It would therefore be useful to convert the product of four generators inside of the trace you're trying to compute into a sum of products of two generators. This can be done by noting the following commutator-anticommutator identity:
$$
...
Top 50 recent answers are included
