New answers tagged group-theory
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 ...
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 ...
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]\...
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 ...
0
When we talk about "symmetries" of some object - we mean that the object "stays the same" under certain transformations. So, let us identify a set of such transformations $G$ and objects $X$. An element of the transformation set $g\in G$ should be able to act on an object $x \in X$. I'll denote this action as function application $g(x) \...
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
Physics uses mathematics as a tool to have theories that describe existing data and can predict new ones. It depends on observations and measurements.
The SU groups came into the study of data slowly. First SU(2), to describe the two baryon states of proton and neutron, that are almost of the same mass. It was useful in nuclear physics studies.
SU(3) first ...
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 ...
0
The adjoint of $\phi$ is defined as $\phi^\dagger = (\phi^\ast_1, \phi^\ast_2)$,
so that
$$\phi^\dagger \phi = (\phi^\ast_1, \phi^\ast_2) \cdot \binom{\phi_1}{\phi_2} = \phi^\ast_1 \phi_1 + \phi^\ast_2 \phi_2$$
The same works for the derivatives.
This is valid of $\phi $ as a "vector" of c-numbers. Is $\phi$ an operator the $\ast$ has to be ...
1
Terms like $$\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}{...
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}^+(...
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.
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 ...
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.
2
As it is often the case the issue lies in conventions. For mathematicians the Lie Algebra generators are any basis that can span the algebra as a vector space, for physicist we usually require the generators themselves to be hermitian (e.g. think about the Pauli matrices), because of their interpretation as observables.
Let us also make clear the difference ...
1
You are certainly correct that the Lie algebra of $U(d)$ consists of skew-Hermitian $d \times d$ matrices. However, physicists will often implicitly complexify Lie algebras, without ever bothering to mention that they are doing it. The complexification of $u(d)$ is indeed $gl(d, \mathbb{C})$. That's because multiplying by $i$ we get Hermitian matrix, and any ...
0
As far as I remember, it is only a matter of definition of the term "generator". If $H$ is hermitian
$$H^+=H$$
then $G=iH$ is skew hermitian:
$$H^+=(iG)^+=-iG=-H$$
So after all, the parameter inside the exponential differs only by an imaginary unit, which doesn't change a lot, at least notation-wise. Instead of generating a one-parameter sub-group ...
1
It has something to do with the Wiener-Hopf method.
Basically, if you recall quantum mechanics and how the Green's function for a Schrodinger equation is found there, it is a Fourier transform of an evolution operator. However, the evolution operator has the initial condition, and, typically, you do not know pre-history, which prevents you from making the ...
1
Te intuitive way to prove it is to notice that those open string oscillators have the same quantum numbers as a Yang-Mills gluon. See the answers to the question 2 coinciding D-branes leads to a U(2)
gauge theory.
The idea is that both (first excited open string oscillators and YM gluons) have $d-2$ polarization states, come in $N^{2}$ varieties, two color/...
2
If you want to learn about both Lie groups and how to solve the Hydrogen atom using those, I would recommend "Lie groups, physics, and geometry" by Gilmore.
It is well written, with physics and engineering students in mind, with lots of examples on simple cases. The first six or seven chapters are very nice (then it can go too deep into ...
Top 50 recent answers are included
