# Tag Info

22

There are a number of imprecisions in your question, mostly having to do with confusing the Lie group and its Lie algebra. I suppose this will make it hard to read the mathematical literature. Having said that, the first volume of Kobayashi and Nomizu is probably the canonical reference. Let me try to summarise. Let me assume that $H$ is connected. The ...

10

In this relatively simple example, one can observe that the subalgebras $\{\sigma_a \otimes \frac{1\mp\eta_1}{2}\}$ are the two commuting copies of $su(2)$. For more complicated situations, one actually has an algorithm to veify the simplicity of a Lie algebra. This is because (the root systems of) simple Lie algebras are classified by Cartan, thus one just ...

10

I don't know if these rise to the level of "useful," but: Yang-Mills theory with gauge group $G_2$ is interesting because $G_2$ has trivial center. So people simulate it on a lattice, try to understand in what sense it might be confining, how string tensions scale, if it has a deconfinement phase transition, and so on. The idea is that looking at a group ...

9

Yes. G2 shows up often, starting with atomic physics (perhaps Racah is the first; see R. E. Behrends, J. Dreitlein, C. Fronsdal, and B. W. Lee, “Simple groups and strong interaction symmetries,” Rev. Mod. Phys. 34, 1 (1962).). You will find some refences in my 1976 Phys rev paper on cns.physics.gatech.edu/GroupTheory/refs . I have whole folder of physics ...

9

Here we will sketch a possible derivation. Let $\eta\in {\rm Mat}_{n\times n}(\mathbb{R})$ be a real symmetric matrix of signature $(p,q)$, where $n=p+q$. Define the Lie group $$O(p,q)~:=~ \{ \Lambda\in {\rm Mat}_{n\times n}(\mathbb{R}) \mid \Lambda^T\eta \Lambda = \eta \},$$ where $\Lambda^T$ denotes the transposed $\Lambda$ matrix. Prove for fun that ...

8

By definition, the metric tensor $\eta_{ij}$ transforms trivially under the defining rep of $SO(n,m)$. $$\eta_{ij}=[D(g^{-T})]_{i}^{\ k}[D(g^{-T})]_{j}^{\ l}\eta_{kl} =[D(g^{-1})]^{k}_{\ i}[D(g^{-1})]^{l}_{\ j}\eta_{kl}$$ and this holds for all $g\in SO(n,m)$. Consider a one-parameter subgroup of the defining rep with matrices $D(g)=e^{tJ}$ where ...

7

I) The Casimir invariants of a Lie algebra $L$ over a field $\mathbb{F}$ are the central elements of the universal enveloping algebra $U(L)$. Example: The angular momentum square $\vec{J}^2$ is a quadratic Casimir invariant of the Lie algebra $L=sl(2,\mathbb{C})$. II) Given a bilinear associative/invariant form $B:L\times L\to \mathbb{F}$, one can create ...

7

I'll give you enough hints to complete the proof yourself. If you're desperate, I'm following the notes by Zuber, which are available online, IIRC. Let's start with some notation: pick some basis $\{t_a\}$ of your Lie algebra, then $$[t_a,t_b] = C_{ab}{}^c t_c$$ defines the structure constants. If you define $$g_{ab} = C_{ad}{}^e C_{be}{}^d,$$ then this ...

6

Well, if something is invariant under the action of some group it means that every element leaves it invariant. In particular, every element that belongs to some subgroup. Also, you can get an obvious representation of the subgroup by restriction. If $\rho: G \to {\rm Aut}(V)$ is a representation then so is $\rho_{H}: H \to {\rm Aut}(V)$ for $H < G$. ...

6

First of all, note that the real Abelian Lie group $U(1)$ comes in two (multiplicatively written) versions: Compact $U(1)~\cong~e^{i\mathbb{R}}~\cong~S^1$, and Non-compact $U(1)~\cong~e^{\mathbb{R}}\cong~\mathbb{R}_+\backslash\{0\}$. Also note that in the physics literature, we often identifies charge operators with Lie algebra generators for a Cartan ...

6

The different definitions you mentioned are NOT definitions. In fact, what you are describing are different representations of the Lorentz Algebra. Representation theory plays a very important role in physics. As far as the Lie algebra are concerned, the generators $L_{\mu\nu}$ are simply some operators with some defined commutation properties. The choices ...

5

One answer is that most calculations in modern physics do not actually depend on the explicit realization of the Pauli matrices $\sigma_a$, $a=1,2,3$, but rather on the relations $$\sigma_{a}^{\dagger}~=~ \sigma_{a},\qquad\qquad {\rm tr}(\sigma_{a})~=~0,\qquad\qquad a=1,2,3,$$ $$\sigma_a \sigma_b ~=~ \delta_{ab} {\bf 1}_{2\times 2} + ... 5 Dear Leandro, it's because the Pauli matrices, together with the 2\times 2 identity matrix, form a full real basis of all the Hermitian 2\times 2 matrices (note that 2\times 2 Hermitian matrices depend on four real parameters), and the identity matrix is irrelevant in a Hamiltonian because it's just a conventional energy shift that acts on all vectors ... 4 An infinitesimal generator X is a vector field, which satisfies Leibniz rule$$X[fg]~=~ fX[g]+g X[f].$$In the 1-dimensional case, the generator is of the form X=p(x)\frac{\partial}{\partial x}, where p=p(x) is some function. Assume furthermore that there exists a bijective smooth function h=h(x), such that$$p(x) h'(x)~=~1.$$In other ... 4 For practical reasons, physicists like to label the states of the system by a set of "quantum numbers". Technically this means that you are looking for a set of mutually commuting Hermitian operators such that: (i) Every vector from the basis of common eigenvectors of these operators is uniquely characterized by the set of eigenvalues (i.e. the above ... 4 It means that the charge operator Q is a Lie algebra generator for some Lie group G. the field \phi\in V takes values/transform in a representation V of the Lie group. (Note that any Lie group representation V is also Lie algebra representation of the corresponding Lie algebra.) the charge operator \rho(Q) in the representation \rho: G \to ... 4 Because of u\leftrightarrow d isospin symmetry. For a more detailed explanation, see e.g. chapter 8 of 't Hooft's lecture notes. The pdf file is available here. 4 There are many methods to compute the matrix elements of a simple Lie algebra generators in a given representation. For the problem at hand, I'll try to describe two methods quite in detail and sketch two other methods. The individual computations actually involve elementary linear algebra and combinatorics, but they are quite lengthy. For example to obtain ... 4 The irreducible SU(3) representations of Dynkin indices (n,0) are the n- symmetric tensor powers of the fundamental representation. Therefore let e_1, e_2, e_3 be an orthonormal basis of the fundamental representation space, then  v_{ii} = e_i \otimes e_i   v_{ij} = \frac{e_i \otimes e_j + e_j \otimes e_i }{\sqrt2}, i \ne j ... 4 A group is a set of transformations that don't change the internal relationship in "something", whether it's another mathematical structure or a set of equations describing a physical object or the physical object itself. So putting a symmetry into equations means to find a set of functions or maps f such that$$ x \text{ is OK} \Leftrightarrow f(x) ...

4

To add to what @Qmechanic says, note also that for a representation $\rho:\mathfrak g\to \mathfrak{gl}(V)$ of a Lie algebra $\mathfrak g$ acting on a vector space $V$, a vector subspace $W$ of $V$ is called an invariant subspace of the representation provided $\rho(X)w\in W$ for all $X\in\mathfrak g$ and $w\in W$. A representation $\rho$ is said to be ...

4

According to quantum mechanics, if you make a measurement of the magnitude squared of the orbital angular momentum of the electron in the hydrogen atom, then possible outcomes of the measurement are the discrete values $$L^2 = \ell(\ell+1)\hbar^2, \qquad \ell\geq 0$$ Given that this is the outcome of the measurement of the magnitude squared, there is a ...

4

I believe you can, if you try to follow the path of finding representations of the $SO(n)$ group over a given Hilbert space. I really haven't done the calculation, but if it is the same, you would have something like this: $H=L_2(\mathbb R^n,\mathbb C)$ would be the Hilbert Space that would correspond to spin 0 particles, and the representation of the ...

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 ...

3

If the symbols $\hat A,\hat B$ mean some functionals of $a,a^\dagger$ which don't depend on $\rho$, your problem has no solution. You simply cannot rewrite objects such as $\rho H$ in the form $H' \rho$; matrices and operators don't commute with each other and their refusal to commute isn't a formality – it's a fundamental fact. A proof that it can't be ...

3

1) The Lie group $G$ behind the meson octet is the $SU(3)_{\rm Flavor}$ Lie group over the three lightest quark flavors $u$, $d$, and $s$. More precisely, the fundamental $SU(3)$ representation $$V~=~\text{Fund} ~=~{\bf 3}$$ is a linear span of the $u$, $d$, and $s$ quarks. The three complex coefficients are collected in a $3\times 1$ column vector $v\in ... 3 My answer will focus just on the mathematical parts pertaining to partial differential equations. Scale invariance is the fact that some partial differential equations stay the same if you appropriately scale the variables. For example the heat equation (where$\boldsymbol{x}$is the position vector in 1, 2 or 3D, doesn't matter) $$\partial_t ... 3 In terms of classical general relativity: Einstein's equations$$ G_{ab} = 8\pi T_{ab} $$can be formulated, in local coordinates, as a system of second order partial differential equations for the metric unknown g_{ab}. The matter field equations further generate some family of partial differential equations. Given a continuous symmetry (as guaranteed ... 3 The basis states on the left are given by$$|{\uparrow\uparrow}\rangle, |{\uparrow\downarrow}\rangle, |{\downarrow\uparrow}\rangle,\text{ and }|{\downarrow\downarrow}\rangle.$$On the right, you are supposed to symmetrize these states with respect to exchanging the first and second spin (that is what sym and antisym stand for). There is only a single ... 3 Concerning a Lie algebra representation$\rho: L \to {\rm End}(V,\mathbb{F})$, where$L$is a Lie algebra, where$\mathbb{F}$is a field (typically$\mathbb{F}=\mathbb{R}$or$\mathbb{F}=\mathbb{C}$), where$V$is a$\mathbb{F}$-vector space, and where$\rho$is a Lie algebra homomorphism, be aware that physicists refer to both the map$\rho\$ and the vector ...

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