Tag Info

New answers tagged

6

Let me first make a general remark about internal symmetry groups, unrelated to our problem of the correct symmetry group for QCD. The symmetry must act on Hilbert space as a unitary operator for the conservation of probability. Now let us turn to the strong interaction. The most important experimental facts were that Observed hadron spectrum was ...


3

Let $\overline{\mathbb{R}^{p,q}}$ denote the conformal compactification of $\mathbb{R}^{p,q}$. Let $n:=p+q$. [If $n=1$, then any transformation is automatically a conformal transformation, so let's assume $n\geq 2$.] On one hand, there is the (global) conformal group consisting of the set globally defined conformal transformations on ...


5

Here's my two cents worth. Why Lie Algebras? First I'm just going to talk about Lie algebras. These capture almost all information about the underlying group. The only information omitted is the discrete symmetries of the theory. But in quantum mechanics we usually deal with these separately, so that's fine. The Lorentz Lie Algebra It turns out that the ...


2

Firstly, what book is this? It will help greatly if I can reference it myself. It is highly likely that when he says $\mbox{SO}(1,3)$ [or $\mbox{SO}(3,1)$!] that he means $\mbox{SO}(1,3)-\uparrow$, which is absolutely not the same! But most people are very lazy about this. Here you're picking out the simply-connected region of $\mbox{O}(1,3)$ ...


3

So I take it you are clearly aware that the big A Adjoint representation is the homomorphism you're after in this case, so you're seeking a more general method. Also, I'm assuming you know that the homomorphism of Lie algebras can only lift to a group homomorphism if the homomorphism's domain is simply connected, in which case there is a unique group ...


4

First notice that the generators are $-i\sigma_k/2$ and $-iL_k$, since the groups are real Lie groups and thus the structure tensor must be real. The answer to your question is positive. In principle it is enough to take the exponential of the Lie algebra isomorphism and a surjective Lie group homomorphism arises this way $\phi : SU(2)\to SO(3)$: ...


3

The correct electroweak gauge group is $SU(2)_L \times U(1)_Y$ where $Y$ denotes the weak hypercharge. After the Higgs field spontaneously breaks this exact symmetry, third generator of $SU(2)_L$ (weak isospin) and weak hypercharge combine to give the remaining unbroken $U(1)_{em}$. Gauge bosons and fermions fall under different representations of this ...


4

It looks like this loophole is not explicitly discussed in the "axioms", but it is mentioned in the paragraph before equation (2) which I copy here: A symmetry transformation is said to be an internal symmetry transformation if it commutes with P. This implies that it acts only on particle-type indices, and has no matrix elements between particles ...


3

Actually we have the following Lie algebra isomorphism $$u(1)\oplus su(2)\cong u(2),$$ and there exists the following Lie group isomorphism $$[U(1)\times SU(2)]/\mathbb{Z}_2 ~\cong~ U(2).$$ In other words, there is a two-to-one map between $U(1)\times SU(2)$ and $U(2)$. So in that sense the Glashow-Salam-Weinberg $U(1)\times SU(2)$ model already contains ...


8

Nice question! The short answer is that the group is not $SU(2)\times U(1)$, it is $SU(2)_L \times U(1)_{em}$. In other words the two groups act on different standard model particles differently. For example the left handed neutrino does interact weakly and so transforms under the $SU(2)_L$, but is electrically neutral so it doesn't transform under the ...


3

There are 3 actions of the Galilean group on the free particle: On the configuration space, on the phase space and on the quantum state space (wave functions). The Galilean Lie algebra is faithfully realized on the configuration space by means of vector fields, but its lifted action on Poisson algebra of functions on the phase apace and on the wave functions ...


2

Consider a theory of fields $\phi:M\to T$ where $M$ is a manifold, and $T$ is a set. In physics, $T$ is often either a vector space or a manifold. We call $M$ the domain of the theory, and we call $T$ the target space. of the theory. We call a function from $M$ to $T$ a field configuration, and the set of all field configurations is denoted $\mathcal F$. ...


5

Using the Littlewood-Richardson (LR) rules for Young tableaux, one may show that $$ \begin{array}{c} [~~]\cr [~~] \end{array} \quad\otimes\quad \begin{array}{c} [a]\cr [b] \end{array} \quad=\quad\begin{array}{c} [~~]\cr [~~] \cr [a]\cr [b]\end{array} \quad\oplus\quad\begin{array}{rl} [~~]&[a]\cr [~~]&\cr [b] \end{array} ...


4

The central extensions are classified by the second cohomology group: http://en.wikipedia.org/wiki/Group_extension . If this group is trivial then each central extension is semidirect (and hence in some sense trivial). In particular, this is the case for the Poincare group but not for the Galilei group. However, if you want to take a nonrelativistic limit ...


2

Well, this might not be exactly what OP is looking for, but the statement in Ref. 1 is in general not correct. That infinitesimal (global) symmetries (of an action) satisfy a Lie algebra does not imply that the corresponding Noether charges must also form a Lie algebra. There could be (classical) anomalies. Example: One example is free Schrödinger theory, ...


1

The confusion possibly comes from the casual notation, for example the last term in equation (3) in its full form ought to be $J_i\epsilon^a t^a_{ij}\phi_j$, which is just a number; while in the original notation $J \epsilon^a t^a \phi$ it might lead you to think it is a matrix because of the presence of $t^a$. One quick way to check the mistake is to ...



Top 50 recent answers are included