# Tag Info

## New answers tagged representation-theory

0

Perhaps the correct perspective is that the combination $L_{\mu\nu}\Phi(0)$ needs to be translated as follows: $$L_{\mu\nu}\Phi(0) \to \left(e^{ix\cdot P} L_{\mu\nu} e^{-ix\cdot P}\right) e^{ix\cdot P}\Phi(0)$$ which now correctly matches the notation in Di Francesco.

1

Perhaps, I do not quite understand the question, but the properties of the two matrix are rather obvious: $$\mathbf{a}^T\mathbb{I}\mathbf{b}=\sum_{i,j}a_i\delta_{i,j}b_j=\sum_i a_i b_i=\mathbf{a}\cdot\mathbf{b}$$ $$\mathbf{a}^T\mathbb{J}\mathbf{b}=\sum_{i,j}a_ib_j=\sum_i a_i \sum_jb_j$$ This latter is sometimes useful for concize notation.

1

Yes (for each value of $S^2$ separately): i) construct $S^{\pm} = (S_X \pm i S_Y)$ and get that they act as raising/lowering operators on $S_Z$ (by noticing that these are eigenoperatos of $S_Z$) ii) construct $S^2 = S_X^2+S_Y^2+S_Z^2$ and see that it commutes with all of these operators, and deduce that it can be diagonalized together with $S_Z$, for ...

0

I do not have that text, and I would rather not shadow-box with its logic. There are better ways to introduce the SU(2) conjugate doublet representation, $\tilde \xi\equiv \zeta \xi^*$, linked to, e.g. this or this. He is using the special form of U, (3), dictated by unitarity, to demonstrate what he claims, namely that $\xi \sim \tilde \xi$, (10). However, ...

1

We cannot reduce the degeneracy even more using only tools from angular momentum theory. You need to find other symmetries, and even then there's no guarantee you will remove all degeneracies. Consider the situation where you have three particles with angular momentum $\ell=1$. In the decomposition of $1\otimes 1\otimes 1$, you will find three sets of ...

3

Never forget to think about statistics when considering quantum mechanics. Your question is related to the three correlations between pairs of three variables. Famously, "is correlated with" isn't as transitive as we'd like to think. Let $|\chi\rangle:=\sum_m|m\rangle$. Write $\sim$ between complex numbers of the same modulus. We can choose three ...

2

We don't know without further information. The $su(2)$ representation $\rho: su(2) \to {\rm End}(V)$ of the angular momentum $\vec{J}$ could be (completely) reducible $$V~=~\oplus_i V_i.$$ Here $V$ denotes the Hilbert space of the system. E.g. the projection operator $P_i$ for each irreducible subspace $V_i$ would commute with $\rho(J^2)$ and $\rho(J_z)$.

2

Wider net is wise, given the responses on MSE. I do not understand why the sum in the above equation only runs over 𝑚′. 𝑙 is associated with the inclination angle 𝜃 and 𝑚 is associated with the azimuthal angle 𝜙, both of which can be affected by a general rotation. I suppose another way of phrasing this is: why is the 𝑙 index of the spherical harmonic ...

6

It is possible to construct singlet states for two particles of any identical spin or angular momentum $j$. In general: \begin{align} \vert 00\rangle = \sum_{m_1 m_2} C^{00}_{jm_1;j,-m_1} \vert j,m_1\rangle \vert j,-m_1\rangle \end{align} where $C^{JM}_{j_1m_1;j_2m_2}$ is a Clesbsh-Gordan coefficient. Furthermore note that singlet states are not necessarily ...

1

Yes, you understand right. You represent the quark with $q^i_\alpha$, a simultaneous weak SU(2) doublet and color SU(3) triplet. It is a hypercharge singlet with eigenvalue 1/3 throughout, so that can be skipped. The color and weak isospin group are in Cartesian product, so the representation you are looking at is a tensor product. So the quark is a 6-vector ...

1

This has nothing to do with $SU(3)$, the doublet representation is that of the electroweak $SU(2)$. Both $u$ and $d$ are themselves triplets transforming in the fundamental representation $\mathbb{3}$ of $SU(3)$. Your last sentence sounds correct to me.

2

If $0\leq l<m$, then $l+1<m+1$, so multiplying these inequalities yields \begin{align} l(l+1)<m(m+1) \end{align} (we can do this since everything is non-negative), so this violates the second inequality you wrote down. Next, suppose $m<-l$, which means $0\leq l<-m$, and thus by the same reasoning as above, we have \begin{align} l(l+1)&< ...

4

Hints: Firstly, $$SO(10) ~\supseteq~ SO(4)\times SO(6),\tag{1}$$ so we get the branching rules $${\bf 10}~\stackrel{(1)}{\cong}~({\bf 4},{\bf 1}) \oplus ({\bf 1},{\bf 6}),\tag{2}$$ and $${\bf 10}\wedge{\bf 10}~\stackrel{(2)}{\cong}~ ({\bf 4}\wedge{\bf 4},{\bf 1})\oplus ({\bf 4},{\bf 6})\oplus ({\bf 1},{\bf 6}\wedge{\bf 6}).\tag{3}$$ Here the space of 2-...

Top 50 recent answers are included