# Tag Info

4

"The same rule" is associativity, but everything else is quite different. That is, the group commutator is, in some convention, $$[R_{x}(\alpha),R_{z}(\beta)]= R_{x}(\alpha)R_{z}(\beta) R_{x}(-\alpha)R_{z}(-\beta),$$ where $R_{x}(\alpha)= e^{-i\alpha J_x}$, etc... For notational simplicity, let's call $-i\alpha J_x =A$ and $-i\beta J_z =B$, antihermitean, ...

3

Your orthogonal matrix $$R(\phi,\theta)= \begin{bmatrix} \cos(\theta) &\sin(\theta) & 0 \\ \sin(\theta)\cos(\phi) & -\cos(\theta)\cos(\phi) & \sin(\phi)\\ \sin(\theta)\sin(\phi) & -\cos(\theta)\sin(\phi) &-\cos(\phi) \end{bmatrix}$$ must have antisymmetric generators. To find ...

2

Set $i=l$ and $j=k$ in 80.17 and sum over the repeated indices. The LHS becomes $\sum_a {\rm tr}\{T^a T^a\}$ the RHS becomes $N^2-1$. There are $N^2-1$ traceless hermitian matrices, so if ${\rm tr}\{T^aT^b\}=\delta_{ab}$ the LHS is $N^2-1$ also. So everything is consistent. It would not be consistent with the 1/2.

2

If an object, say, a field $\phi\in V$ lives in a Lie algebra representation $\rho: \mathfrak{g}\to {\rm End}(V)$, then it is implicitly understood that the gauge covariant derivative $$D_{\mu}~=~\partial_{\mu}-ig~A_{\mu}\tag{1}$$ should be interpreted as $$D_{\mu}\phi~=~\partial_{\mu}\phi-ig~\rho(A_{\mu})\phi.\tag{2}$$ If $\Phi\in {\rm End}(V)$ is (...

2

This complexifying is useful only in the case of Lorentz algebra $so(1,3)$ (or $so(3,1)$). For a general Lie algebra, the complexifying unnecessarily complicates things for no apparent benefit. (it is understood that the assertion is in the context of representation of real world particles. If anyone can elaborate on the virtue of complexification of say, ...

1

The parametrization you've given just isn't of the exponential form you're after. The matrix you've written down is parametrized by the location of the $x$ axis after the rotation, which has polar angle $\theta$ and azimuthal angle $\phi$ in polar spherical coordinates drawn around the old $x$ axis. It is not a rotation by angle $\phi$ about an axis at ...

1

This is indeed a matter of normalization, as A_user_with_NoName correctly commented. If the normalization is $Tr(T^{a}T^{b}) = \frac{1}{2}\delta^{ab}$, then the formula for $SU(N)$ is valid. The proof is given in the link provided by A_user_with_NoName.

1

It is well-known that the finite dimensional irreps $V_{\ell}$ of the Lie algebra $so(3)$ are classified by spin $\ell\in\frac{1}{2}\mathbb{N}_0$. To rule out half-integer representations for the orbital angular momentum (OAM) Lie algebra $${\rm span}_{\mathbb{R}}(L_1,L_2,L_3)~\cong~ so(3),$$ one should use the fact the OAM operators  L_j~=~\sum_{k,\ell=1}^...

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