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1

For the translation operator, according to Sakurai (2nd ed, pp. 42-46), there are four conditions we want the infinitesimal translations to accomplish: 1) If $|\alpha\rangle$ is normalized, then so is the translated ket $\hat{T}(d\mathbf{x})|\alpha\rangle$ 2) $\hat{T}(d\mathbf{x'})\hat{T}(d\mathbf{x''})=\hat{T}(d\mathbf{x'}+d\mathbf{x''})$ 3) \hat{T}(-d\... 1 Here is an explanation without exponentials. For the translation operator, for example, we want (by definition of what the translation operator should do) \begin{align} \left<x\right|T(\epsilon) \left| \psi \right> &= \left<x + \epsilon\right|\left. \psi \right>\\ &= \psi(x + \epsilon)\\ &\approx \psi(x) + \epsilon \,\psi'(x)\\ &... 1 The exponential of an operator is defined by its series so \begin{align} e^A= 1+ A +\frac{1}{2!}A^2+\frac{1}{3!}A^3 \end{align} so we have, respectively \begin{align} R_z(\phi)&=e^{-i \hat J_z \phi/\hbar}&\approx 1-\frac{i}{\hbar} \hat J_z \phi +\ldots \, ,\tag{1}\\ U(t)&=e^{-i \hat H t/\hbar}&\approx 1-\frac{i}{\hbar} \hat H t+\ldots \, ,\\... 1 The tensor product \begin{align} (1,1)\otimes(1,1)=(2,2)+2(1,1)+(0,0)+(3,0)+(0,3) \end{align} and in fact(3,1)$has dimension$24$where$(3,0)$has dimension 10 so it's likely$\Delta$is in$(3,0)$, not (3,1). You need to work out the remaining quantum numbers for your particles - let's call them$\alpha,\beta$and$\gamma$respectively. Once you ... 0 Okay so basically with the change below everything is correct. An adjoint representation of SU(2) is the following:$ T_1 = \frac{1}{\sqrt{2}}\begin{pmatrix}0&1&0\\1&0&1\\0&1&0\end{pmatrix}  \;\;\;\;\;\;T_2 =\frac{1}{\sqrt{2}} \begin{pmatrix}0&-i&0\\i&0&-i\\0&i&0\end{pmatrix}  \;\;\;\;\;\;T_3 =\begin{...

0

A note to @WunderNatur's "'accidental' isomorphism between low dimensional simple lie algebras": "Accidental" is the key word. Since we failed to find anything "accidental" for $\mathfrak{so}(1,4)$, the superstring saga is confined to AdS (thanks to the "accidental" isomorphism between $\mathfrak{so}(2,3)$ and $\mathfrak{sp}(4,\mathbb{R})$) and confronted ...

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 ...

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.

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, ...

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.

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 $$L_j~=~\sum_{k,\ell=1}^... 2 If an object, say, a field \phi\in V lives in a Lie algebra representation \rho: \mathfrak{g}\to {\rm End}(V), then it is implicitly understood that the gauge covariant derivative$$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 (...

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