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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 ...