Exponential of tensor product, excess scalar exponential

Question

I know that in a hilbert space $\epsilon=\epsilon_1\otimes\epsilon_2$ it is true that the rotation operator can be decomposed into rotation operators acting on the single hilbert spaces $$\mathcal{D}=\mathcal{D_1}\mathcal{D_2}:=(\mathcal{D_1}\otimes I_2)*(I_1\otimes\mathcal{D}_2)$$ It holds also true that the angular momenta of the single spaces add $$J=J_1+J_2:=(J_1\otimes I_2+I_1\otimes J_2)$$ Bothe results are connected via $$\mathcal{D}=\exp(-i/\hbar\cdot\beta J)$$ This leads to a problem though, as the following calculation shows: First we note that $[(J_1\otimes I_2),(I_1\otimes J_2)]=0$ which we use to justify decomposing the exponential into two exponentials. \begin{align} \mathcal{D}&=\exp(-i/\hbar\cdot\beta J)=\exp(-i/\hbar\beta\cdot (J_1\otimes I_2+I_1\otimes J_2))\\&=\exp(-i/\beta\cdot (J_1\otimes I_2))\cdot\exp(-i/\hbar\beta(I_1\otimes J_2))\\&=\sum(-i/\hbar\beta)^n\cdot (J_1\otimes I_2)^n\cdot\sum(-i/\hbar\beta)^n\cdot (I_1\otimes J_2)^n \end{align} Now examine (Note I droppend the factorial for convenience) $$\sum(-i/\hbar\beta)^n\cdot (J_1\otimes I_2)^n=\sum(-i/\hbar\beta)^n\cdot (J_1^n\otimes I_2^n)=\sum(-i/\hbar\beta)^n\cdot (J_1^n\otimes I_2)=\sum(-i/\hbar\beta)^n\cdot J_1^n\otimes \sum(-i/\hbar\beta)^n\cdot I_2=\exp(-i/\hbar\beta J_1)\otimes\exp(-i/\hbar\beta)$$ So in total \begin{align}\mathcal{D}&=(\exp(-i/\hbar\beta J_1)\otimes\exp(-i/\hbar\beta))\cdot(\exp(-i/\hbar\beta)\otimes\exp(-i/\hbar\beta J_2))\\&=(\mathcal{D_1}\otimes\exp(-i/\hbar\beta))\cdot (\exp(-i/\hbar\beta)\otimes\mathcal{D_2})\end{align}

Where is the mistake, how do I get rid of these excess scalar exponential functions, that don'tappear in the first equation

Answer 1

Let's do the case when $A$ acts on a space with basis ${\bf e}_i$ and $I$ is the two-by-two identity matrix for the space with basis ${\bf f}_1$, ${\bf f}_2$. Then in the basis ${\bf e}_i\otimes {\bf f}_1$, ${\bf e}_i\otimes {\bf f}_2$ we can write the matrix corresponding to $\alpha(A\otimes I)$ as $$ \alpha(A\otimes I)=\alpha \left[\matrix{A &0\cr 0 &A}\right]=\left[\matrix{\alpha A &0\cr 0 &\alpha A}\right] $$ so $$ \exp \alpha(A\otimes I)=\exp \left[\matrix{\alpha A &0\cr 0 &\alpha A}\right]= \left[\matrix{e^{\alpha A} &0\cr 0 &e^{\alpha A}}\right]\\= (\exp{\alpha A})\otimes I $$