# Exponential of tensor product, excess scalar exponential

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

• The way you add angular moment is weird. Normally the Hilbert space of two particles (with angular momenta) is the tensor product of the individual Hilbert spaces not the direct sum as you write.
– lcv
Jan 9, 2020 at 11:54
• I used the wrong symbol. Everywhere. I will correct that. @Icv
– user224659
Jan 9, 2020 at 12:27
• I don’t think this is fixed. The Hilbert space for the 2-particle system is $D_1\otimes D_2$. I dunno what the * means in $(D_1\otimes 1)*(1\otimes D_2)$. Jan 9, 2020 at 14:41
• The * ist just a multiplication in the sense that you apply the operators one after one. both of the terms you wrote down mean the same thing.
– user224659
Jan 9, 2020 at 15:36

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 $$\exp \alpha(A\otimes I)=\exp \left[\matrix{\alpha A &0\cr 0 &\alpha A}\right]= \left[\matrix{e^{\alpha A} &0\cr 0 &e^{\alpha A}}\right]\\= (\exp{\alpha A})\otimes I$$