Exponentiation in tensor product of Hilbert spaces Consider a Hilbert space $\mathcal{H}$, defined as a tensor product of other 2-dimensional Hilbert spaces.
If you have $$H\equiv\beta\left(\sigma^{(1)}_1 -\sigma^{(2)}_1 \right)^2\in\mathcal{H}\equiv\mathcal{H}_{(1)}\otimes\mathcal{H}_{(2)},$$
where $$\sigma_1^{(1)}\equiv\sigma_1\otimes 1\!\!1, \;\;\; \sigma_1^{(2)}\equiv 1\!\!1\otimes\sigma_1,$$ $\sigma_1$ (Pauli matrix)  being $$\sigma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix},$$
how would you calculate $$\exp{\left\{-i H t \right\}}\,?$$
My attempt: knowing that $$H=\beta\left(\left(\sigma_1^{(1)}\right)^2 + \left(\sigma_1^{(2)}\right)^2 - 2\sigma_1\otimes\sigma_1\right)=2\beta\left( 1\!\!1_4 -\sigma_1\otimes\sigma_1 \right),$$ where $1\!\!1_4$ is the 4-dimensional identity matrix, I write $$\exp{\left\{-i H t \right\}}=\exp{\left\{-i\,2\beta\left(1\!\!1_4 -\sigma_1\otimes\sigma_1 \right)t\right\}}=$$ $$=\left[1\!\!1_4 \exp{\left\{-i\,2\beta t\right\}}\right] \cdot \exp{\left\{i\,2\beta\left(\sigma_1\otimes\sigma_1\right)t \right\}}=$$ $$=\left[1\!\!1_4 \exp{\left\{-i\,2\beta t\right\}}\right] \cdot \left[\exp{\left\{i\,2\beta\sigma_1 t\right\}}\otimes\exp{\left\{i\,2\beta\sigma_1 t\right\}} \right].$$
Is it correct?
 A: Very nice question! In the following, the idea will be to (unitarily) diagonalize $H$ as
$$
\exp(U DU^\dagger)=\sum_{n=0}^\infty \frac{(U DU^\dagger)^n}{n!}=\sum_{n=0}^\infty \frac{U D^nU^\dagger}{n!}=U\sum_{n=0}^\infty \frac{ D^n}{n!}U^\dagger=U \exp(D)U^\dagger
$$
where $D=\operatorname{diag}(d_1,\ldots,d_n)$ so $\exp(D)=\operatorname{diag}(e^{d_1},\ldots,e^{d_n})$ as is readily verified.
You already calculated out that $H=2\beta (\operatorname{id}_4-\sigma_1\otimes\sigma_1)$ so as $\operatorname{id}_4-\sigma_1\otimes\sigma_1$ is hermitian (since $\sigma_1=\sigma_1^\dagger$) we can unitarily diagonalize it. The eigenvalues and possible eigenvectors of $\operatorname{id}_4-\sigma_1\otimes\sigma_1$ are given by
$$
\lambda_1=2,\ v_1=(-1,0,0,1)\qquad\qquad \lambda_2=2,\ v_2=(0,-1,1,0)\\
\lambda_3=0,\ v_3=(0,1,1,0)\qquad\qquad \lambda_4=0,\ v_4=(1,0,0,1)
$$
so we already know that we can write $H=2\beta U DU^\dagger$ with $D=\operatorname{diag}(2,2,0,0)$ (or any permutation of that, which doesn't matter as they all differ only by a unitary transformation). The unitary matrix now consists of a set of orthonormal eigenvectors of the corresponding eigenvalues. One quickly sees that $\lbrace v_1,\ldots,v_4\rbrace$ is an orthogonal set but $\Vert v_1\Vert=\ldots=\Vert v_4\Vert=\sqrt{2}$ which means
$$
U=\Big(\frac{v_1}{\Vert v_1\Vert},\ldots,\frac{v_4}{\Vert v_4\Vert}\Big)=\frac{1}{\sqrt{2}}\begin{pmatrix} -1&0&0&1\\0&-1&1&0\\0&1&1&0\\1&0&0&1 \end{pmatrix}=U^\dagger
$$
is a possible choice for $U$. With this we get
$$
\exp(-iHt)=\exp(U \operatorname{diag}(-4i\beta t,-4i\beta t,0,0)U^\dagger)
=U \operatorname{diag}(e^{-4i\beta t},e^{-4i\beta t},1,1)U^\dagger
$$
as elaborated in the beginning which by simple calculation then results in
$$
\exp(-iHt)=\frac12(1+e^{-4i\beta t})\operatorname{id}_4+\frac12(1-e^{-4i\beta t})(\sigma_1\otimes\sigma_1).
$$
