There is a well-known formula for rotations of Pauli vectors
$e^{i\theta \vec{n}\cdot\vec{\sigma}}=\cos{\theta}+i\sin{\theta} \vec{n}\cdot\vec{\sigma}$ with $\vec{\sigma}=(\sigma_x,\sigma_y,\sigma_z)$.
Now I am dealing with a similar formula with an added difficulty: two spin vectors, $\vec{\sigma_1}$ and $\vec{\sigma_2}$, such that my rotation is given by
$e^{i\theta \vec{\sigma_1}\cdot\vec{\sigma_2}}$. It is not possible to, a priori, writte the same expression as for one spin because the dot product will produce non-commuting terms, i.e. $[\sigma_x^{1}\sigma_x^{2},\sigma_y^{1}\sigma_y^{2}]\neq 0$ which prevents one from using the BKH formula.
Is there any known expression for such a two-spin rotation?
Thank you!
Edit: My goal is to apply this transformation to a state and see how it transforms. Specifically, I want to evaluate $e^{i\theta \vec{\sigma_1}\cdot\vec{\sigma_2}}|\phi\rangle$, where $|\phi\rangle$ is the bell state $\frac{1}{\sqrt{2}}(|00\rangle+|11\rangle)$
Best Answer
Review spin addition for two spin doublets. Your formula is actually simpler, in a sense! No expansion of the exponential is warranted. Expanding it is a hallmark of some misconception.
Anything that happens to the exponent happens to the exponential. You are liable to get confused by the complacent notation you express your exponential in. The exponent of it is proportional to a 4×4 matrix, $$ (\vec\sigma_1\otimes {\mathbb I}+ {\mathbb I} \otimes \vec \sigma_2 )^2/2 -3 {\mathbb I}_4 \\ = \vec \sigma_1\otimes \cdot \vec \sigma_2 +(\vec \sigma_1^2\otimes {\mathbb I}+ {\mathbb I}\otimes \vec \sigma_2^2 )/2 -3 {\mathbb I}_4\\ = \vec \sigma_1\otimes \cdot \vec \sigma_2 ~, $$ where the dimensionality of the unlabelled identity matrices, 2, is obvious from context and omitted for simplicity.
Now, you know from the linked answer that the above 4×4 matrix may be reduced by a 4×4 Clebsch similarity transform change of basis to a diagonal matrix, $$ O ~\vec \sigma_1\otimes \cdot \vec \sigma_2 ~ O^{-1}= 5{\mathbb I}_3 \oplus (-3), $$ with evident 3×3 and 1×1 blocks.
The very same similarity transformation will transform the exponential of your matrix then, the trivial exponential of a diagonal matrix, $$ O~ \exp (i\theta ~ \vec \sigma_1\otimes \cdot \vec \sigma_2 ) ~O^ {-1}= e^{i5\theta}{\mathbb I}_3 \oplus e^{-3i\theta}\\ = \operatorname{diag} (e^{i5\theta},e^{i5\theta},e^{i5\theta} , e^{-3i\theta} ). $$ If you insisted on returning to your original, 1-tensor-2, dysfunctional basis, you trivially reverse the above orthogonal Clebsch equivalence transformation.