Expression involving Pauli spin matrices Jayaseetha Rau, in her paper Relaxation Phenomena in Spin and Harmonic Oscillator Systems starts from an expression (equation 3.6 in her paper)
\begin{equation}\tag{1}
U^C(\tau) = \exp(i\lambda\sigma_{1j}\sigma_{2j}\tau) =
\exp\left(\frac{i\lambda\tau}{2}[(\sigma_{1j} + \sigma_{2j})^2 - \sigma_{1j}^2 - \sigma_{2j}^2]\right),
\end{equation}
where $\sigma_{1j}$ and $\sigma_{2j}$ are the Pauli matrices corresponding to two spin systems, to get
\begin{equation}\tag{2}
U^C(\tau) = \frac{e^{-3i\lambda\tau}}{4}\left[1 + 3e^{4i\tau\lambda} + \sigma_{1j}\sigma_{2j}(e^{4i\tau\lambda} - 1)\right].
\end{equation}
I don't know how to get (2) from (1). Equation (1), in vector form, is just
\begin{equation}\tag{3}
U^C(\tau) = e^{i\lambda\tau\vec{\sigma}_1\cdot\vec{\sigma}_2} = \exp\left(\frac{i\lambda\tau}{2}[(\vec{\sigma}_{1} + \vec{\sigma}_{2})^2 - \sigma_{1}^2 - \sigma_{2}^2]\right).
\end{equation}
Since the square of each Pauli matrix is an identity,
\begin{equation}\tag{4}
\sigma_1^2 = \sigma_2^2 = 3I_2,
\end{equation}
$I_2$ being the $2 \times 2$ identity matrix. Therefore, equation (3) becomes
\begin{equation}\tag{5}
U^C(\tau) = e^{-3i\lambda\tau}\exp\left(\frac{i\lambda\tau}{2}(\vec{\sigma}_{1} + \vec{\sigma}_{2})^2\right).
\end{equation}
I am unable to get (2) from (5).
 A: An alternative that may simplify things along the way is using their matrix representation and a similarity transformation. With this I mean using the fact
\begin{equation}
\sum_{j=1}^3 \sigma_{j} \otimes \sigma_{j} = u \cdot d \cdot u \, , 
\end{equation}
with\begin{eqnarray}
d &=& \mathrm{diag}(1,1,-3,1) \, ,
\\
u &=& \begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} & 0 \\
0 & \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} & 0 \\
0 & 0 & 0 & 1
\end{pmatrix} \, .
\end{eqnarray}
You can check that $u \cdot u = \mathbf{1}$. This simplifies exponentiation, as you find
\begin{equation}
\exp \left[i \lambda \tau \sum_{j=1}^3 \sigma_{j} \otimes \sigma_{j} \right] = u \cdot \exp \left[i \lambda \tau d \right] \cdot u \, .
\end{equation}
Since $d$ is diagonal, exponentiation is trivial. Then you can check that there are numbers $\alpha$ and $\beta$ that satisfy
\begin{equation}
u \cdot \exp \left[i \lambda \tau \sum_{j=1}^3 \sigma_{j} \otimes \sigma_{j} \right] \cdot u = \exp \left[i \lambda \tau d \right] = \alpha \, \mathbf{1} + \beta \, d \, .
\end{equation}
This is an equation for diagonal matrices, which makes it simple to solve, and will give you the result for your equation $(2)$.
