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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).

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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 $(2)$.

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  • $\begingroup$ Hi @secavara, thanks for your answer. Can you clarify why $\exp(i\lambda\tau \sigma_j \otimes \sigma_j) = u \cdot \exp(i\lambda\tau d) \cdot u$? I am using the summation convention to keep the expression simpler. $\endgroup$ – Amey Joshi Jan 17 at 13:52
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    $\begingroup$ You can see this by using the definition of matrix exponentiation: $\exp\left( i \lambda \tau \, u \cdot d \cdot u \right) = \sum_{k=0}^\infty \frac{1}{k!} \left( i \lambda \tau \, u \cdot d \cdot u \right)^k = \sum_{k=0}^\infty \frac{1}{k!} \left( i \lambda \tau \right)^k \left(u \cdot d \cdot u \right)^k$. Since $u \cdot u = \mathbf{1}$, we have $\left(u \cdot d \cdot u \right)^k = (u \cdot d \cdot u) \cdot ...\cdot (u \cdot d \cdot u) = u \cdot d^k \cdot u $. Plugging that back into the sum, gives you the result. $\endgroup$ – secavara Jan 17 at 14:02

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