Exponentiating an anti-diagonal matrix

Question

I want to understand an equation I saw in a paper. If $\theta$ is some angle, and $\sigma_y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}$, then it is stated that $e^{-i \theta \sigma_y } = R_{\theta}$ (the rotation matrix of angle $\theta$).

Now set $A:= -i \theta \sigma_y = \begin{bmatrix} 0 & \theta \\ -\theta & 0 \end{bmatrix}$, so that $e^{A}$ is defined as $\sum_{j=0}^{\infty} \frac {1} {k!} A^k$. Now I know how to exponentiate a diagonal matrix, but I am not sure how to do it in this case (anti-diagonal). Insights appreciated.

Answer 1

Observe the following. If $A = -i\theta \sigma_y$ then we have that \begin{eqnarray*} A & = & \left [ \begin{array}{cc} 0 & -\theta \\ \theta & 0 \\ \end{array} \right ] \\ A^2 & = & \left [ \begin{array}{cc} -\theta^2 & 0 \\ 0 & -\theta^2 \\ \end{array} \right ] \\ A^3 & = & \left [ \begin{array}{cc} 0 & \theta^3 \\ -\theta^3 & 0 \\ \end{array} \right ] \\ A^4 & = & \left [ \begin{array}{cc} \theta^4 & 0 \\ 0 & \theta^4 \\ \end{array} \right ] \\ \end{eqnarray*}

demonstrating that that the odd powers have nonzero off-diagonal entries, while the even powers have nonzero diagonal entries. We can therefore break up the sum for the exponential into an even part and an odd part:

\begin{eqnarray*} \exp(A) & = & \sum_{k=0}^\infty \frac{1}{k!}A^k \\ & = & \sum_{k=0}^\infty \frac{1}{(2k)!} A^{2k} + \sum_{k=0}^\infty \frac{1}{(2k+1)!}A^{2k+1} \end{eqnarray*}

Now notice for the even series we get \begin{eqnarray*} \sum_{k=0} \frac{1}{(2k)!} \left [ \begin{array}{cc} -\theta^2 & 0 \\ 0 & -\theta^2 \\ \end{array} \right ]^{k} & = & \left [ \begin{array}{cc} \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!} \theta^{2k} & 0 \\ 0 & \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!} \theta^{2k} \\ \end{array} \right ] \\ & = & \left [ \begin{array}{cc} \cos\theta & 0 \\ 0 & \cos\theta \\ \end{array} \right ]. \end{eqnarray*}

Convince yourself that the odd series yields $\pm \sin\theta$ in the off-diagonal entries.

Answer 2

As mentioned above, the diagonal terms are $\sum_{j=0}^\infty (-1)^j \frac{1}{(2j)!}\theta^{2j}$. This is same as $\frac{e^{i\theta} + e^{-i\theta}}{2} = \cos(\theta)$.

Similarly, the two diagonal entries are negative of each other, and they can be shown as $-\sin(\theta)$ and $\sin(\theta)$ respectively thus showing that the overall matrix is a rotation matrix.