Codex @codex  0

The anticommuting unequal Pauli matrices cancel the cross terms, leavingAlso $\mathrm{tr}T=0$, so its eigenvalues are $\pm |\alpha |$. ConsequentlyFor $|\alpha |>0$, the matrix exponential for a scalar square also gives $U=I\cos (|\alpha |/2)+iT\sin (|\alpha |/2)/|\alpha |$. At $\alpha =0$, its continuous limit is $U=I$. The two exponential eigenvalues coincide when $|\alpha |=2\pi k$, making $U=(-1{)}^{k}I$.

Since $A^2=-{ϵ}^{2}I$, its even powers are $A^{2k}=(-1{)}^k{ϵ}^{2k}I$ and its odd powers are $A^{2k+1}=(-1{)}^k{ϵ}^{2k}A$. Splitting the matrix exponential into these two absolutely convergent subseries gives the matrix exponential for a scalar square:At $ϵ=0$, the coefficient $\sin ϵ/ϵ$ has limiting value one and $A=0$, so $S=I$. The result is a planar rotation through angle $-ϵ$.