= Solution
At $t=0$ the <power series> gives $\mathbf v(0)=\mathbf v_0$. Equate the coefficient of $t^n$ after differentiating the series and substituting into the <linear system of ordinary differential equations>:
$$
(n+1)\mathbf v_{n+1}=-iH\mathbf v_n.
$$
<Mathematical induction> yields $\mathbf v_n=(-iH)^n\mathbf v_0/n!$. Summing gives the <matrix exponential> solution $\mathbf v(t)=e^{-iHt}\mathbf v_0$.
For the specified diagonal <Hermitian matrix>, its powers and exponential act separately on the two components. Thus
$$
\boxed{U(t)=e^{-iHt}=\begin{pmatrix}e^{-it}&0\\0&e^{it}\end{pmatrix},\qquad\mathbf v(t)=U(t)\mathbf v_0.}
$$
This <unitary matrix> preserves $\mathbf v^\dagger\mathbf v$: the two components acquire opposite phases of unit <modulus>. The coefficient recurrence applies to any constant finite square matrix $H$; unitarity here additionally uses its <Hermitian> property.
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