= Solution
For a nonzero <quantum state> $\psi$ and a <self-adjoint operator> $A$, the <expectation value> and <standard deviation> of its measurement are
$$
\boxed{a=\mathbb E_\psi(A)=\frac{\langle\psi,A\psi\rangle}{\langle\psi,\psi\rangle},\qquad
\Delta_\psi(A)=\frac{\|(A-aI)\psi\|}{\|\psi\|}=\sqrt{\mathbb E_\psi(A^2)-a^2}.}
$$
The squared deviation is also called the <variance> of the <observable>. All displayed products are understood on <quantum states> for which the relevant moments exist. Self-adjointness gives $\overline{\langle\psi,A\psi\rangle}=\langle A\psi,\psi\rangle=\langle\psi,A\psi\rangle$, so $a$ is real. Expanding the squared <norm> gives the <variance> expression. A <norm> is nonnegative and is zero exactly for the zero <vector>, hence $\Delta_\psi(A)=0$ precisely when $A\psi=a\psi$. Thus \b[zero dispersion is equivalent to being an <eigenstate>].
For bounded $H$, the <operator exponential> is the <norm>-convergent power series
$$
U_t=\sum_{n=0}^\infty\frac{(-itH/\hbar)^n}{n!}.
$$
For a general unbounded <self-adjoint> <Hamiltonian operator>, the <spectral theorem> defines the same notation by
$$
U_t=\int_{\mathbb R}e^{-itE/\hbar}\,dP_H(E).
$$
This definition does not require the power series to converge on every <quantum state>. <complex conjugation> of the spectral multiplier gives $U_t^*=U_{-t}$, and multiplication of the phases gives $U_{-t}U_t=I$. Thus \b[$U_t$ is <unitary>], preserves the <quantum state> <norm>, and has $U_0=I$.
For initial data in the domain of $H$, differentiating the spectral expression yields $\partial_tU_t\psi_0=-(i/\hbar)HU_t\psi_0$. Consequently
$$
\boxed{\psi_t=U_t\psi_0}
$$
solves the <Schrödinger equation> with the prescribed initial <quantum state>. Uniqueness follows by differentiating $U_{-t}\psi_t$: the two terms cancel, leaving the constant $\psi_0$.
In the two-dimensional energy <basis>, expand the initial <quantum state> as $\psi_0=c_1\Psi_1+c_2\Psi_2$. Each energy <eigenvector> receives its own phase under the <operator exponential>, giving
$$
\boxed{\psi_t=c_1e^{-itE_1/\hbar}\Psi_1+c_2e^{-itE_2/\hbar}\Psi_2.}
$$
This is the general solution because every initial <vector> has a unique <basis> expansion.
In that <orthonormal basis> the other <observable> has the real symmetric <matrix>
$$
[A]=\begin{pmatrix}\cos2\alpha&\sin2\alpha\\\sin2\alpha&-\cos2\alpha\end{pmatrix}.
$$
Its conjugate transpose equals itself, so it is <self-adjoint>. Also $A^2=I$ and $\operatorname{tr}A=0$, giving <eigenvalues> $+1$ and $-1$. Direct substitution gives the normalized <orthogonal> <eigenvectors>
$$
\boxed{v_+=\cos\alpha\,\Psi_1+\sin\alpha\,\Psi_2,\qquad
v_-=-\sin\alpha\,\Psi_1+\cos\alpha\,\Psi_2.}
$$
An irrelevant global phase may multiply either <vector>. With the chosen lower <eigenstate> initially,
$$
\boxed{\psi_t=-\sin\alpha\,e^{-itE_1/\hbar}\Psi_1+\cos\alpha\,e^{-itE_2/\hbar}\Psi_2.}
$$
The <Born rule> says that the <probability> for the higher <eigenstate> is $|\langle v_+,\psi_t\rangle|^2$. Its amplitude is $\sin\alpha\cos\alpha(e^{-itE_2/\hbar}-e^{-itE_1/\hbar})$. Using $|e^{-iu}-e^{-iv}|^2=4\sin^2((u-v)/2)$ gives
$$
\boxed{P_+(t)=\sin^2(2\alpha)\sin^2\left(\frac{(E_2-E_1)t}{2\hbar}\right).}
$$
This is the <transition probability for a rotated two-level observable>. Its time dependence comes from the relative phase between two energy <eigenstates>. If their energies coincide, or if the measured <observable> has the energy <eigenvectors> as its <eigenvectors>, this transition <probability> is zero at all times.
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