= Solution
Inside the infinite well, the time-independent <Schrödinger equation> is $-\hbar^2\Psi''/(2m)=E\Psi$ with Dirichlet endpoints. Negative or zero <energy> gives no nonzero solution satisfying both endpoints. For $E>0$, the sine solution requires $ka=n\pi$. Thus
$$
\boxed{E_n=\frac{n^2\pi^2\hbar^2}{2ma^2},\qquad\Psi_n(x)=\sqrt{\frac2a}\sin\frac{n\pi x}a,\quad n\geq1.}
$$
The <normalization> follows from $\int_0^a\sin^2(n\pi x/a)\,dx=a/2$. These functions form an <orthonormal basis>. Substitute the expansion into $i\hbar\partial_t\psi=H\psi$ and take the <inner product> with each basis vector: orthonormality gives $i\hbar\dot c_n=E_nc_n$, equivalently
$$
\boxed{c_n(t)=c_n(0)e^{-iE_nt/\hbar}.}
$$
For a general infinite expansion this is interpreted in the usual <Hamiltonian> domain or its <unitary> evolution; the particular state below has only two terms.
Put $\theta=\pi x/a$. The identity $\sin\theta(16\cos^2\theta-1)=4\sin3\theta+3\sin\theta$ gives $c_1=3/5$, $c_3=4/5$, all others zero. Their squared magnitudes sum to one. Hence
$$
\boxed{\langle H\rangle=\frac9{25}E_1+\frac{16}{25}E_3=\frac{153\pi^2\hbar^2}{50ma^2}.}
$$
The evolved state is $(3e^{-iE_1t/\hbar}\Psi_1+4e^{-iE_3t/\hbar}\Psi_3)/5$. Its position <probability> includes an <interference> term with frequency $\Omega=(E_3-E_1)/\hbar=4\pi^2\hbar/(ma^2)$. Specifically,
$$
P(t)=\frac2{25\pi}\int_0^{\pi/4}\left[9\sin^2\theta+16\sin^23\theta+24\sin\theta\sin3\theta\cos(\Omega t)\right]d\theta.
$$
The three required integrals are $\pi/8-1/4$, $\pi/8+1/12$ and $1/4$, respectively. Therefore
$$
\boxed{P(0\leq X\leq a/4;t)=\frac14-\frac{11}{150\pi}+\frac{12}{25\pi}\cos\left(\frac{4\pi^2\hbar t}{ma^2}\right).}
$$
The <energy> <expectation> is constant while the spatial <probability> oscillates, because relative <energy> phases affect <interference> but not the <energy> probabilities.
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