= Solution
{c}
The identity $\cos^2x=\tfrac12+\tfrac14e^{2ix}+\tfrac14e^{-2ix}$ gives the <momentum eigenstate> expansion
$$
\Psi(x,0)=A\sqrt{2\pi}\left(\frac12\psi_0+\frac14\psi_2+\frac14\psi_{-2}\right).
$$
Using the <orthonormal basis> from part (a), the <wavefunction normalization> is
$$
1=|A|^2(2\pi)\left(\frac14+\frac1{16}+\frac1{16}\right)=|A|^2\frac{3\pi}{4}.
$$
Hence $|A|=2/\sqrt{3\pi}$. The <wavefunction normalization> leaves a <global phase> free; the positive real choice used in the requested expansion gives
$$
\boxed{A=\frac2{\sqrt{3\pi}},\qquad \Psi(x,0)=\sqrt{\frac23}\left(\psi_0(x)+\frac12\psi_2(x)+\frac12\psi_{-2}(x)\right).}
$$
The two nonzero <eigenvalues> of the <momentum operator> have the same <energy eigenvalue> $E_\star=2\hbar^2/m$. Under <unitary time evolution>, their coefficients acquire the same <dynamical phase>. With $\Omega=E_\star/\hbar=2\hbar/m$,
$$
\boxed{\begin{aligned}
\Psi(x,t)&=\sqrt{\frac23}\left(\psi_0(x)+\frac{e^{-i\Omega t}}2\psi_2(x)+\frac{e^{-i\Omega t}}2\psi_{-2}(x)\right)\\
&=\frac{1+e^{-i\Omega t}\cos(2x)}{\sqrt{3\pi}}.
\end{aligned}}
$$
At $t=0$, $1+\cos(2x)=2\cos^2x$ recovers the initial <wavefunction>. Its <probability density>, useful below, is
$$
|\Psi(x,t)|^2=\frac{1+\cos^2(2x)+2\cos(2x)\cos(\Omega t)}{3\pi}.
$$
Back to article page
