= Solution
The <orthonormal basis> makes the squared <norm> equal to the sum of squared expansion coefficients:
$$
1=|\mathcal N|^2(2\pi R)\left(\frac14+\frac1{16}+\frac1{16}\right)
=|\mathcal N|^2\frac{3\pi R}{4}.
$$
Choose a positive real <wavefunction normalization> constant, fixing an irrelevant <global phase>:
$$
\boxed{\mathcal N=\frac2{\sqrt{3\pi R}},\qquad
\Psi(0)=\sqrt{\frac23}\,\phi_0-\frac1{\sqrt6}(\phi_2+\phi_{-2}).}
$$
As a direct check, $\int_0^{2\pi R}\sin^4(x/R)\,dx=3\pi R/4$.
Back to article page
