= Solution
The second infinite-well state has
$$
\boxed{\rho_2(x)=\frac2L\sin^2\frac{2\pi x}{L}}\quad(0<x<L).
$$
Its maxima are at $L/4$ and $3L/4$, with nodes at the walls and center. The density is zero outside.
If the second bound state exists in a symmetric <finite square well>, odd parity preserves its central node and two equal lobes. However, the density is nonzero at the walls and has exterior tails proportional to $e^{-2\kappa|x-x_{\rm wall}|}$. Normalization includes those tails, and internal peak positions and heights change. \b[Finite barriers change penetration and energy; symmetry preserves the central node.] The original wells figure compares the two normalized second-state densities.
Back to article page
