 Solution

The second infinite-well state hasIts 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 $e^{-2\kappa |x-x_{\mathrm{wall}}|}$. Normalization includes those tails, and internal peak positions and heights change. Finite barriers change penetration and energy; symmetry preserves the central node. The original wells figure compares the two normalized second-state densities.