I have a question from the very basics of Quantum Mechanics. Given this theorem:
For the discrete bound-state spectrum of a one-dimensional potential let the allowed energies be $E_1<E_2< E_3< ...$ with $E_1$ the ground state energy. Let the associated energy eigenstates be $ψ_1,ψ_2,ψ_3,...$. The wavefunction $ψ_1$ has no nodes,$ψ_2$ has one node, and each consecutive wavefunction has one additional node. In conclusion $ψ_n$ has $n−1$ nodes.
What is the physical interpretation for the number of nodes in the concrete energy eigenstate? I understand that the probability of finding the particle in the node point is $0$ for the given energy. However, why does the ground state never have a node? or why does every higher energy level increments number of nodes precisely by 1?