Number of Nodes in energy eigenstates

Question

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?

Answer 1

The physical interpretation behind the increase of energy with the number of nodes can be understood in a very crude manner as follows:

Nodes are points of zero probability densities. Since the wavefunction is continuous, the probability density is also a continuous function. So the regions in the neighbourhood of nodes will have small probability densities. Physically, this means that the particle has less space to move around. That is, the particle is more confined and uncertainty in position $(\Delta x)$ decreases. This increases $\Delta p$ (due to the uncertainty principle), causing an increase in energy. Hence, the energy increases as the number of nodes increase. So the ground state should not have any node.

Answer 2

I guess there is not that much to grasp, unless you can really understand dark spots on an electron diffraction pattern. Very roughly explanation would be to interpret wave functions of a particle in a potential well as "standing waves", or as two interfering waves reflected from the walls of the well. Increasing the energy leads to higher harmonics, which leads to additional nodes. Nodes' numbering is the same as in the case of a classical string.