# Number of Nodes in energy eigenstates

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?

• Commented Jan 12, 2017 at 18:01

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.