Lowest energy level of a quantum system
The ground state of a quantum-mechanical system is its stationary state of lowest energy; the energy of the ground state is known as the zero-point energy of the system. An excited state is any state with energy greater than the ground state. In quantum field theory, the ground state is usually called the vacuum state or the vacuum.
If more than one ground state exists, they are said to be degenerate. Many systems have degenerate ground states. Degeneracy occurs whenever there exists a unitary operator that acts non-trivially on a ground state and commutes with the Hamiltonian of the system.
According to the third law of thermodynamics, a system at absolute zero temperature exists in its ground state; thus, its entropy is determined by the degeneracy of the ground state. Many systems, such as a perfect crystal lattice, have a unique ground state and therefore have zero entropy at absolute zero. It is also possible for the highest excited state to have absolute zero temperature for systems that exhibit negative temperature.
Absence of nodes in one dimension
In one dimension, the ground state of the Schrödinger equation can be proven to have no nodes.
Consider the average energy of a state with a node at x = 0; i.e., ψ(0) = 0. The average energy in this state would be
where V(x) is the potential.
With integration by parts:
Hence in case that is equal to zero, one gets:
Now, consider a small interval around ; i.e., . Take a new (deformed) wave function ψ'(x) to be defined as , for ; and , for ; and constant for . If is small enough, this is always possible to do, so that ψ'(x) is continuous.
Assuming around , one may write
where is the norm.
Note that the kinetic-energy densities hold everywhere because of the normalization. More significantly, the average kinetic energy is lowered by by the deformation to ψ'.
Now, consider the potential energy. For definiteness, let us choose . Then it is clear that, outside the interval , the potential energy density is smaller for the ψ' because there.
On the other hand, in the interval we have
which holds to order .
However, the contribution to the potential energy from this region for the state ψ with a node is
lower, but still of the same lower order as for the deformed state ψ', and subdominant to the lowering of the average kinetic energy.
Therefore, the potential energy is unchanged up to order , if we deform the state with a node into a state ψ' without a node, and the change can be ignored.
We can therefore remove all nodes and reduce the energy by , which implies that ψ' cannot be the ground state. Thus the ground-state wave function cannot have a node. This completes the proof. (The average energy may then be further lowered by eliminating undulations, to the variational absolute minimum.)
As the ground state has no nodes it is spatially non-degenerate, i.e. there are no two stationary quantum states with the energy eigenvalue of the ground state (let's name it ) and the same spin state and therefore would only differ in their position-space wave functions.
The reasoning goes by contradiction: For if the ground state would be degenerate then there would be two orthonormal stationary states and — later on represented by their complex-valued position-space wave functions and — and any superposition with the complex numbers fulfilling the condition would also be a be such a state, i.e. would have the same energy-eigenvalue and the same spin-state.
Now let be some random point (where both wave functions are defined) and set:
(according to the premise no nodes).
Therefore the position-space wave function of is
for all .
But i.e., is a node of the ground state wave function and that is in contradiction to the premise that this wave function cannot have a node.
Note that the ground state could be degenerate because of different spin states like and while having the same position-space wave function: Any superposition of these states would create a mixed spin state but leave the spatial part (as a common factor of both) unaltered.