Ground state

The ground state of a quantum-mechanical system is its lowest-energy state; 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 the quantum field theory, the ground state is usually called the vacuum state or the vacuum. The ground state of a quantum-mechanical system is its lowest-energy state; 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 the 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. 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. Now, consider a small interval around x = 0 {displaystyle x=0} ; i.e., x ∈ [ − ϵ , ϵ ] {displaystyle xin } . Take a new (deformed) wave function ψ'(x) to be defined as ψ ′ ( x ) = ψ ( x ) {displaystyle psi '(x)=psi (x)} , for x < − ϵ {displaystyle x<-epsilon } ; and ψ ′ ( x ) = − ψ ( x ) {displaystyle psi '(x)=-psi (x)} , for x > ϵ {displaystyle x>epsilon } ; and constant for x ∈ [ − ϵ , ϵ ] {displaystyle xin } . If ϵ {displaystyle epsilon } is small enough, this is always possible to do, so that ψ'(x) is continuous. Assuming ψ ( x ) ≈ − c x {displaystyle psi (x)approx -cx} around x = 0 {displaystyle x=0} , one may write where N = 1 1 + 4 3 | c | 2 ϵ 3 {displaystyle N={frac {1}{sqrt {1+{frac {4}{3}}|c|^{2}epsilon ^{3}}}}} is the norm.