Ground state

Energy levels for an electron in an atom: ground state and excited states. After absorbing energy, an electron may jump from the ground state to a higher energy excited 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. The ground state of a quantum field theory 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 which 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.

1D ground state has no nodes

In 1D, the ground state of the Schrödinger equation has no nodes. This can be proved considering the average energy of a state with a node at x=0, i.e. \psi(0)=0. Consider the average energy in this state


\left\langle\psi| H|\psi\right\rangle=\int dx\; \left(-\frac{\hbar^2}{2m}\psi^* \frac{d^2\psi}{dx^2}+V(x)|\psi(x)|^2\right)
where V(x) is the potential. Now consider a small interval around x=0, i.e. x\in[-\epsilon,\epsilon]. Take a new wave function \psi'(x) to be defined as \psi'(x)=\psi(x), x<-\epsilon and \psi'(x)=-\psi(x), x>\epsilon and constant for x\in[-\epsilon,\epsilon]. If \epsilon is small enough then this is always possible to do so that \psi'(x) is continuous. Assuming \psi(x)\approx-cx around x=0, we can write


\psi'(x)=N\left\{\begin{array}{ll}
|\psi(x)| & |x|>\epsilon\\
c\epsilon & |x|\le\epsilon
\end{array}\right.

where N=\frac{1}{\sqrt{1+4|c|^2\epsilon^3/3}} is the norm. Note that the kinetic energy density |d\psi'/dx|^2<|d\psi/dx|^2 everywhere because of the normalization. Now consider the potential energy. For definiteness let us choose V(x)\ge 0. Then it is clear that outside the interval x\in[-\epsilon,\epsilon] the potential energy density is smaller for the \psi' because |\psi'|<|\psi| there. On the other hand, in the interval x\in[-\epsilon,\epsilon] we have


{V^\epsilon_{avg}}'=\int_{-\epsilon}^\epsilon dx\; V(x)|\psi'|^2=\frac{\epsilon^2|c|^2}{1+4|c|^2\epsilon^3/3}\int_{-\epsilon}^\epsilon V(x)\simeq 2\epsilon^3|c|^2 V(0)+\cdots,

which is correct to order \epsilon^3. On the other hand, the contribution to the potential energy from this region for the state with a node, \psi(x), is


V^\epsilon_{avg}=\int_{-\epsilon}^\epsilon dx\; V(x)|\psi|^2= |c|^2\int_{-\epsilon}^\epsilon dx\; x^2V(x)\simeq \frac{2\epsilon^3|c|^2}{3}V(0)+\cdots,

which is of the same order as for the state \psi'. Therefore, the potential energy is unchanged up to order \epsilon^2 if we deform the state with a node \psi into a state without a node \psi'. We can therefore remove all nodes and reduce the energy, which implies that the ground-state wave function cannot have a node. This completes the proof.

Examples

Initial wave functions for the first four states of a one-dimensional particle in a box

Notes

  1. "Unit of time (second)". SI Brochure. BIPM. Retrieved 2013-12-22.

Bibliography

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