词条 | Ground state |
释义 |
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. Ground state has no nodes in one-dimensionIn one dimension, the ground state of the Schrödinger equation can be proven to have no nodes.[1] Consider the average energy of a state with a node at {{mvar|x}} = 0; i.e., {{math|ψ(0)}} = 0. The average energy in this state would be where {{math|V(x)}} is the potential. Now, consider a small interval around ; i.e., . Take a new (deformed) wave function {{math|ψ{{'}}(x)}} to be defined as , for ; and , for ; and constant for . If is small enough, this is always possible to do, so that {{math|ψ{{'}}(x)}} is continuous. Assuming around , one may write where is the norm. Note that the kinetic-energy density everywhere because of the normalization. More significantly, the average kinetic energy is lowered by by the deformation to {{math|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 {{math|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 {{math|ψ}} with a node is lower, but still of the same lower order as for the deformed state {{math|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 {{math|ψ{{'}}}} without a node, and the change can be ignored. We can therefore remove all nodes and reduce the energy by , which implies that {{math|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.) Examples
Notes1. ^ See, for example, {{cite thesis |last= Cohen|first= M. |date=1956 |title=The energy spectrum of the excitations in liquid helium |type= Ph.D. |publisher=California Institute of Technology |url=http://thesis.library.caltech.edu/1007/1/Cohen_m_1956.pdf |chapter=Appendix A: Proof of non-degeneracy of the ground state}} Published as {{cite journal |doi=10.1103/PhysRev.102.1189 |title=Energy Spectrum of the Excitations in Liquid Helium |journal=Physical Review |volume=102 |issue=5 |pages=1189 |year=1956 |last1=Feynman |first1=R. P. |last2=Cohen |first2=Michael |bibcode=1956PhRv..102.1189F}} 2. ^{{cite web |title=Unit of time (second) |work=SI Brochure |url=http://www.bipm.org/en/si/si_brochure/chapter2/2-1/second.html |publisher=International Bureau of Weights and Measures |accessdate=2013-12-22}} Bibliography
1 : Quantum mechanics |
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