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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク ground state ： ウィキペディア英語版 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. 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 an average energy in the state with a node at $x=0$, i.e. $\backslash psi(0)=0$. Consider the average energy in this state $$ \left\langle\psi| H|\psi\right\rangle=\int dx\; \left(-\frac\psi^ * \frac+V(x)|\psi(x)|^2\right) where $V(x)$ is the potential. Now consider a small interval around $x=0$, i.e. $x\backslash in()$. Take a new wavefunction $\backslash psi\text{'}(x)$ to be defined as and $\backslash psi\text{'}(x)=-\backslash psi(x),\; x>\backslash epsilon$ and constant for $x\backslash in()$. If epsilon is small enough then this is always possible to do so that $\backslash psi\text{'}(x)$ is continuous. So assuming $\backslash psi(x)\backslash approx-cx$ around $x=0$, we can write the new function as $$ \psi'(x)=N\left\ |\psi(x)| & |x|>\epsilon\\ c\epsilon & |x|\le\epsilon \end\right. where $N=\backslash frac\}\text{'}=\backslash int\_^\backslash epsilon\; dx\backslash ;\; V(x)|\backslash psi\text{'}|^2=\backslash frac\backslash int\_^\backslash epsilon\; V(x)\backslash approx\; \backslash fracV(0)+\backslash dots\backslash ;.$ which is correct to this order of $\backslash epsilon$ and $\backslash dots$ indicate higher order corrections. On the other hand, the potential energy in the $\backslash psi$ state is $$ V^\epsilon_=\int_^\epsilon dx\; V(x)|\psi|^2=\int_^\epsilon dx\; |c|^2|x|^2V(x)\approx\fracV(0)+\dots\;. which is the same as that of the $\backslash psi\text{'}$ state to the order shown. Therefore, the potential energy unchanged to leading order in $\backslash epsilon$ by deforming the state with a node $\backslash psi$ into a state without a node $\backslash psi\text{'}$. We can do this by removing all nodes thereby reducing the energy, which implies that the ground state energy must not have a node. This completes the proof. 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「ground state」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.