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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 good quantum number ： ウィキペディア英語版 good quantum number In quantum mechanics, given a particular Hamiltonian $H$ and an operator $O$ with corresponding eigenvalues and eigenvectors given by $O|q\_j\backslash rangle=q\_j|q\_j\backslash rangle$, then the numbers (or the eigenvalues) $q\_j$ are said to be "good quantum numbers" if every eigenvector $|q\_j\backslash rangle$ remains an eigenvector of $O$ ''with the same eigenvalue'' as time evolves. Hence, if: $O|q\_j\backslash rangle=O\backslash sum\_k\; c\_k(0)\; |e\_k\backslash rangle\; =\; q\_j\; |q\_j\backslash rangle$ then we require ::$O\backslash sum\_k\; c\_k(0)\; \backslash exp(-i\; e\_k\; t/\backslash hbar)\backslash ,|e\_k\backslash rangle=q\_j\backslash sum\_k\; c\_k(0)\; \backslash exp(-i\; e\_k\; t/\backslash hbar)\backslash ,|e\_k\backslash rangle$ for all eigenvectors $|q\_j\backslash rangle$ in order to call $q$ a good quantum number (where $e\_k$s represent the eigenvectors of the Hamiltonian) Theorem: A necessary and sufficient condition for q (which is an eigenvalue of an operator O) to be good is that $O$ commutes with the Hamiltonian $H$ Proof: Assume $()=0$. :: If $|\backslash psi\_0\backslash rangle$ is an eigenvector of $O$, then we have (by definition) that $O\; |\backslash psi\_0\backslash rangle=q\_j\; |\; \backslash psi\_0\backslash rangle$, and so : ::$O|\backslash psi\_t\backslash rangle=O\backslash ,T(t)\backslash ,|\backslash psi\_0\backslash rangle$ ::$=O\; e^|\backslash psi\_0\backslash rangle$ ::$=\; O\; \backslash sum\_^\; \backslash frac\; (-i\; H\; t/\backslash hbar)^\; |\backslash psi\_0\backslash rangle$ ::$=\; \backslash sum\_^\; \backslash frac\; (-i\; H\; t/\backslash hbar)^\; O\; |\backslash psi\_0\backslash rangle$ ::$=q\_j\; |\backslash psi\_t\backslash rangle$ ==Ehrenfest Theorem and Good Quantum Numbers== Ehrenfest Theorem gives the rate of change of the expectation value of operators. It reads as follows: :$\backslash frac\backslash langle\; A(t)\backslash rangle\; =\; \backslash left\backslash langle\backslash frac\backslash right\backslash rangle\; +\; \backslash frac\backslash langle()\backslash rangle$ Commonly occurring operators don't depend explicitly on time. If such operators commute with the Hamiltonian, then their expectation value remains constant with time. Now, if the system is in one of the common eigenstates of the operator$A$ (and $H$ too), then system remains in this eigenstate as time progresses. Any measurement of the quantity $A$ will give us the eigenvalue (or the good quantum number) associated with the eigenstates in which the particle is. This is actually a statement of conservation in Quantum Mechanics. In non-relativistic treatment,$l$and $s$ are good quantum numbers but in relativistic quantum mechanics they are no longer good quantum numbers as $L$ and $S$ do not commute with $H$ (in Dirac theory). $J=L+S$ is a good quantum number in relativistic quantum mechanics as $J$ commutes with $H$. 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「good quantum number」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.