翻訳と辞書 Words near each other ・ Sum of angles of a triangle ・ Sum of Christianity ・ Sum of Logic ・ Sum of normally distributed random variables ・ Sum of Parts ・ Sum of perpetuities method ・ Sum of public power ・ Sum of radicals ・ Sum of squares ・ Sum of the Parts ・ Sum of two squares theorem ・ Sum Practysis of Medecyne and other Short Works ・ Sum rule ・ Sum rule in differentiation ・ Sum rule in integration ・ Sum rule in quantum mechanics ・ Sum to Infinity ・ Sum to Infinity (book) ・ Sum Ying Fung ・ Sum-free sequence ・ Sum-free set ・ Sum-frequency generation ・ Sum-of-squares optimization ・ Sum-of-the-parts analysis ・ Sum-product number ・ Sum/One ・ Suma ・ Suma (co-operative) ・ Suma (moth) ・ Suma Bhattacharya Dictionary Lists mini英和辞書 mini和英辞書 Webster 1913 Latin-English FOLDOC Wikipedia English ウィキペディア

翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Sum rule in quantum mechanics ： ウィキペディア英語版 Sum rule in quantum mechanics In quantum mechanics, a sum rule is a formula for transitions between energy levels, in which the sum of the transition strengths is expressed in a simple form. Sum rules are used to describe the properties of many physical systems, including solids, atoms, atomic nuclei, and nuclear constituents such as protons and neutrons. The sum rules are derived from general principles, and are useful in situations where the behavior of individual energy levels is too complex to be described by a precise quantum-mechanical theory. In general, sum rules are derived by using Heisenberg's quantum-mechanical algebra to construct operator equalities, which are then applied to the particles or energy levels of a system. ==Derivation of sum rules〔Sanwu Wang, ''Generalization of the Thomas-Reiche-Kuhn and the Bethe sum rules'', Physical Review A 60, 262 (1999). http://prola.aps.org/abstract/PRA/v60/i1/p262_1〕== Assume that the Hamiltonian $\backslash hat$ has a complete set of eigenfunctions $|n\backslash rangle$ with eigenvalues $E\_n$: :$$ \hat |n\rangle = E_n |n\rangle. For the Hermitian operator $\backslash hat$ we define the repeated commutator $\backslash hat^$ by: :$$ \begin \hat^ & \equiv \hat\\ \hat^ & \equiv (\hat ) = \hat\hat-\hat\hat\\ \hat^ & \equiv (\hat^ ), \ \ \ k=1,2,\ldots \end The operator $\backslash hat^$ is Hermitian since $\backslash hat$ is defined to be Hermitian. The operator $\backslash hat^$ is anti-Hermitian: :$$ \left(\hat^\right)^\dagger = (\hat\hat)^\dagger-(\hat\hat)^\dagger = \hat\hat - \hat\hat = -\hat^. By induction one finds: :$$ \left(\hat^\right)^\dagger = (-1)^k \hat^ and also :$$ \langle m | \hat^ | n \rangle = (E_m-E_n)^k \langle m | \hat | n \rangle. For a Hermitian operator we have :$$ |\langle m | \hat | n \rangle|^2 = \langle m | \hat | n \rangle \langle m | \hat | n \rangle^\ast = \langle m | \hat | n \rangle \langle n | \hat | m \rangle. Using this relation we derive: :$$ \begin \langle m | (\hat^ ) | m \rangle &= \langle m | \hat \hat^ | m \rangle - \langle m | \hat^\hat | m \rangle\\ &= \sum_n \langle m | \hat |n\rangle\langle n| \hat^ | m \rangle - \langle m | \hat^ |n\rangle\langle n| \hat | m \rangle\\ &= \sum_n \langle m | \hat |n\rangle \langle n| \hat| m \rangle (E_n-E_m)^k - (E_m-E_n)^k \langle m | \hat |n\rangle\langle n| \hat | m \rangle \\ &= \sum_n (1-(-1)^k) (E_n-E_m)^k |\langle m | \hat | n \rangle|^2. \end The result can be written as :$$ \langle m | (\hat^ ) | m \rangle = \begin 0, & \mboxk\mbox \\ 2 \sum_n (E_n-E_m)^k |\langle m | \hat | n \rangle|^2, & \mboxk\mbox. \end For $k=1$ this gives: :$$ \langle m | [\hat, [\hat,\hat]] | m \rangle = 2 \sum_n (E_n-E_m) |\langle m | \hat | n \rangle|^2. ==Example== See oscillator strength. 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「Sum rule in quantum mechanics」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.