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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 Hellmann–Feynman theorem ： ウィキペディア英語版 Hellmann–Feynman theorem In quantum mechanics, the Hellmann–Feynman theorem relates the derivative of the total energy with respect to a parameter, to the expectation value of the derivative of the Hamiltonian with respect to that same parameter. According to the theorem, once the spatial distribution of the electrons has been determined by solving the Schrödinger equation, all the forces in the system can be calculated using classical electrostatics. The theorem has been proven independently by many authors, including Paul Güttinger (1932), Wolfgang Pauli (1933), Hans Hellmann (1937) and Richard Feynman (1939). The theorem states =\int}}}\psi_\lambda\ \mathrmV},|}} where *$\backslash hat\_$ is a Hamiltonian operator depending upon a continuous parameter $\backslash lambda\backslash ,$, *$\backslash psi\_\backslash lambda\backslash ,$ is an eigen-wavefunction (eigenfunction) of the Hamiltonian, depending implicitly upon $\backslash lambda\backslash ,$, *$E\backslash ,$ is the energy (eigenvalue) of the wavefunction, *$\backslash mathrmV\backslash ,$ implies an integration over the domain of the wavefunction. ==Proof== This proof of the Hellmann–Feynman theorem requires that the wavefunction be an eigenfunction of the Hamiltonian under consideration; however, one can also prove more generally that the theorem holds for non-eigenfunction wavefunctions which are stationary (partial derivative is zero) for all relevant variables (such as orbital rotations). The Hartree–Fock wavefunction is an important example of an approximate eigenfunction that still satisfies the Hellmann–Feynman theorem. Notable example of where the Hellmann–Feynman is not applicable is for example finite-order Møller–Plesset perturbation theory, which is not variational. The proof also employs an identity of normalized wavefunctions – that derivatives of the overlap of a wavefunction with itself must be zero. Using Dirac's bra–ket notation these two conditions are written as :$\backslash hat\_|\backslash psi\_\backslash lambda\backslash rangle\; =\; E\_|\backslash psi\_\backslash lambda\backslash rangle,$ :$\backslash langle\backslash psi\_\backslash lambda|\backslash psi\_\backslash lambda\backslash rangle\; =\; 1\; \backslash Rightarrow\; \backslash frac\backslash lambda\}\backslash langle\backslash psi\_\backslash lambda|\backslash psi\_\backslash lambda\backslash rangle\; =0.$ The proof then follows through an application of the derivative product rule to the expectation value of the Hamiltonian viewed as a function of λ: :$$ \begin \frac} &= \frac\lambda}\langle\psi_\lambda|\hat_|\psi_\lambda\rangle \\ &=\bigg\langle\frac\bigg|\hat_\bigg|\psi_\lambda\bigg\rangle + \bigg\langle\psi_\lambda\bigg|\hat_\bigg|\frac\bigg\rangle + \bigg\langle\psi_\lambda\bigg|\frac_}\bigg|\psi_\lambda\bigg\rangle \\ &=E_\bigg\langle\frac\bigg|\psi_\lambda\bigg\rangle + E_\bigg\langle\psi_\lambda\bigg|\frac\bigg\rangle + \bigg\langle\psi_\lambda\bigg|\frac_}\bigg|\psi_\lambda\bigg\rangle \\ &=E_\frac\lambda}\langle\psi_\lambda|\psi_\lambda\rangle + \bigg\langle\psi_\lambda\bigg|\frac_}\bigg|\psi_\lambda\bigg\rangle \\ &=\bigg\langle\psi_\lambda\bigg|\frac_}\bigg|\psi_\lambda\bigg\rangle. \end For a deep critical view of the proof see 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「Hellmann–Feynman theorem」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.