TIPT - Phys 852 Quantum mechanics II Spring 2008...

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Unformatted text preview: Phys 852, Quantum mechanics II, Spring 2008 Time-Independent Perturbation Theory 1/14/2008 Prof. Michael G. Moore, Michigan State University 1 The central problem in time-independent perturbation theory: Let H be the unperturbed (a.k.a. ‘background’, ‘bare’) Hamiltonian whose eigenvalues and eigenvectors are known. Let E (0) n be the n th unperturbed energy eigenvalue, and | n (0) ) be the n th unperturbed energy eigenstate. They satisfy H | n (0) ) = E (0) n | n (0) ) (1) and ( n (0) | n (0) ) = 1 . (2) Let V be a Hermitian operator which ‘perturbs’ the system, such that the full Hamiltonian is H = H + V. (3) The standard approach is to instead solve H = H + λV , (4) and use λ as for book-keeping during the calculation, but set λ = 1 at the end of the calculation. However, if there is a readily identifiable small parameter in the definition of the perturbation operator V , then use it in the place of λ , and do not set it to 1. In these notes we will treat λ as a small parameter, and will not set it equal to unity. The goal is to find the eigenvalues and eigenvectors of the full Hamiltonian (4). Let E n and | n ) be the n th eigenvalue and its corresponding eigenstate. They satisfy H | n ) = E n | n ) (5) and ( n | n ) = 1 . (6) Because λ is a small parameter, it is assumed that accurate results can be obtained by expanding E n and | n ) in powers of λ , and keeping only the leading term(s). Formally expanding the perturbed quantities gives E n = E (0) n + λE (1) n + λ 2 E (2) n + ..., (7) and | n ) = | n (0) ) + λ | n (1) ) + λ 2 | n (2) ) + ..., (8) where E ( j ) n and | n ( j ) ) are yet-to-be determined expansion coefficients Inserting these expansions into the eigenvalue equation (5) then gives ( H + λV )( | n (0) ) + λ | n (1) ) + λ 2 | n (2) ) + ... . ) = ( E (0) n + λE (1) n + λ 2 E (2) n + ... )( | n (0) ) + λ | n (1) ) + λ 2 | n (2) ) + ... . ) . (9) 1 Because of the linear independence of terms in a power series, this equation can only be satisfied for arbitrary λ if all terms with the same power of λ cancel independently. Equating powers of λ thus gives λ : H | n (0) ) = E (0) n | n (0) ) , (10) λ 1 : ( H − E (0) n ) | n (1) ) = − V | n (0) ) + E (1) n | n (0) ) (11) λ 2 : ( H − E (0) n ) | n (2) ) = − V | n (1) ) + E (2) n | n (0) ) + E (1) n | n (1) ) (12) etc ... (13) This generalizes to λ j : ( H − E (0) n ) | n ( j ) ) = − V | n ( j − 1) ) + j summationdisplay k =1 E ( k ) n | n ( j − k ) ) . (14) 2 The non-degenerate case We will now describe how to solve these equations in the case where none of the unperturbed energy levels are degenerate. Trick #1 : To obtain the j th correction to the n th energy eigenvalue, simply hit the λ j equation from the left with the bra ( n (0) | . For j = 1 this gives E (1) n = ( n (0) | V | n (0) ) , (15) where we have used ( n (0) | ( H − E (0) n ) = 0 and ( n (0) | n (0) ) = 1....
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TIPT - Phys 852 Quantum mechanics II Spring 2008...

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