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Unformatted text preview: Phys 852, Quantum mechanics II, Spring 2008 TimeIndependent Perturbation Theory 1/14/2008 Prof. Michael G. Moore, Michigan State University 1 The central problem in timeindependent 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 bookkeeping 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 yettobe 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 nondegenerate 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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This note was uploaded on 11/26/2010 for the course PHYSICS PHYS 852 taught by Professor Michaelmoore during the Spring '10 term at Michigan State University.
 Spring '10
 MichaelMoore
 mechanics, The Bible

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