Download Time Independent Perturbation Theory - Quantum Mechanics I | PHYS 516 and more Study notes Quantum Mechanics in PDF only on Docsity! Time-Independent Perturbation Theory Robert Gilmore Physics Department, Drexel University, Philadelphia, Pennsylvania 19104, USA (Dated: January 30, 2007, My Physics Class: Quantum Mechanics I) Perturbation theory is introduced by diagonalizing a 3 × 3 matrix. Generalization to a larger basis is immediate. This treatment is simpler than the usual treatment and leads immediately to results one higher order than usual in both the perturbed eigenvalues and eigenfunctions. I. INTRODUCTION Computing eigenvalues/vectors is all about diagonal- izing matrices: finite × finite or ∞ × ∞. We treat the problem H0 + ǫH1 = E1Ia + ∗ ∗ ∗ ∗ E2Ib + ∗ ∗ ∗ ∗ ∗ (1) If some eigenvalues of H0 are degenerate (e.g., the 2p lev- els in hydrogen) then corresponding submatrices should first be diagonalized: for example E1Ia + ǫ(H1)ij , where (H1)ij is the submatrix of ǫH1 in the subspace of states that have degenerate eigenvalue E1 of H0. II. A SIMPLE PERTURBATION PROBLEM The method we propose is valid for H0 with nondegen- erate eigenvalues. All the results we need can be deter- mined by using a simple 3× 3 matrix and then applying the rules of “general covariance” at the end of the calcu- lation. We want to compute the eigenvalues and eigenvectors of the 3× 3 matrix H0 + ǫH1. We begin by constructing the secular equation from E1 + ǫh11 − λ ǫh12 ǫh13 ǫh21 E2 + ǫh22 − λ ǫh23 ǫh31 ǫh32 E3 + ǫh33 − λ → (E1 + ǫh11 − λ)(E2 + ǫh22 − λ)(E3 + ǫh33 − λ) +ǫ3(h23h31h12 + h21h13h32) −ǫ2(E1 + ǫh11 − λ)h23h32 −ǫ2(E2 + ǫh22 − λ)h31h13 −ǫ2(E3 + ǫh33 − λ)h12h21 (2) where (ǫ(H1)ij → ǫhij). To determine the eigenvalues we set the determinant equal to zero. We shall solve for the perturbation of the eigenvalue E2. To do this we set the determinant equal to zero, divide by (E1 + ǫh11 − λ)(E3 + ǫh33 − λ), and rearrange the equation to find (E2 + ǫh22 − λ) = −Aǫ 2 − Bǫ2 − Cǫ3 (3) so that λ = E2 + ǫh22 + ǫ 2(A + B) + ǫ3C (4) The terms A, B, C in this expression are: A = − (E2 + ǫh22 − λ)h31h13 (E1 + ǫh11 − λ)(E3 + ǫh33 − λ) B = − h21h12 (E1 + ǫh11 − λ) − h23h32 (E3 + ǫh33 − λ) C = h23h31h12 + h21h13h32 (E1 + ǫh11 − λ)(E3 + ǫh33 − λ) (5) The coefficients A, B, C are functions of ǫ. Since λ − (E2 + ǫh22) is of order ǫ 2 (c.f, Eq. (3)), the term Aǫ2 is fourth order and can be neglected if we wish to compute corrections to E2 only to third order. The coefficients B and C have Taylor series in ǫ beginning with a constant term. To construct the correction to the energy E2 to third order it is sufficient to replace λ → E2 + ǫh22 in the denominators of B and C. We find E (3) 2 = E2 + ǫh22 + ǫ 2 3 ∑ j 6=2 h2jhj2 (E2 − Ej) + ǫ(h22 − hjj) +ǫ3 3 ∑ j 6=2,k 3 ∑ k 6=2 h2jhjkhk2 (E2 − Ej)(E2 − Ek) (6)
