Numerical Methods in Quantum Mechanics - Dipartimento di Fisica

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quadratic) interaction potential. This will allow to study problems that, unlike the harmonic oscillator, do not have a simple analytical solution. 1.1.1 Units The Schrödinger equation for a one-dimensional harmonic oscillator is, in usual notations: d 2 ( ψ dx 2 = −2m ¯h 2 E − 1 ) 2 Kx2 ψ(x) (1.2) where K the force constant (the force on the mass being F = −Kx, proportional to the displacement x and directed towards the origin). Classically such an oscillator has a frequency (angular frequency) √ K ω = m . (1.3) It is convenient to work in adimensional units. These are the units that will be used by the codes presented at the end of this chapter. Let us introduce adimensional variables ξ, defined as ξ = ( mK ¯h 2 ) 1/4 x = (using Eq.(1.3) for ω), and ɛ, defined as ( ) mω 1/2 x (1.4) ¯h ε = Ē hω . (1.5) By inserting these variables into the Schrödinger equation, we find d 2 ( ) ψ dξ 2 = −2 ε − ξ2 ψ(ξ) (1.6) 2 which is written in adimensional units. 1.1.2 Exact solution One can easily verify that for large ξ (such that ε can be neglected) the solutions of Eq.(1.6) must have an asymptotic behavior like ψ(ξ) ∼ ξ n e ±ξ2 /2 (1.7) where n is any finite value. The + sign in the exponent must however be discarded: it would give raise to diverging, non-physical solutions (in which the particle would tend to leave the ξ = 0 point, instead of being attracted towards it by the elastic force). It is thus convenient to extract the asymptotic behavior and assume ψ(ξ) = H(ξ)e −ξ2 /2 (1.8) where H(ξ) is a well-behaved function for large ξ (i.e. the asymptotic behavior is determined by the second factor e −ξ2 /2 ). In particular, H(ξ) must not grow like e ξ2 , or else we fall back into a undesirable non-physical solution. 6
Under the assumption of Eq.(1.8), Eq.(1.6) becomes an equation for H(ξ): H ′′ (ξ) − 2ξH ′ (ξ) + (2ε − 1)H(ξ) = 0. (1.9) It is immediate to notice that ε 0 = 1/2, H 0 (ξ) = 1 is the simplest solution. This is the ground state, i.e. the lowest-energy solution, as will soon be clear. In order to find all solutions, we expand H(ξ) into a series (in principle an infinite one): ∞∑ H(ξ) = A n ξ n , (1.10) n=0 we derive the series to find H ′ and H ′′ , plug the results into Eq.(1.9) and regroup terms with the same power of ξ. We find an equation ∞∑ [(n + 2)(n + 1)A n+2 + (2ε − 2n − 1)A n ] ξ n = 0 (1.11) n=0 that can be satisfied for any value of ξ only if the coefficients of all the orders are zero: (n + 2)(n + 1)A n+2 + (2ε − 2n − 1)A n = 0. (1.12) Thus, once A 0 and A 1 are given, Eq.(1.12) allows to determine by recursion the solution under the form of a power series. Let us assume that the series contain an infinite number of terms. For large n, the coefficient of the series behave like A n+2 A n → 2 n , that is: A n+2 ∼ 1 (n/2)! . (1.13) Remembering that exp(ξ 2 ) = ∑ n ξ2n /n!, whose coefficient also behave as in Eq.(1.13), we see that recursion relation Eq.(1.12) between coefficients produces a function H(ξ) that grows like exp(ξ 2 ), that is, produces unphysical diverging solutions. The only way to prevent this from happening is to have in Eq.(1.12) all coefficients beyond a given n vanish, so that the infinite series reduces to a finite-degree polynomial. This happens if and only if ε = n + 1 2 (1.14) where n is a non-negative integer. Allowed energies for the harmonic oscillator are thus quantized: ( E n = n + 1 ) ¯hω n = 0, 1, 2, . . . (1.15) 2 The corresponding polynomials H n (ξ) are known as Hermite polynomials. H n (ξ) is of degree n in ξ, has n nodes, is even [H n (−ξ) = H n (ξ)] for even n, odd [H n (−ξ) = −H n (ξ)] for odd n. Since e −ξ2 /2 is node-less and even, the complete wave function corresponding to the energy E n : ψ n (ξ) = H n (ξ)e −ξ2 /2 (1.16) 7
Page 1 and 2: Lecture notes Numerical Methods in
Page 3 and 4: 3.3 Code: crossection . . . . . . .
Page 5 and 6: Introduction The aim of these lectu
Page 7 and 8: is an important and well-known libr
Page 9: Chapter 1 One-dimensional Schrödin
Page 13 and 14: point its value is still ∼ 60% of
Page 15 and 16: an equation involving finite differ
Page 17 and 18: eigenvalue) to force the code to pe
Page 19 and 20: 1.3.3 Laboratory Here are a few hin
Page 21 and 22: The Hamiltonian can thus be written
Page 23 and 24: In the following we will use q 2 e
Page 25 and 26: The dominating term close to the nu
Page 27 and 28: whose goal is to give an estimate,
Page 29 and 30: Chapter 3 Scattering from a potenti
Page 31 and 32: 3.2 Scattering of H atoms from rare
Page 33 and 34: 1. Solution of the radial Schrödin
Page 35 and 36: • In the limit of a weak potentia
Page 37 and 38: By modifying ψ as ψ + δψ, the e
Page 39 and 40: This demonstrates Eq.(4.19), since
Page 41 and 42: that the same equations, for a fini
Page 43 and 44: eported here are immediately genera
Page 45 and 46: BLAS 5 (collected here: dgemm.f 6 )
Page 47 and 48: known as generalized eigenvalue pro
Page 49 and 50: The Hamiltonian operator for this p
Page 51 and 52: Chapter 6 Self-consistent Field A w
Page 53 and 54: (the variations of all other normal
Page 55 and 56: nucleus. Even in open-shell atoms,
Page 57 and 58: Chapter 7 The Hartree-Fock approxim
Page 59 and 60: Now that we have the expectation va
Page 61 and 62:
number of possible Slater determina
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Let us now introduce a further vari
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The index R is a reminder that both
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Molecular orbitals theory can expla
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Verify how sensitive the result is
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potential, we can hope to obtain a
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with wave vector k will be purely k
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Fast Fourier Transform In the simpl
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truncate it, it is convenient to co
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A reciprocal lattice of vectors G m
Page 81 and 82:
The origin of the coordinate system
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Chapter 11 Exact diagonalization of
Page 85 and 86:
The only nonzero matrix elements fo
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The code requires the number N of s
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Appendix A From two-body to one-bod
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Appendix B Accidental degeneracy an
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and [J 2 1 , J z] = [J 2 2 , J z] =
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part and symmetric coordinate part)
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and the corresponding energy is E =
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D.3.1 Laboratory • observe the ef
Page 101:
• Relative error: |a − b| < ɛa
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