Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Homework No. 10 (Spring 2016) PHYS 530A: Quantum Mechanics II Due date: Thursday, 2016 May 5, 4.30pm 1. (110 points.) The Hamiltonian for an hydrogenic atom is H= p2 Ze2 − . 2µ r (1) (a) Using h̄ ∂ h̄ ∇ and H=− (2) i i ∂t in conjunction with the fact that Hamiltonian is a constant of motion, construct the ‘time-independent Schrödinger equation’ for hydrogenic wavefuntions to be h̄2 2 Ze2 − ∇ − ψn (r) = En ψn (r). (3) 2µ r p= (b) We shall confine our discussion to bound states (En < 0). Define µZ 2 e4 1 En = − 2h̄2 n2 and h̄2 a0 = 2 , µe later being the (first) Bohr radius. Thus, derive 2Z Z2 1 2 ∇ + ψn (r) = 2 2 ψn (r). a0 r a0 n (c) The Laplacian in spherical polar coordinates is 1 ∂ 1 ∂2 1 ∂ 1 ∂ 2∂ 2 . + sin θ + ∇ = 2 r r ∂r ∂r r 2 sin θ ∂θ ∂θ sin2 θ ∂φ2 (4) (5) (6) Since angular momentum is a constant of motion we can expand the wavefunctions in the form n−1 X l X ψn (r) = Rnl (r)Ylm (θ, φ), (7) l=0 m=−l where the spherical harmonics satisfy ∂ 1 ∂2 1 ∂ 2 2 sin θ + = h̄2 l(l + 1)Ylm (θ, φ). L Ylm (θ, φ) = −h̄ 2 2 sin θ ∂θ ∂θ sin θ ∂φ (8) Thus, derive the differential equation for the radial part of the wavefunction as 1 d 2d l(l + 1) 2Z Z2 r − + − Rnl (r) = 0. (9) r 2 dr dr r2 a0 r a20 n2 1 (d) In terms of the dimensionless variable x= derive 2Zr a0 (10) d2 2 d l(l + 1) 1 1 + − + − Rnl (x) = 0. dx2 x dx x2 x (2n)2 (e) For x ≫ 1 argue that we have 1 d2 Rnl (x) = 0. − dx2 (2n)2 (11) (12) Solve this equation to learn the asymptotic behaviour of the radial wavefunction to be x Rnl (x) ∼ e− 2n for x ≫ 1. (13) (f) For x ≪ 1 argue that we have l(l + 1) 1 d 2 d Rnl (x) = 0. x − x2 dx dx x2 (14) Solve this equation to learn the limiting behaviour of the radial wavefunction to be Rnl (x) ∼ xl for x ≪ 1. (15) (g) Use the above limiting forms to define x Rnl (x) = xl e− 2n L(x), (16) and derive, in terms of x , n 2 d d y 2 + {(2l + 1) + 1 − y} + (n − l − 1) L(y) = 0. dy dy y= (17) (18) Compare this to the differential equation satisfied by the Laguerre polynomials, (α) Ln (y), the Laguerre equation, 2 d d y 2 + {α + 1 − y} + n L(α) (19) n (y) = 0. dy dy Thus, derive Rnl (r) = N 2Zr na0 where N is a normalization constant. 2 l Zr − na e (2l+1) 0L n−l−1 2Zr na0 , (20) (h) The normalization constant N is, in principle, determined using Z ∞ r 2 dr|Rnl (r)|2 = 1. (21) 0 Verify that, the above statement does not, immediately, lead to the orthogonality relation for Laguerre polynomials, Z ∞ (n + α)! (α) dx xα e−x L(α) . (22) n (x)Ln′ (x) = δnn′ n! 0 (i) Let us derive the Hellmann-Feynman theorem. Consider the energy eigenvalue equation [H(λ) − E(λ)]|E, λi = 0 (23) and its adjoint hE, λ|[H(λ) − E(λ)] = 0. Differentiate with respect to λ: ∂E ∂H |E, λi + (E − H)|E, λi = 0, − ∂λ ∂λ and multiply with hE, λ| to obtain ∂H ∂E ∂H = hE, λ| |E, λi = , ∂λ ∂λ ∂λ (24) (25) (26) which is the statement of the Hellmann-Feynman theorem. (j) Use the Hellmann-Feynman theorem, by regarding λ as Z, in the hydrogenic atom to evaluate Z ∞ 1 Z 1 = . (27) r 2 dr |Rnl (r)|2 = r n r a0 n2 0 (k) Use the orthogonality relation of Laguerre polynomials in Eq. (22) in Eq. (27) to derive the normalization constant N as 3 s 2 Z 2 (n − l − 1)! . (28) N= 2 n a0 (n + l)! Thus, derive the hydrogenic wavefunction l 3 s Zr 2Zr 2 Z 2 (n − l − 1)! 2Zr − na (2l+1) ψnlm (r) = 2 Ylm (θ, φ). (29) e 0 Ln−l−1 n a0 (n + l)! na0 na0 3