Download Homework No. 10 (Spring 2016) PHYS 530A: Quantum Mechanics II

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