Elements of Quantum Mechanics
3.9 The Two-Body Problem
3.9.1 The Hydrogen-Like Atom Problem
The radial part of the wave function satisfies the following equation:
1 r2
d dr
r2dR
dr
+2μ 2
E+ Zq2
4πε0r −l(l+1)2 2μr2
R(r)=0 (3.143) For R(r) to be well behaved at r=0 and also as r → ∞, we would obtain the following discrete energy eigenvalues of the problem see Appendix B:
En= −|E1|
n2 (3.144)
where n=1, 2, 3,. . .
represents the total quantum number and
|E1| = 1
2μZ2α2c2 (3.145)
3.9 The Two-Body Problem 55 represents the magnitude of the ground state energy. Further,
α= q2
4πε0c ≈ 1
137.036 (3.146)
represents the fine structure constant and c(≈2.998×108m/s) represents the speed of light in free space. For the hydrogen atom
mN=mp ≈ 1.6726×10−27kg giving
μH ≈ 9.1045×10−31kg
where we have taken me ≈ 9.1094 × 10−31kg. On the other hand, for the deuterium atom
mN=mD ≈ 3.3436×10−27kg giving
μD ≈ 9.1070×10−31kg
Now, for the n=n1 →n =n2transition, the wavelength of the emitted radiation is given by
λ= hc En1 −En2
(3.147) or
λ= 2 h μZ2α2c2
1 n22 − 1
n21 −1
(3.148)
When n2=1, 2, and 3 we have what is known as Lyman series, the Balmer series, and the Paschen series, respectively. For the n=3→n=2 transition, the wavelength of the emitted radiation comes out to be
6565.2 Å and 6563.4 Å
for hydrogen and deuterium, respectively. The corresponding wavelength for the n=4→n=2 transition is
4863.1 Å and 4861.7 Å
Such a small difference in the wavelength was first observed by Urey in 1932 which led to the discovery of deuterium.
In spectroscopy the energy levels are usually written in wavenumber units which are obtained by dividing by hc:
Tn= En
hc = −Z2
n2R (3.149)
where
R=2π2μ ch3
q2 4πε0
2
=μcα2
2 h (3.150)
is known as the Rydberg constant. Values of the Rydberg constant for different hydrogen like atoms are given below:
R=109677.58 cm−1 (for the hydrogen atom) 109707.56 cm−1 (for the deuterium atom) 109722.40 cm−1 (for the He+-atom) 109728.90 cm−1 (for the Li++ -atom)
The slight difference in the values is because of the difference in the values of the reduced massμ.
The normalized radial part of the wave function is given by see Appendix B:
Rnl(ρ)=N e−ρ/2ρl1F1(−nr, 2 l+2,ρ) (3.151) where nr is known as the radial quantum number and for the1F1function to be a polynomial, nrcan take only the following values:
nr =0, 1, 2, 3,. . ...
The total quantum number is given by
n=l+1+nr
Thus
n=1, 2, 3,. . .with l=0, 1, 2,. . .n−1 The normalization constant is given by
N= γ3/2 (2 l+1)!
&
(n+l)! 2n(n−l−1)!
'1/2
(3.152) In the above equation
ρ=γr; γ = 2Z na0
a0= 2 μ(q2/4πε0)
⎫⎪
⎪⎬
⎪⎪
⎭
(3.153)
3.9 The Two-Body Problem 57 where a0is the Bohr radius. Further
1F1(a, c,ρ)=1+a
cρ+a(a+1) c (c+1)
ρ2
2! + · · · (3.154) represents the confluent hypergeometric function. For given values of n and l, a would be a negative integer or zero and the above function would be a polynomial.
For example, for n=2, l=0, we will have nr=1:
N=γ3/2 2!
4×1 1/2
= 1
√2 Z
a0 3/2
and
1F1(−1, 2,ρ)=1−ρ Thus 2
R2 0(r)= 1
√2 Z
a0 3/2
1−1
2ξ e−ξ/2 (3.155)
where
ξ = r a0
(3.156) Similarly, one can calculate other wave functions. We give below the first few Rnl(r)
R1 0(r)=2 Z
a0 3/2
e−ξ (3.157)
R21(r)= 1 2√
6 Z
a0 3/2
ξe−ξ/2 (3.158)
R30(r)= 2 3√
3 Z
a0 3/2
1−2 3ξ+ 2
27ξ2 e−ξ/3 (3.159a) R31(r)= 8
27√ 6
Z a0
3/2 ξ−1
6ξ2 e−ξ/3 (3.159b) R32(r)= 4
81√ 30
Z a0
3/2
ξ2e−ξ/3 (3.160)
The wave functions are normalized so that ∞
0
|Rnl(r)|2 r2 dr=1 (3.161)
The complete wave function is given by
ψnlm(r,θ,φ)=Rnl(r) Ylm(θ,φ) (3.162) where Ylm(θ,φ) are the spherical harmonics tabulated below.
Looking at Eq. (3.162) we see that the energy depends on the total quantum number n. Since for each value of n we have values of l ranging from 0 to n–1 and for each value of l, the m values range from –l to +l there are
n−1
l=0
(2 l+1)=n2
statesψnlmbelonging to a particular energy. The degeneracy with respect to m is due to spherical symmetry of the potential energy function. But the l-degeneracy is peculiar to the Coulomb field and is, in general, removed for non-Coulomb potentials.
Further,
Y0,0=(4π)−1/2 (3.163)
Y1,1= − 3
8π
1/2
sinθeiφ (3.164)
Y1,0= 3
4π
1/2
cosθ (3.165)
Y1,−1= 3
8π
1/2
sinθe−iφ (3.166)
Y2,2 = 15
32π
1/2
sin2θe2iφ (3.167)
Y2,1 = − 15
8π
1/2
sinθ cosθeiφ (3.168)
Y2,0= 5
16π
1/2
3 cos2θ−1
(3.169) Y2,−1=
15 8π
1/2
sinθcosθe−iφ (3.170) Y2,−2=
15 32π
1/2
sin2θe−2iφ (3.171)
and so on. The ground state eigenfunction is ψ1,0,0(n=1, l=0, m=0), the first excited state(n=2)is fourfold degenerateψ2,0,0,ψ2,1,−1,ψ2,1,0, andψ2,1,1. Similarly n = 3 state is ninefold degenerate. In general, the states character- ized by the quantum number n are n2-fold degenerate. The wave functions are orthonormal, i.e.,
ψnlm∗ ψnlmr2dr sinθdθdφ=δnnδllδmm (3.172) It may be a worthwhile exercise for the reader to see that the above wave functions satisfy Eq. (3.172) with E given by Eq. (3.144).
Problems 59
Problems
Problem 3.1 Consider a potential energy function given by the following equation (see Fig.3.3) V(x)= ∞ x<0
=0 0<x<a
=V0 x>a
(3.173)
Assumeψ and dψdx to be continuous at x=a and that the wave function vanishes at x=0 and as x → +∞.
(a) Using the above boundary conditions, solve the one-dimensional Schrödinger equation to obtain the following transcendental equations which would determine the discrete values of energy:
−ξcotξ=
α2−ξ2 (3.174)
where
ξ=
2μEa2
2 and α=
2μV0a2
2 (3.175)
(b) Assumingα=3π
2μV0a2 2 =9π2 show that there will be three bound states (see Fig.3.3) with
ξ=2.83595, 5.64146, and 8.33877
V (x) h2/ 2μa2
a
V0
∞ Fig. 3.3 The first three
eigenvalues and
eigenfunctions of an isolated well forα=3π;α=
2mV0a2 2
Problem 3.2 Consider a symmetric potential energy function so that V(– x)=V(x). Show that the solu- tions are either symmetric or antisymmetric functions of x, i.e., eitherψ(– x)=+ψ(x) orψ(– x)
=–ψ(x)
[Hint: In Eq. (3.22), make the transformation x →–x and assuming V (– x)= V(x) show that ψ(– x) satisfies the same equation asψ(x); hence, we must haveψ(– x)=λψ(x). Make the transformation x→–x again to proveλ= ±1]
Problem 3.3 Consider a potential energy function given by the following equation
V(x)=
,0; |x|<d2
V0; |x|>d2 (3.176)
Since the potential energy variation is symmetric about x=0, the solutions are either symmetric or antisymmetric functions of x. (see Problem 3.2). For E < V0solve the Schrödinger equation (in the two regions). Assumingψand dψdx continuous at x=±d/2 and that the wave function must vanish as x→ ± ∞obtain the following transcendental equations which would determine the discrete values of energy
ξtanξ=
α2−ξ2 for symmetric states (3.177)
−ξcotξ=
α2−ξ2 for antisymmetric states (3.178) where
ξ=
2μE d2
42 and α=
2μV0d2
42 (3.179)
For a given value ofα, the solutions of Eqs. (3.177) and (3.178) will give the bound states for the potential well problem given by Eq. (3.176). Obviously, forα<π/2 we will have only bound state. For given values of V0,μand d, as→0, the value ofαwill become large and we will have a continuum of states implying that all energy levels are possible. Thus in the limit of→0, we have the results of classical mechanics.
Problem 3.4 Using the results of the previous problem, obtain the energy eigenvalues for a single well corresponding to the following values of various parameters:μ=me, V0=20 eV; d=5 Å
[Ans: E1≈1.088 eV; E2≈4.314 eV, E3≈9.527 eV, E4≈16.253 eV]
Problem 3.5 Consider the three-dimensional harmonic oscillator V=1
2μ
ω12x2+ω22y2+ω23z2
(3.180) Use the method of separation of variables to solve the Schrödinger equation (in Cartesian coordinates) and show that the energy eigenvalues are given by
E= n1+1
2 ω1+ n2+1
2 ω2+ n3+1
2 ω3 (3.181)
n1, n2, n3 = 0, 1, 2,. . .. The corresponding wave functions are products of the Hermite–Gauss functions.
Problem 3.6 Calculate the wavenumbers corresponding to the Hα(n=3→n=2) and the Hβ(n=4
→n=2) lines of the Balmer series for the hydrogen atom. What will be the corresponding wavelengths [Ans:≈6563 Å and 4861 Å]
Problem 3.7 Calculate the wavelengths for the n=4→n=3 transition in the He+atom [Ans:≈4686 Å]
Problems 61 Problem 3.8 Calculate the wavelengths corresponding to the n=2→n=1; n=3→n=1; n=4→ n=1, and n=5→n=1 transitions of the Lyman series of the hydrogen atom
[Ans:≈1216 Å, 1026 Å, 973 Å, 950 Å]
Problem 3.9 Using the expressions for spherical harmonics, write all wave functions corresponding to the n=2 and n=3 states of the hydrogen atom. Show that they are fourfold and ninefold degenerate.
(Actually, if we take into account the spin states, they are 8-fold and 18-fold degenerate)
Problem 3.10 Show that∞ 0
R10(r)R20(r) r2dr=0