Download CHAPTER 2: PARTICLE IN A CENTRAL POTENTIAL. THE

CHAPTER 2: PARTICLE IN A CENTRAL POTENTIAL.
THE HYDROGEN ATOM
(From Cohen-Tannoudji, Chapter VII)
A. STATIONARY STATES OF A PARTICLE IN A CENTRAL POTENTIAL
We will be dealing with the following three topics:
A) Stationary states of a particle in a central potential
V(~r) is invariant under any rotation about the origin, that is H, Lk = 0, and thus
the eigenfunctions of L̂2 and L̂z are also eigenfunctions of H .
B) Motion of the center of mass and relative motion for a system of two interacting particles
(i) a two particle system in which interaction energy depends only on the particles’
relative position can be replaced by a simpler problem of one fictitious particle;
(ii) in addition, when the interaction depends only on the distance between particles, then the fictitious particle’s motion is governed by a central potential.
C) Exactly solvable problems
(i) V(~r) is a Coulomb potential: hydrogen, deuterium, tritium, He+,Li+;
(ii) V(~r) is a quadratic potential: isotropic three-dimensional harmonic oscillator.
1. Outline of the problem
a. REVIEW OF SOME CLASSICAL RESULTS
Force on the particle located at the point M
dV ~r
~
F~ = −∇V(r)
=−
dr r
(2.1)
is always directed to the origin O. In this case the angular momentum theorem
~ = ~r × ~p is a constant of motion:
implies that the angular momentum L
~
dL
= ~0
dt
(2.2)
~.
and the particle trajectory is on the plane through the origin and perpendicular to L
Let us consider now the position and velocity of the particle at time t. The velocity
can be decomposed into the radial vr and tangential velocity ~v⊥ defined through the
relations
dr
dt
~r × ~v = r ~v⊥
vr =
(2.3)
(2.4)
so that the modulus of the angular momentum is given as
L
~ =
~r × µ~v = µr ~v⊥
(2.5)
The total energy of the particle
1
1
1 2
µ~v + V(r) = µ~v2r + µ~v2⊥ + V(r)
2
2
2
(2.6)
~2
1 2
L
E = µvr +
+ V(r)
2
2
2µr
(2.7)
2
~ 2
2
2
2
~
L = L = µ r ~v⊥ = r2 p2⊥
(2.8)
E =
can be rewritten as
which follows from
The classical Hamiltonian of the system is then
~2
p2r
L
+
+ V(r)
H =
2µ 2µr2
(2.9)
dr
pr = µ
dt
(2.10)
where the momentum
~ 2 can be expressed using the spherical
is conjugate to the radial coordinate r, and L
coordinates r, θ and φ as
~ 2 = p2 +
L
θ
1
2
p
φ
sin2 θ
(2.11)
~ 2 is a constant of motion as V(r) depends only on r and not on θ and φ that is
L
the kinetic energy associated with tangential motion T ⊥ = 21 µv2⊥ is a constant in the
~ 2 plays the role of a parameter
Hamiltonian. That is L
~ 2 dV
∂H
d pr
d2r
L
= µ 2 =−
= 3−
dt
∂r
dr
dt
µr
(2.12)
and this allows us to regard this problem as equivalent to a one-dimensional problem
(with 0 ≤ r < ∞) with a particle of mass µ in the effective potential
~2
L
Veff (r) = V(r) +
2µr2
(2.13)
b. THE QUANTUM MECHANICAL HAMILTONIAN
The eigenvalue problem in the coordinate representation is given as

 2

 ~
− ∆ + V(r) ϕ ~r = Eϕ ~r
2µ
(2.14)
where the Laplacian ∆ in spherical coordinates is
1  ∂2
1 ∂
1 ∂2 
1 ∂2

r + 2  2 +
+ 2
2
2
r ∂r
tan θ ∂θ sin θ ∂φ
r ∂θ

∆ =

(2.15)
Recall that
 2

2
 ∂
1 ∂
1 ∂ 

+ 2
L̂2 = −~2  2 +
2
tan θ ∂θ sin θ ∂φ
∂θ
(2.16)
and you can write down the quantum mechanical Hamiltonian in the form analogous
to the one we found in the classical case
~ 1 ∂ r + 1 L̂2 + V(r)
Ĥ = − 2µ
r ∂r2
2µr2
2
2
The eigenvalue equation then becomes

 2

 ~ 1 ∂2
1 2
−
r+
L̂ + V(r) ϕ(r, θ, φ) = Eϕ(r, θ, φ)
2
2
2µ r ∂r
2µr
(2.17)
2. Separation of Variables
a. ANGULAR DEPENDENCE OF THE EIGENFUNCTIONS
ˆ
The three components of the angular momentum operator ~
L act only on the angular
variables θ and φ and thus they commute with all operators that depend on r only. In
addition they commute with L̂2, so the three components of the angular momentum
are constants of motion:
h
ˆ
Ĥ, ~L
= 0
i
(2.18)
Obviously also Ĥ, L̂2 = 0, so we can choose the complete set of commuting observables to be represented by the operators
Ĥ , L̂2, and L̂z
We therefore require the eigenfunctions ϕ(r, θ, φ) to be simultaneous eigenfunctions
of this C.S.C.O. Their eigenvalue equations
Ĥϕ
L̂2ϕ
L̂zϕ
~r = Eϕ ~r
~r = l(l + 1)~2ϕ ~r
~r = m~ϕ ~r
(2.19)
(2.20)
(2.21)
now constitute a system of differential equations that must be solved simultaneously.
We know the general form of the eigenfunctions of L̂2and L̂z for a fixed value of the
quantum numbers l and m which is given by the spherical harmonics Ylm (θ, φ). Thus
the general solution will be of the following form
ϕ ~r = R(r)Ylm (θ, φ)
(2.22)
and the remaining problem is to determine the radial function R(r) such that ϕ ~r is
an eigenfunction of Ĥ .
b. THE RADIAL EQUATION
Inserting ϕ ~r = R(r)Ylm (θ, φ) into the eigenvalue equation
 2

2
 ~ 1 ∂

1 2
−
r+
L̂ + V(r) ϕ(r, θ, φ) = Eϕ(r, θ, φ)
2
2
2µ r ∂r
2µr
(2.23)
we can formulate the eigenvalue equation for the function of the radial coordinate
R(r) only
 2

2
 ~ 1 d2

l(l + 1)~
−
r+
+ V(r) R(r) = ER(r)
2
2
2µ r dr
2µr
(2.24)
We note at this point that the Laplacian is not necessarily defined at r = 0. In order
for ϕ ~r = R(r)Ylm (θ, φ) to be the solution of the eigenvalue problem, we need to
make sure that the behavior of the solutions R(r) at r = 0 is sufficiently regular.
Now we face the eigenvalue problem, i.e. represented by a differential equation,
which depends only on r and l as parameters, i.e. we are looking for eigenvalues
and eigenvectors of Ĥl which is different for different values of l.
In the state space E~r , we consider subspaces E(l, m) for fixed values of l and m, and
study the eigenvalue equation of Ĥ in each of these subspaces:
- the equation is the same in the (2l + 1) subspaces E(l, m) associated with l;
- Ek,l are eigenvalues of Ĥl in a given E(l, m), and they can be discrete or continuous
depending on k which represents various eigenvalues with the same value of l;
- Rk,l(r) are eigenfunctions of Ĥl in E(l, m).
The eigenvalue equation (i.e. the differential equation)
 2

 ~ 1 d2

l(l + 1)~2
−
r+
+ V(r) Rk,l(r) = Ek,lRk,l(r)
2
2
2µ r dr
2µr
(2.25)
can be satisfied by taking
Rk,l(r) =
1
uk,l(r)
r
and by multiplying both sides of the equation by r
2
d
− 2µ 2
dr
~2
l(l+1)~2
+
+ V(r) uk,l(r) = Ek,luk,l(r)
2µr2
(2.26)
Note that the last two terms in the square bracket constitute the effective potential
l(l + 1)~2
Veff (r) = V(r) +
2µr2
(2.27)
As r ≥ 0 then l(l + 1)~2/2µr2 ≥ 0 and thus the term l(l + 1)~2/2µr2 always repels the
particle from the origin. We call this term centrifugal potential.
Examples of Veff for l = 0, 1, 2:
c. BEHAVIOR OF THE SOLUTIONS OF THE RADIAL EQUATION AT THE ORIGIN
We have to examine the behavior of R(r) at r = 0 to know whether they really are
solutions of the eigenvalue problem. We assume
• V(r) is for r → 0 either finite or approaches infinity less rapidly than 1r ;
• for r → 0 the radial function Rk,l(r) ∼ Cr s
Substituting Rk,l(r) into the eigenvalue equation
 2

2
 ~ 1 d2

l(l + 1)~
−
r+
+ V(r) Rk,l(r) = Ek,lRk,l(r)
2
2µ r dr2
2µr
(2.28)
we get
~2 1 d2
l(l + 1)~2
s
s ) + V(r)(Cr s ) = E (Cr s )
−
r(Cr
)
+
(Cr
k,l
2µ r dr2
2µr2
~2
l(l + 1)~2 s−2
s−2
+
r
+ V(r)r s = Ek,lr s
− (s + 1) s r
2µ
2µ
(2.29)
(2.30)
and setting the coefficients of the dominant term to zero we get
−s(s + 1) + l(l + 1) = 0
(2.31)
The equation −s(s + 1) + l(l + 1) = 0 has two solutions
(
either s = l
or
s = −(l + 1)
(2.32)
that is, for a given Ek,l, there are therefore two linearly independent solutions of the
eigenvalue equation above behaving at r → 0 like
1. rl, or
2.
r
1 .
l+1
1 as this diverges for r = 0 and is thus
However we have to reject the solution l+1
r
physically unacceptable.
The acceptable solution of the eigenvalue problem converges to zero at r = 0 for all
values of l (recall that Rk,l(r) = 1r uk,l(r)):
uk,l(r)
∼
r→0
Crl+1
Consequently, to the eigenvalue equation we have to add a condition
uk,l(0) = 0.
(2.33)
3. Stationary states of a particle in a central potential
a. QUANTUM NUMBERS
Summary:
• The eigenfunctions of Ĥ are required to be the simultaneous eigenfunctions of
L̂2 and L̂z which determines their angular dependence
ϕk,l,m ~r = Rk,l(r)Ylm (θ, φ) = 1r uk,l(r)Ylm (θ, φ).
• The eigenvalue equation of Ĥ is replaced by a differential equation involving only
the radial coordinate r and depending on a parameter l.
In addition:
In principle the functions ϕ (r, θ, φ) must be square integrable
Z 2 2
ϕ
k,l,m (r, θ, φ) r dr dΩ = 1
(2.34)
and we can separately integrate their radial and angular component
Z Z ∞
ϕ
2 r2dr dΩ =
(r,
θ,
φ)
r2dr
k,l,m
0
m
2
2 Z
Rk,l(r)
dΩ Yl (θ, φ)
(2.35)
The spherical harmonics Ylm (θ, φ) are already normalized, so the normalization condition reduces to
Z ∞
0
Z ∞
2
r2dr Rk,l(r) =
dr
0
2
uk,l(r) = 1
(2.36)
We note that if the spectrum has a continuous part, i.e. k is a continuous index, then
Z ∞
0
r2dr R∗k0,l(r)Rk,l(r) =
Z ∞
0
dr u∗k0,l(r)uk,l(r) = δ k0 − k
(2.37)
where δ k0 − k is δ-function. Since this is not a square integrable function, the normalization integral diverges for k = k0.
An eigenfunction
ϕk,l,m ~r = Rk,l(r)Ylm (θ, φ)
depends on three parameters as it is a simultaneous eigenfunction of Ĥ , L̂2 and L̂z
with eigenvalues Ek , l(l + 1)~2, and m~.
We call these parameters - k, l and m - quantum numbers:
k - a radial quantum number,
l - an azimuthal quantum number,
m - a magnetic quantum number.
1
The radial part Rk,l(r) = r uk,l(r) of ϕk,l,m ~r are independent of m and are given by
the eigenvalue equation
2
d
− 2µ 2
dr
~2
l(l+1)~2
+
+ V(r) uk,l(r) = Ek,luk,l(r)
2µr2
The angular part depends only on l and m, and it does NOT depend on the form of
the potential V(r).
b. DEGENERACY OF THE ENERGY LEVELS
(2l + 1) orthogonal functions ϕk,l,m ~r with k and l fixed and m varying from −l to +l
are eigenfunctions of Ĥ with the same eigenvalue Ek,l.
That is Ek,l is at least (2l + 1)-fold degenerate. This is called essential degeneracy
as it exists for all V(r) and is due to the fact that Ĥ depends on L̂2 but not on L̂z.
It is also possible that Ek,l = Ek0,l0 for l , l0. This occurs for certain potentials and it
is called accidental degeneracy.
Remarks:
For a fixed l, the radial equation has at most one solution which is physically acceptable. This is ensured by the condition uk,l(0) = 0.
The behavior of the solutions as r → ∞ follows from the behavior of the potential
V(r → ∞) → 0. The negative values of Ek,l form a discrete set.
B. MOTION OF THE CENTER OF MASS AND RELATIVE MOTION FOR A SYSTEM OF TWO INTERACTING PARTICLES
1. Motion of the center of mass and relative motion in classical mechanics
Let us consider two spinless particles with masses m1 and m2 and positions ~r1 and
~r2 respectively, and assume that the force between the particles is derived from the
potential V(r) = V(~r1 − ~r2).
in classical mechanics the system is described by the Lagrangian
1
2 1
2
˙
˙
£ ~r1, ~r1;~r2, ~r2 = T − V = m1~r˙ 1 + m2~r˙ 2 − V ~r1 − ~r2 .
2
2
(2.38)
The momental of the particles are
~p1 = m1~r˙ 1
~p2 = m2~r˙ 2
(2.39)
(2.40)
The study of the motion of the two particles is simplified by introducing
(i) center of mass coordinates
m1~r1 + m2~r2
~rG =
m1 + m2
(2.41)
~r = ~r1 − ~r2
(2.42)
(ii) relative coordinates
where
m2
~r
m1 + m2
m1
~r2 = ~rG −
~r
m1 + m2
~r1 = ~rG +
(2.43)
(2.44)
The Lagrangian can now be rewritten as follows
"
#2
"
#2
1
m2 ˙
1
m1 ˙
~r + m2 ~r˙G −
~r − V ~r
£ ~rG , ~r˙G ;~r, ~r˙ = m1 ~r˙G +
2
m1 + m2
2
m1 + m2
1 ˙2 1 ˙2
= M~rG + µ~r − V ~r
(2.45)
2
2
where M is the total mass of the system and µ is the reduced mass (i.e. the geometric
mean of m1 and m2):
M = m1 + m2
m1m2
µ =
m1 + m2
1
1
1
=
+
µ
m1 m2
(2.46)
(2.47)
(2.48)
Conjugate momenta are the total momentum and the relative momentum defined as
~pG = M~r˙G = m1~r˙ 1 + m2~r˙ 2 = ~p1 + ~p2
~p = µ~r˙ =
m2 ~p1 − m1 ~p2
m1 + m2
~p
~p1 ~p2
=
−
µ
m1 m2
(2.49)
(2.50)
(2.51)
The classical Hamiltonian and the equations of motion are
2
~pG
~p2
+
+ V ~r
2M 2µ
~p˙ G = ~0
~ ~r
~p˙ = −∇V
H ~rG , ~pG ;~r, ~p =
(2.52)
(2.53)
(2.54)
2
~pG
where 2M is the kinetic energy of a fictitious particle with mass M and position in the
center of mass which is, as indicated by ~p˙ G = ~0, in uniform rectilinear motion.
In the center of mass frame (which is in uniform rectilinear motion ~p˙ G = ~0) the total
momentum ~pG = 0 so the Hamiltonian reduces to
~p2
Hr =
+ V ~r
2µ
(2.55)
which characterizes the energy associated with the relative motion of the two particles.
Here a new fictitious particle, relative particle, is introduced with the mass µ (the
reduced mass of the original particles) and the relative coordinate ~r and momentum
~p. This relative particle behaves as being a subject of a potential V(r) equal to the
potential energy of interaction between two real particles.
The study of the relative motion of two interacting particles reduces to study
of a motion of a single fictitious particle characterized by µ, ~r and ~p.
2. Separation of variables in quantum mechanics
a. OBSERVABLES ASSOCIATED WITH THE CENTER OF MASS AND THE RELATIVE PARTICLE
ˆ
ˆ
ˆ
ˆ
~ 1, R
~ 2, P
~ 1, and P
~ 2 satisfy the canonical commutation relations, e.g.
The observables R
the x components (similarly relations hold for the other components)
h
h
X̂1, P̂1x
i
= i~
(2.56)
X̂2, P̂2x
i
= i~
(2.57)
We can define new coordinates and momenta
~ˆ 1 + m2R
~ˆ 2
m1R
ˆ~
RG =
m1 + m2
~ˆ = R
~ˆ 1 − R
~ˆ 2
R
~ˆ G = P
~ˆ 1 + P
~ˆ 2
P
ˆ
ˆ
~
~
m
P
−
m
P
2 1
1 2
~ˆ =
P
m1 + m2
(2.58)
(2.59)
(2.60)
(2.61)
which satisfy the following commutation relations, e.g. the x components (and similarly for the other components)
h
i
X̂G , P̂Gx = i~
h
i
X̂, P̂ x = i~
(2.62)
(2.63)
b. EIGENVALUES AND EIGENFUNCTIONS OF THE HAMILTONIAN
The Hamiltonian
ˆ~ 2
ˆ~ 2
P2
P1
ˆ~
ˆ~
+
+ V R1 − R2
Ĥ =
2m1 2m2
(2.64)
can be rewritten in terms of the new coordinates as
2
ˆ2
ˆ
~
~
PG P
~ˆ
Ĥ =
+
+V R
2M 2µ
(2.65)
This can be expressed as the sum
Ĥ = ĤG + Ĥr
(2.66)
where
ˆ2
~
PG
ĤG =
2M
ˆ~ 2
P
~ˆ
+V R
Ĥr =
2µ
(2.67)
(2.68)
ĤG and Ĥr commute
h
ĤG , Ĥr
i
= 0
(2.69)
and thus satisfy the eigenvalue equations with the common eigenvector
ĤG |ϕi = EG |ϕi
Ĥr |ϕi = Er |ϕi
(2.70)
Ĥ|ϕi = E|ϕi
(2.72)
E = EG + E r
(2.73)
(2.71)
which imply that
where
Consider {|~rG ,~ri} representation with the basis vectors common to the observables
~ˆ G and R
~ˆ , φ(~rG ,~r):
R
~ˆ G φ(~rG ,~r) = ~rG φ(~rG ,~r)
R
~ˆ rG ,~r) = ~rφ(~rG ,~r)
Rφ(~
(2.74)
(2.75)
The conjugate momenta are
~G
~ˆ G = ~ ∇
P
i
~~
ˆ
~
P = ∇
i
(2.76)
(2.77)
The state space factorizes into the tensor product E = E~rG ⊗ E~r and the Hamiltonians
ĤG and Ĥr appear as extensions to E acting on E~rG and E~r respectively.
This implies
|ϕi = |χG i ⊗ |ωr i
(2.78)
where
(
ĤG |χG i = EG |χG i
|χG i ∈ E~rG
(2.79)
Ĥr |ωr i = Er |ωr i
|ωr i ∈ E~r
(2.80)
and
(
(i) The eigenvalue equation
~2
−
∆G χG ~rG = EG χG ~rG
2M
(2.81)
describes the motion of a free fictitious particle and has the solution
χG ~rG
i~
1
~
=
e pG ·~rG
(2π~)3/2
(2.82)
whose energy is
EG =
2
~pG
2M
≥0
(2.83)
(ii) The eigenvalue equation
 2

 ~

− ∆ + V ~r  ωr ~r = Er ωr ~r
2µ
(2.84)
describes the behavior of the two interacting particles in the center of mass frame
(which is equivalent to a single fictitious relative particle in central potential).
The interaction depends only on the distance between the particles ~r1 − ~r2 and not
on the direction of ~r1 − ~r2.
The relative particle is subject to a central potential V(~r).
The total angular momentum of the system of two real particles is given as
ˆ
ˆ
ˆ
J~ = ~L1 + ~L2
~Lˆ 1 = R
~ˆ 1 × P
~ˆ 1
~Lˆ 2 = R
~ˆ 2 × P
~ˆ 2
(2.85)
(2.86)
(2.87)
and it can be written as
ˆ
ˆ
ˆ
J~ = ~LG + ~L
(2.88)
where
~Lˆ G = R
~ˆ G × P
~ˆ G
~Lˆ = R
~ˆ × P
~ˆ
(2.89)
(2.90)
ˆ~ ~ˆ
such that LG , L = 0 and their components satisfy the standard angular momentum
commutation relations.
C. THE HYDROGEN ATOM
1. Introduction
The hydrogen atom is the system of two particles: proton, with mass m p and charge
q, and electron, with mass me and charge −q
m p = 1.7 × 10−27 kg
me = 0.9 × 10−30 kg
q = 1.6 × 10−19 Coulomb
(2.91)
(2.92)
(2.93)
interacting via the Coulomb potential
e2
q2 1
V(r) = −
= −
4π0 r
r
which depends on the distance r between proton and electron
(2.94)
The Hydrogen atom in the center of mass coordinate system:
Classically the Hamiltonian for the relative motion of the particles is
~p2 e2
H ~r, ~p =
−
2µ r
(2.95)
with the reduced mass
mem p
me
µ=
' me 1 −
me + m p
mp
!
(2.96)
where me/m p ≈ 1/1800, so the center of mass is in almost in the same place as the
proton and thus the relative particle can be identified with the electron.
2. The Bohr model
Niels Bohr postulated fixed classical orbits for motion of electron around proton
1 2 e2
E = µv −
2
r
(2.97)
The force, given as µ × acceleration of circular motion, equals to the Coulomb force
µv2
e2
= 2
r
r
(2.98)
The quantization condition introduced empirically by Bohr is
µvr = n~; where n is a positive integer
(2.99)
implies (homework: derive these relations)
1
En = − 2 E I
n
(2.100)
(2.101)
rn = n2a0
(2.102)
(2.103)
vn =
1
v0
n
(2.104)
The values of the introduced constants are
µe4
EI =
2~2
~2
a0 =
µe2
e2
v0 =
~
(2.105)
(2.106)
(2.107)
E I ' 13.6 eV
(2.108)
a0 ' 0.52 Å
(2.109)
The Bohr model yielded the correct values for the energy levels of the hydrogen
atom. Moreover, it provides correct value of the ionization energy E I and the Bohr
radium a0 and thus correctly characterizes atomic dimensions.
However, its classical character prevents it to be the ultimate theory of the hydrogen
atom which would be free of internal inconsistencies.
Quantum theory of the hydrogen atom is free of these inconsistencies.
3. Quantum mechanical theory of the hydrogen atom
Relative motion of the proton and electron is studied in the center of mass in which
the eigenvalue equation has the following form

 2
2
 ~
e 
− ∆ −  ϕ ~r = Eϕ ~r
2µ
r
(2.110)
1
ϕk,l,m ~r = uk,l(r)Ylm (θ, φ)
r
(2.111)
The eigenfunctions
where uk,l(r) satisfies
 2 2

2
2
 ~ d
l(l + 1)~
e 
−
+
−  uk,l(r) = Euk,l(r)
2
2
2µ dr
r
2µr
(2.112)
uk,l(0) = 0
(2.113)
and the condition
The spectrum include dicrete and continuous parts, for example:
a. CHANGE OF VARIABLES
It is convenient to work with dimensionless quantities which we define as
ρ = r/a0
q
λk,l =
−Ek,l/E I
(2.114)
(2.115)
The radial eigenvalue equation becomes
 2

 d
l(l + 1) 2
2  u (ρ) = 0

−
+
−
λ
k,l  k,l
ρ
dρ2
ρ2
(2.116)
b. SOLVING THE RADIAL EQUATION
We will seek the solution by expanding uk,l(ρ) in power series but first we will have a
look at its asymptotic behavior.
α. Asymptotic behavior
For ρ → ∞ the terms in the eigenvalue equation above proportional to 1/ρ and 1/ρ2
are much smaller than λ2k,l, so we can neglect them obtaining
 2

 d

2

− λk,l uk,l(ρ) = 0
2
dρ
At this limit the solution therefore is e±ρλk,l .
(2.117)
If we consider the full eigenvalue equation, including the terms proportional to 1/ρ
and 1/ρ2, we expect the solution to be given by e±ρλk,l multiplied by a power of ρ.
We note the solution e+ρλk,l must be rejected as it is not bounded at ρ → ∞, and we
are left with e−ρλk,l as the only physically acceptable asymptotic solution.
The solution we are looking for is thus given as
uk,l(ρ) = e−ρλk,l yk,l(ρ)
(2.118)
where yk,l(ρ) must satisfy the equation
 2
"
#


d
2
d
l(l
+
1)


−
2λ
+
−
yk,l(ρ) = 0


k,l
 dρ2

2
dρ
ρ
ρ
(2.119)
yk,l(0) = 0.
(2.120)
and the condition
β. Solutions in the form of power series
We expect the solution in the following form
yk,l(ρ) = ρ s
∞
X
c q ρq
(2.121)
q=0
where by definition
c0 , 0
(2.122)
The condition yk,l(0) = 0 implies that s > 0. We first calculate the derivatives
∞
X
d
yk,l(ρ) =
(q + s)cq ρq+s−1
dρ
q=0
∞
X
d2
q+s−2
y
(ρ)
=
(q
+
s)(q
+
s
−
1)c
ρ
q
k,l
dρ2
q=0
(2.123)
(2.124)
and we substitute these into the eigenvalue equation
 2
"
#

2 l(l + 1) 
d
d

+
−
−
2λ
yk,l(ρ) = 0


k,l
 dρ2

2
dρ
ρ
ρ
(2.125)
to get
∞
X
(q + s)(q + s − 1)cq ρq+s−2 − 2λk,l
q=0
"
+
2 l(l + 1)
−
ρ
ρ2
∞
X
(q + s)cq ρq+s−1
q=0
#X
∞
cq ρq+s = 0
(2.126)
q=0
In order for this equation to be zero, all its coefficients (at each power of ρ) must be
zero.
The lowest order term is ρ s−2. By taking its coefficient as zero we get (homework)
[−l(l + 1) + s(s − 1)] c0 = 0
(2.127)
which is satisfied, since c0 , 0, by the following relations
(
s=l+1
s = −l
and we choose as physically acceptable the condition
s=l+1
(the other solution would be diverging for ρ → 0).
(2.128)
By setting the coefficients of the general term in ρq+s−2 we obtain the recurrence
(homework)
h
i
q(q + 2l + 1)cq = 2 (q + l)λk,l − 1 cq−1
(2.129)
If we fix c0, we can calculate using this recursion all the other coefficients of the
series c1, c2, ... cq.
Since cq/cq−1 → 0 as q → ∞, the series is convergent for all q.
Thus we have determined, for any value of λk,l, the solution of the eigenvalue
equation for yk,l(ρ) with the condition yk,l(0) = 0.
c. ENERGY QUANTIZATION. RADIAL FUNCTIONS
We require the preceding solution to have correct, i.e. physically acceptable, asymptotic behavior for q → ∞. In this limit, the equation
h
i
q(q + 2l + 1)cq = 2 (q + l)λk,l − 1 cq−1
(2.130)
q2cq ∼ 2qλk,lcq−1
(2.131)
reduces to
which implies that
cq
cq−1
∼
q→∞
2λk,l
q
(2.132)
Now consider
e2ρλk,l =
∞
X
dqρq
(2.133)
q=0
where
dq =
q
2λk,l
q!
(2.134)
This implies that
2λk,l
dq
=
dq−1
q
(2.135)
That is the series for yk,l(ρ) being considered behaves like e2ρλk,l and thus
uk,l(ρ) = e−ρλk,l yk,l(ρ) = e+ρλk,l
(2.136)
which is not physically acceptable.
Consequently, we must reject all cases in which the expansion
yk,l(ρ) = ρ s
X
c q ρq
(2.137)
q=0
is an infinite series.
That is the only possible values of λk,l are those for which the expansion above
reduces to a polynomial.
uk,l is than physically acceptable since its asymptotic behavior is dominated by e−ρλk,l .
Therefore we need an integer k such that r.h.s. of the equation
h
i
q(q + 2l + 1)cq = 2 (q + l)λk,l − 1 cq−1
(2.138)
goes to zero as q = k. Then cq = 0 for q ≥ k.
For fixed l, we label the corresponding values of λk,l by this integer (k ≥ 1 as c0 never
goes to zero).
By the equation above we have
λk,l =
1
k+l
(2.139)
For a given l, the only negative energies possible are
Ek,l =
−E I
;
2
(k + l)
k = 1, 2, 3, . . .
(2.140)
yk,l(ρ) is therefore a polynomial whose term of the lowest order is ρl+1 and whose
term of the highest order is ρk+l.
The various coefficients can be calculated in terms of c0 by the recursion
2(k − q)
cq = −
c
q(q + 2l + 1)(k + l) q−1
(2.141)
or alternatively by the formula
!q
2
(k − 1)!
(2l + 1)!
cq = (−1)q
c0
k + l (k − q − 1)! q!(q + 2l + 1)!
(2.142)
uk,l(ρ) is then given as e−ρλk,l yk,l(ρ) and c0 is determined by the normalization condition
Z ∞
0
2
dr uk,l(r) = 1
Finally we obtain the true functions Rk,l(r) by
(2.143)
uk,l (r)
r .
Examples:
−3/2 −r/a0
Rk=1,l=0(r) = 2 a0
e
!
−3/2
r
e−r/2a0
Rk=2,l=0(r) = 2 2a0
1−
2a0
1 r −r/2a0
Rk=1,l=1(r) = 2a0 −3/2 √
e
3 a0
(2.144)
(2.145)
(2.146)
4. Discussion of the results
a. ORDER OF MAGNITUDE OF ATOMIC PARAMETERS
The ionization energy E I and the Bohr radius a0 play important roles in giving an
order of magnitude of the energies and spatial extensions of the wavefunctions
where
1 2 2
E I = α µc
2
1
a0 =
o
α C
e2
q2
1
α =
=
'
~c 2π0~c 137
~
~
'
' 3.8 × 10−3 Å
oC =
µc mec
(2.147)
(2.148)
(2.149)
(2.150)
The Bohr radius a0 is about 100× the Compton wavelength oC of electron.
The ionization potential E I ≈ 10−5 − 10−4 µc2 where the rest-mass energy
µc2 ≈ mec2 ' 0.51 × 106 eV
(2.151)
E I mec2
(2.152)
which implies
This condition justifies our choice of non-relativistic Schrödinger equation to describe
the hydrogen atom.
b. ENERGY LEVELS
α. Possible values of the quantum numbers; degeneracies
For a fixed l, there exist an inifinite number of possible energy values with k = 1, 2, 3...
and each of these is at least (2l+1)-fold degenerate. This is the essential degeneracy.
recall that this follows from the fact that the radial wavefunction depends on l but not
on m.
There exist accidental degeneracies when Ek,l = Ek0,l0 (l , l0) if k + l = k0 + l0.
Ek,l depends on (k + l) rather than k and l separately (only for hydrogen). We set
n = k+l
(2.153)
which we call principal quantum number.
The energy levels are then
1
En = − 2 E I
n
(2.154)
Since k ≥ 1, there is only a finite number of the values of l associated with the same
n, that is, if n is fixed l = 0, 1, ..., n − 1.
The shell n contains n subshells, each with (2l + 1) orthogonal states.
The total degeneracy of the level n is given as
gn =
n−1
X
(2l + 1) = 2
l=0
(n − 1)n
+ n = n2
2
(2.155)
β. Spectroscopic notation
The energy shells are labelled by n = 1, 2..or by the letters K, L, .... The subshells are
labelled by the corresponding number n followed by l in the spectroscopic notation:
l=0
l=1
l=2
l=3
l=4
...
↔
↔
↔
↔
↔
s
p
d
f
g
...
alphabetical order
(2.156)
(2.157)
Example:
The ground state is K shell with 1s subshell, the first excited state is L shell with the
subshells 2s and 2p, etc.
The hydrogen atom energy levels
c. WAVE FUNCTIONS
Wavefunctions are labeled by the quantum numbers n, l, and m which uniquely characterize the eigenfunctions φn,l,m(~r) of the C.S.C.O. Ĥ , L̂2, and L̂z.
α. Angular dependence
the angular dependence is given by the spherical harmonics Ylm (θ, φ)
β. Radial dependence
Since n = k + l, we can label the radial wavefunctions by n and l (rather than k and l)
Rk=1,l=0 ≡ Rn=1,l=0;
Rk=2,l=0 ≡ Rn=2,l=0;
(2.158)
Rk=1,l=1 ≡ Rn=2,l=1
(2.160)
(2.159)
the behavior of Rn,l(r) around r = 0 is rl, so only states belonging to s subshell (l = 0)
give a nonzero probability of presence at the origin.
The greater l is, the larger is the region around proton with negligible probability of
finding an electron. This has interesting physical consequences (hyperfine structure,...).
γ. Formula for successive Bohr radii rn = n2a0
Let us consider various states with l = n − 1, and calculate variation of probability
density with r for each of the preceding levels in an infinitesimal solid angle dΩ about
a fixed direction of polar angles θ and φ
2
d3P(r, θ, φ) = ϕn,l,m(r, θ, φ) r2drdΩ
2 2
m
2
= Rn,l(r) r dr × Yl (θ, φ) dΩ
2
where the term Rn,l(r) r2dr is the probability density between r and r + dr.
(2.161)
(2.162)
The condition l = n − 1 implies k = n − l = 1, so Rn,l(r) contains only one term in
(r/a0)n−1 and the probability density
fn(r) =
=
 !n−1
2
2


r  r
−r/na
0

e


a20 a0
!2n
r
e−2r/na0
a0
(2.163)
(2.164)
this function has a maximum for
r = rn = n2a0
which is the Bohr radii of the Bohr orbit corresponding to the energy En.
(2.165)
The wavefunctions of the lowest energy levels:
1s level
2s level
2p level
ϕn=1,l=0,m=0 = q1 e−r/a0
πa30
ϕn=2,l=0,m=0 = q 1
1 − 2ar e−r/2a0
0
8πa30
ϕn=2,l=1,m=1 = − q1 ar e−r/2a0 sin θeiφ
8 πa30 0
ϕn=2,l=1,m=0 = q1 ar e−r/2a0 cos θ
4 2πa30 0
ϕn=2,l=1,m=−1 = q1 ar e−r/2a0 sin θe−iφ
8 πa30 0
THE ISOTROPIC THREE-DIMENSIONAL HARMONIC OSCILLATOR
(Complement BVII)
Our objectives are:
1. Solving the radial equation
2. Finding energy levels and stationary wave functions
Consider a spinless particle of mass µ subjected to the potential
i
1 h 2 2
2
2
2
2
V(x, y, z) = µ ω x x + ωy y + ωz z
2
(2.166)
ω x = ωy = ωz = ω
(2.167)
ˆ2
~
P
~ˆ
Ĥ =
+V R
2µ
(2.168)
where
The Hamiltonian is given as
The standard solution relies on separating x, y and z coordinates in the coordinate
representation and solving the system in a way analogous to solving the harmonic
oscillator in one dimension. The energy eigenvalues found are
!
3
n + ~ω
2
(2.169)
1
(n + 1)(n + 2)
2
(2.170)
En =
their degree of degeneracy is
gn =
and the eigenfunctions associated with the eigenvalue En, n = n x = ny = nz, are
 2 3/4
β2 2 2 2  β 
1
− x +y +z
ϕn x,ny,nz (x, y, z) =  
e 2
q
π
2n x+ny+nz n x!ny!nz!
× Hn x (βx)Hny (βy)Hnz (βz)
(2.171)
(2.172)
where
r
β =
µω
,
~
n = n x + ny + nz
(2.173)
(2.174)
Since the oscillator is isotropic we can use the spherical coordinates; the potential
becomes a central potential
1 2 2
V(~r) → V(r) = µω r
2
C.S.C.O. is given by Ĥ , L̂2 and L̂z.
(2.175)
1. Solving the radial equation
 2

2
2
 ~ 1 d
1 2 2 l(l + 1)~ 
−
 Rk,l(r) = Ek,lRk,l(r)
r + µω r +
2
2
2µ r dr
2
2µr
Rk,l(r) =
εk,l =
1
uk,l(r)
r
2µEk,l
~2
 2

 d
4 r2 − l(l + 1) + ε  u (r) = 0

−
β
k,l  k,l
dr2
r2
(2.176)
(2.177)
(2.178)
(2.179)
We also have to add the condition
uk,l(0) = 0
(2.180)
to get the physically acceptable behavior of the wavefunction at the origin r = 0.
Considering the asymptotic behavior for r → ∞ simplifies the eigenvalue equation:
 2

 d

4
2

− β r  uk,l(r)
2
dr
'
r→∞
0
(2.181)
whose two solutions are eβ r /2 and e−β r /2, however only the latter is physically
acceptable as it does not diverges for large r.
2 2
2 2
We therefore seek the eigenfunctions in the form
2 r2 /2
−β
uk,l(r) = e
yk,l(r)
(2.182)
yk,l(r) must satisfy
d2
dr
2r
y
−
2β
k,l
2
d
l(l + 1)
yk,l(r) = 0
yk,l + εk,l − β2 −
2
dr
r
#
"
yk,l(0) = 0
(2.183)
(2.184)
We expand yk,l(r) into a power series
yk,l(r) = r s
∞
X
aqrq
(2.185)
q=0
where a0 is by definition nonzero
a0 , 0
(2.186)
We substitute yk,l(r) into the eigenvalue equation. The term of the lowest order is
r s−2. Its coefficient is zero if
[s(s − 1) − l(l + 1)] a0 = 0
(2.187)
which implies the physically acceptable (to satisfy yk,l(0) = 0) solution
s = l+1
(2.188)
The next term in the expansion is r s−1 and its coefficient equals to
[s(s + 1) − l(l + 1)] a1
(2.189)
Since s is fixed already, the coefficient can go to zero only if
a1 = 0
(2.190)
The general term in rq+s is zero if
(q + s + 2)(q + s + 1) − l(l + 1) aq+2
h
i
2
2
+ εk,l − β − 2β (q + s) aq = 0
(2.191)
(2.192)
or
h
i
(q + 2) (q + 2l + 3) aq+2 = (2q + 2l + 3) β2 − εk,l aq
This recursion implies that all coefficients aq of rank q = odd are zero.
Coefficients of even rank are proportional to a0.
(2.193)
If no coefficients on r.h.s. go to zero, then yk,l(r) has the form of an implicit power
series for which
aq+2
∼
q→∞
aq
2β2
q
(2.194)
The behavior of the coefficients is the same as for the exponential function
e
β2 r2
=
∞
X
c2pr2p
(2.195)
p=0
with
β2p
c2p =
p!
(2.196)
Then
c2p+2
c2p
∼
p→∞
2β2
p
(2.197)
2 2
and in this case the yk,l(r) would be dominated by eβ r which is not physically acceptable.
Physically acceptable cases are those in which there exists an even integer k, positive or zero, such that
εk,l = (2k + 2l + 3)β2
(2.198)
so the coefficients of the recursion of even rank greater than k are zero. ConseP
q must be finite and thus reduces to a
quently the series for yk,l(r) = r s ∞
a
r
q
q=0
polynomial, and the radial function uk,l(r) decreases exponentially as r goes to infinity.
2. Energy levels and stationary wave functions
The energy eigenvalues are
Ek,l = ~ω k + l +
3
2
!
(2.199)
where k is an even positive integer or zero. Since Ek,l actually depends on
n = k+l
(2.200)
we can rewrite it as
Ek,l = ~ω n +
3
2
!
where k is an even positive integer or zero and l is any non-negative integer.
(2.201)
Thus n can take only integral values, positive or zero:
(k, l) = (0, n), (2, n − 2), . . . , (n − 2, 2), (n, 0)
(k, l) = (0, n), (2, n − 2), . . . , (n − 3, 3), (n − 1, 1)
if n is even
(2.202)
if n is odd
(2.203)
This implies the values of l:
n=0:
n=1:
n=2:
n=3:
n=4:
l=0
l=1
l = 0, 2
l = 1, 3
l = 0, 2, 4
(2.204)
En = (n + 23 )~ω = 23 ~ω, 25 ~ω, 72 ~ω, 92 ~ω, 11
2 ~ω, ...
For each pair (k, l) there exist one and only one radial wavefunction uk,l(r).
There is (2l + 1) eigenfunctions of Ĥ , L̂2 and L̂z:
1
ϕk,l,m ~r = uk,l(r)Ylm (θ, φ)
r
(2.205)
and consequently the degree of degeneracy
gn =
X
(2l + 1)
if n is even
(2.206)
if n is odd
(2.207)
l=0,2,...,n
gn =
X
l=1,3,...,n
(2l + 1)
This specifically yields
for even n : gn =
for odd n : gn =
n/2
X
1
(4p + 1) = (n + 1)(n + 2)
2
p=0
(n−1)/2
X
p=0
1
(4p + 3) = (n + 1)(n + 2)
2
(2.208)
(2.209)
For each of the pair (k, l), we have one and only one uk,l(r) (to within a factor a0), and
thus (2l + 1) common eigenfunctions of Ĥ and L̂2 with eigenvalues En and l(l + 1)~2.
Example: wavefunctions for the three lowest energy levels
(i) E0 = 32 ~ω
k=l = 0
(2.210)
y0,0(r) reduces to a0r where a0 is determined by normalization
 2 3/4
 β 
2 r2 /2
−β


ϕ0,0,0 ~r =   e
π
The ground state is non-degenerate.
(2.211)
(ii) the first excited state E1 = 52 ~ω
This state is three-fold degenerate
(
k=0
l=1
(2.212)
with y0,1(r) = a0r2.
The three eigenfunctions are
r
ϕ0,1,m ~r =
2 r2 /2 m
8 β3/2
−β
βre
Y1 (θ, φ)
3 π1/4
m = 1, 0, −1
(2.213)
The spherical harmonics Ylm (θ, φ) are such that
r
3
z
4π r
i
r h −1
3
1
Y
−
Y
=
x
√
1
1
4π
2
r
i
r h −1
3
1
y
√ Y1 + Y1 = −i
4π
2
rY10(θ, φ) =
(2.214)
(2.215)
(2.216)
and the Hermite polynomial of order 1 is
H1(u) = 2u
(2.217)
so that the functions ϕ0,1,m are related to the functions ϕn x,ny,nz of the basis given by
the Hermite polynomials (Eq. (2.172)) by the equations
ϕn x=0,ny=0,nz=1 = ϕk=0,l=1,m=0
i
1 h
ϕn x=1,ny=0,nz=0 = √ ϕk=0,l=1,m=−1 − ϕk=0,l=1,m=1
2
i
i h
ϕn x=0,ny=1,nz=0 = √ ϕk=0,l=1,m=−1 + ϕk=0,l=1,m=1
2
(2.218)
(2.219)
(2.220)
(iii) the second excited state E2 = 27 ~ω
This state is six-fold degenerate, and k and l can take the values
k = 0,
l=2
(2.221)
k = 2,
l=0
(2.222)
The function y0,2(r) is a0r3, and the function y2,0(r)
#
"
2
y2,0(r) = a0r 1 − β2r2
3
(2.223)
the six basis functions in the eigenspace associated with E2 are of the form
r
16 β3/2 2 2 −β2r2/2 m
β r e
Y2 (θ, φ)
1/4
r 15 π
!
3 β3/2
2 2 2 2 2 −β2r2/2
1− β r β r e
ϕ2,0,0 ~r =
2 π1/4
3
ϕ0,2,m ~r =
(2.224)
(2.225)
Through the explicite expressions for the spherical harmonics and the Hermite polynomials we can prove the relations:
i
1 h
ϕk=2,l=0,m=0 = − √ ϕn x=2,ny=0,nz=0 + ϕn x=0,ny=2,nz=0 + ϕn x=0,ny=0,nz=2
3
i
h
i
1 h
1
√ ϕk=0,l=2,m=2 + ϕk=0,l=2,m=−2 = − √ ϕn x=2,ny=0,nz=0 − ϕn x=0,ny=2,nz=0
2
2
i
1 h
√ ϕk=0,l=2,m=2 − ϕk=0,l=2,m=−2 = iϕn x=1,ny=1,nz=0
2
i
1 h
√ ϕk=0,l=2,m=1 − ϕk=0,l=2,m=−1 = −ϕn x=1,ny=0,nz=1
2
i
1 h
√ ϕk=0,l=2,m=1 + ϕk=0,l=2,m=−1 = −iϕn x=0,ny=1,nz=1
2
"
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2
1
1
ϕk=0,l=2,m=0 = √ ϕn x=0,ny=0,nz=2 − ϕn x=2,ny=0,nz=0 − ϕn x=0,ny=2,nz=0
2
2
3