量子力学
Quantum mechanics
School of Physics and Information Technology
QUANTUM MECHANICS IN
THREE DIMENSIONS
Chapter 4
4.1 Schrödinger Equation in Spherical Coordinates 131
4.2 The Hydrogen atom 145
4.1 Schrödinger Equation in Spherical Cordinations
(1) The generalization of the Schrödinger Equation from one-dimensional to three-one-dimensional is straightward. The SE says
;
Ψ
=
∂
Ψ
∂
H
t
i
the Hamiltonian operator H is obtained from the classical energy
by the standard prescription (applied now to y,z as well as x)
)
,
,
(
)
(
2
1
2
1
2 2 2 2z
y
x
V
p
p
p
m
V
mv
+
=
x+
y+
z+
,
x
i
p
x∂
∂
→
,
y
i
p
y∂
∂
→
.
z
i
p
zor, for short As
,
∂
∂
+
∂
∂
+
∂
∂
=
→
+
+
=
k
z
j
y
i
x
i
p
k
p
j
p
i
p
p
x y z,
∇
=
i
p
∂
∂
+
∂
∂
+
∂
∂
=
∇
k
z
j
y
i
x
where Thus,
2
2
1
2 2Ψ
+
Ψ
∇
−
=
Ψ
+
⋅
=
∂
Ψ
∂
V
m
V
p
p
m
t
i
where 2 2 2 2 2 2 2z
y
x
∂
∂
+
∂
∂
+
∂
∂
=
∇
⋅
∇
=
∇
is the Laplacian, in),
,
,
,
(
)
,
(
)
,
(
x
t
V
r
t
V
x
y
z
t
V
→
=
(2) The probability of finding the particle in the infinitesimal volume d3r, And in 3-dimensional space
as well as
),
,
,
,
(
)
,
(
r
t
=
Ψ
x
y
z
t
Ψ
,
3
dxdydz
r
d
=
isΨ
(
r
,
t
)
2d
3r
.
(3) Therefore the normalization condition reads
.
1
)
,
(
2 3=
Ψ
r
t
d
r
(4) If the potential is time-independent, the time-independent Schrodinger equation reads
,
2
2 2
Ψ
=
Ψ
+
Ψ
∇
−
V
E
m
and there will be a complete set of stationary states
.
)
(
)
,
(
r
t
=
r
e
−iEnt/Ψ
n(
r
,
t
)
=
ψ
n(
r
)
e
n.
Ψ
ψ
The general solution to the (time-dependent) Schrodinger equation is
.
)
(
)
,
(
)
,
(
=
Ψ
=
− /Ψ
n
t iE n
n n
n n
n
e
r
c
t
r
c
t
4.1.1 Separation of Variables
(1) Spherical coordinates
Cartesian coordinates:
(
x
,
y
,
z
)
(
r
,
θ
,
φ
)
Spherical coordinates:
,
sin
2φ
θ
θ
τ
r
drd
d
d
=
P
x
y
z
o
) , , (x y z Mθ
r
• •ϕ
ϕ
ϕ
ϕ
z
y xA
=
=
sin
cos
,
φ
θ
φ
θ
r
x
In spherical coordinates the Laplacian takes the form
∂
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
=
∇
2 2 2 2 2 2 2 2sin
1
sin
sin
1
1
φ
θ
θ
θ
θ
θ
r
r
r
r
r
r
,
sin
2φ
θ
θ
τ
r
drd
d
The time-independent Schrodinger equation in Cartesian coordinates
In spherical coordinates
ψ
ψ
ψ
ψ
ψ
E
V
z
y
x
m
∂
+
=
∂
+
∂
∂
+
∂
∂
−
2 2 2 2 2 2 22
ψ
ψ
φ
ψ
θ
θ
ψ
θ
θ
θ
ψ
E
V
r
r
r
r
r
r
m
∂
+
=
∂
+
∂
∂
∂
∂
+
∂
∂
∂
∂
−
2 2 2 2 2 2 2 2sin
1
sin
sin
1
1
2
We begin by looking for solutions that are separable into products:
)
,
(
)
(
)
,
,
(
θ
φ
θ
φ
ψ
r
=
R
r
Y
Putting this into above equation, we have
ERY
VRY
Y
r
R
Y
r
R
dr
dR
r
dr
d
r
Y
m
∂
+
=
∂
+
∂
∂
∂
∂
+
−
2 2 2 2 2 2 2sin
sin
sin
2 2
/
2
mr
−
Dividing by RY and multiplying by
(
( ))
0 2 sin 1 sin sin 1 1 2 2 2 2 2 2 = − − ∂ ∂ + ∂ ∂ ∂ ∂+ Y mr V r E
Y Y Y dr dR r dr d
R θ θ θ θ θ φ
(
)
∂ ∂ + ∂ ∂ ∂ ∂ − = − − 2 2 2 2 2 2 sin 1 sin sin 1 ) ( 2 1φ
θ
θ
θ
θ
θ
Y Y Y Y E r V mr dr dR r dr d RThe term on the left hand depends only on r, whereas the right depends only on ; accordingly, each must be a constant, which is in the form l(l+1):
(
(
)
)
(
1
);
2
1
2 2 2+
=
−
−
mr
V
r
E
l
l
dr
dR
r
dr
d
R
).
1
(
sin
1
sin
sin
1
2 22
=
−
+
∂
∂
+
∂
∂
∂
∂
l
l
Y
Y
Y
Y
θ
θ
θ
θ
θ
φ
4.1.2 The Angular Equation
Solution of Y : Equation 4.17 determines Y function as
Multiplying above equation by Ysin2 , it becomes:
).
1
(
sin
1
sin
sin
1
2 2
2
=
−
+
∂
∂
+
∂
∂
∂
∂
l
l
Y
Y
Y
Y
θ
θ
θ
θ
θ
φ
.
sin
)
1
(
sin
sin
2 22
Y
l
l
Y
Y
θ
φ
θ
θ
θ
θ
=
−
+
∂
∂
+
∂
∂
∂
∂
As always, we try separation of variables:
)
(
)
(
)
,
(
θ
φ
=
Θ
θ
⋅
Φ
φ
Y
(
1
)
sin
1
0
.
sin
sin
1
2 2 2
=
Φ
Φ
+
+
+
Θ
Θ
θ
θ
θ
θ
θ
d
φ
d
l
l
d
d
d
d
The first term is a function only of , and the second is a function only of , so each must be a constant. This time I will call the separation constant m2:
φ
(
1
)
sin
;
sin
sin
1
2 2m
l
l
d
d
d
d
=
+
+
Θ
Θ
θ
d
θ
θ
d
θ
(
)
θ
Θ
θ
θ
.
1
22 2
m
d
d
−
=
Φ
Φ
φ
(1) The equation is easy to solve:
.
)
(
2 2
2
φ
φ
φ
im
e
m
d
d
=
Φ
Φ
−
=
Φ
Now, when advances by 2 , we return to the same point in space, so it is natural to require that
φ
).
(
)
2
(
π
+
φ
=
φ
In other words,
(
)
1
2 2
=
=
+ im i m
im
e
e
e
φ π φ πFrom this it follows that m must be an integer:
,
3
,
2
,
1
,
0
±
±
±
=
m
(2) The equation is not simple.
θ
(
)
[
1
sin
]
0
.
sin
sin
Θ
+
l
l
+
2−
m
2Θ
=
d
d
d
d
θ
θ
θ
θ
Turn into
θ
x by:x
=
cos
θ
(
)
0
.
1
1
2
)
1
(
22 2
2 2
=
Θ
−
−
+
+
Θ
−
Θ
−
x
m
l
l
d
d
x
d
d
x
θ
θ
Above equation is lth-order associated Legendre equation.
Therefore, the solution of is
),
(cos
)
(
)
(
θ
=
AP
lmx
=
AP
lmθ
Θ
where Pm
l is the associated Legendre function, defined by
),
(
)
1
(
)
(
| | 2
/ | | 2
x
P
dx
d
x
x
P
lm m
m
and Pl(x) is the lth Legendre polynomial, defined by the Rodrigues formula:
,
)
1
(
!
2
1
)
(
2 ll l l
x
dx
d
l
x
P
≡
−
),
1
3
(
1
)
(
,
)
(
,
1
)
(
2 1 0−
=
=
=
x
x
P
x
x
P
x
P
)
3
30
35
(
8
1
)
(
),
3
5
(
2
1
)
(
),
1
3
(
2
)
(
2 4 4 3 3 2 2+
−
=
−
=
−
=
x
x
x
P
x
x
x
P
x
x
P
Pl(x) is a polynomial (of degree l) in x, and is even or odd according to the parity
is not, in general, a polynomial——if m is odd it carries a factor of
),
(
)
1
(
)
(
| | 2 / | | 2x
P
dx
d
x
x
P
l m m ml
≡
−
But for associated Legendre function Pm l:
.
1
−
x
2l
m
≤
,
1
)
(
)
(
0 00
x
=
P
x
=
P
,
)
(
)
(
1 01
x
P
x
x
P
=
=
(
1
)
(
)
1
,
)
(
2 1/2 1 21
1
P
x
x
dx
d
x
x
P
=
−
=
−
),
1
3
(
2
1
)
(
)
(
2 20
2
x
=
P
x
=
x
−
P
,
1
3
)
(
)
1
(
)
(
2 1/2 2 21
2
P
x
x
x
dx
d
x
x
P
=
−
=
−
(
1
)
,
3
)
(
)
1
(
)
(
2 2 22 2
2
2
P
x
x
dx
d
x
x
P
=
−
=
−
As
x
=
cos
θ
,
,
1
=
r
r
=
cos
θ
,
P
x
y
z
o
) , , (x y z M
θ
r
•
•
ϕ
ϕ
ϕ
ϕ
z
y x
A
,
sin
θ
=
r
)
(cos
θ
m l
P
r
=
Plot:
,
1
)
(
0
0
x
=
P
P
10(
x
)
=
cos
θ
,
,
sin
)
(
1
1
x
=
θ
),
1
cos
3
(
2
1
)
(
20
2
x
=
θ
−
P
,
cos
sin
3
)
(
1
2
x
=
θ
θ
P
,
sin
3
)
(
22
2
x
=
θ
P
)
(cos
θ
m l
P
r
=
Plot:
P
x
y
z
o
) , , (x y z M
θ
r
•
•
ϕ
ϕ
ϕ
ϕ
z
y x
A
), 1 cos
3 (
1 2
−
=
θ
r (3cos 1),
r
=
3
sin
θ
cos
θ
,
r
=
3
sin
2θ
,
2 −
=
θ
),
cos
3
cos
5
(
2
1
)
(
30
3
x
=
θ
−
θ
P
(
5
cos
1
)
,
sin
2
3
)
(
21
3
x
=
θ
θ
−
P
,
cos
sin
15
)
(
22
3
x
=
θ
θ
P
(
1
cos
)
,
sin
15
)
(
23
3
x
=
θ
−
θ
P
)
(cos
θ
m l
P
r
=
Plot:
( )
(
5
cos
3
cos
),
2
1
3θ
θ
θ
=
−
(1) Notice that l must be a nonnegative integer, for the Rodrigues formula to make any sense.
Some Notes:
),
(
)
1
(
)
(
| | 2
/ | | 2
x
P
dx
d
x
x
P
lm m
m
l
≡
−
m
≤
l
If |m|>l, then Plm=0. Therefore, for any given l, there are (2l+1) possible
values of m: values of m:
,
2
,
1
,
0
=
l
l
l
l
l
m
=
−
,
−
+
1
,
,
−
1
,
0
,
1
,
−
1
,
( )
)
(
),
(
,
),
(
),
(
x
P
1x
P
1x
P
0x
(2) Now, the volume element in spherical coordinates is
so the normalization condition becomes
1
sin
sin
2 2 22 2
=
⋅
=
θ
θ
φ
φ
θ
θ
ψ
r
drd
d
R
r
dr
Y
d
d
,
sin
2 3
φ
θ
θ
drd
d
r
r
d
=
It is convenient to normalize R and Y separately: It is convenient to normalize R and Y separately:
,
1
0
2 2
=
∞
dr
r
R
2sin
1
,
0 0
2
=
π π
φ
θ
θ
d
d
Y
where R determined by V(r) and Y can be obtained.
(
(
)
)
(
1
);
2
1
2 2 2
+
=
−
−
mr
V
r
E
l
l
dr
dR
r
The normalized angular wave functions are called spherical harmonics:
where
),
(cos
|)!
|
(
|)!
|
(
4
)
1
2
(
)
,
(
θ
π
ε
φ
θ
φ ml im m
l
e
P
m
l
m
l
l
Y
+
−
+
=
( )
≤
≥
−
=
.
0
for
,
1
,
0
for
,
1
m
m
m
ε
Now we list here some few spherical harmonics: See book for more
φ
θ
θ
π
φ
θ
ie
Y
±=
sin
(
5
cos
−
1
)
±64
21
)
,
(
2Actually, the Ys are automatically orthogonal, so
[
(
,
)
]
(
,
)
sin
,
2
0 0
*
m m l l m
l m
l
Y
d
d
Y
θ
φ
′ ′θ
φ
θ
θ
φ
=
δ
′δ
′π π
Notice that
( )
1
(
,
)
)
,
(
θ
φ
m lm*θ
φ
m
l
Y
Y
−=
−
For historical reasons, l is called the azimuthal quantum number, and m
Notice that the angular part of the wave function, Y( , ), is the same for all spherically symmetric potentials; the actual shape of the potential, V(r), affects only the radial part of the wave function, R(r), which is determined by Equation 4.16:
4.1.3 The Radial Equation
(
(
)
)
(
1
)
.
2
2 2 2
R
l
l
R
E
r
V
mr
dr
dR
r
dr
d
+
=
−
−
so that
This equation can be simplified if we change variables as
),
(
)
(
r
rR
r
u
=
,
r
u
R
=
=
1
2r
−
u
,
dr
du
r
dr
dR
,
2 2 2
dr
u
d
r
dr
dR
r
dr
d
and hence
.
)
1
(
2
)
(
2
22
2 2 2
Eu
u
r
l
l
m
r
V
dr
u
d
m
=
+
+
+
−
This is called the radial equation; it is identical in form to the one-dimensional Schrödinger Equation , except that the effective potential,
)
1
(
2
l
l
+
contains an extra piece, the so-called centrifugal term, . It tends to throw the particle outward (away from the origin), just like the centrifugal (pseudo-) force in classical mechanics. Meanwhile, the normalization condition becomes
,
)
1
(
2
)
(
22
r
l
l
m
r
V
V
eff=
+
+
2 2
)
1
(
2
r
l
l
m
+
,
1
0
2 2
=
∞
dr
r
R
01
.
2
=
∞
The
infinite spherical well
:
Find the wave function and the allowed energies.
Solution:
1. Outside the well, the wave function is zero: u(r,r=a or r>a)=0.
>
∞
<
=
.
if
,
,
if
,
0
)
(
a
r
a
r
r
V
a
∞
1. Outside the well, the wave function is zero: u(r,r=a or r>a)=0.
,
)
1
(
22 2
2
u
k
r
l
l
dr
u
d
−
+
=
2. Inside the well, the radial equation reads
where
k
=
2
mE
.
(1) The case l=0 is easy:
).
cos(
)
sin(
)
(
2 2
2
kr
B
kr
A
r
u
u
k
dr
u
d
+
=
Then the solution is
(
)
sin(
)
cos(
)
.
r
kr
B
r
kr
A
r
R
=
+
As the second term blows up, so we must choose
r
→
0
B=0.r
kr
A
r
R
(
)
=
sin(
)
The boundary condition then requires that
0
)
sin(
0
)
sin(
)
(
=
=
ka
=
r
ka
A
a
R
ka
=
n
π
,
n
=
1
,
2
,
3
,
The allowed energies are evidently
mE
k
=
2
,
(
1
,
2
,
3
,
),
2
22 2 2
0
=
n
=
ma
n
E
nπ
which is the same for the one-dimensional infinite square well.
,
3
,
2
,
1
,
=
=
n
a
n
The normalization condition:
,
1
0
2
=
∞
dr
u
2
;
a
A
=
yields
Tacking on the angular part (l=0,m=0)
,
4
1
)
(cos
4
1
)
,
(
000 0
π
θ
π
φ
θ
=
P
=
Y
r
r
k
a
r
R
n(
)
=
2
sin(
n)
a
n
k
n=
π
we conclude that
ψ
nlm(
r
,
θ
,
φ
)
=
R
n(
r
)
Y
lm(
θ
,
φ
)
r
a
r
n
a
Y
R
nn
)
/
sin(
2
1
0 0 00
π
π
ψ
=
=
Notice that the stationary states are labeled by three quantum
number, n, l and m:
nlm. The energy, however, depends only on
(2) The case l is in any integer:
The general solution of above equation is:
.
)
1
(
22 2
2
u
k
r
l
l
dr
u
d
−
+
=
),
(
)
(
)
(
r
A
rj
kr
B
rn
kr
u
=
⋅
l+
⋅
lwhere jl(kr) is the spherical Bessel function of order l, and nl(kr) is the
spherical Neumann function of order l. They are defined as follows:
,
sin
1
)
(
)
(
x
x
dx
d
x
x
x
j
l l
l
≡
−
,
cos
1
)
(
)
(
x
x
dx
d
x
x
x
n
l l
Spherical Bessel function:
,
sin
1
)
(
)
(
x
x
dx
d
x
x
x
j
l l
l
≡
−
,
sin
)
(
0
x
x
x
j
=
,
cos
sin
)
(
21
x
x
x
x
x
j
=
−
,
cos
1
)
(
)
(
x
x
dx
d
x
x
x
n
l l
l
≡
−
−
Spherical Neumann function
,
cos
)
(
0
x
x
x
n
=
−
,
sin
cos
)
(
21
x
x
x
x
x
The asymptotic properties of two functions :
0
lim
x→Generally, for small x, we have
,
1
)
(
lim
0 0=
→
j
x
x
,
3
)
(
lim
1 0x
x
j
x→
=
,
1
)
(
lim
00
n
x
x
x→
=
−
,
1
)
(
lim
1 20
n
x
x
x→
=
−
(
2
1
)
!
!
)
(
lim
0
=
+
→
l
x
x
j
l l x(
)
.
!
!
1
2
)
(
lim
1 0 + →−
−
=
l l xx
l
x
n
Proof: For small x,
in the general solution,
u
(
r
)
=
A
⋅
rj
l(
kr
)
+
B
⋅
rn
l(
kr
),
(
)
.
!
!
1
2
)
(
lim
10
→
∞
−
−
=
+→ l l
x
x
l
x
n
As when x 0, Neumann functions blow up, that is
we must set B=0, and hence
).
(
)
(
)
(
)
(
r
A
rj
kr
R
r
A
j
kr
u
=
⋅
l=
⋅
lThe boundary condition then requires that R(a)=0. Evidently k must be chosen such that
;
0
)
(
ka
=
j
lthat is, (ka) is a zero of the lth-order spherical Bessel function. Now, the Bessel functions are oscillatory; each one has an infinite number of zeros. However, unfortunately for us, they are not regularly located and must be computed numerically. At any rate, if we suppose that
,
0
)
(
nl=
l
where nl is the nth zero of the lth spherical Bessel function. The allowed energies, then, are given by
the boundary condition requires that
,
1
nl
a
k
=
β
),
,
3
,
2
,
1
(
,
2
2 2 2
=
=
n
ma
E
nlβ
nland the wave functions are
),
,
(
)
(
)
,
(
)
(
)
,
,
(
θ
φ
θ
φ
β
θ
φ
ψ
ml nl
l nl m
l nl
nlm
r
Y
a
j
A
Y
r
R
r
=
=
The hydrogen atom consists of a heavy, essentially motionless proton, of charge e, together with a much lighter electron (charge –e) that orbits around it, bound by the mutual attraction of opposite charges.
From Coulomb’s law, the potential energy (in SI units) is
1
2
e
4.2 The Hydrogen Atom
r
e
e
−
.
1
4
)
(
0 2
r
e
r
V
πε
−
=
e
Then the radial equation for hydrogen atom says
.
)
1
(
2
1
4
2
22
0 2
2 2 2
Eu
u
r
l
l
m
r
e
dr
u
d
m
=
+
+
−
+
−
Our problem is to solve this equation for u(r), and determine the allowed energies, E. Now we consider this problem in detail by using analytical method.
Incidentally, Coulomb potential, 1 , admits two different states,
4 )
(
0 2
r e r
V
πε
− =
continuous states and bound states, which are separately corresponds to the following situations:
e
e
−
continuous states E>0, bound states E<0,
r
e
e
−
describing electron-proton
The radial equation for Hydrogen atom is
(1) Simplify it (tidy up):
4.2.1 The Radial Wave Function
.
)
1
(
2
1
4
2
22
0 2
2 2 2
Eu
u
r
l
l
m
r
e
dr
u
d
m
=
+
+
−
+
−
πε
(
E<0
)
1. Radial Solution:
(1) Simplify it (tidy up):
As E<0, then we let
κ
≡
−
2
mE
.
Dividing above equation by E, we have
( )
( )
( )
.
)
1
(
1
2
1
2 20 2
2 2
u
r
l
l
r
me
r
d
u
d
+
+
−
=
κ
κ
κ
πε
κ
κ
πε
ρ
κ
ρ
20 2 0
2
and
,
me
r
≡
≡
So that
.
)
1
(
1
0 22 2
u
l
l
d
u
d
+
+
−
=
ρ
ρ
ρ
ρ
(2) The asymptotic properties of the solution:
∞
→
ρ
)
a
In this case, the constant term in the bracket of above equation dominates, so (approximately)
.
2 2
u
d
u
d
=
ρ
The general solution of it is
,
ρ ρBe
Ae
0
)
ρ
→
b
but the second term blows up as , so
e
ρ→
∞
ρ
→
∞
B=0. Evidently,,
)
1
(
ρ
>>
≈
Ae
−ρu
for large .ρ
In this case, the centrifugal term dominates; approximately, then:
u
l
l
d
u
d
+
+
−
=
0 22 2
)
1
(
1
ρ
ρ
ρ
ρ
u
l
l
d
u
d
2 2
2
)
1
(
ρ
ρ
+
=
The general solution of it is
,
1 l
l
D
C
u
=
ρ
++
ρ
−d
ρ
ρ
ρ
d
ρ
ρ
( 1)
)
1
(
+
−
− +=
C
l
lDl
ld
du
ρ
ρ
ρ
( 2)
1 2
2
)
1
(
)
1
(
+
−+
+
− +=
Cl
l
lDl
l
ld
u
d
ρ
ρ
But for 0, the term -l blows up, so D=0. Thus
,
)
1
(
<<
≈
C
l+1u
ρ
ρ
for small .ρ
(3) Introduce new function v( ) to simplify solution:
The next step is to peel off the asymptotic behavior, introducing the new function v( ):
),
(
)
(
ρ
ρ
1e
ρv
ρ
u
(
ρ
)
=
ρ
l+1e
−ρv
(
ρ
),
u
=
l+ −in the hope that v( ) will turn out to be simpler than u( ). Then
(
)
ρ
ρ
ρ
ρ
ρ
ρ
ρ
ρ ρ
ρ
d
dv
e
v
e
v
e
l
d
du
l − l+ − l+ −+
−
+
=
1
(
)
1(
)
1(
+
−
)
+
=
−ρ
ρ
ρ
ρ
ρd
dv
v
l
e
l
u
l
l
d
u
d
+
+
−
=
0 22 2
)
1
(
1
ρ
ρ
ρ
ρ
(
+
−
)
+
+
+
+
+
−
−
=
− 22 2 2
1
2
)
1
(
2
2
ρ
ρ
ρ
ρ
ρ
ρ
ρ
ρ
ρd
v
d
d
dv
l
l
l
l
e
d
u
d
lIn terms of v( ), then, the radial equation of u( ) reads
(
1
)
[
2
(
1
)
]
0
.
2
0 2 2=
+
−
+
−
+
+
l
v
d
dv
l
d
v
d
ρ
ρ
ρ
ρ
ρ
(4) Solve above equation by power series method:
Finally, we assume the solution, v( ), can be expressed as a power series in :
.
)
(
0 ∞ ==
j j jc
v
ρ
ρ
Now replace v( ) into equation and our problem is to determine the coefficients of the series, c1,c2,c3, … . Differentiating term by term:
(
1
)
.
0 1 0 1 0 ∞ = + ∞ = − ∞ =
+
=
=
=
j j j j j j j jj
c
j
j
c
c
d
d
d
dv
ρ
ρ
ρ
ρ
ρ
Differentiating again,(
1
)
1 1.
2
2 ∞
− +
+
=
j
j
c
j jv
d
ρ
(
1
)
.
0 1 2 = +
+
=
j jc
j
j
d
ρ
ρ
Inserting these into Equation 4.61, we have
(
+
)
+
∞ = − + 0 1 11
j j jc
j
j
ρ
ρ
(
1
)
[
2
(
1
)
]
0
.
2
0 2 2=
+
−
+
−
+
+
l
v
d
dv
l
d
v
d
ρ
ρ
ρ
ρ
ρ
(
+
1
−
ρ
)
2
l
(
)
∞ = +
+
0 11
j j jc
j
ρ
(
)
[
0−
2
+
1
]
+
ρ
l
0
(
1
)
2
(
1
)
0
1
+
+
+
∞
=
+
l
c
j
j
j
j
j
ρ
(
)
∞
=
+
+
0
1
1
j
j j
c
j
ρ
(
)
[
0−
2
+
1
]
+
ρ
l
0
0
=
∞
=
j
j j
c
ρ
∞
=
−
0
2
j
j j
jc
ρ
(
+
)
=
−
∞
+ +
1 1
1
2
j
c
jρ
j∞
−
2
jc
jρ
jwhere
=0
j j=0
Equating the coefficient of like powers yields
(
j
+
1
)
c
j+1+
2
(
l
+
1
)
(
j
+
1
)
c
j+1−
2
jc
j+
[
0−
2
(
l
+
1
)
]
c
j=
0
,
j
ρ
or:
(
)
(
1
)
(
2
2
)
.
1
2
01 j
j
c
l
j
j
l
j
c
+
+
+
−
+
+
=
+
This recursion formula determines the coefficients, and hence the function
v( ) : We start with c0, and recursion formula gives us c1; putting this back in, we obtain c2, and so on.
At last, after c0 being fixed eventually by normalization, the solution of v( ) and u( ) will be got.
2. Energies of the solutions:
=
=
+ −(
)
)
(
ρ
ρ
1e
ρv
ρ
u
lρ
l+1e
−ρ.
0
∞
=
j
j j
c
ρ
If j is very large, that is j , the recursion formula says
(
)
(
)
(
)
1
.
2
1
2
)
2
2
(
1
1
2
01 j j j
j
c
j
c
j
j
j
c
l
j
j
l
j
c
+
=
+
≅
+
+
+
−
+
+
=
+
ρ
Then
1
2
−
≅
jj
c
j
c
−
≅
−21
2
2
j
c
j
j
!
,
2
0
c
j
c
so
,
!
2
)
(
0 20 0 0
ρ
ρ
ρ
ρ
c
e
j
c
c
v
j
j j
j
j
j
≅
=
=
∞
= ∞
=
and hence
ρ
ρ
ρ
=
+e
−u
(
)
l 1c
0e
2ρ=
c
0ρ
l+1e
ρ,
which blows up at large and is not permitted because the solution will not be properly normalized. In order to satisfy the normalization condition, there is only one way out of this dilemma: The series must terminate. There must occur some maximal integer, jmax, such that
.
0
1
max +
=
j
c
Evidently, from recursion formula
(
)
(
1
)
(
2
2
)
,
1
2
01 j
j
c
l
j
j
l
j
c
+
+
+
−
+
+
=
+
we get
(
1
)
0
.
2
j
max+
l
+
−
ρ
0=
(
)
(
)
jj
c
l
j
j
l
j
c
)
2
2
(
1
1
2
max 01
max
+
+
+
−
+
+
=
+
ρ
Defining , which is the so-called principle quantum number, we have
1
max
+
+
≡
j
l
n
we have
.
2
0
=
n
ρ
mE
2
−
≡
κ
ρ
πε
2κ
0 2 0
2
me
≡
2 0 2 2 0 2
4 2
2
8
2
π
ε
ρ
κ
me
m
so the allowed energies are
3
,
2
,
1
,
1
1
4
2
2 2 12
0 2
2
=
=
−
=
E
n
n
n
e
m
E
nπε
This is the famous Bohr formula ——by any measure the most important
result in all of quantum mechanics. Bohr obtained it in 1913 by a serendipitous mixture of inapplicable classical physics and premature quantum theory.
mixture of inapplicable classical physics and premature quantum theory.
And we also find that
n
me
2
2
0 22
0
≡
=
κ
πε
ρ
1
1
,
4
0 22
an
n
me
=
=
πε
κ
where
m
me
a
2 102 0
10
529
.
0
4
−×
=
.
an
r
r
=
≡
κ
ρ
It follows that
3. The overall solutions of Hydrogen atom:
Finally, the spatial wave functions of hydrogen are labeled by three quantum numbers (n,l, and m):
),
,
(
)
(
)
,
,
(
θ
φ
θ
φ
ψ
ml nl
nlm
(
r
,
θ
,
φ
)
=
R
(
r
)
Y
(
θ
,
φ
),
ψ
nlmr
=
R
nlr
Y
lwhere
,
1
)
(
1
)
(
max
0 1
1
= − + −
+
=
=
=
j
j
j j l
l
nl
e
c
r
v
e
r
r
u
r
R
ρ
ρρ
ρ
ρρ
and v( ) is a polynomial of degree jmax= n-l-1 in , whose coefficients are determined by the recursion formula
(
)
(
1
)
(
2
2
)
.
1
2
1 j
j
c
l
j
j
n
l
j
c
+
+
+
−
+
+
=
+
an
.
r
=
(1) The ground state:
The ground state (that is, the state of lowest energy) is the case n=1; putting in the accepted values for the physical constants, we get
.
6
.
13
4
2
2
0 2
2
1
eV
e
m
E
=
−
=
−
πε
Evidently the binding energy of hydrogen (the amount of energy you would have to impart to the electron in the ground state in order to ionize the atom) is have to impart to the electron in the ground state in order to ionize the atom) is
13.6 eV. As the principle quantum number n=1=jmax+l+1, the angular quantum number must be zero (l=0), whence also m=0, so the wave function is
).
,
(
)
(
)
,
,
(
10 00100
θ
φ
θ
φ
ψ
r
=
R
r
Y
= − +
=
max
0 1
1
)
(
j
j
j j l
nl
e
c
r
r
R
ρ
ρρ
a
r
an
r
=
=
ρ
0 10
1
)
(
e
c
r
r
R
=
ρ
−ρa r
e
a
c
r
Normalizing R10 by
,
1
4
|
|
4
|
|
|
|
2 0 3 2 2 0 0 2 / 2 2 2 0 0 2 210
=
=
=
=
∞ − ∞
a
c
a
a
c
dr
r
e
a
c
dr
r
R
r awe have 0
2
.
a
c
=
a r a re
a
e
r
R
2 /3 /
2
2
)
(
=
−=
− −.
1
)
,
(
)
(
)
,
,
(
/ 3 0 0 10 100 a re
a
Y
r
R
r
=
=
−π
φ
θ
φ
θ
ψ
Meanwhile, Y00= 1/ 4π , and hence the ground state of hydrogen is
e
a
e
a
r
R
2 3(2) The first excited states n=2:
If n=2 the energy is
.
4
3
4
6
13
2
21
2
.
eV
eV
.
E
E
=
=
−
=
−
This is the first excited states, since we can have either l=0 (in which case
m=0) or l=1 (in which case m=-1, 0, or 1); Evidently there are four states that share the same energy E2.
If l=0, the recursion relation gives
0
=
j
1
=
j
(
)
(
0
1
)
(
0
0
2
)
;
2
1
0
0
2
0 0
1
c
c
c
=
−
+
+
+
−
+
+
=
(
)
(
1
1
)
(
1
0
2
)
0
;
2
1
0
1
2
1
2
=
+
+
+
−
+
+
=
c
c
1
1
0
2
1
max
=
n
−
l
−
=
−
−
=
and therefore so
),
1
(
)
(
0 0 01
0
max
ρ
ρ
ρ
ρ
=
=
−
=
−
=
=
c
c
c
c
v
j
j
j j
a r
e
a
r
a
c
v
e
r
r
R
20 0)
/22
1
(
2
)
(
1
)
(
=
ρ
−ρρ
=
−
−Normalization
and we find
If l=1, the recursion formula terminates the series after a single term;
0
1
1
2
1
max
=
n
−
l
−
=
−
−
=
j
,
)
(
00
0
c
c
v
j
j j
=
=
=
ρ
ρ
a r
re
a
c
r
R
21 02 /24
)
(
=
−( )
ρ
ρ
e
ρv
r
r
R
nl(
)
=
1
l+1 −(3) The excited states for arbitrary n:
For arbitrary n, the possible values of l are
,
1
,
...
,
2
,
1
,
0
−
=
n
l
and for each l there are (2l+1) possible values of m, so the total degeneracy of the energy level En is
.
)
1
2
(
)
(
21
0
n
l
n
d
n
l
=
+
=
−
=0
l=
The polynomial v( ) (defined by the recursion formula Eq.4.76 is a function well known to applied mathematicians; apart from normalization, it can be written as
),
2
(
)
(
2 110
ρ
ρ
ρ
−+−∞
=
=
=
nl lj
j
j
L
c
v
where
( )
1
(
)
)
(
L
x
dx
d
x
L
qp p
p p
is the associated Laguerre polynomial, and
is the qth Laguerre polynomial.
(
x q)
q x
q
e
x
dx
d
e
x
L
(
)
≡
−Therefore the radial wave function is
1
0
=
L
1
1
=
−
x
+
L
1
4
2
2
=
x
−
x
+
L
)
2
(
1
)
(
=
ρ
l+1 −ρ⋅
2(nl++1l)−(2l+1)ρ
nl
e
L
r
r
R
na
r
=
ρ
⋅
=
− −+−na
r
L
e
na
r
na
l l n na
r l
2
1
2 11 /
=
)
(
10
r
R
1
− /⋅
102
=
1
e
− /⋅
(
N
10×
1
)
a
a
r
L
e
a
a r a
r
1
1 0
=
L
)
2
1
(
4
1
2
1
2
1
20 2 / 1 1 2 / 20a
r
N
e
a
a
r
L
e
a
R
=
−r a⋅
=
−r a⋅
−
a
r
a
r
L
11=
4
−
2
Normalization Generally, we can normalize Rnl as
1
)
(
0 2 2=
∞dr
r
r
R
nlto give normalized Rnl as follows
)
/
2
(
2
)
(
/L
2 11r
na
na
r
e
N
r
R
nl ll na r nl nl + − − −
⋅
=
with normalization constant Nnl being
Then, finally, the normalized hydrogen wave function are
)
,
(
)
/
2
(
2
)
,
,
(
θ
φ
/ 2 11θ
φ
ψ
ml l
l n l na
r nl
nlm
L
r
na
Y
na
r
e
N
r
=
⋅
− −+−⋅
Notice that whereas the wave functions depend on all three quantum numbers, the energies are determined by n alone. This is a peculiarity of the Coulomb potential; generally, the energies depend also on l.
The wave functions are mutually orthogonal
m m l l n n m
l n
nlm
ψ
′ ′ ′r
θ
drd
θ
d
φ
=
δ
′δ
′δ
′ψ
* 2sin
See the figures of solutions of
In practice such perturbations are always present; transitions (or, as they are In principle, if you put a hydrogen atom into some stationary state nlm, it should stay there forever. However, if you tickle it slightly (by collision with another atom, say, or by shining light on it), the electron may undergo a
transition to some other stationary state——either by absorbing energy, and
moving up to a higher-energy state, or by giving off energy (typically in the form of electromagnetic radiation), and moving down.
4.2.2 The Spectrum of Hydrogen
sometimes called, “quantum jumps ”) are constantly occurring, and the result is that a container of hydrogen gives off light (photons), whose energy corresponds to the difference in energy between the initial and final states:
eV
n
n
E
E
E
f i
f
i
−
=
−
−
⋅
=
13
.
6
1
21
2γ
,
1
1 2
E
n
E
n=
E
1=
−
13
.
6
eV
.
i
E
f
E
1
E
2
ν
λ
=
c
Now according to the Planck formula, the energy of a photon is proportional to its frequency:
Meanwhile, the wavelength is given by , so
,
ν
γ
h
E
=
ν
γ
h
n
n
E
E
f i
=
−
−
=
11
21
2λ
ν
1
1
1
2 2
1
h
c
h
n
n
c
E
f i
=
=
−
−
1 7
2
0 2
3 1
10
097
.
1
4
4
−
×
=
=
=
e
m
c
m
hc
E
R
πε
π
n
n
i fc
n
in
fc
λ
−
=
−
=
11
21
21
21
21
i f
i
f
n
n
R
n
n
hc
E
λ
is known as the Rydberg constant. Above equation is the Rydberg formula for the spectrum of hydrogen; it was discovered empirically in 19th century, and
the greatest triumph of Bohr’s theory was its ability to account for this result— —and to calculate R in terms of the fundamental constants of nature.
Transitions to the ground state (nf=1) lie in the ultraviolet; they are known to spectroscopists as the Lyman series.
Spectrum of Hydrogen:
Transitions to the first excited state (nf=2) fall in the visible region; they constitute Balmer series.
Transitions to nf=3 (Paschen series) are in the infrared region; and so on.
莱曼系
莱曼系莱曼系
莱曼系
巴尔末系
巴尔末系巴尔末系
巴尔末系
OXYGEN
HYDROGEN
These spectral lin
