量子力学
Quantum mechanics
School of Physics and Information Technology
Shaanxi Normal University
Chapter 4
QUANTUM MECHANICS IN
THREE DIMENSIONS
4.1 Schrödinger Equation in Spherical Coordinates 131
4.2 The Hydrogen atom
145
4.3 Angular momentum
160
4.4 Spin
171
4.1 Schrödinger Equation in Spherical Cordinations
(1) The generalization of the Schrödinger Equation from onedimensional to three-dimensional is straightward. The SE says
∂Ψ
ih
= HΨ;
∂t
the Hamiltonian operator H is obtained from the classical energy
1 2
1
mv + V =
( p x2 + p y2 + p z2 ) + V ( x, y, z )
2
2m
by the standard prescription (applied now to y,z as well as x)
h ∂
px →
,
i ∂x
h ∂
py →
,
i ∂y
h ∂
pz →
.
i ∂z
As
v
v
v
r
r h  ∂ v ∂ v ∂ v
p = p x i + p y j + p z k → p =  i +
j + k ,
i  ∂x
∂y
∂z 
or, for short
r h
p = ∇,
i
where
 ∂ v ∂ v ∂ v
∇ =  i +
j + k 
∂z 
∂y
 ∂x
Thus
∂Ψ  1 r r
h2 2

ih
=
p ⋅ p + V Ψ = −
∇ Ψ + VΨ ,
∂t  2m
2m

where
2
2
2
∂
∂
∂
∇2 = ∇ ⋅ ∇ = 2 + 2 + 2
∂x
∂y
∂z
Cartesian coordinates.
is the Laplacian, in
And in 3-dimensional space
r
V ( x, t ) → V (r , t ) = V ( x, y, z , t ),
as well as
r
Ψ (r , t ) = Ψ ( x, y, z , t ),
(2) The probability of finding the particle in the infinitesimal volume d3r,
r
d r = dxdydz,
3
is
r 2 3r
Ψ (r , t ) d r .
(3) Therefore the normalization condition reads
∫
r 2 3r
Ψ ( r , t ) d r = 1.
r
d r
3
(4) If the potential is time-independent, the time-independent Schrodinger
equation reads
h2 2
−
∇ Ψ + VΨ = EΨ,
2m
and there will be a complete set of stationary states
r
r − iEnt / h
Ψn (r , t ) = ψ n (r )e
.
The general solution to the (time-dependent) Schrodinger equation is
r
r
r −iEnt / h
Ψ (r , t ) = ∑ cn Ψn (r , t ) = ∑ cnψ n (r )e
.
n
n
z
4.1.1 Separation of Variables
(1) Spherical coordinates
Cartesian coordinates:
Spherical coordinates:
θ
( x, y , z )
(r ,θ , φ )
A
M ( x , y, z )
z
o
x
y
ϕ
x
 x = r sin θ cos φ ,

 y = r sin θ sin φ ,
 z = r cos θ .

r
•
•
y
P
dτ = r 2 sin θdrd θdφ ,
In spherical coordinates the Laplacian takes the form
∂ 
∂ 
1 ∂  2 ∂
1
1
2
∇ = 2 r
+ 2
 sin θ
+ 2 2
∂θ  r sin θ
r ∂r  ∂r  r sin θ ∂θ 
 ∂2 
 2 
 ∂φ 
The time-independent Schrodinger equation in Cartesian coordinates
h 2  ∂ 2ψ ∂ 2ψ ∂ 2ψ
 2 + 2 + 2
−
2m  ∂x
∂y
∂z

 + Vψ = Eψ

In spherical coordinates
h 2  1 ∂  2 ∂ψ 
1
∂ 
∂ψ 
1
−
+ 2
+ 2 2
r
 sin θ
 2
2m  r ∂r  ∂r  r sin θ ∂θ 
∂θ  r sin θ
 ∂ 2ψ
 2
 ∂φ

 + Vψ = Eψ

We begin by looking for solutions that are separable into products:
ψ ( r , θ , φ ) = R ( r )Y (θ , φ )
Putting this into above equation, we have
h 2  Y d  2 dR 
R
∂ 
∂Y 
R
−
r
+
sin
+
θ





2m  r dr  dr  r 2 sin θ ∂θ 
∂θ  r 2 sin 2 θ
 ∂ 2Y 
 2  + VRY = ERY
 ∂φ 
Dividing by RY and multiplying by − 2mr 2 / h 2
 1 d  2 dR 
1
1
∂ 
∂Y 
r
+
sin
+
θ





∂θ  Y sin 2 θ
 R dr  dr  Y sin θ ∂θ 
 ∂ 2Y  2mr 2
 2  − 2 (V (r ) − E ) = 0
h
 ∂φ 
 1
 1 d  2 dR  2mr 2

∂ 
∂Y 
1
(
)
r
−
V
r
−
E
=
−
+
(
)
sin
θ







2
2
R
dr
dr
h
Y
sin
θ
∂
θ
∂
θ
Y
sin
θ







 ∂ 2Y 
 2 
 ∂φ 
The term on the left hand depends only on r, whereas the right depends only on θ
φ; accordingly, each must be a constant, which is in the form l(l+1):
 1 d  2 dR  2mr 2

r
 − 2 (V (r ) − E ) = l (l + 1);

h
 R dr  dr 

V = V (r )
 1
∂ 
∂Y 
1
θ
sin
+



2
Y
sin
∂
∂
Y
sin
θ
θ
θ
θ



Spherically
symmetric
potential
 ∂ Y 
 2  = −l (l + 1).
 ∂φ 
2
4.1.2 The Angular Equation
Solution of Y : Equation 4.17 determines Y function as
1
∂ 
∂Y 
1
 sin θ
+
Y sin θ ∂θ 
∂θ  Y sin 2 θ
 ∂ 2Y 
 2  = −l (l + 1).
 ∂φ 
Multiplying above equation by Ysin2θ, it becomes:
∂ 
∂Y  ∂ 2Y
2
sin θ
 sin θ
 + 2 = −l (l + 1) sin θY .
∂θ 
∂θ  ∂φ
As always, we try separation of variables:
Y (θ , φ ) = Θ(θ ) ⋅ Φ (φ )
Plugging this in, and dividing by Y, we find
2
1 

1
Φ
d 
dΘ 
d
2
= 0.
 sin θ
 + l (l + 1)sin θ  +
 sin θ
2
dθ 
dθ 
Θ 
 Φ dφ
The first term is a function only of θ, and the second is a function only of φ , so
each must be a constant. This time I will call the separation constant m2:
1
d 
dΘ 
2
2
(
)
sin
θ
sin
θ
+
l
l
+
1
sin
θ
=
m
;



Θ
dθ 
dθ  
1 d 2Φ
2
=
−
m
.
2
Φ dφ
(1) The φ equation is easy to solve:
d 2Φ
2
imφ
=
−
m
Φ
⇒
Φ
(
φ
)
=
e
.
2
dφ
Now, when φ advances by 2π, we return to the same point in space, so it is
natural to require that
Φ (2π + φ ) = Φ (φ ).
In other words,
e
im (φ + 2π )
=e
imφ
⇒e
i 2πm
=1
From this it follows that m must be an integer:
m = 0 , ± 1, ± 2 , ± 3 , L
(2) The θ equation is not simple.
d
sin θ
dθ
dΘ 

2
2
(
)
sin
θ
+
l
l
+
1
sin
θ
−
m
Θ = 0.


dθ 

[
]
Turn θ into x by: x = cos
θ
2
2


d
Θ
d
Θ
m
2
(1 − x )
Θ = 0.
− 2x
+ l (l + 1) −
2
2
1− x 
dθ
dθ 
Above equation is lth-order associated Legendre equation.
Therefore, the solution of Θ is
Θ (θ ) = APl m ( x ) = APl m (cos θ ),
where Pml is the associated Legendre function, defined by
| m|
m
2 |m| / 2
Pl ( x ) ≡ (1 − x )
 d 
  Pl ( x ),
 dx 
and Pl(x) is the lth Legendre polynomial, defined by the Rodrigues formula:
l
1  d  2
Pl ( x ) ≡ l   ( x − 1) l ,
2 l!  dx 
P0 ( x ) = 1,
P1 ( x ) = x,
1
P2 ( x ) = (3 x 2 − 1),
2
1
P3 ( x ) = (5 x 3 − 3 x ),
2
1
P4 ( x ) = (35 x 4 − 30 x 2 + 3)
8
Pl(x) is a polynomial (of degree l) in x, and is even or odd according to the parity
of l.
But for associated Legendre function Pml:
|m|
2 |m| / 2
m
Pl ( x ) ≡ (1 − x )
 d 
  Pl ( x ),
 dx 
m≤l
is not, in general, a polynomial——if m is odd it carries a factor of
1− x2 .
P00 ( x ) = P0 ( x ) = 1,
P10 ( x ) = P1 ( x ) = x,
1
1
(
P ( x) = 1 − x
)
2 1/ 2
d
P1 ( x ) = 1 − x 2 ,
dx
1
P ( x ) = P2 ( x ) = (3 x 2 − 1),
not a polynomial
2
1
2 1/ 2 d
P2 ( x ) = (1 − x )
P2 ( x ) = 3 x 1 − x 2 ,
dx
2
d
P22 ( x ) = (1 − x 2 ) 2 P2 ( x ) = 3(1 − x 2 ),
dx
0
2
z
As x = cos θ ,
0
0
P ( x ) = 1,
P ( x ) = cos θ ,
P11 ( x ) = sin θ ,
Plot:
m
r = Pl (cos θ )
r = 1,
θ
0
1
A
M ( x , y, z )
z
o
x
x
r = cos θ ,
r
•
y
ϕ
•
y
P
r = sin θ ,
Plot:
z
m
r = Pl (cos θ )
1
P ( x ) = (3 cos 2 θ − 1),
2
P21 ( x ) = 3 sin θ cos θ ,
θ
0
2
P22 ( x ) = 3 sin 2 θ ,
1
r = (3 cos 2 θ − 1),
2
A
M ( x , y, z )
z
o
x
x
r = 3 sin θ cos θ ,
r
•
y
ϕ
•
y
P
r = 3 sin 2 θ ,
r = Pl m (cos θ )
1
0
P3 ( x ) = (5 cos 3 θ − 3 cos θ ),
2
3
P31 ( x ) = sin θ (5 cos 2 θ − 1),
2
Plot:
1
r (θ ) = (5 cos 3 θ − 3 cos θ ),
2
P32 ( x ) = 15 sin 2 θ cos θ ,
(
)
P33 ( x ) = 15 sin θ 1 − cos 2 θ ,
r (0 ) = 1, r (π ) = −1
Some Notes:
(1) Notice that l must be a nonnegative integer, for the Rodrigues formula
to make any sense.
| m|
m
2 |m| / 2
Pl ( x ) ≡ (1 − x )
 d 
  Pl ( x ),
 dx 
m≤l
If |m|>l, then Plm=0. Therefore, for any given l, there are (2l+1) possible
values of m:
l = 0 ,1 , 2 , L
m = − l , − l + 1 , L , − 1 , 0 ,1 , L l − 1 , l
Pl ± l ( x ), Pl ± (l −1 ) ( x ), L , Pl ± 1 ( x ), Pl 0 ( x )
(2) Now, the volume element in spherical coordinates is
d 3 r = r 2 sin θdrd θdφ ,
so the normalization condition becomes
2
2
ψ
r
sin
θ
drd
θ
d
φ
=
R
r
dr ⋅ ∫ Y sin θ dθ dφ = 1
∫
∫
2
2
2
It is convenient to normalize R and Y separately:
∫
∞
0
2
2
R r dr = 1,
2π
π
0
0
∫ ∫
2
Y sin θ dθ d φ = 1,
where R determined by V(r) and Y can be obtained.
 1 d  2 dR  2mr 2

r
 − 2 (V (r ) − E ) = l (l + 1);

h
 R dr  dr 

The normalized angular wave functions are called spherical harmonics:
m
Yl (θ , φ ) = ε
( 2 l + 1) ( l − | m |)! im φ m
e Pl (cos θ ),
4π ( l + | m |)!
 (− 1 )m ,
ε =
1,
where
for m ≥ 0 ,
for m ≤ 0 .
Now we list here some few spherical harmonics: See book for more
Y
±1
3
21
(θ , φ ) = m
sin θ ( 5 cos 2 θ − 1) e ± iφ
64 π
Notice that
Yl
−m
(θ , φ ) = (− 1) Yl
m
m*
(θ , φ )
Actually, the Ys are automatically orthogonal, so
∫ ∫ [Y
2π
0
π
0
m
l
]
*
(θ , φ ) Yl ′m ′ (θ , φ ) sin θ d θ d φ = δ ll ′δ m m ′ ,
For historical reasons, l is called the azimuthal quantum number, and m
the magnetic quantum number.
4.1.3 The Radial Equation
Notice that the angular part of the wave function, Y(θ,Φ), 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:
d  2 dR  2mr 2
r
 − 2 (V (r ) − E )R = l (l + 1) R.
dr  dr 
h
This equation can be simplified if we change variables as
u (r ) = rR(r ),
so that
u
R= ,
r
dR 1  du

= 2  r − u ,
dr r  dr

d  2 dR 
d 2u
r
=r 2 ,
dr  dr 
dr
and hence
h 2 d 2u 
h 2 l (l + 1) 
−
+ V (r ) +
u = Eu.
2
2
2m dr
2m r 

This is called the radial equation; it is identical in form to the one-dimensional
Schrödinger Equation , except that the effective potential,
Veff
h 2 l (l + 1)
= V (r ) +
,
2
2m r
h 2 l (l + 1)
. It tends
2m r 2
contains an extra piece, the so-called centrifugal term,
to throw the particle outward (away from the origin), just like the centrifugal
(pseudo-) force in classical mechanics. Meanwhile, the normalization condition
becomes
∫
∞
0
2
2
R r dr = 1,
∫
∞
0
2
u dr = 1 .
The infinite spherical well:
0,
V (r ) = 
∞ ,
if r < a ,
a
if r > a .
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.
2. Inside the well, the radial equation reads
d 2u  l (l + 1)
2mE
2
where
=  2 − k u ,
k=
.
2
dr
h
 r

(1) The case l=0 is easy:
d 2u
2
=
−
k
u ⇒ u (r ) = A sin(kr ) + B cos(kr ).
2
dr
sin(kr )
cos(kr )
+B
.
Then the solution is R ( r ) = A
r
r
As r → 0 the second term blows up, so we must choose B=0.
sin(kr )
R(r ) = A
r
The boundary condition then requires that
sin(ka)
sin(ka
R(a) = A
= 0 ⇒ sin(ka) = 0
r
The allowed energies are evidently
2mE
k=
h
n 2π 2 h 2
En 0 =
,
2
2ma
ka = nπ , n = 1,2,3, L
nπ
, n = 1,2,3,L
kn =
a
(n = 1,2,3, L),
which is the same for the one-dimensional infinite square well.
The normalization condition:
∫
∞
0
2
u dr = 1,
yields
A=
2
;
a
2 sin(k n r )
Rn (r ) =
a
r
Tacking on the angular part (l=0,m=0)
0
0
Y (θ , φ ) =
we conclude that
ψ n 00
1 0
P0 (cos θ ) =
4π
1
,
4π
nπ
kn =
a
ψ nlm ( r , θ , φ ) = Rn ( r )Yl m (θ , φ )
1 sin(nπr / a )
=RY =
r
2πa
0
n 0
Notice that the stationary states are labeled by three quantum
number, n, l and m: ψnlm. The energy, however, depends only on
n and l:Enl.
(2) The case l is in any integer:
d 2 u  l (l + 1)
2 
=
−
k

u .
2
2
dr
 r

The general solution of above equation is:
u (r ) = A ⋅ rjl (kr ) + B ⋅ rnl (kr ),
where 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:
l
 1 d  sin x
jl ( x) ≡ (− x) l 
,

 x dx  x
l
 1 d  cos x
nl ( x) ≡ −(− x) l 
,

 x dx  x
Spherical Bessel function:
l
 1 d  sin x
jl ( x) ≡ (− x) l 
,

 x dx  x
sin x
j0 ( x ) =
,
x
sin x cos x
j1 ( x) = 2 −
,L
x
x
Spherical Neumann function
l
l  1 d  cos x
nl ( x) ≡ −(− x) 
,

 x dx  x
cos x
,
n0 ( x) = −
x
cos x sin x
n1 ( x) = 2 −
,L
x
x
The asymptotic properties of two functions :
lim j0 ( x) = 1,
x →0
lim j1 ( x) =
x →0
x
,L
3
lim x → 0
1
lim n0 ( x) = − ,
x →0
x
1
lim n1 ( x) = − 2 , L
x →0
x
Generally, for small x, we have
xl
lim jl ( x) =
x →0
(2l + 1)!!
(
2l − 1)!!
lim n ( x) = −
.
x →0
l
Proof: For small x,
x3 x5 x7
sin x ≈ x −
+
−
+L
3! 5! 7!
x2 x4 x6
cos x ≈ 1 −
+
−
+L
2! 4! 6!
x l +1
As when x0, Neumann functions blow up, that is
(
2l − 1)!!
lim n ( x) = −
→ ∞.
x →0
in the general solution,
and hence
l
x l +1
u (r ) = A ⋅ rjl (kr ) + B ⋅ rnl (kr ), we must set B=0,
u (r ) = A ⋅ rjl (kr ) ⇒ R(r ) = A ⋅ jl (kr ).
The boundary condition then requires that R(a)=0. Evidently k must be chosen
such that
jl (ka) = 0;
that 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
jl ( β nl ) = 0,
the boundary condition requires that
1
k = β nl ,
a
where βnl is the nth zero of the lth spherical Bessel function. The allowed
energies, then, are given by
h2
2
Enl =
β
nl ,
2
2ma
(n = 1,2,3,L),
and the wave functions are
m
ψ nlm ( r , θ , φ ) = Rnl ( r )Yl (θ , φ ) = Anl jl (
β nl
a
r )Yl m (θ , φ ),
with the constant Anl to be determined by normalization. Each energy level
is (2l+1)-fold degenerate, since there are different values of m for each
value of l.
4.2 The Hydrogen Atom
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.
−e
r
From Coulomb’s law, the potential
energy (in SI units) is
e
e2 1
V (r ) = −
.
4πε 0 r
Then the radial equation for hydrogen atom says
h 2 d 2u  e 2 1 h 2 l (l + 1) 
−
+ −
+
 u = Eu.
2
2
2m dr
 4πε 0 r 2m r 
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.
e2 1
Incidentally, Coulomb potential, V ( r ) = −
, admits two different states,
4πε 0 r
continuous states and bound states, which are separately corresponds to the
following situations:
−e
r
−e
e
continuous states E>0,
describing electron-proton
scattering
e
bound states E<0,
representing Hydrogen atom
4.2.1 The Radial Wave Function
1. Radial Solution:
The radial equation for Hydrogen atom is
h 2 d 2u  e 2 1 h 2 l (l + 1) 
−
+ −
+
 u = Eu.
2
2
2m dr
 4πε 0 r 2m r 
(1) Simplify it (tidy up):
As E<0, then we let
κ≡
− 2mE
.
h
Dividing above equation by E, we have

d 2u
me 2
1
l (l + 1) 
u.
= 1 −
+
2
2 
2
d (κr )
 2πε 0 h κ (κr ) (κr ) 
This suggests that we introduce
(E<0)
ρ ≡ κr ,
So that
me 2
and ρ 0 ≡
2πε 0 h 2κ
d 2u  ρ 0 l (l + 1) 
= 1 −
+
u.

2
2
ρ
ρ 
dρ

(2) The asymptotic properties of the solution:
a) ρ → ∞
In this case, the constant term in the bracket of above equation dominates, so
(approximately)
2
d u
= u.
2
dρ
The general solution of it is
u = Ae − ρ + Be ρ ,
but the second term e ρ → ∞ blows up as ρ → ∞ , so B=0. Evidently,
u ( ρ >> 1) ≈ Ae − ρ ,
for large
ρ
.
b) ρ → 0
In this case, the centrifugal term dominates; approximately, then:
d u  ρ 0 l (l + 1) 
= 1 −
+
u

2
2
dρ
ρ
ρ 

2
d 2u l (l + 1)
=
u
2
2
dρ
ρ
The general solution of it is
u = Cρ l +1 + Dρ − l ,
du
= C (l + 1) ρ l − Dlρ −(l +1)
dρ
d 2u
l −1
− (l + 2 )
=
Cl
(
l
+
1
)
ρ
+
Dl
(
l
+
1
)
ρ
dρ 2
But for ρ→0, the term ρ-l blows up, so D=0. Thus
u ( ρ << 1) ≈ Cρ l +1 ,
for small ρ .
(3) Introduce new function v(ρ) to simplify solution:
The next step is to peel off the asymptotic behavior, introducing the new
function v(ρ):
u ( ρ ) = ρ l +1e − ρ v ( ρ ),
in the hope that v(ρ) will turn out to be simpler than u(ρ). Then
du
l −ρ
l +1 − ρ
l +1 − ρ dv
= (l + 1)ρ e v ( ρ ) − ρ e v ( ρ ) + ρ e
dρ
dρ
dv 
l −ρ 
= ρ e (l + 1 − ρ )v + ρ

d
ρ


2



(
+
1
)
d 2u
l
l
dv
d
v
l −ρ
 + 2 (l + 1 − ρ )
= ρ e  − 2l − 2 + ρ +
+ρ
2
2 
dρ
ρ
d
ρ
d
ρ



In terms of v(ρ), then, the radial equation of u(ρ) reads
d 2u  ρ 0 l (l + 1) 
= 1 −
+
u

2
2
dρ
ρ
ρ 

d 2v
dv
(
)
ρ
+
2
l
+
1
−
ρ
+ [ρ 0 − 2(l + 1)]v = 0 .
2
dρ
dρ
[4.61]
(4) Solve above equation by power series method:
Finally, we assume the solution, v(ρ), can be expressed as a power series in ρ:
∞
v( ρ ) = ∑ c j ρ j .
j =0
Now replace v(ρ) into equation and our problem is to determine the coefficients
of the series, c1,c2,c3, … . Differentiating term by term:
∞
∞

dv
d  ∞
 ∑ c j ρ j  = ∑ c j jρ j −1 = ∑ ( j + 1)c j +1 ρ j .
=

d ρ d ρ  j = 0
j =0
 j =0
Differentiating again,
∞
d 2v
j −1
(
)
=
j
j
+
1
c
ρ
.
∑
j +1
2
dρ
j =0
Inserting these into Equation 4.61, we have
d 2v
dv
ρ
+ 2 (l + 1 − ρ )
+ [ρ 0 − 2(l + 1)]v = 0 .
2
dρ
dρ
∞
∞
ρ ∑ j ( j + 1)c j +1 ρ
j =0
j −1
j
(
)
j
+
1
c
ρ
+ 2(l + 1 − ρ ) ∑
j +1
j =0
∞
j
+ [ρ 0 − 2(l + 1)] ∑ c j ρ = 0
j =0
∞
∑
∞
j ( j + 1)c j +1 ρ j + 2 (l + 1) ∑ ( j + 1)c j +1 ρ
j
j =0
j =0
∞
− 2∑
∞
j
c
ρ
=0
jc j ρ + [ρ 0 − 2(l + 1)] ∑ j
j
j =0
j =0
∞
where
− 2 ∑ ( j + 1)c j +1 ρ
∞
j +1
= − 2 ∑ jc j ρ j
j =0
=0
j =0
=0
Equating the coefficient of like powers yields
j ( j + 1)c j +1 + 2 (l + 1) ( j + 1)c j +1 − 2 jc j + [ρ 0 − 2 (l + 1) ]c j = 0,
or:
c j +1
2 ( j + l + 1) − ρ 0
=
c j.
( j + 1)( j + 2l + 2)
This recursion formula determines the coefficients, and hence the function
v(ρ) : 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(ρ)
and u(ρ) will be got.
∞
u(ρ ) = ρ
l +1 − ρ
e v( ρ ) = ρ
l +1 − ρ
e
j
c
ρ
∑ j .
j =0
2. Energies of the solutions:
If j is very large, that is j→∞, the recursion formula says
c j +1
Then
2 ( j + l + 1) − ρ 0
2j
2
=
cj ≅
cj =
c j.
( j + 1)( j + 2l + 2)
( j + 1) j
j +1
2j
2 2
2
c0 ,
c j ≅ c j −1 ≅
c j −2 ⇒ c j ≅
j!
j
j j −1
so
∞
∞
j =0
j =0
v ( ρ ) = ∑ c j ρ j ≅ c0 ∑
2j j
ρ = c0 e 2 ρ ,
j!
and hence
u ( ρ ) = ρ l +1e − ρ c0 e 2 ρ = c0 ρ l +1e ρ ,
which blows up at large ρ→∞ 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
c jmax +1 = 0 .
Evidently, from recursion formula
c j +1
2 ( j + l + 1) − ρ 0
=
cj,
( j + 1)( j + 2l + 2)
we get
c jmax +1
2 ( j max + l + 1) − ρ 0
=
cj
( j + 1)( j + 2l + 2)
2 ( j max + l + 1) − ρ 0 = 0 .
Defining
we have
n ≡ j max + l + 1
, which is the so-called principle quantum number,
ρ 0 = 2 n.
− 2mE
κ≡
h
me 2
ρ0 ≡
2πε 0 h 2κ
h 2κ 2
me 4
E=−
=− 2 2 2 2
2m
8π ε 0 h ρ 0
so the allowed energies are
 m
En = −  2
 2 h
 e

 4πε 0
2



2
 1
1
 2 = 2 E1 ,
n
 n
n = 1, 2,3 L
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.
And we also find that
me 2
ρ0 ≡
= 2n
2
2πε 0 h κ
where
 me 2
κ = 
2
4
πε
h
0

4πε 0 h 2
−10
a=
=
0
.
529
×
10
m
2
me
1
1
 =
,
 n an
is the so-called Bohr radius.
It follows that
r
ρ ≡ κr =
.
an
3. The overall solutions of Hydrogen atom:
Finally, the spatial wave functions of hydrogen are labeled by three quantum
numbers (n,l, and m):
ψ nlm ( r , θ , φ ) = Rnl ( r )Yl m (θ , φ ),
where
u 1 l +1 − ρ
1 l +1 − ρ jmax
Rnl (r ) = = ρ e v( ρ ) = ρ e ∑ c j ρ j ,
r r
r
j =0
and v(ρ) is a polynomial of degree jmax= n-l-1 in ρ, whose coefficients are
determined by the recursion formula
c j +1
2( j + l + 1 − n )
=
c j.
( j + 1)( j + 2l + 2)
r
ρ=
.
an
(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
 m
E1 = −  2
 2h
 e

 4πε 0
2



2

 = − 13 .6 eV .

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
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
ψ 100 ( r , θ , φ ) = R10 ( r )Y00 (θ , φ ).
1 −ρ
1 l +1 − ρ jmax
j
Rnl (r ) = ρ e ∑ c j ρ ⇒ R10 (r ) = ρe c0
r
r
j =0
c0 − r / a
R10 (r ) = e
a
r
r
=
ρ=
an a
Normalizing R10 by
∫
∞
0
R10
2
2
|
c
|
r 2 dr = 02
a
∫
we have
R10 (r ) =
∞
0
2
3
|
c
|
a
−2r / a 2
2 a
0
e
r dr = 2
=| c0 |
= 1,
a
4
4
2
c0 =
.
a
2
a3
−
3
2 −r / a
e − r / a = 2a e
Meanwhile, Y00= 1/ 4π , and hence the ground state of hydrogen is
ψ 100 ( r , θ , φ ) = R10 ( r )Y00 (θ , φ ) =
1
πa 3
e −r / a .
(2) The first excited states n=2:
If n=2 the energy is
13 .6 eV
E1
E2 = 2 = −
= − 3.4 eV .
2
4
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
jmax = n − l − 1 = 2 − 0 − 1 = 1
j=0
2 (0 + 0 + 1 − 2 )
c1 =
c0 = − c0 ;
(0 + 1)(0 + 0 + 2)
j =1
2 (1 + 0 + 1 − 2 )
c2 =
c1 = 0;
(1 + 1)(1 + 0 + 2)
so
jmax =1
v( ρ ) =
j
c
ρ
∑ j = c0 − c0 ρ = c0 (1 − ρ ),
j =0
Normalization
and therefore
c0
1 −ρ
r −r / 2a
R20 (r ) = ρe v( ρ ) =
(1 − )e
r
2a
2a
If l=1, j max = n − l − 1 = 2 − 1 − 1 = 0 the recursion formula terminates the
series after a single term;
0
v( ρ ) = ∑ c j ρ j = c0 ,
j =0
Normalization
and we find
1 l +1 − ρ
Rnl (r ) = ρ e v(ρ )
r
R21 (r ) =
c0
−r / 2 a
re
4a 2
(3) The excited states for arbitrary n:
For arbitrary n, the possible values of l are
l = 0, 1, 2, ... , n − 1,
and for each l there are (2l+1) possible values of m, so the total degeneracy
of the energy level En is
n −1
d (n) = ∑ (2l + 1) = n 2 .
l=
=0
0
The polynomial v(ρ) (defined by the recursion formula Eq.4.76 is a function
well known to applied mathematicians; apart from normalization, it can be
written as
∞
v( ρ ) = ∑ c j ρ j = L2nl−+l1−1 (2 ρ ),
j =0
where
p
p d 
p
Lq − p ( x) ≡ (− 1)   Lq ( x)
 dx 
is the associated Laguerre polynomial, and
q
 d  −x q
Lq ( x) ≡ e   e x
 dx 
x
(
L0 = 1
)
is the qth Laguerre polynomial.
L1 = − x + 1
L2 = x 2 − 4 x + 1
L
Therefore the radial wave function is
1 l +1 − ρ 2l +1
Rnl (r ) = ρ e ⋅ L( n +l ) −( 2l +1) (2 ρ )
r
l
Examples:
1  r  − r / na 2l +1  2r 
⋅ Ln −l −1  
=
 e
na  na 
 na 
1
 2r  1
R10 (r ) = e − r / a ⋅ L10   = e − r / a ⋅ ( N10 ×1)
a
 a  a
r
ρ=
na
L10 = 1
Normalization
R20 =
1 −r / 2a 1  r  1 −r / 2a
1
r
⋅ L1   =
⋅ N 20 (1 − )
e
e
2a
4
2a
 a  2a
r
r
Normalization L   = 4 − 2
a
a
1
1
Generally, we can normalize Rnl as
∫
∞
0
2
Rnl ( r ) r 2 dr = 1
to give normalized Rnl as follows
l
Rnl (r ) = N nl ⋅ e
− r / na
 2r  2l +1
  Ln −l −1 (2r / na)
 na 
with normalization constant Nnl being
 2 
N nl =  
 na 
3/ 2
1/ 2
 (n − l − 1)! 


 2n[(n + l )!]3 


Then, finally, the normalized hydrogen wave function are
l
ψ nlm (r ,θ , φ ) = N nl ⋅ e
− r / na
 2r  2l +1
m
  Ln −l −1 (2r / na) ⋅ Yl (θ , φ )
 na 
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
*
2
ψ
ψ
r
∫ nlm n′l′m′ sin θdrdθdφ = δ nn′δ ll′δ mm′
Visualizing the hydrogen wave functions is not easy. See book!
See the figures of solutions of
Hydrogen atom
4.2.2 The Spectrum of Hydrogen
In principle, if you put a hydrogen atom into some stationary stateψnlm, 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.
In practice such perturbations are always present; transitions (or, as they are
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:
Ei
Ef
E2
E1
1
E n = 2 E1 ,
n
E1 = − 13 .6 eV .
 1

1
Eγ = E i − E f = − 13 .6 2 − 2  ⋅ eV
n

n
f 
 i
Now according to the Planck formula, the energy of a photon is proportional
to its frequency:
Eγ = h ν ,
c
Meanwhile, the wavelength is given by
λ=
 1

1
Eγ = − E1  2 − 2  = h ν
n

n
i
f


E1  1
1 
ν
1
−
− 2 =h =h
2

c  ni n f 
c
λ
E1  1
1 
=
− 2 =
2

λ hc  n f ni 
1
ν
, so
 1

1
R 2 − 2 
n

n
f
i


where
E1
m
R=
=
hc 4πch 3
 e

 4πε 0
2
2

 = 1 .097 × 10 7 m −1

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.
Spectrum of Hydrogen:
Transitions to the ground state (nf=1)
lie in the ultraviolet; they are known
to spectroscopists as the Lyman series.
莱曼系
Transitions to the first excited state
(nf=2) fall in the visible region; they
constitute Balmer series.
巴尔末系
Transitions to nf=3 (Paschen series)
are in the infrared region; and so on.
帕刑系
OXYGEN
HYDROGEN
prism
diffraction grating
These spectral lines are produces by
"exciting" gas atoms and molecules with
high voltage (about 5000 volts). This
energy kicks electrons to higher energy
levels where they are unstable and drop
back towards the ground state (lower
energy levels). As the electrons make this
downward transition, they release energy
in the form of visible light.
The emission and absorption spectrum of hydrogen in the
visible range is the following
End