Notice that the angular part of the wave function, Y ,
, is the same for all spherically symmetric potentials; the actual shape of the potential,
Vr , affects
only the radial part of the wave function, Rr
, 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
2
− =
u r
dr du
r dr
dR ,
2 2
2
dr u
d r
dr dR
r dr
d =
and hence
. 1
2 2
2 2
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
2 2
r l
l m
r V
V
eff
+ +
=
2 2
1 2
r l
l m
+
, 1
2 2
=
∞
dr r
R .
1
2
=
∞
dr u
The
infinite spherical well
:
Find the wave function and the allowed energies.
Solution: 1. Outside the well, the wave function is
zero: ur,r=a or ra=0 .
∞ =
. if
, ,
if ,
a r
a r
r V
a
∞
1. Outside the well, the wave function is zero: ur,r=a or ra=0
.
, 1
2 2
2 2
u k
r l
l dr
u d
− +
=
2. Inside the well, the radial equation reads
where
. 2mE
k =
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
. cos
sin r
kr B
r kr
A r
R +
=
As the second term blows up, so we must choose B=0
.
→ r
r kr
A r
R sin
=
The boundary condition then requires that
sin sin
= =
= ka
r ka
A a
R ,
3 ,
2 ,
1 ,
= =
n n
ka
π
The allowed energies are evidently
mE k
2 =
, ,
3 ,
2 ,
1 ,
2
2 2
2 2
= =
n ma
n E
n
π
which is the same for the one-dimensional infinite square well.
, 3
, 2
, 1
, =
= n
a n
k
n
π
The normalization condition:
, 1
2
=
∞
dr u
; 2
a A
=
yields
Tacking on the angular part l=0,m=0
, 4
1 cos
4 1
,
π θ
π φ
θ
= =
P Y
r r
k a
r R
n n
sin 2
=
a n
k
n
π
=
we conclude that
, ,
,
φ θ
φ θ
ψ
m l
n nlm
Y r
R r
=
r a
r n
a Y
R
n n
sin 2
1
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 n and l:E
nl
.
2 The case
l is in any integer:
The general solution of above equation is:
. 1
2 2
2 2
u k
r l
l dr
u d
− +
=
, kr
rn B
kr rj
A r
u
l l
⋅ +
⋅ =
where j
l
kr is the spherical
Bessel function of order
l , and
n
l
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
l
− −
≡
Spherical Bessel function:
, sin
1 x
x dx
d x
x x
j
l l
l
− ≡
, sin
x x
x j
=
, cos
sin
2 1
x x
x x
x j
− =
, cos
1 x
x dx
d x
x x
n
l l
l
− −
≡
Spherical Neumann function
, cos
x x
x n
− =
, sin
cos
2 1
x x
x x
x n
− =
The asymptotic properties of two functions :
lim
→ x
Generally, for small x, we have
, 1
lim =
→
x j
x
, 3
lim
1
x x
j
x
=
→
, 1
lim x
x n
x
− =
→
, 1
lim
2 1
x x
n
x
− =
→
1 2
lim +
=
→
l x
x j
l l
x
. 1
2 lim
1 +
→
− −
=
l l
x
x l
x n
Proof: For small x,
+ −
+ −
≈ 7
5 3
sin
7 5
3
x x
x x
x
+ −
+ −
≈ 6
4 2
1 cos
6 4
2
x x
x x
in the general solution,
, kr
rn B
kr rj
A r
u
l l
⋅ +
⋅ =
. 1
2 lim
1
∞ →
− −
=
+ →
l l
x
x l
x n
As when x 0
, Neumann functions
blow up, that is
we must set B=0
, and hence
. kr
j A
r R
kr rj
A r
u
l l
⋅ =
⋅ =
The boundary condition then requires that Ra=0
. Evidently k
must be chosen such that
; =
ka j
l
that is, ka
is a zero of the l
th-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
, =
nl l
j
β
where
nl
is the n
th zero of the l
th 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
nl nl
β
and the wave functions are
, ,
, ,
,
φ θ
β φ
θ φ
θ ψ
m l
nl l
nl m
l nl
nlm
Y r
a j
A Y
r R
r =
=
with the constant A
nl
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
.
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
