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Spherical Coordinates Review Schroedinger's equation:

2 2 
∂Ψ( r ,t)


−
∇ Ψ( r ,t) + V ( r )Ψ( r ,t) = i
2m
∂t
Space
Time
2M

∇ ψ + 2 E − V (r ) ψ = 0

2
(
)
T=
E
−i t
Ae 
Assume spherical symmetry:
⎡1 ∂ ⎛ 2 ∂⎞
1
∂ ⎛
∂⎞
1
∂2 ⎤
r
+ 2
sin θ ⎟ + 2 2
ψ + k 2 (r)ψ = 0
⎢ 2
2 ⎥
⎜
⎟
⎜
∂θ ⎠ r sin θ ∂φ ⎦
⎣ r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⎝
k 2 (r) =
2M
E − V (r)
2

(
)
Separation of variables.
For spherical symmetry, the solution is separable in spherical coordinates:
(
)
() () ()
ψ r,θ , φ = R r P θ F φ
⎡1 ∂⎛ 2 ∂⎞
⎤
1
∂ ⎛
∂⎞
1
∂2
2
r
+ 2
sin θ ⎟ + 2 2
+ k (r) ⎥ R r P(θ )F(φ ) = 0
⎢ 2
⎜
⎟
⎜
2
∂r
∂r
∂
θ
∂
θ
⎝
⎠
⎝
⎠
r sin θ
r sin θ ∂φ
⎣r
⎦
()
Equations.
Boundary conditions.
∂ ⎛ 2 ∂R ⎞
2 2
r
+
r
k (r) − l(l + 1) R = 0
∂r ⎜⎝ ∂r ⎟⎠
(
)
l = 0, 1, 2 i i i
1 ∂ ⎛
∂P ⎞
m2
sin θ ⎟ − 2 P + l(l + 1)P = 0
⎜
sin θ ∂θ ⎝
∂θ ⎠ sin θ
∂2 F
2
+
m
F =0
2
∂φ
m = 0, ±1, ±2 i i i ± l − 1
imφ
ψ (r, θ , φ ) = ∑ Rl (r)Plm (cos θ )e
lm
= ∑ Rl (r)Ylm (θ , φ )
lm
l
m Ylm
0
0
Y00 = (1 / 4π )
1
0
Y10 = ( 3 / 4π )
1
±1
2
0
2
±1
Y2±1 =  (15 / 16π )
2
±2
Y2±2 =  (15 / 32π )
⎧⎪ l=0,1,2...
⎨
⎪⎩ m=0,±1,±2...±l
1/2
1/2
cos θ
Y1±1 =  ( 3 / 8π )
sin θ e±iφ
1/2
Y20 = ( 5 / 16π )
1/2
( 3cos θ − 1)
2
1/2
1/2
sin θ cos θ e±iφ
sin 2 θ e±i2φ
This will also hold in general for any quantum mechanical problem where the potential is
spherical symmetry, so long as the potential energy depends only on r, for example an atom.
This is the reason why it is so important for atomic, nuclear and particle physics.
Meaning of the quantum numbers l and m
Orbital angular momentum:
  
L = r × p.
2
L = l(l + 1) 2
z componant of orbital angular momentum: Lz =
Lz
Lz = m 

L = l ( l + 1) 
Ly
Lx
 
r × pz
= m
Meaning of the quantum numbers l and m
Orbital angular momentum:
  
L = r × p.
2
L = l(l + 1) 2
z componant of orbital angular momentum: Lz =
Lz
 
r × pz
Lz = m 

L = l ( l + 1) 
Ly
Lx
= m
Angular Probability Functions.
s states:
p states:
d states:
f states:
4 quantum numbers
n : radial quantum number (energy)
l : orbital angular momentum, l = 0 (s), l = 1( p), l = 2 (d),
ml : "magnetic" ml ≤ l. eg. l = 1: ml = −1, 0, + 1
ms : "spin" ms ≤ s. eg. s = 1 / 2 : ms = −1 / 2 , + 1 / 2
l = 3 ( f ),
l = 4 (g) • • •
Solution for R(r) depends on exact form of V (r).
Free particle: V (r) = 0
l
⎛ 1 d ⎞ ⎛ sin x ⎞
jl (x) = x l ⎜ −
⎝ x dx ⎟⎠ ⎜⎝ x ⎟⎠
where x ≡ kr.
jl (x) are called spherical Bessel functions.
First few Spherical Bessel Functions.
( )
j0 kr =
( )
sin kr
kr
cos ( kr )
(
)
j ( kr ) =
−
kr
( kr )
sin kr
1
2
( )
j2 kr =
( ) − 3cos ( kr ) − sin ( kr )
kr
( kr )
( kr )
3sin kr
3
2
For bound system, boundry connditions put constraints on k → kn .
Rl (kr) → Rnl (kr)
Homework. Due Friday, Nov. 6
Show that ψ 00 = R0Y00 and ψ 11 = R1Y11 are solutions to the Schroedinger equation.
⎡1 ∂⎛ 2 ∂⎞
1
∂ ⎛
∂⎞
1
∂2 ⎤
2
r
+
sin
θ
+
ψ
(r,
θ
,
φ
)
+
k
ψ (r,θ , φ ) = 0
⎢ 2
⎥
⎜
⎟
⎜
⎟
2
2
2
2
∂θ ⎠ r sin θ ∂φ ⎦
⎣ r ∂r ⎝ ∂r ⎠ r sin θ ∂θ ⎝
Par5cle in a spherical box with rigid walls. Boundary Condition:
R(r) → R(a) → 0
( )
jnl kn a = 0
First few Spherical Bessel Functions.
( )
jn0 ka =
( )
sin ka
k10 a = π
ka
( ) − cos ( ka )
( )
ka
( ka )
jn1 ka =
( )
jn 2 ka =
sin ka
2
( ) − 3cos ( ka ) − sin ( ka )
ka
( ka )
( ka )
3sin ka
3
2
k20 a = 2π
k30 a = 3π
k11 a=4.49
k21 a = 7.73
k31 a=10.90
k12 a=5.76
k22 a=9.10
k32 a=12.32
Zeros of Spherical Bessel Func5ons n
1
l 2
3
4
0
3.14
6.28
9.43
12.57
1
4.49
7.73
10.90
14.07
2
5.76
9.10
12.32
15.57
3
6.99
10.42
13.70
16.92
4
8.18
11.71
15.04
18.30
Applica5on to Nuclear Physics In the extreme nuclear shell model the nucleons are moving freely in a spherical nucleus with infinitely rigid walls at a radius a. Exercise:
Calculate the energies of the three lowest quantum states, 1s, 1p,1d, 2s
of a neutron moving in such a box of radius 5 fm.
hc = 1240 eV-nm = 1240 × 10 −6 MeV-nm = 1240 MeV-fm
1240
c =
MeV-fm =197.3 MeV-fm mM c 2 ≈ 939 MeV
2π
 2 c 2 ( ka )
p2
2 k 2
2 c2 k 2
E=
=
=
=
2m
2m
2mc 2
2mc 2 a 2
2
hc = 1240 eV-nm = 1240 × 10 −6 MeV-nm = 1240 MeV-fm
1240
c =
MeV-fm =197.3 MeV-fm
2π
k10 a = π
k11a=4.493
k12 a=5.76
k20 a = 2π
(197.3) MeV 2 -fm 2
2 c2
=
= 0.8 MeV
2mc 2 a 2 2 ( 939 ) 5 2 MeV-fm 2
E10 = 0.8π 2 = 8 MeV
(
E11 = 0.8 4.49
(
2
E12 = 0.8 5.76
(
) = 16 MeV
2
) = 26 MeV
)
E12 = 0.8 2 2 π 2 = 32 MeV
2
( )
32 MeV 26 MeV 16 MeV 8 MeV Shell structure of protons in a spherical box.
2s shell - 40Ca
n = 2 l = 0 ml = 0
ms = ±1 / 2
1d shell - 36Ar
n = 1 l = 2 ml = −2, − 1, 0, + 1, + 2
1p shell - 16O
n = 1 l = 1 ml = +1 0 − 1
1s shell - 4He
n = 1 l = 0 ml = 0
ms = ±1 / 2
ms = ±1 / 2
ms = ±1 / 2
Nucleons (protons and neutrons) obey the Pauli exclusion principle. Protons and neutrons fill shells seperately because they are not iden5cal par5cles.