SCPY152 Lecture 22 Radial Functions in Central Potential Problems

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SCPY152
Lecture 22 Radial Functions in Central
Potential Problems
Udom Robkob
Department Physics, MUSC
March 18, 2015
Udom RobkobDepartment Physics, MUSC
SCPY152Lecture 22 Radial Functions in Central Pote
Today Topics
Radial Schrodinger’s equation
Free particle in spherical coordinates
Rigid potential sphere
Isotropic harmonic oscillator
Coulomb potential well
Radial Schrodinger’s Equation
From the quantized angular momentum operator
L2 Ylm = ~2 l(l + 1)Ylm , we have radial Schrodinger’s
equation in the form;
2m
~2 l(l + 1)
1 d
2 dREl
r
+ 2 E − V (r) −
REl = 0.
r2 dr
dr
~
2mr2
(1)
This equation will be solved for various potential V (r) in
the subsequence slides.
Free Particle
For a case of free particle, V (r) = 0, then we have radial
equation in the form
2mE l(l + 1)
1 d
2 dREl
r
+
−
REl = 0
r2 dr
dr
~2
r2
d2 REl 2 dREl
l(l + 1)
+
+ −
REl = 0. (2)
dr2
r dr
r2
Let us define =
ρ2
2mE
~2
= k 2 and ρ = kr, then we have
dRkl
dRkL
+ ρ2 − l(l + 1) Rkl = 0
+ 2ρ
2
dρ
dρ
(3)
where REl → Rkl (ρ).
Free Particle
This equation in known as sherical Bessel differential
equation.
Its solution is spherical Bessel functions of the first kind,
jl (kr), and the second kind, nl (kr),so that our solution will
be
Rkl (kr) = C1 jl (kr) + C2 nl (kr)
(4)
Only the first kind functions are finite at origin, r = 0,
hence we set C2 = 0 to eliminate the second kind function.
Rigid Potential Sphere
The potential function of rigid potential sphere is
0, 0 ≤ r < a
V (r) =
∞,
r>a
(5)
Its solution inside sphere is
ϕ(r, θ, φ) = Rkl (r)Ylm (θ, φ)
Rkl (r) = Cjl (kr)
~2 k 2
Ek =
2m
(6)
(7)
(8)
A particle is free inside the sphere of radius a and vanish at
boundary;
0 = Rkl (a) = Cjl (ka) = 0, → ka = zn,l
(9)
where zn,l is the zero point of jl (ka).
Rigid Potential Sphere
Where is the zero points?
The quantized energies will be
En,l =
2
~2 zn,l
2ma2
(10)
Isotropic Harmonic Oscillator
The corresponding radial equation is
1 d
2 dRE,l (r)
r
r2 dr
dr
2
2m
1
~ l(l + 1)
+ 2 E − mω 2 r2 −
RE,l (r) = 0
(11)
~
2
2mr2
√
Let us define = 2E/~ω, α = mω/~, and ρ = αr, then
we have
d2 R,l (ρ) 2 dR,l (ρ)
+
+
dρ2
ρ dρ
l(l + 1)
− ρ2 −
R,l (ρ) = 0
ρ2
(12)
This equation will have solution when
= 4n + 2l + 3, n = 0, 1, 2, ...
(13)
Isotropic Harmonic Oscillator
Then we have the quantized energies
3
~ω
En,l = 2n + l +
2
(14)
List of these quantized energies (in unit of ~ω),
n
0
0
0
1
0
1
l En,l (~ω)
0
3/2
1
5/2
2
7/2
0
7/2
3
9/2
1
9/2
(15)
Note the appearance of degenerated excited states.
Coulomb Potential Well
Coulomb potential function is
k
Ze2
V (r) = − , k =
r
4π0
(16)
Radial equation becomes
d2 REl 2 dREl 2m
k ~2 l(l + 1)
+
+ 2 E+ −
REl = 0 (17)
dr2
r dr
~
r
2mr2
We define ”effective potential” as
k ~2 l(l + 1)
Vef f (r) = − +
r
2mr2
(18)
In case of l = 1, we have
Bounded states occur at negative energy.
We define the following parameters
2 = −
~
8mE
4π0 ~2
=
,
a
=
0
2
2
~
e m
mcα
where a0 = 0.529˚
A and α = 1/137 is fine structure
constant.
Next we define new variable ρ = r, then radial equation
becomes
d2 Rl (ρ) 2 dRl (ρ)
1
2
l(l + 1)
+
−
+
+
Rl (ρ) = 0
dρ2
ρ dρ
4 a0 ρ
ρ2
(19)
This equation will have solution when
2
= n, n = 1, 2, 3, ..., and l = 0, 1, 2, .., n − 1
a0
(20)
Then we get the quantized energies in the form
a0 =
2
4
8mE
4
→ 2 a20 = 2 → − 2 a20 = 2
n
n
~
n
or
E → En = −
1
~2 1
1
= − mc2 α2 2
2
n
2ma20 n2
(21)
In case of Hydrogen atom. m = me and me c2 = 0.511M eV ,
then we have
13.6eV
En = −
(22)
n2
We have three quantum numbers;
n = 1, 2, 3, ...; principle quantum numbers
l = 0, 1, 2, ..., n − 1; angular momentum quantum number
m = −l, −(l − 1), ..., (l − 1), l; magnetic quantum numbers
Back to the radial function, Rnl (r), we have
Graphs of the radial function, Rnl (r);
Graphs of the radial distribution function, |Rnl (r)|2 r2 ;
A full expression of sate function of particle waves in
Columonb potential well will be
ϕnlm (r, θ, φ) = Rnl (r)Ylm (θ, φ)
(23)
Exercise - give expressions of ϕ100 and ϕ210 for a case of
Z = 1.

SCPY152 Lecture 22 Radial Functions in Central Potential Problems

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