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Today’s Outline - October 21, 2013
• Hydrogen atom solution
• Hydrogen atom wavefunctions
Reading Assignment: Chapter 4.3
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Wednesday, October 30, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 1 / 17
Today’s Outline - October 21, 2013
• Hydrogen atom solution
• Hydrogen atom wavefunctions
Reading Assignment: Chapter 4.3
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Wednesday, October 30, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 1 / 17
Today’s Outline - October 21, 2013
• Hydrogen atom solution
• Hydrogen atom wavefunctions
Reading Assignment: Chapter 4.3
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Wednesday, October 30, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 1 / 17
Today’s Outline - October 21, 2013
• Hydrogen atom solution
• Hydrogen atom wavefunctions
Reading Assignment: Chapter 4.3
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Wednesday, October 30, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 1 / 17
Today’s Outline - October 21, 2013
• Hydrogen atom solution
• Hydrogen atom wavefunctions
Reading Assignment: Chapter 4.3
Homework Assignment #08:Chapter 4: 10,13,14,15,16,38due Wednesday, October 30, 2013
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 1 / 17
Hydrogen atom
The potential of the hydrogen atomis simply the Coulomb potential,which is spherically symmetric
where we assume that the nucleus(proton) is stationary because it ismuch more massive than the elec-tron
V (r) = − e2
4πε0
1
r
we can substitute this potential intothe radial equation
dividing by Eand rearranging
rewriting it withcommon terms
− ~2
2m
d2u
dr2+
[− e2
4πε0
1
r+
~2
2m
l(l + 1)
r2
]u = Eu
− ~2
2mE
d2u
dr2=
[1 +
e2
4πε0E
1
r− ~2
2mE
l(l + 1)
r2
]u
− ~2
2mE
d2u
dr2=
[1 +
me2
2πε0~2~2
2mE
1
r− ~2
2mE
l(l + 1)
r2
]u
initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =
√−2mE/~
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 2 / 17
Hydrogen atom
The potential of the hydrogen atomis simply the Coulomb potential,which is spherically symmetric
where we assume that the nucleus(proton) is stationary because it ismuch more massive than the elec-tron
V (r) = − e2
4πε0
1
r
we can substitute this potential intothe radial equation
dividing by Eand rearranging
rewriting it withcommon terms
− ~2
2m
d2u
dr2+
[− e2
4πε0
1
r+
~2
2m
l(l + 1)
r2
]u = Eu
− ~2
2mE
d2u
dr2=
[1 +
e2
4πε0E
1
r− ~2
2mE
l(l + 1)
r2
]u
− ~2
2mE
d2u
dr2=
[1 +
me2
2πε0~2~2
2mE
1
r− ~2
2mE
l(l + 1)
r2
]u
initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =
√−2mE/~
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 2 / 17
Hydrogen atom
The potential of the hydrogen atomis simply the Coulomb potential,which is spherically symmetric
where we assume that the nucleus(proton) is stationary because it ismuch more massive than the elec-tron
V (r) = − e2
4πε0
1
r
we can substitute this potential intothe radial equation
dividing by Eand rearranging
rewriting it withcommon terms
− ~2
2m
d2u
dr2+
[− e2
4πε0
1
r+
~2
2m
l(l + 1)
r2
]u = Eu
− ~2
2mE
d2u
dr2=
[1 +
e2
4πε0E
1
r− ~2
2mE
l(l + 1)
r2
]u
− ~2
2mE
d2u
dr2=
[1 +
me2
2πε0~2~2
2mE
1
r− ~2
2mE
l(l + 1)
r2
]u
initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =
√−2mE/~
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 2 / 17
Hydrogen atom
The potential of the hydrogen atomis simply the Coulomb potential,which is spherically symmetric
where we assume that the nucleus(proton) is stationary because it ismuch more massive than the elec-tron
V (r) = − e2
4πε0
1
r
we can substitute this potential intothe radial equation
dividing by Eand rearranging
rewriting it withcommon terms
− ~2
2m
d2u
dr2+
[− e2
4πε0
1
r+
~2
2m
l(l + 1)
r2
]u = Eu
− ~2
2mE
d2u
dr2=
[1 +
e2
4πε0E
1
r− ~2
2mE
l(l + 1)
r2
]u
− ~2
2mE
d2u
dr2=
[1 +
me2
2πε0~2~2
2mE
1
r− ~2
2mE
l(l + 1)
r2
]u
initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =
√−2mE/~
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 2 / 17
Hydrogen atom
The potential of the hydrogen atomis simply the Coulomb potential,which is spherically symmetric
where we assume that the nucleus(proton) is stationary because it ismuch more massive than the elec-tron
V (r) = − e2
4πε0
1
r
we can substitute this potential intothe radial equation
dividing by Eand rearranging
rewriting it withcommon terms
− ~2
2m
d2u
dr2+
[− e2
4πε0
1
r+
~2
2m
l(l + 1)
r2
]u = Eu
− ~2
2mE
d2u
dr2=
[1 +
e2
4πε0E
1
r− ~2
2mE
l(l + 1)
r2
]u
− ~2
2mE
d2u
dr2=
[1 +
me2
2πε0~2~2
2mE
1
r− ~2
2mE
l(l + 1)
r2
]u
initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =
√−2mE/~
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 2 / 17
Hydrogen atom
The potential of the hydrogen atomis simply the Coulomb potential,which is spherically symmetric
where we assume that the nucleus(proton) is stationary because it ismuch more massive than the elec-tron
V (r) = − e2
4πε0
1
r
we can substitute this potential intothe radial equation
dividing by Eand rearranging
rewriting it withcommon terms
− ~2
2m
d2u
dr2+
[− e2
4πε0
1
r+
~2
2m
l(l + 1)
r2
]u = Eu
− ~2
2mE
d2u
dr2=
[1 +
e2
4πε0E
1
r− ~2
2mE
l(l + 1)
r2
]u
− ~2
2mE
d2u
dr2=
[1 +
me2
2πε0~2~2
2mE
1
r− ~2
2mE
l(l + 1)
r2
]u
initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =
√−2mE/~
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 2 / 17
Hydrogen atom
The potential of the hydrogen atomis simply the Coulomb potential,which is spherically symmetric
where we assume that the nucleus(proton) is stationary because it ismuch more massive than the elec-tron
V (r) = − e2
4πε0
1
r
we can substitute this potential intothe radial equation
dividing by Eand rearranging
rewriting it withcommon terms
− ~2
2m
d2u
dr2+
[− e2
4πε0
1
r+
~2
2m
l(l + 1)
r2
]u = Eu
− ~2
2mE
d2u
dr2=
[1 +
e2
4πε0E
1
r− ~2
2mE
l(l + 1)
r2
]u
− ~2
2mE
d2u
dr2=
[1 +
me2
2πε0~2~2
2mE
1
r− ~2
2mE
l(l + 1)
r2
]u
initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =
√−2mE/~
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 2 / 17
Hydrogen atom
The potential of the hydrogen atomis simply the Coulomb potential,which is spherically symmetric
where we assume that the nucleus(proton) is stationary because it ismuch more massive than the elec-tron
V (r) = − e2
4πε0
1
r
we can substitute this potential intothe radial equation
dividing by Eand rearranging
rewriting it withcommon terms
− ~2
2m
d2u
dr2+
[− e2
4πε0
1
r+
~2
2m
l(l + 1)
r2
]u = Eu
− ~2
2mE
d2u
dr2=
[1 +
e2
4πε0E
1
r− ~2
2mE
l(l + 1)
r2
]u
− ~2
2mE
d2u
dr2=
[1 +
me2
2πε0~2~2
2mE
1
r− ~2
2mE
l(l + 1)
r2
]u
initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =
√−2mE/~
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 2 / 17
Hydrogen atom
The potential of the hydrogen atomis simply the Coulomb potential,which is spherically symmetric
where we assume that the nucleus(proton) is stationary because it ismuch more massive than the elec-tron
V (r) = − e2
4πε0
1
r
we can substitute this potential intothe radial equation
dividing by Eand rearranging
rewriting it withcommon terms
− ~2
2m
d2u
dr2+
[− e2
4πε0
1
r+
~2
2m
l(l + 1)
r2
]u = Eu
− ~2
2mE
d2u
dr2=
[1 +
e2
4πε0E
1
r− ~2
2mE
l(l + 1)
r2
]u
− ~2
2mE
d2u
dr2=
[1 +
me2
2πε0~2~2
2mE
1
r− ~2
2mE
l(l + 1)
r2
]u
initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =
√−2mE/~
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 2 / 17
Hydrogen atom
The potential of the hydrogen atomis simply the Coulomb potential,which is spherically symmetric
where we assume that the nucleus(proton) is stationary because it ismuch more massive than the elec-tron
V (r) = − e2
4πε0
1
r
we can substitute this potential intothe radial equation
dividing by Eand rearranging
rewriting it withcommon terms
− ~2
2m
d2u
dr2+
[− e2
4πε0
1
r+
~2
2m
l(l + 1)
r2
]u = Eu
− ~2
2mE
d2u
dr2=
[1 +
e2
4πε0E
1
r− ~2
2mE
l(l + 1)
r2
]u
− ~2
2mE
d2u
dr2=
[1 +
me2
2πε0~2~2
2mE
1
r− ~2
2mE
l(l + 1)
r2
]u
initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =
√−2mE/~
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 2 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ +��Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ +��Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ +��Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ +��Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ +��Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ +��Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ +��Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ +��Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ + Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ +��Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
1
κ2d2u
dr2=
[1− me2
2πε0~2κ1
κr+
l(l + 1)
(κr)2
]u
d2u
dρ2=
[1− ρ0
ρ+
l(l + 1)
ρ2
]u
As ρ→∞, the constant term dominates
and the solution is of the form
but the second term is unbounded in thelimit of ρ→∞, thus B = 0
in the limit of ρ→ 0, the centrifugal termis dominant
if we substitute
ρ ≡ κr , ρ0 ≡me2
2πε0~2κ
just as with the harmonicoscillator, we start with theasymptotic solution and thengeneralize
d2u
dρ2≈ u
u(ρ) = Ae−ρ +��Beρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 3 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Hydrogen atom (cont.)
for ρ → 0 the solution satis-fies
this has a solution
this can be shown to satisfythe equation
but the second term blows upas ρ→ 0, so D = 0
d2u
dρ2=
l(l + 1)
ρ2u
u(ρ) = Cρl+1 +D
ρl
du
dρ= (l + 1)Cρl − l
D
ρ(l+1)
d2u
dρ2= l(l + 1)Cρl−1 + l(l + 1)
D
ρ(l+2)
u(ρ) ∼ Cρl+1
the solution we seek, including the asymptotic portions is thus
u(ρ) = ρl+1e−ρv(ρ)
where v(ρ) is a polynomial in ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 4 / 17
Asymptotic behavior
u(ρ) = ρl+1e−ρv(ρ)
The exponential term serves tolimit the asymptotic behavior of thewavefunction
It remains only to determine thepolynomial “wavy part” of the solu-tion, v(ρ). This is done in the sameway as was the analytical solutionof the harmonic oscillator.
0
1
2
3
4
5
0 2 4 6 8 10
u(ρ
)
ρ
l=0
l=1
l=2
l=3
e-ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 5 / 17
Asymptotic behavior
u(ρ) = ρl+1e−ρv(ρ)
The exponential term serves tolimit the asymptotic behavior of thewavefunction
It remains only to determine thepolynomial “wavy part” of the solu-tion, v(ρ). This is done in the sameway as was the analytical solutionof the harmonic oscillator.
0
1
2
3
4
5
0 2 4 6 8 10
u(ρ
)
ρ
l=0
l=1
l=2
l=3
e-ρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 5 / 17
Asymptotic behavior
u(ρ) = ρl+1e−ρv(ρ)
The exponential term serves tolimit the asymptotic behavior of thewavefunction
It remains only to determine thepolynomial “wavy part” of the solu-tion, v(ρ). This is done in the sameway as was the analytical solutionof the harmonic oscillator.
0
1
2
3
4
5
0 2 4 6 8 10
u(ρ
)
ρ
l=0
l=1
l=2
l=3
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 5 / 17
Asymptotic behavior
u(ρ) = ρl+1e−ρv(ρ)
The exponential term serves tolimit the asymptotic behavior of thewavefunction
It remains only to determine thepolynomial “wavy part” of the solu-tion, v(ρ). This is done in the sameway as was the analytical solutionof the harmonic oscillator.
0
1
2
3
4
5
0 2 4 6 8 10
u(ρ
)
ρ
l=0
l=1
l=2
l=3
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 5 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ=
(l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρdv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2=
lρl−1e−ρ[
(l + 1− ρ)v + ρdv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]
= ρle−ρ{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ=
(l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρdv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2=
lρl−1e−ρ[
(l + 1− ρ)v + ρdv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]
= ρle−ρ{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv
− ρl+1e−ρv + ρl+1e−ρdv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2=
lρl−1e−ρ[
(l + 1− ρ)v + ρdv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]
= ρle−ρ{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv
+ ρl+1e−ρdv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2=
lρl−1e−ρ[
(l + 1− ρ)v + ρdv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]
= ρle−ρ{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ
dv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2=
lρl−1e−ρ[
(l + 1− ρ)v + ρdv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]
= ρle−ρ{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ
dv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]
d2u
dρ2=
lρl−1e−ρ[
(l + 1− ρ)v + ρdv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]
= ρle−ρ{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ
dv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2=
lρl−1e−ρ[
(l + 1− ρ)v + ρdv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]= ρle−ρ
{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ
dv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2= lρl−1e−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]
− ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]= ρle−ρ
{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ
dv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2= lρl−1e−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]
+ ρle−ρ[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]= ρle−ρ
{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ
dv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2= lρl−1e−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]
= ρle−ρ{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ
dv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2= lρl−1e−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]= ρle−ρ
{
ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ
dv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2= lρl−1e−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]= ρle−ρ
{ρd2v
dρ2
+ 2(l + 1− ρ)dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ
dv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2= lρl−1e−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]= ρle−ρ
{ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ
+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The wavy part
The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)
u(ρ) = ρl+1e−ρv(ρ)
du
dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ
dv
dρ
= ρle−ρ[
(l + 1− ρ)v + ρdv
dρ
]d2u
dρ2= lρl−1e−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]− ρle−ρ
[(l + 1− ρ)v + ρ
dv
dρ
]+ ρle−ρ
[−v + (l + 1− ρ)
dv
dρ+
dv
dρ+ ρ
d2v
dρ2
]= ρle−ρ
{ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 6 / 17
The Schrodinger equation for v(ρ)
Substituting into the Schrodinger equation for u(ρ)
0 =d2u
dρ2−[
1− ρ0ρ
+l(l + 1)
ρ2
]u
= ρle−ρ{ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}− ρle−ρ
[ρ− ρ0 +
l(l + 1)
ρ
]v
= ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
we will solve this in the same way as the harmonic oscillator, assumingthat v(ρ) is an infinite polynomial in ρ
v(ρ) =∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 7 / 17
The Schrodinger equation for v(ρ)
Substituting into the Schrodinger equation for u(ρ)
0 =d2u
dρ2−[
1− ρ0ρ
+l(l + 1)
ρ2
]u
= ρle−ρ{ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}− ρle−ρ
[ρ− ρ0 +
l(l + 1)
ρ
]v
= ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
we will solve this in the same way as the harmonic oscillator, assumingthat v(ρ) is an infinite polynomial in ρ
v(ρ) =∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 7 / 17
The Schrodinger equation for v(ρ)
Substituting into the Schrodinger equation for u(ρ)
0 =d2u
dρ2−[
1− ρ0ρ
+l(l + 1)
ρ2
]u
= ρle−ρ{ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}− ρle−ρ
[ρ− ρ0 +
l(l + 1)
ρ
]v
= ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
we will solve this in the same way as the harmonic oscillator, assumingthat v(ρ) is an infinite polynomial in ρ
v(ρ) =∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 7 / 17
The Schrodinger equation for v(ρ)
Substituting into the Schrodinger equation for u(ρ)
0 =d2u
dρ2−[
1− ρ0ρ
+l(l + 1)
ρ2
]u
= ρle−ρ{ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}− ρle−ρ
[ρ− ρ0 +
l(l + 1)
ρ
]v
= ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
we will solve this in the same way as the harmonic oscillator, assumingthat v(ρ) is an infinite polynomial in ρ
v(ρ) =∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 7 / 17
The Schrodinger equation for v(ρ)
Substituting into the Schrodinger equation for u(ρ)
0 =d2u
dρ2−[
1− ρ0ρ
+l(l + 1)
ρ2
]u
= ρle−ρ{ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+
[ρ− 2(l + 1) +
l(l + 1)
ρ
]v
}− ρle−ρ
[ρ− ρ0 +
l(l + 1)
ρ
]v
= ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
we will solve this in the same way as the harmonic oscillator, assumingthat v(ρ) is an infinite polynomial in ρ
v(ρ) =∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 7 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1
=∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj
+ 2(l + 1)∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1
=∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj
+ 2(l + 1)∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1
=∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj
+ 2(l + 1)∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1 =
∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj
+ 2(l + 1)∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1 =
∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj
+ 2(l + 1)∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1 =
∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj
+ 2(l + 1)∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1 =
∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj
+ 2(l + 1)∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1 =
∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj
+ 2(l + 1)∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1 =
∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1 =
∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj
+ [ρ0 − 2(l + 1)]∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
taking the derivatives
shifting the indices by onesince the first term vanishesanyway
substituting back into theSchrodinger equation
v =∞∑j=0
cjρj
dv
dρ=∞∑j=0
jcjρj−1 =
∞∑j=0
(j + 1)cj+1ρj
d2v
dρ2=∞∑j=0
j(j + 1)cj+1ρj−1
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
=∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 8 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj
=2
j + 1cj −→ cj =
2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj
= c0e2ρ −→ u(ρ) = c0ρ
l+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj
=2
j + 1cj −→ cj =
2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj
= c0e2ρ −→ u(ρ) = c0ρ
l+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj
=2
j + 1cj −→ cj =
2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj
= c0e2ρ −→ u(ρ) = c0ρ
l+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj
=2
j + 1cj −→ cj =
2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj
= c0e2ρ −→ u(ρ) = c0ρ
l+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj
=2
j + 1cj −→ cj =
2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj
= c0e2ρ −→ u(ρ) = c0ρ
l+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj
=2
j + 1cj −→ cj =
2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj
= c0e2ρ −→ u(ρ) = c0ρ
l+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj =
2
j + 1cj
−→ cj =2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj
= c0e2ρ −→ u(ρ) = c0ρ
l+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj =
2
j + 1cj −→ cj =
2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj
= c0e2ρ −→ u(ρ) = c0ρ
l+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj =
2
j + 1cj −→ cj =
2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj
= c0e2ρ −→ u(ρ) = c0ρ
l+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj =
2
j + 1cj −→ cj =
2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj = c0e
2ρ
−→ u(ρ) = c0ρl+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
The recursion relation for v(ρ)
0 =∞∑j=0
j(j + 1)cj+1ρj + 2(l + 1)
∞∑j=0
(j + 1)cj+1ρj
− 2∞∑j=0
jcjρj + [ρ0 − 2(l + 1)]
∞∑j=0
cjρj
equating coefficients of like powers of j and solving for cj+1
0 = j(j + 1)cj+1 + 2(l + 1)(j + 1)cj+1 − 2jcj + [ρ0 − 2(l + 1)]cj
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
looking at the asymptotic behavior for large j
cj+1 '2j
j(j + 1)cj =
2
j + 1cj −→ cj =
2j
j!c0
v(ρ) = c0
∞∑j=0
2j
j!ρj = c0e
2ρ −→ u(ρ) = c0ρl+1eρ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 9 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m
= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2
=E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m
= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2
=E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m
= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2
=E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m
= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2
=E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m
= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2
=E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m
= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2
=E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m
= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2
=E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m
= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2
=E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2
=E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2
=E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Limiting the sum
The sum needs to be limited by set-ting a maximum value jmax suchthat cjmax+1 ≡ 0
by inspection of the recursion rela-tion, one can see that jmax can beconstrained by setting the numera-tor equal to zero
If we define an integer n, called theprincipal quantum number we get
the energy now becomes
cj+1 =
[2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
]cj
0 = 2(jmax + l + 1)− ρ0
n ≡ jmax + l + 1
ρ0 = 2n
E = −~2κ2
2m= − me4
8π2ε20~2ρ20
En = −
[m
2~2
(e2
4πε0
)2]
1
n2=
E1
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 10 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1
→ l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1
m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1
m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1
m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1
m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n=
1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n=
1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n=
1
an
a =4πε0~2
me2
= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n=
1
an
a =4πε0~2
me2= 0.529× 10−10m
ρ =r
an
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Quantum numbers and the Bohr radius
From the defining relation for jmax , it is possible to obtain constraints forthe azimuthal quantum number, l
Thus, if n = 0, both l and jmax must be zero as well. The lower bound forl is always zero and it can at most be n − 1 when jmax = 0. In general
n = jmax + l + 1 → l = n − 1− jmax
n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n=
1
an
a =4πε0~2
me2= 0.529× 10−10m
ρ =r
anC. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 11 / 17
Hydrogen wave functions
It is now possible to assemble the entire hydrogen atom wave function
ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)
v(r/na) =∞∑j=0
cj
( r
na
)jbut the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials
and Lq(x) are Laguerre polynomials
Rnl(r) =1
r
( r
na
)l+1e−r/nav(r/na)
cj+1 =
{2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
}cj
v(ρ) = L2l+1n−l−1(2ρ)
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
Lq(x) = ex(
d
dx
)q (e−xxq
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 12 / 17
Hydrogen wave functions
It is now possible to assemble the entire hydrogen atom wave function
ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)
v(r/na) =∞∑j=0
cj
( r
na
)jbut the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials
and Lq(x) are Laguerre polynomials
Rnl(r) =1
r
( r
na
)l+1e−r/nav(r/na)
cj+1 =
{2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
}cj
v(ρ) = L2l+1n−l−1(2ρ)
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
Lq(x) = ex(
d
dx
)q (e−xxq
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 12 / 17
Hydrogen wave functions
It is now possible to assemble the entire hydrogen atom wave function
ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)
v(r/na) =∞∑j=0
cj
( r
na
)jbut the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials
and Lq(x) are Laguerre polynomials
Rnl(r) =1
r
( r
na
)l+1e−r/nav(r/na)
cj+1 =
{2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
}cj
v(ρ) = L2l+1n−l−1(2ρ)
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
Lq(x) = ex(
d
dx
)q (e−xxq
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 12 / 17
Hydrogen wave functions
It is now possible to assemble the entire hydrogen atom wave function
ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)
v(r/na) =∞∑j=0
cj
( r
na
)j
but the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials
and Lq(x) are Laguerre polynomials
Rnl(r) =1
r
( r
na
)l+1e−r/nav(r/na)
cj+1 =
{2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
}cj
v(ρ) = L2l+1n−l−1(2ρ)
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
Lq(x) = ex(
d
dx
)q (e−xxq
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 12 / 17
Hydrogen wave functions
It is now possible to assemble the entire hydrogen atom wave function
ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)
v(r/na) =∞∑j=0
cj
( r
na
)j
but the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials
and Lq(x) are Laguerre polynomials
Rnl(r) =1
r
( r
na
)l+1e−r/nav(r/na)
cj+1 =
{2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
}cj
v(ρ) = L2l+1n−l−1(2ρ)
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
Lq(x) = ex(
d
dx
)q (e−xxq
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 12 / 17
Hydrogen wave functions
It is now possible to assemble the entire hydrogen atom wave function
ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)
v(r/na) =∞∑j=0
cj
( r
na
)jbut the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials
and Lq(x) are Laguerre polynomials
Rnl(r) =1
r
( r
na
)l+1e−r/nav(r/na)
cj+1 =
{2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
}cj
v(ρ) = L2l+1n−l−1(2ρ)
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
Lq(x) = ex(
d
dx
)q (e−xxq
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 12 / 17
Hydrogen wave functions
It is now possible to assemble the entire hydrogen atom wave function
ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)
v(r/na) =∞∑j=0
cj
( r
na
)jbut the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials
and Lq(x) are Laguerre polynomials
Rnl(r) =1
r
( r
na
)l+1e−r/nav(r/na)
cj+1 =
{2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
}cj
v(ρ) = L2l+1n−l−1(2ρ)
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
Lq(x) = ex(
d
dx
)q (e−xxq
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 12 / 17
Hydrogen wave functions
It is now possible to assemble the entire hydrogen atom wave function
ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)
v(r/na) =∞∑j=0
cj
( r
na
)jbut the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials
and Lq(x) are Laguerre polynomials
Rnl(r) =1
r
( r
na
)l+1e−r/nav(r/na)
cj+1 =
{2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
}cj
v(ρ) = L2l+1n−l−1(2ρ)
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
Lq(x) = ex(
d
dx
)q (e−xxq
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 12 / 17
Hydrogen wave functions
It is now possible to assemble the entire hydrogen atom wave function
ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)
v(r/na) =∞∑j=0
cj
( r
na
)jbut the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials
and Lq(x) are Laguerre polynomials
Rnl(r) =1
r
( r
na
)l+1e−r/nav(r/na)
cj+1 =
{2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
}cj
v(ρ) = L2l+1n−l−1(2ρ)
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
Lq(x) = ex(
d
dx
)q (e−xxq
)
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 12 / 17
Hydrogen wave functions
It is now possible to assemble the entire hydrogen atom wave function
ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)
v(r/na) =∞∑j=0
cj
( r
na
)jbut the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials
and Lq(x) are Laguerre polynomials
Rnl(r) =1
r
( r
na
)l+1e−r/nav(r/na)
cj+1 =
{2(j + l + 1)− ρ0
(j + 1)(j + 2l + 2)
}cj
v(ρ) = L2l+1n−l−1(2ρ)
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
Lq(x) = ex(
d
dx
)q (e−xxq
)C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 12 / 17
Laguerre polynomials
Laguerre polynomials
Lq(x) = ex(
d
dx
)q (e−xxq
)
L0 = 1
L1 = −x + 1
L2 = x2 − 4x + 2
associated Laguerre polynomials
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
L00 = 1
L10 = 1
L20 = 2
L01 = −x + 1
L11 = −2x + 4
L21 = −6x + 18
L02 = x2 − 4x + 2
L12 = 3x3 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 13 / 17
Laguerre polynomials
Laguerre polynomials
Lq(x) = ex(
d
dx
)q (e−xxq
)L0 = 1
L1 = −x + 1
L2 = x2 − 4x + 2
associated Laguerre polynomials
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
L00 = 1
L10 = 1
L20 = 2
L01 = −x + 1
L11 = −2x + 4
L21 = −6x + 18
L02 = x2 − 4x + 2
L12 = 3x3 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 13 / 17
Laguerre polynomials
Laguerre polynomials
Lq(x) = ex(
d
dx
)q (e−xxq
)L0 = 1
L1 = −x + 1
L2 = x2 − 4x + 2
associated Laguerre polynomials
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
L00 = 1
L10 = 1
L20 = 2
L01 = −x + 1
L11 = −2x + 4
L21 = −6x + 18
L02 = x2 − 4x + 2
L12 = 3x3 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 13 / 17
Laguerre polynomials
Laguerre polynomials
Lq(x) = ex(
d
dx
)q (e−xxq
)L0 = 1
L1 = −x + 1
L2 = x2 − 4x + 2
associated Laguerre polynomials
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
L00 = 1
L10 = 1
L20 = 2
L01 = −x + 1
L11 = −2x + 4
L21 = −6x + 18
L02 = x2 − 4x + 2
L12 = 3x3 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 13 / 17
Laguerre polynomials
Laguerre polynomials
Lq(x) = ex(
d
dx
)q (e−xxq
)L0 = 1
L1 = −x + 1
L2 = x2 − 4x + 2
associated Laguerre polynomials
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
L00 = 1
L10 = 1
L20 = 2
L01 = −x + 1
L11 = −2x + 4
L21 = −6x + 18
L02 = x2 − 4x + 2
L12 = 3x3 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 13 / 17
Laguerre polynomials
Laguerre polynomials
Lq(x) = ex(
d
dx
)q (e−xxq
)L0 = 1
L1 = −x + 1
L2 = x2 − 4x + 2
associated Laguerre polynomials
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
L00 = 1
L10 = 1
L20 = 2
L01 = −x + 1
L11 = −2x + 4
L21 = −6x + 18
L02 = x2 − 4x + 2
L12 = 3x3 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 13 / 17
Laguerre polynomials
Laguerre polynomials
Lq(x) = ex(
d
dx
)q (e−xxq
)L0 = 1
L1 = −x + 1
L2 = x2 − 4x + 2
associated Laguerre polynomials
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
L00 = 1
L10 = 1
L20 = 2
L01 = −x + 1
L11 = −2x + 4
L21 = −6x + 18
L02 = x2 − 4x + 2
L12 = 3x3 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 13 / 17
Laguerre polynomials
Laguerre polynomials
Lq(x) = ex(
d
dx
)q (e−xxq
)L0 = 1
L1 = −x + 1
L2 = x2 − 4x + 2
associated Laguerre polynomials
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
L00 = 1
L10 = 1
L20 = 2
L01 = −x + 1
L11 = −2x + 4
L21 = −6x + 18
L02 = x2 − 4x + 2
L12 = 3x3 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 13 / 17
Laguerre polynomials
Laguerre polynomials
Lq(x) = ex(
d
dx
)q (e−xxq
)L0 = 1
L1 = −x + 1
L2 = x2 − 4x + 2
associated Laguerre polynomials
Lpq−p(x) ≡ (−1)p(
d
dx
)p
Lq(x)
L00 = 1
L10 = 1
L20 = 2
L01 = −x + 1
L11 = −2x + 4
L21 = −6x + 18
L02 = x2 − 4x + 2
L12 = 3x3 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 13 / 17
Complete hydrogen solution
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
we can use this general solution to generate all the eigenfunctions of thehydrogen atom
ψ100 =
√(2
a
)3 (0)!
2[(1)!]3e−r/a
(2r
a
)0 [L10(2r/a)
]Y 00 (θ, φ)
=
√4
a3e−r/a
1√4π
=1√πa3
e−r/a
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 14 / 17
Complete hydrogen solution
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
we can use this general solution to generate all the eigenfunctions of thehydrogen atom
ψ100 =
√(2
a
)3 (0)!
2[(1)!]3e−r/a
(2r
a
)0 [L10(2r/a)
]Y 00 (θ, φ)
=
√4
a3e−r/a
1√4π
=1√πa3
e−r/a
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 14 / 17
Complete hydrogen solution
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
we can use this general solution to generate all the eigenfunctions of thehydrogen atom
ψ100 =
√(2
a
)3 (0)!
2[(1)!]3e−r/a
(2r
a
)0 [L10(2r/a)
]Y 00 (θ, φ)
=
√4
a3e−r/a
1√4π
=1√πa3
e−r/a
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 14 / 17
Complete hydrogen solution
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
we can use this general solution to generate all the eigenfunctions of thehydrogen atom
ψ100 =
√(2
a
)3 (0)!
2[(1)!]3e−r/a
(2r
a
)0 [L10(2r/a)
]Y 00 (θ, φ)
=
√4
a3e−r/a
1√4π
=1√πa3
e−r/a
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 14 / 17
Complete hydrogen solution
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
we can use this general solution to generate all the eigenfunctions of thehydrogen atom
ψ100 =
√(2
a
)3 (0)!
2[(1)!]3e−r/a
(2r
a
)0 [L10(2r/a)
]Y 00 (θ, φ)
=
√4
a3e−r/a
1√4π
=1√πa3
e−r/a
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 14 / 17
More hydrogen solutions
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
ψ200 =
√(2
2a
)3 (1)!
4[(2)!]3e−r/2a
(2r
2a
)0 [L11(2r/2a)
]Y 00 (θ, φ)
=1√
32a3e−r/2a
(4− 2
r
a
) 1√4π
=1√8a3
(1− r
2a
)e−r/2a
ψ210 =
√(2
2a
)3 (0)!
4[(3)!]3e−r/2a
(2r
2a
)1 [L30(2r/2a)
]Y 01 (θ, φ)
=1
12√
6a36e−r/2a
( ra
) 1
2
√3
πcos θ =
1
4√
2πa3r
ae−r/2a cos θ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 15 / 17
More hydrogen solutions
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
ψ200 =
√(2
2a
)3 (1)!
4[(2)!]3e−r/2a
(2r
2a
)0 [L11(2r/2a)
]Y 00 (θ, φ)
=1√
32a3e−r/2a
(4− 2
r
a
) 1√4π
=1√8a3
(1− r
2a
)e−r/2a
ψ210 =
√(2
2a
)3 (0)!
4[(3)!]3e−r/2a
(2r
2a
)1 [L30(2r/2a)
]Y 01 (θ, φ)
=1
12√
6a36e−r/2a
( ra
) 1
2
√3
πcos θ =
1
4√
2πa3r
ae−r/2a cos θ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 15 / 17
More hydrogen solutions
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
ψ200 =
√(2
2a
)3 (1)!
4[(2)!]3e−r/2a
(2r
2a
)0 [L11(2r/2a)
]Y 00 (θ, φ)
=1√
32a3e−r/2a
(4− 2
r
a
) 1√4π
=1√8a3
(1− r
2a
)e−r/2a
ψ210 =
√(2
2a
)3 (0)!
4[(3)!]3e−r/2a
(2r
2a
)1 [L30(2r/2a)
]Y 01 (θ, φ)
=1
12√
6a36e−r/2a
( ra
) 1
2
√3
πcos θ =
1
4√
2πa3r
ae−r/2a cos θ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 15 / 17
More hydrogen solutions
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
ψ200 =
√(2
2a
)3 (1)!
4[(2)!]3e−r/2a
(2r
2a
)0 [L11(2r/2a)
]Y 00 (θ, φ)
=1√
32a3e−r/2a
(4− 2
r
a
) 1√4π
=1√8a3
(1− r
2a
)e−r/2a
ψ210 =
√(2
2a
)3 (0)!
4[(3)!]3e−r/2a
(2r
2a
)1 [L30(2r/2a)
]Y 01 (θ, φ)
=1
12√
6a36e−r/2a
( ra
) 1
2
√3
πcos θ =
1
4√
2πa3r
ae−r/2a cos θ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 15 / 17
More hydrogen solutions
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
ψ200 =
√(2
2a
)3 (1)!
4[(2)!]3e−r/2a
(2r
2a
)0 [L11(2r/2a)
]Y 00 (θ, φ)
=1√
32a3e−r/2a
(4− 2
r
a
) 1√4π
=1√8a3
(1− r
2a
)e−r/2a
ψ210 =
√(2
2a
)3 (0)!
4[(3)!]3e−r/2a
(2r
2a
)1 [L30(2r/2a)
]Y 01 (θ, φ)
=1
12√
6a36e−r/2a
( ra
) 1
2
√3
πcos θ =
1
4√
2πa3r
ae−r/2a cos θ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 15 / 17
More hydrogen solutions
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
ψ200 =
√(2
2a
)3 (1)!
4[(2)!]3e−r/2a
(2r
2a
)0 [L11(2r/2a)
]Y 00 (θ, φ)
=1√
32a3e−r/2a
(4− 2
r
a
) 1√4π
=1√8a3
(1− r
2a
)e−r/2a
ψ210 =
√(2
2a
)3 (0)!
4[(3)!]3e−r/2a
(2r
2a
)1 [L30(2r/2a)
]Y 01 (θ, φ)
=1
12√
6a36e−r/2a
( ra
) 1
2
√3
πcos θ
=1
4√
2πa3r
ae−r/2a cos θ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 15 / 17
More hydrogen solutions
ψnlm =
√(2
na
)3 (n − l − 1)!
2n[(n + l)!]3e−r/na
(2r
na
)l [L2l+1n−l−1(2r/na)
]Yml (θ, φ)
ψ200 =
√(2
2a
)3 (1)!
4[(2)!]3e−r/2a
(2r
2a
)0 [L11(2r/2a)
]Y 00 (θ, φ)
=1√
32a3e−r/2a
(4− 2
r
a
) 1√4π
=1√8a3
(1− r
2a
)e−r/2a
ψ210 =
√(2
2a
)3 (0)!
4[(3)!]3e−r/2a
(2r
2a
)1 [L30(2r/2a)
]Y 01 (θ, φ)
=1
12√
6a36e−r/2a
( ra
) 1
2
√3
πcos θ =
1
4√
2πa3r
ae−r/2a cos θ
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 15 / 17
Radial functions by n
We can plot the radial wavefunc-tions as a function of r/a for eachvalue of n
As with all other wavefunctions wehave seen, as the quantum num-ber rises, the number of nodes in-creases
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 5 10 15 20
a3
/2R
r/a
n=1
l=0
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 16 / 17
Radial functions by n
We can plot the radial wavefunc-tions as a function of r/a for eachvalue of n
As with all other wavefunctions wehave seen, as the quantum num-ber rises, the number of nodes in-creases
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 5 10 15 20
a3
/2R
r/a
n=1
l=0
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 16 / 17
Radial functions by n
We can plot the radial wavefunc-tions as a function of r/a for eachvalue of n
As with all other wavefunctions wehave seen, as the quantum num-ber rises, the number of nodes in-creases
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 5 10 15 20
a3
/2R
r/a
n=2
l=0
l=1
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 16 / 17
Radial functions by n
We can plot the radial wavefunc-tions as a function of r/a for eachvalue of n
As with all other wavefunctions wehave seen, as the quantum num-ber rises, the number of nodes in-creases
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 5 10 15 20
a3
/2R
r/a
n=3
l=0
l=1
l=2
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 16 / 17
Radial functions by n
We can plot the radial wavefunc-tions as a function of r/a for eachvalue of n
As with all other wavefunctions wehave seen, as the quantum num-ber rises, the number of nodes in-creases
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 5 10 15 20
a3
/2R
r/a
n=3
l=0
l=1
l=2
l=3
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 16 / 17
Radial functions by l
Similarly, we can plot them for eachvalue of l
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 5 10 15 20
a3
/2R
r/a
l=0
n=1
n=2
n=3
n=4
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 17 / 17
Radial functions by l
Similarly, we can plot them for eachvalue of l
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 5 10 15 20
a3
/2R
r/a
l=0
n=1
n=2
n=3
n=4
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 17 / 17
Radial functions by l
Similarly, we can plot them for eachvalue of l
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 5 10 15 20
a3
/2R
r/a
l=1
n=2
n=3
n=4
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 17 / 17
Radial functions by l
Similarly, we can plot them for eachvalue of l
-0.1
0
0.1
0.2
0.3
0.4
0.5
0 5 10 15 20
a3
/2R
r/a
l=2
n=3
n=4
C. Segre (IIT) PHYS 405 - Fall 2013 October 21, 2013 17 / 17