today’s outline - october 09, 2018segre/phys405/18f/lecture_14.pdf · today’s outline - october...
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Today’s Outline - October 09, 2018
• Example 4.1
• Hydrogen atom solution
• Hydrogen atom wavefunctions
• Angular momentum
Homework Assignment #06:Chapter 3: 16,18,26,27,33,38due Thursday, October 11, 2018
Homework Assignment #07:Chapter 4: 1,2,5,7,9,10due Tuesday, October 23, 2018
Next Tuesday lecture recording: Friday, October 12, 13:50-15:05 in 204SB
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 1 / 42
Today’s Outline - October 09, 2018
• Example 4.1
• Hydrogen atom solution
• Hydrogen atom wavefunctions
• Angular momentum
Homework Assignment #06:Chapter 3: 16,18,26,27,33,38due Thursday, October 11, 2018
Homework Assignment #07:Chapter 4: 1,2,5,7,9,10due Tuesday, October 23, 2018
Next Tuesday lecture recording: Friday, October 12, 13:50-15:05 in 204SB
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 1 / 42
Today’s Outline - October 09, 2018
• Example 4.1
• Hydrogen atom solution
• Hydrogen atom wavefunctions
• Angular momentum
Homework Assignment #06:Chapter 3: 16,18,26,27,33,38due Thursday, October 11, 2018
Homework Assignment #07:Chapter 4: 1,2,5,7,9,10due Tuesday, October 23, 2018
Next Tuesday lecture recording: Friday, October 12, 13:50-15:05 in 204SB
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 1 / 42
Today’s Outline - October 09, 2018
• Example 4.1
• Hydrogen atom solution
• Hydrogen atom wavefunctions
• Angular momentum
Homework Assignment #06:Chapter 3: 16,18,26,27,33,38due Thursday, October 11, 2018
Homework Assignment #07:Chapter 4: 1,2,5,7,9,10due Tuesday, October 23, 2018
Next Tuesday lecture recording: Friday, October 12, 13:50-15:05 in 204SB
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 1 / 42
Today’s Outline - October 09, 2018
• Example 4.1
• Hydrogen atom solution
• Hydrogen atom wavefunctions
• Angular momentum
Homework Assignment #06:Chapter 3: 16,18,26,27,33,38due Thursday, October 11, 2018
Homework Assignment #07:Chapter 4: 1,2,5,7,9,10due Tuesday, October 23, 2018
Next Tuesday lecture recording: Friday, October 12, 13:50-15:05 in 204SB
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 1 / 42
Today’s Outline - October 09, 2018
• Example 4.1
• Hydrogen atom solution
• Hydrogen atom wavefunctions
• Angular momentum
Homework Assignment #06:Chapter 3: 16,18,26,27,33,38due Thursday, October 11, 2018
Homework Assignment #07:Chapter 4: 1,2,5,7,9,10due Tuesday, October 23, 2018
Next Tuesday lecture recording: Friday, October 12, 13:50-15:05 in 204SB
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 1 / 42
Today’s Outline - October 09, 2018
• Example 4.1
• Hydrogen atom solution
• Hydrogen atom wavefunctions
• Angular momentum
Homework Assignment #06:Chapter 3: 16,18,26,27,33,38due Thursday, October 11, 2018
Homework Assignment #07:Chapter 4: 1,2,5,7,9,10due Tuesday, October 23, 2018
Next Tuesday lecture recording: Friday, October 12, 13:50-15:05 in 204SB
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 1 / 42
Today’s Outline - October 09, 2018
• Example 4.1
• Hydrogen atom solution
• Hydrogen atom wavefunctions
• Angular momentum
Homework Assignment #06:Chapter 3: 16,18,26,27,33,38due Thursday, October 11, 2018
Homework Assignment #07:Chapter 4: 1,2,5,7,9,10due Tuesday, October 23, 2018
Next Tuesday lecture recording: Friday, October 12, 13:50-15:05 in 204SB
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 1 / 42
Today’s Outline - October 09, 2018
• Example 4.1
• Hydrogen atom solution
• Hydrogen atom wavefunctions
• Angular momentum
Homework Assignment #06:Chapter 3: 16,18,26,27,33,38due Thursday, October 11, 2018
Homework Assignment #07:Chapter 4: 1,2,5,7,9,10due Tuesday, October 23, 2018
Next Tuesday lecture recording: Friday, October 12, 13:50-15:05 in 204SB
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 1 / 42
3D solution in spherical coordinates
The angular solutions are spherical harmonics
Yml (θ, φ) = ε
√(2l + 1)(l − |m|)!
4π(l + |m|)!e imφPm
l (cos θ), ε =
{(−1)m, m ≥ 0
1, m ≤ 0
l = 0, 1, 2, 3, · · · azimuthal quantum numberm = −l ,−(l − 1), · · ·, l − 1, l magnetic quantum number
The radial solutions, R(r) = u(r)/r depend on the potential and mustsatisfy the equation:
− ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u = Eu
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 2 / 42
3D solution in spherical coordinates
The angular solutions are spherical harmonics
Yml (θ, φ) = ε
√(2l + 1)(l − |m|)!
4π(l + |m|)!e imφPm
l (cos θ), ε =
{(−1)m, m ≥ 0
1, m ≤ 0
l = 0, 1, 2, 3, · · · azimuthal quantum numberm = −l ,−(l − 1), · · ·, l − 1, l magnetic quantum number
The radial solutions, R(r) = u(r)/r depend on the potential and mustsatisfy the equation:
− ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u = Eu
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 2 / 42
3D solution in spherical coordinates
The angular solutions are spherical harmonics
Yml (θ, φ) = ε
√(2l + 1)(l − |m|)!
4π(l + |m|)!e imφPm
l (cos θ), ε =
{(−1)m, m ≥ 0
1, m ≤ 0
l = 0, 1, 2, 3, · · · azimuthal quantum number
m = −l ,−(l − 1), · · ·, l − 1, l magnetic quantum number
The radial solutions, R(r) = u(r)/r depend on the potential and mustsatisfy the equation:
− ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u = Eu
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 2 / 42
3D solution in spherical coordinates
The angular solutions are spherical harmonics
Yml (θ, φ) = ε
√(2l + 1)(l − |m|)!
4π(l + |m|)!e imφPm
l (cos θ), ε =
{(−1)m, m ≥ 0
1, m ≤ 0
l = 0, 1, 2, 3, · · · azimuthal quantum numberm = −l ,−(l − 1), · · ·, l − 1, l magnetic quantum number
The radial solutions, R(r) = u(r)/r depend on the potential and mustsatisfy the equation:
− ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u = Eu
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 2 / 42
3D solution in spherical coordinates
The angular solutions are spherical harmonics
Yml (θ, φ) = ε
√(2l + 1)(l − |m|)!
4π(l + |m|)!e imφPm
l (cos θ), ε =
{(−1)m, m ≥ 0
1, m ≤ 0
l = 0, 1, 2, 3, · · · azimuthal quantum numberm = −l ,−(l − 1), · · ·, l − 1, l magnetic quantum number
The radial solutions, R(r) = u(r)/r depend on the potential and mustsatisfy the equation:
− ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u = Eu
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 2 / 42
3D solution in spherical coordinates
The angular solutions are spherical harmonics
Yml (θ, φ) = ε
√(2l + 1)(l − |m|)!
4π(l + |m|)!e imφPm
l (cos θ), ε =
{(−1)m, m ≥ 0
1, m ≤ 0
l = 0, 1, 2, 3, · · · azimuthal quantum numberm = −l ,−(l − 1), · · ·, l − 1, l magnetic quantum number
The radial solutions, R(r) = u(r)/r depend on the potential and mustsatisfy the equation:
− ~2
2m
d2u
dr2+
[V (r) +
~2
2m
l(l + 1)
r2
]u = Eu
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 2 / 42
Example 4.1 (review)
The potential for the infinite spher-ical well is
inside the well the wave equation is
with k =√
2mE/~, it becomes
V (r) =
{0, r ≤ a
∞, r ≥ a
Eu = − ~2
2m
d2u
dr2+
~2
2m
l(l + 1)
r2u
d2u
dr2=
[l(l + 1)
r2− k2
]u
We must solve this for each value of l separately. The l = 0 solution is
ψn00 =1√2πa
sin(nπr/a)
rEn0 =
n2π2~2
2ma2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 3 / 42
Example 4.1 (review)
The potential for the infinite spher-ical well is
inside the well the wave equation is
with k =√
2mE/~, it becomes
V (r) =
{0, r ≤ a
∞, r ≥ a
Eu = − ~2
2m
d2u
dr2+
~2
2m
l(l + 1)
r2u
d2u
dr2=
[l(l + 1)
r2− k2
]u
We must solve this for each value of l separately. The l = 0 solution is
ψn00 =1√2πa
sin(nπr/a)
rEn0 =
n2π2~2
2ma2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 3 / 42
Example 4.1 (review)
The potential for the infinite spher-ical well is
inside the well the wave equation is
with k =√
2mE/~, it becomes
V (r) =
{0, r ≤ a
∞, r ≥ a
Eu = − ~2
2m
d2u
dr2+
~2
2m
l(l + 1)
r2u
d2u
dr2=
[l(l + 1)
r2− k2
]u
We must solve this for each value of l separately. The l = 0 solution is
ψn00 =1√2πa
sin(nπr/a)
rEn0 =
n2π2~2
2ma2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 3 / 42
Example 4.1 (review)
The potential for the infinite spher-ical well is
inside the well the wave equation is
with k =√
2mE/~, it becomes
V (r) =
{0, r ≤ a
∞, r ≥ a
Eu = − ~2
2m
d2u
dr2+
~2
2m
l(l + 1)
r2u
d2u
dr2=
[l(l + 1)
r2− k2
]u
We must solve this for each value of l separately. The l = 0 solution is
ψn00 =1√2πa
sin(nπr/a)
rEn0 =
n2π2~2
2ma2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 3 / 42
Example 4.1 (review)
The potential for the infinite spher-ical well is
inside the well the wave equation is
with k =√
2mE/~, it becomes
V (r) =
{0, r ≤ a
∞, r ≥ a
Eu = − ~2
2m
d2u
dr2+
~2
2m
l(l + 1)
r2u
d2u
dr2=
[l(l + 1)
r2− k2
]u
We must solve this for each value of l separately. The l = 0 solution is
ψn00 =1√2πa
sin(nπr/a)
rEn0 =
n2π2~2
2ma2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 3 / 42
Example 4.1 (review)
The potential for the infinite spher-ical well is
inside the well the wave equation is
with k =√
2mE/~, it becomes
V (r) =
{0, r ≤ a
∞, r ≥ a
Eu = − ~2
2m
d2u
dr2+
~2
2m
l(l + 1)
r2u
d2u
dr2=
[l(l + 1)
r2− k2
]u
We must solve this for each value of l separately. The l = 0 solution is
ψn00 =1√2πa
sin(nπr/a)
rEn0 =
n2π2~2
2ma2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 3 / 42
Example 4.1 (review)
The potential for the infinite spher-ical well is
inside the well the wave equation is
with k =√
2mE/~, it becomes
V (r) =
{0, r ≤ a
∞, r ≥ a
Eu = − ~2
2m
d2u
dr2+
~2
2m
l(l + 1)
r2u
d2u
dr2=
[l(l + 1)
r2− k2
]u
We must solve this for each value of l separately. The l = 0 solution is
ψn00 =1√2πa
sin(nπr/a)
rEn0 =
n2π2~2
2ma2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 3 / 42
Example 4.1 (review)
The potential for the infinite spher-ical well is
inside the well the wave equation is
with k =√
2mE/~, it becomes
V (r) =
{0, r ≤ a
∞, r ≥ a
Eu = − ~2
2m
d2u
dr2+
~2
2m
l(l + 1)
r2u
d2u
dr2=
[l(l + 1)
r2− k2
]u
We must solve this for each value of l separately. The l = 0 solution is
ψn00 =1√2πa
sin(nπr/a)
r
En0 =n2π2~2
2ma2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 3 / 42
Example 4.1 (review)
The potential for the infinite spher-ical well is
inside the well the wave equation is
with k =√
2mE/~, it becomes
V (r) =
{0, r ≤ a
∞, r ≥ a
Eu = − ~2
2m
d2u
dr2+
~2
2m
l(l + 1)
r2u
d2u
dr2=
[l(l + 1)
r2− k2
]u
We must solve this for each value of l separately. The l = 0 solution is
ψn00 =1√2πa
sin(nπr/a)
rEn0 =
n2π2~2
2ma2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 3 / 42
Example 4.1 (cont.)
What about the solutions for l 6= 0?
they involve Bessel functions; thegeneral solution being
jl(x) are spherical Bessel functionsof order l and nl(x) are sphericalNeumann functions of order l .
while spherical Bessel functions arefinite at the origin, spherical Neu-mann functions are infinite andthus we again set Bl = 0
u(r) = Al r jl(kr) + Bl r nl(kr)
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
nl(x) = −(−x)l(
1
x
d
dx
)l cos x
x
R(r) = Al jl(kr)
we still must apply the boundary condition that jl(ka) = 0 but this is a bitmore complex than for the l = 0 case
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 4 / 42
Example 4.1 (cont.)
What about the solutions for l 6= 0?
they involve Bessel functions; thegeneral solution being
jl(x) are spherical Bessel functionsof order l and nl(x) are sphericalNeumann functions of order l .
while spherical Bessel functions arefinite at the origin, spherical Neu-mann functions are infinite andthus we again set Bl = 0
u(r) = Al r jl(kr) + Bl r nl(kr)
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
nl(x) = −(−x)l(
1
x
d
dx
)l cos x
x
R(r) = Al jl(kr)
we still must apply the boundary condition that jl(ka) = 0 but this is a bitmore complex than for the l = 0 case
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 4 / 42
Example 4.1 (cont.)
What about the solutions for l 6= 0?
they involve Bessel functions; thegeneral solution being
jl(x) are spherical Bessel functionsof order l and nl(x) are sphericalNeumann functions of order l .
while spherical Bessel functions arefinite at the origin, spherical Neu-mann functions are infinite andthus we again set Bl = 0
u(r) = Al r jl(kr) + Bl r nl(kr)
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
nl(x) = −(−x)l(
1
x
d
dx
)l cos x
x
R(r) = Al jl(kr)
we still must apply the boundary condition that jl(ka) = 0 but this is a bitmore complex than for the l = 0 case
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 4 / 42
Example 4.1 (cont.)
What about the solutions for l 6= 0?
they involve Bessel functions; thegeneral solution being
jl(x) are spherical Bessel functionsof order l
and nl(x) are sphericalNeumann functions of order l .
while spherical Bessel functions arefinite at the origin, spherical Neu-mann functions are infinite andthus we again set Bl = 0
u(r) = Al r jl(kr) + Bl r nl(kr)
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
nl(x) = −(−x)l(
1
x
d
dx
)l cos x
x
R(r) = Al jl(kr)
we still must apply the boundary condition that jl(ka) = 0 but this is a bitmore complex than for the l = 0 case
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 4 / 42
Example 4.1 (cont.)
What about the solutions for l 6= 0?
they involve Bessel functions; thegeneral solution being
jl(x) are spherical Bessel functionsof order l and nl(x) are sphericalNeumann functions of order l .
while spherical Bessel functions arefinite at the origin, spherical Neu-mann functions are infinite andthus we again set Bl = 0
u(r) = Al r jl(kr) + Bl r nl(kr)
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
nl(x) = −(−x)l(
1
x
d
dx
)l cos x
x
R(r) = Al jl(kr)
we still must apply the boundary condition that jl(ka) = 0 but this is a bitmore complex than for the l = 0 case
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 4 / 42
Example 4.1 (cont.)
What about the solutions for l 6= 0?
they involve Bessel functions; thegeneral solution being
jl(x) are spherical Bessel functionsof order l and nl(x) are sphericalNeumann functions of order l .
while spherical Bessel functions arefinite at the origin, spherical Neu-mann functions are infinite andthus we again set Bl = 0
u(r) = Al r jl(kr) + Bl r nl(kr)
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
nl(x) = −(−x)l(
1
x
d
dx
)l cos x
x
R(r) = Al jl(kr)
we still must apply the boundary condition that jl(ka) = 0 but this is a bitmore complex than for the l = 0 case
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 4 / 42
Example 4.1 (cont.)
What about the solutions for l 6= 0?
they involve Bessel functions; thegeneral solution being
jl(x) are spherical Bessel functionsof order l and nl(x) are sphericalNeumann functions of order l .
while spherical Bessel functions arefinite at the origin, spherical Neu-mann functions are infinite andthus we again set Bl = 0
u(r) = Al r jl(kr) + Bl r nl(kr)
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
nl(x) = −(−x)l(
1
x
d
dx
)l cos x
x
R(r) = Al jl(kr)
we still must apply the boundary condition that jl(ka) = 0 but this is a bitmore complex than for the l = 0 case
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 4 / 42
Example 4.1 (cont.)
What about the solutions for l 6= 0?
they involve Bessel functions; thegeneral solution being
jl(x) are spherical Bessel functionsof order l and nl(x) are sphericalNeumann functions of order l .
while spherical Bessel functions arefinite at the origin, spherical Neu-mann functions are infinite andthus we again set Bl = 0
u(r) = Al r jl(kr) + Bl r nl(kr)
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
nl(x) = −(−x)l(
1
x
d
dx
)l cos x
x
R(r) = Al jl(kr)
we still must apply the boundary condition that jl(ka) = 0 but this is a bitmore complex than for the l = 0 case
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 4 / 42
Example 4.1 (cont.)
What about the solutions for l 6= 0?
they involve Bessel functions; thegeneral solution being
jl(x) are spherical Bessel functionsof order l and nl(x) are sphericalNeumann functions of order l .
while spherical Bessel functions arefinite at the origin, spherical Neu-mann functions are infinite andthus we again set Bl = 0
u(r) = Al r jl(kr) + Bl r nl(kr)
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
nl(x) = −(−x)l(
1
x
d
dx
)l cos x
x
R(r) = Al jl(kr)
we still must apply the boundary condition that jl(ka) = 0 but this is a bitmore complex than for the l = 0 case
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 4 / 42
Spherical Bessel functions
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
j0(x) =sin x
x
j1(x) = (−x)1
x
(cos x
x− sin x
x2
)=
sin x
x2− cos x
x
j2(x) = (−x)2(
1
x
d
dx
)x cos x − sin x
x3
= x
(−x sin x
x3− 3
x cos x − sin x
x4
)=
(3 sin x − 3x cos x − x2 sin x
x3
)0
+1
0 5 10
j l(x)
x
Clearly the roots are not at nice, simple, locations!
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 5 / 42
Spherical Bessel functions
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
j0(x) =sin x
x
j1(x) = (−x)1
x
(cos x
x− sin x
x2
)=
sin x
x2− cos x
x
j2(x) = (−x)2(
1
x
d
dx
)x cos x − sin x
x3
= x
(−x sin x
x3− 3
x cos x − sin x
x4
)=
(3 sin x − 3x cos x − x2 sin x
x3
)
0
+1
0 5 10j 0
(x)
x
Clearly the roots are not at nice, simple, locations!
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 5 / 42
Spherical Bessel functions
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
j0(x) =sin x
x
j1(x) = (−x)1
x
(cos x
x− sin x
x2
)
=sin x
x2− cos x
x
j2(x) = (−x)2(
1
x
d
dx
)x cos x − sin x
x3
= x
(−x sin x
x3− 3
x cos x − sin x
x4
)=
(3 sin x − 3x cos x − x2 sin x
x3
)
0
+1
0 5 10j 0
(x)
x
Clearly the roots are not at nice, simple, locations!
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 5 / 42
Spherical Bessel functions
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
j0(x) =sin x
x
j1(x) = (−x)1
x
(cos x
x− sin x
x2
)=
sin x
x2− cos x
x
j2(x) = (−x)2(
1
x
d
dx
)x cos x − sin x
x3
= x
(−x sin x
x3− 3
x cos x − sin x
x4
)=
(3 sin x − 3x cos x − x2 sin x
x3
)
0
0 5 10j 1
(x)
x
Clearly the roots are not at nice, simple, locations!
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 5 / 42
Spherical Bessel functions
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
j0(x) =sin x
x
j1(x) = (−x)1
x
(cos x
x− sin x
x2
)=
sin x
x2− cos x
x
j2(x) = (−x)2(
1
x
d
dx
)x cos x − sin x
x3
= x
(−x sin x
x3− 3
x cos x − sin x
x4
)=
(3 sin x − 3x cos x − x2 sin x
x3
)
0
0 5 10j 1
(x)
x
Clearly the roots are not at nice, simple, locations!
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 5 / 42
Spherical Bessel functions
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
j0(x) =sin x
x
j1(x) = (−x)1
x
(cos x
x− sin x
x2
)=
sin x
x2− cos x
x
j2(x) = (−x)2(
1
x
d
dx
)x cos x − sin x
x3
= x
(−x sin x
x3− 3
x cos x − sin x
x4
)
=
(3 sin x − 3x cos x − x2 sin x
x3
)
0
0 5 10j 1
(x)
x
Clearly the roots are not at nice, simple, locations!
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 5 / 42
Spherical Bessel functions
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
j0(x) =sin x
x
j1(x) = (−x)1
x
(cos x
x− sin x
x2
)=
sin x
x2− cos x
x
j2(x) = (−x)2(
1
x
d
dx
)x cos x − sin x
x3
= x
(−x sin x
x3− 3
x cos x − sin x
x4
)=
(3 sin x − 3x cos x − x2 sin x
x3
)0
0 5 10j 2
(x)
x
Clearly the roots are not at nice, simple, locations!
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 5 / 42
Spherical Bessel functions
jl(x) = (−x)l(
1
x
d
dx
)l sin x
x
j0(x) =sin x
x
j1(x) = (−x)1
x
(cos x
x− sin x
x2
)=
sin x
x2− cos x
x
j2(x) = (−x)2(
1
x
d
dx
)x cos x − sin x
x3
= x
(−x sin x
x3− 3
x cos x − sin x
x4
)=
(3 sin x − 3x cos x − x2 sin x
x3
)0
+1
0 5 10j l(
x)
x
Clearly the roots are not at nice, simple, locations!
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 5 / 42
Solutions for all l
jl(ka) = 0
k =1
aβnl , βnl are the roots
Enl =~2
2ma2β2nl
βn1 βn2 βn3
j0 π 2π 3πj1 4.493 7.726 10.904j2 5.762 9.906 12.325j3 6.988 10.420 13.698
The wavefunctions are thus:
ψnlm(r , θ, φ) = Anl jl
(βnl r
a
)Yml (θ, φ)
these are (2l + 1)-fold degenerate states, that is, the energy does notdepend on the quantum numbers which give rise to the degeneracy
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 6 / 42
Solutions for all l
jl(ka) = 0
k =1
aβnl , βnl are the roots
Enl =~2
2ma2β2nl
βn1 βn2 βn3
j0 π 2π 3πj1 4.493 7.726 10.904j2 5.762 9.906 12.325j3 6.988 10.420 13.698
The wavefunctions are thus:
ψnlm(r , θ, φ) = Anl jl
(βnl r
a
)Yml (θ, φ)
these are (2l + 1)-fold degenerate states, that is, the energy does notdepend on the quantum numbers which give rise to the degeneracy
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 6 / 42
Solutions for all l
jl(ka) = 0
k =1
aβnl , βnl are the roots
Enl =~2
2ma2β2nl
βn1 βn2 βn3
j0 π 2π 3πj1 4.493 7.726 10.904j2 5.762 9.906 12.325j3 6.988 10.420 13.698
The wavefunctions are thus:
ψnlm(r , θ, φ) = Anl jl
(βnl r
a
)Yml (θ, φ)
these are (2l + 1)-fold degenerate states, that is, the energy does notdepend on the quantum numbers which give rise to the degeneracy
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 6 / 42
Solutions for all l
jl(ka) = 0
k =1
aβnl , βnl are the roots
Enl =~2
2ma2β2nl
βn1 βn2 βn3
j0 π 2π 3πj1 4.493 7.726 10.904j2 5.762 9.906 12.325j3 6.988 10.420 13.698
The wavefunctions are thus:
ψnlm(r , θ, φ) = Anl jl
(βnl r
a
)Yml (θ, φ)
these are (2l + 1)-fold degenerate states, that is, the energy does notdepend on the quantum numbers which give rise to the degeneracy
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 6 / 42
Solutions for all l
jl(ka) = 0
k =1
aβnl , βnl are the roots
Enl =~2
2ma2β2nl
βn1 βn2 βn3
j0 π 2π 3πj1 4.493 7.726 10.904j2 5.762 9.906 12.325j3 6.988 10.420 13.698
The wavefunctions are thus:
ψnlm(r , θ, φ) = Anl jl
(βnl r
a
)Yml (θ, φ)
these are (2l + 1)-fold degenerate states, that is, the energy does notdepend on the quantum numbers which give rise to the degeneracy
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 6 / 42
Solutions for all l
jl(ka) = 0
k =1
aβnl , βnl are the roots
Enl =~2
2ma2β2nl
βn1 βn2 βn3
j0 π 2π 3πj1 4.493 7.726 10.904j2 5.762 9.906 12.325j3 6.988 10.420 13.698
The wavefunctions are thus:
ψnlm(r , θ, φ) = Anl jl
(βnl r
a
)Yml (θ, φ)
these are (2l + 1)-fold degenerate states, that is, the energy does notdepend on the quantum numbers which give rise to the degeneracy
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 6 / 42
Solutions for all l
jl(ka) = 0
k =1
aβnl , βnl are the roots
Enl =~2
2ma2β2nl
βn1 βn2 βn3
j0 π 2π 3πj1 4.493 7.726 10.904j2 5.762 9.906 12.325j3 6.988 10.420 13.698
The wavefunctions are thus:
ψnlm(r , θ, φ) = Anl jl
(βnl r
a
)Yml (θ, φ)
these are (2l + 1)-fold degenerate states, that is, the energy does notdepend on the quantum numbers which give rise to the degeneracy
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 6 / 42
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 2018 October 09, 2018 7 / 42
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 2018 October 09, 2018 7 / 42
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 2018 October 09, 2018 7 / 42
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 2018 October 09, 2018 7 / 42
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 2018 October 09, 2018 7 / 42
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 2018 October 09, 2018 7 / 42
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 2018 October 09, 2018 7 / 42
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 2018 October 09, 2018 7 / 42
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 2018 October 09, 2018 7 / 42
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 2018 October 09, 2018 7 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 8 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 9 / 42
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 2018 October 09, 2018 10 / 42
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 2018 October 09, 2018 10 / 42
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 2018 October 09, 2018 10 / 42
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 2018 October 09, 2018 10 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 11 / 42
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 2018 October 09, 2018 12 / 42
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 2018 October 09, 2018 12 / 42
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 2018 October 09, 2018 12 / 42
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 2018 October 09, 2018 12 / 42
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 2018 October 09, 2018 12 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solution
and take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
The recursion relation for v(ρ)
0 = ρd2v
dρ2+ 2(l + 1− ρ)
dv
dρ+ [ρ0 − 2(l + 1)]v
assume a polynomial solutionand take 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 =∞∑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 2018 October 09, 2018 13 / 42
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 2018 October 09, 2018 14 / 42
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 2018 October 09, 2018 14 / 42
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 2018 October 09, 2018 14 / 42
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 2018 October 09, 2018 14 / 42
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 2018 October 09, 2018 14 / 42
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 2018 October 09, 2018 14 / 42
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 2018 October 09, 2018 14 / 42
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 2018 October 09, 2018 14 / 42
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 2018 October 09, 2018 14 / 42
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 2018 October 09, 2018 14 / 42
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 2018 October 09, 2018 14 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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
= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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= −13.6eV
n2, n = 1, 2, 3, . . .
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 15 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ
= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n
=1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n=
1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n=
1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n=
1
an
a =4πε0~2
me2
= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n=
1
an
a =4πε0~2
me2= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 = 1, both l and jmax must be zero. The lower bound for l isalways 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
The state with energy En has de-generacy
Recall the definition of ρ0
solving for κ
where a is the Bohr radius
ρ =r
an
d(n) =n−1∑l=0
(2l + 1) = n2
ρ0 =me2
2πε0~2κ= 2n
κ =
(me2
4πε0~2
)1
n=
1
an
a =4πε0~2
me2= 0.529× 10−10m
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 16 / 42
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 2018 October 09, 2018 17 / 42
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 2018 October 09, 2018 17 / 42
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 2018 October 09, 2018 17 / 42
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 2018 October 09, 2018 17 / 42
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 2018 October 09, 2018 17 / 42
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 2018 October 09, 2018 17 / 42
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 2018 October 09, 2018 17 / 42
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 2018 October 09, 2018 17 / 42
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 2018 October 09, 2018 17 / 42
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 2018 October 09, 2018 17 / 42
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 = 3x2 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 18 / 42
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 = 3x2 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 18 / 42
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 = 3x2 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 18 / 42
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 = 3x2 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 18 / 42
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 = 3x2 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 18 / 42
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 = 3x2 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 18 / 42
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 = 3x2 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 18 / 42
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 = 3x2 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 18 / 42
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 = 3x2 − 18x + 18
L22 = 12x2 − 96x + 144
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 18 / 42
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 2018 October 09, 2018 19 / 42
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 2018 October 09, 2018 19 / 42
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 2018 October 09, 2018 19 / 42
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 2018 October 09, 2018 19 / 42
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 2018 October 09, 2018 19 / 42
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 2018 October 09, 2018 20 / 42
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 2018 October 09, 2018 20 / 42
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 2018 October 09, 2018 20 / 42
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 2018 October 09, 2018 20 / 42
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 2018 October 09, 2018 20 / 42
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 2018 October 09, 2018 20 / 42
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 2018 October 09, 2018 20 / 42
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=4
l=0
l=1
l=2
l=3
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 21 / 42
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 2018 October 09, 2018 21 / 42
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 2018 October 09, 2018 21 / 42
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 2018 October 09, 2018 21 / 42
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=4
l=0
l=1
l=2
l=3
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 21 / 42
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 2018 October 09, 2018 22 / 42
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 2018 October 09, 2018 22 / 42
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 2018 October 09, 2018 22 / 42
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 2018 October 09, 2018 22 / 42
Hydrogen orbital density plots
n = 1
l = 0
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 23 / 42
Hydrogen orbital density plots
n = 2
l = 0
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 24 / 42
Hydrogen orbital density plots
n = 2
l = 1
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 25 / 42
Hydrogen orbital density plots
n = 2
l = 1
m = 1
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 26 / 42
Hydrogen orbital density plots
n = 3
l = 0
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 27 / 42
Hydrogen orbital density plots
n = 3
l = 1
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 28 / 42
Hydrogen orbital density plots
n = 3
l = 1
m = 1
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 29 / 42
Hydrogen orbital density plots
n = 3
l = 2
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 30 / 42
Hydrogen orbital density plots
n = 3
l = 2
m = 1
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 31 / 42
Hydrogen orbital density plots
n = 3
l = 2
m = 2
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 32 / 42
Hydrogen orbital density plots
n = 4
l = 0
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 33 / 42
Hydrogen orbital density plots
n = 4
l = 1
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 34 / 42
Hydrogen orbital density plots
n = 4
l = 1
m = 1
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 35 / 42
Hydrogen orbital density plots
n = 4
l = 2
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 36 / 42
Hydrogen orbital density plots
n = 4
l = 2
m = 1
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 37 / 42
Hydrogen orbital density plots
n = 4
l = 2
m = 2
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 38 / 42
Hydrogen orbital density plots
n = 4
l = 3
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 39 / 42
Hydrogen orbital density plots
n = 4
l = 3
m = 1
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 40 / 42
Hydrogen orbital density plots
n = 4
l = 3
m = 2
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 41 / 42
Hydrogen orbital density plots
n = 4
l = 3
m = 3
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 42 / 42