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
![Page 2: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/2.jpg)
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
![Page 3: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/3.jpg)
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
![Page 4: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/4.jpg)
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
![Page 5: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/5.jpg)
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
![Page 6: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/6.jpg)
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
![Page 7: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/7.jpg)
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
![Page 8: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/8.jpg)
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
![Page 9: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/9.jpg)
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
![Page 10: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/10.jpg)
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
![Page 11: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/11.jpg)
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
![Page 12: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/12.jpg)
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
![Page 13: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/13.jpg)
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
![Page 14: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/14.jpg)
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
![Page 15: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/15.jpg)
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
![Page 16: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/16.jpg)
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, . . .
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![Page 17: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/17.jpg)
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
![Page 18: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/18.jpg)
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
![Page 19: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/19.jpg)
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
![Page 20: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/20.jpg)
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
![Page 21: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/21.jpg)
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
![Page 22: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/22.jpg)
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
![Page 23: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/23.jpg)
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
![Page 24: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/24.jpg)
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
![Page 25: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/25.jpg)
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
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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
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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
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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
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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
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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
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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
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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
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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
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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!
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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!
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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!
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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!
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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!
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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!
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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!
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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!
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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
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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
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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
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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
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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
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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
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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
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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/~
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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
![Page 51: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/51.jpg)
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
![Page 52: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/52.jpg)
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
![Page 53: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/53.jpg)
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
![Page 54: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/54.jpg)
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
![Page 55: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/55.jpg)
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
![Page 56: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/56.jpg)
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
![Page 57: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/57.jpg)
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
![Page 58: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/58.jpg)
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
![Page 59: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/59.jpg)
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
![Page 60: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/60.jpg)
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
![Page 61: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/61.jpg)
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
![Page 62: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/62.jpg)
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
![Page 63: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/63.jpg)
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
![Page 64: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/64.jpg)
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
![Page 65: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/65.jpg)
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
![Page 66: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/66.jpg)
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
![Page 67: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/67.jpg)
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
![Page 68: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/68.jpg)
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
![Page 69: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/69.jpg)
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
![Page 70: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/70.jpg)
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
}
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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
}
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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
}
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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
}
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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
}
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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
}
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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
}
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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
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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
![Page 94: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/94.jpg)
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
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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
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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
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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
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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
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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
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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
![Page 101: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/101.jpg)
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
![Page 102: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/102.jpg)
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
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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
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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
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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
![Page 106: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/106.jpg)
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
![Page 107: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/107.jpg)
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
![Page 108: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/108.jpg)
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
![Page 109: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/109.jpg)
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
![Page 110: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/110.jpg)
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
![Page 111: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/111.jpg)
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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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, . . .
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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, . . .
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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, . . .
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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, . . .
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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
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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
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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
![Page 134: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/134.jpg)
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
![Page 135: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/135.jpg)
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
![Page 136: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/136.jpg)
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
![Page 137: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/137.jpg)
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
![Page 138: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/138.jpg)
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
![Page 139: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/139.jpg)
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
![Page 140: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/140.jpg)
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
![Page 141: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/141.jpg)
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
![Page 142: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/142.jpg)
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
![Page 143: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/143.jpg)
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
![Page 144: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/144.jpg)
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
![Page 145: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/145.jpg)
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
![Page 146: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/146.jpg)
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
![Page 147: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/147.jpg)
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
![Page 148: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/148.jpg)
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
![Page 149: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/149.jpg)
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
![Page 150: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/150.jpg)
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
![Page 151: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/151.jpg)
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
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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
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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
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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
)
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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
)
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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
)
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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
)
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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
)
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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
)
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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
)
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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
)
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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
)
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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Hydrogen orbital density plots
n = 1
l = 0
m = 0
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Hydrogen orbital density plots
n = 2
l = 0
m = 0
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Hydrogen orbital density plots
n = 2
l = 1
m = 0
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Hydrogen orbital density plots
n = 2
l = 1
m = 1
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Hydrogen orbital density plots
n = 3
l = 0
m = 0
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Hydrogen orbital density plots
n = 3
l = 1
m = 0
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Hydrogen orbital density plots
n = 3
l = 1
m = 1
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Hydrogen orbital density plots
n = 3
l = 2
m = 0
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Hydrogen orbital density plots
n = 3
l = 2
m = 1
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Hydrogen orbital density plots
n = 3
l = 2
m = 2
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![Page 204: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/204.jpg)
Hydrogen orbital density plots
n = 4
l = 0
m = 0
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![Page 205: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/205.jpg)
Hydrogen orbital density plots
n = 4
l = 1
m = 0
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![Page 206: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/206.jpg)
Hydrogen orbital density plots
n = 4
l = 1
m = 1
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![Page 207: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/207.jpg)
Hydrogen orbital density plots
n = 4
l = 2
m = 0
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 36 / 42
![Page 208: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/208.jpg)
Hydrogen orbital density plots
n = 4
l = 2
m = 1
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![Page 209: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/209.jpg)
Hydrogen orbital density plots
n = 4
l = 2
m = 2
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![Page 210: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/210.jpg)
Hydrogen orbital density plots
n = 4
l = 3
m = 0
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![Page 211: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/211.jpg)
Hydrogen orbital density plots
n = 4
l = 3
m = 1
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![Page 212: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/212.jpg)
Hydrogen orbital density plots
n = 4
l = 3
m = 2
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![Page 213: Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October 09, 2018 Example 4.1 Hydrogen atom solution Hydrogen atom wavefunctions Angular](https://reader035.vdocuments.site/reader035/viewer/2022062918/5edae95e09ac2c67fa6881e9/html5/thumbnails/213.jpg)
Hydrogen orbital density plots
n = 4
l = 3
m = 3
C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 42 / 42