Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October...

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

• Example 4.1

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

[V (r) +

l(l + 1)

]u = Eu

Yml (θ, φ) = ε

√(2l + 1)(l − |m|)!

l (cos θ), ε =

{(−1)m, m ≥ 0

1, m ≤ 0

− ~2

[V (r) +

l(l + 1)

]u = Eu

Yml (θ, φ) = ε

√(2l + 1)(l − |m|)!

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

− ~2

[V (r) +

l(l + 1)

]u = Eu

Yml (θ, φ) = ε

√(2l + 1)(l − |m|)!

l (cos θ), ε =

{(−1)m, m ≥ 0

1, m ≤ 0

− ~2

[V (r) +

l(l + 1)

]u = Eu

Yml (θ, φ) = ε

√(2l + 1)(l − |m|)!

l (cos θ), ε =

{(−1)m, m ≥ 0

1, m ≤ 0

− ~2

[V (r) +

l(l + 1)

]u = Eu

Yml (θ, φ) = ε

√(2l + 1)(l − |m|)!

l (cos θ), ε =

{(−1)m, m ≥ 0

1, m ≤ 0

− ~2

[V (r) +

l(l + 1)

]u = Eu

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

l(l + 1)

[l(l + 1)

r2− k2

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, . . .

with k =√

2mE/~, it becomes

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− k2

ψn00 =1√2πa

sin(nπr/a)

rEn0 =

n2π2~2

2ma2, n = 1, 2, 3, . . .

with k =√

2mE/~, it becomes

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− k2

ψn00 =1√2πa

sin(nπr/a)

rEn0 =

n2π2~2

2ma2, n = 1, 2, 3, . . .

with k =√

2mE/~, it becomes

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− k2

ψn00 =1√2πa

sin(nπr/a)

rEn0 =

n2π2~2

2ma2, n = 1, 2, 3, . . .

with k =√

2mE/~, it becomes

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− k2

ψn00 =1√2πa

sin(nπr/a)

rEn0 =

n2π2~2

2ma2, n = 1, 2, 3, . . .

with k =√

2mE/~, it becomes

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− k2

ψn00 =1√2πa

sin(nπr/a)

rEn0 =

n2π2~2

2ma2, n = 1, 2, 3, . . .

with k =√

2mE/~, it becomes

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− k2

ψn00 =1√2πa

sin(nπr/a)

rEn0 =

n2π2~2

2ma2, n = 1, 2, 3, . . .

with k =√

2mE/~, it becomes

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− k2

ψn00 =1√2πa

sin(nπr/a)

En0 =n2π2~2

2ma2, n = 1, 2, 3, . . .

with k =√

2mE/~, it becomes

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− k2

ψn00 =1√2πa

sin(nπr/a)

rEn0 =

n2π2~2

2ma2, n = 1, 2, 3, . . .

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(

)l sin x

nl(x) = −(−x)l(

)l cos 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

Example 4.1 (cont.)

jl(x) = (−x)l(

)l sin x

nl(x) = −(−x)l(

)l cos x

R(r) = Al jl(kr)

Example 4.1 (cont.)

jl(x) = (−x)l(

)l sin x

nl(x) = −(−x)l(

)l cos x

R(r) = Al jl(kr)

Example 4.1 (cont.)

jl(x) are spherical Bessel functionsof order l

and nl(x) are sphericalNeumann functions of order l .

jl(x) = (−x)l(

)l sin x

nl(x) = −(−x)l(

)l cos x

R(r) = Al jl(kr)

Example 4.1 (cont.)

jl(x) = (−x)l(

)l sin x

nl(x) = −(−x)l(

)l cos x

R(r) = Al jl(kr)

Example 4.1 (cont.)

jl(x) = (−x)l(

)l sin x

nl(x) = −(−x)l(

)l cos x

R(r) = Al jl(kr)

Example 4.1 (cont.)

jl(x) = (−x)l(

)l sin x

nl(x) = −(−x)l(

)l cos x

R(r) = Al jl(kr)

Example 4.1 (cont.)

jl(x) = (−x)l(

)l sin x

nl(x) = −(−x)l(

)l cos x

R(r) = Al jl(kr)

Example 4.1 (cont.)

jl(x) = (−x)l(

)l sin x

nl(x) = −(−x)l(

)l cos x

R(r) = Al jl(kr)

Spherical Bessel functions

jl(x) = (−x)l(

)l sin x

j0(x) =sin x

j1(x) = (−x)1

(cos x

x− sin x

x2− cos x

j2(x) = (−x)2(

)x cos x − sin x

(−x sin x

x3− 3

x cos x − sin x

(3 sin x − 3x cos x − x2 sin x

0 5 10

j l(x)

Clearly the roots are not at nice, simple, locations!

jl(x) = (−x)l(

)l sin x

j0(x) =sin x

j1(x) = (−x)1

(cos x

x− sin x

x2− cos x

j2(x) = (−x)2(

)x cos x − sin x

(−x sin x

x3− 3

x cos x − sin x

0 5 10j 0

jl(x) = (−x)l(

)l sin x

j0(x) =sin x

j1(x) = (−x)1

(cos x

x− sin x

=sin x

x2− cos x

j2(x) = (−x)2(

)x cos x − sin x

(−x sin x

x3− 3

x cos x − sin x

0 5 10j 0

jl(x) = (−x)l(

)l sin x

j0(x) =sin x

j1(x) = (−x)1

(cos x

x− sin x

x2− cos x

j2(x) = (−x)2(

)x cos x − sin x

(−x sin x

x3− 3

x cos x − sin x

0 5 10j 1

jl(x) = (−x)l(

)l sin x

j0(x) =sin x

j1(x) = (−x)1

(cos x

x− sin x

x2− cos x

j2(x) = (−x)2(

)x cos x − sin x

(−x sin x

x3− 3

x cos x − sin x

0 5 10j 1

jl(x) = (−x)l(

)l sin x

j0(x) =sin x

j1(x) = (−x)1

(cos x

x− sin x

x2− cos x

j2(x) = (−x)2(

)x cos x − sin x

(−x sin x

x3− 3

x cos x − sin x

0 5 10j 1

jl(x) = (−x)l(

)l sin x

j0(x) =sin x

j1(x) = (−x)1

(cos x

x− sin x

x2− cos x

j2(x) = (−x)2(

)x cos x − sin x

(−x sin x

x3− 3

x cos x − sin x

0 5 10j 2

jl(x) = (−x)l(

)l sin x

j0(x) =sin x

j1(x) = (−x)1

(cos x

x− sin x

x2− cos x

j2(x) = (−x)2(

)x cos x − sin x

(−x sin x

x3− 3

x cos x − sin x

0 5 10j l(

Solutions for all l

jl(ka) = 0

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

)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

Solutions for all l

jl(ka) = 0

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

(βnl r

)Yml (θ, φ)

Solutions for all l

jl(ka) = 0

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

(βnl r

)Yml (θ, φ)

Solutions for all l

jl(ka) = 0

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

(βnl r

)Yml (θ, φ)

Solutions for all l

jl(ka) = 0

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

(βnl r

)Yml (θ, φ)

Solutions for all l

jl(ka) = 0

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

(βnl r

)Yml (θ, φ)

Solutions for all l

jl(ka) = 0

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

(βnl r

)Yml (θ, φ)

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

we can substitute this potential intothe radial equation

dividing by Eand rearranging

rewriting it withcommon terms

− ~2

[− e2

4πε0

l(l + 1)

]u = Eu

− ~2

4πε0E

r− ~2

l(l + 1)

− ~2

2πε0~2~2

r− ~2

l(l + 1)

initially we are only interested in bound states with E < 0 and so we canmake the usual substitution κ =

√−2mE/~

Hydrogen atom

V (r) = − e2

4πε0

− ~2

[− e2

4πε0

l(l + 1)

]u = Eu

− ~2

4πε0E

r− ~2

l(l + 1)

− ~2

2πε0~2~2

r− ~2

l(l + 1)

√−2mE/~

Hydrogen atom

V (r) = − e2

4πε0

− ~2

[− e2

4πε0

l(l + 1)

]u = Eu

− ~2

4πε0E

r− ~2

l(l + 1)

− ~2

2πε0~2~2

r− ~2

l(l + 1)

√−2mE/~

Hydrogen atom

V (r) = − e2

4πε0

− ~2

[− e2

4πε0

l(l + 1)

]u = Eu

− ~2

4πε0E

r− ~2

l(l + 1)

− ~2

2πε0~2~2

r− ~2

l(l + 1)

√−2mE/~

Hydrogen atom

V (r) = − e2

4πε0

− ~2

[− e2

4πε0

l(l + 1)

]u = Eu

− ~2

4πε0E

r− ~2

l(l + 1)

− ~2

2πε0~2~2

r− ~2

l(l + 1)

√−2mE/~

Hydrogen atom

V (r) = − e2

4πε0

− ~2

[− e2

4πε0

l(l + 1)

]u = Eu

− ~2

4πε0E

r− ~2

l(l + 1)

− ~2

2πε0~2~2

r− ~2

l(l + 1)

√−2mE/~

Hydrogen atom

V (r) = − e2

4πε0

− ~2

[− e2

4πε0

l(l + 1)

]u = Eu

− ~2

4πε0E

r− ~2

l(l + 1)

− ~2

2πε0~2~2

r− ~2

l(l + 1)

√−2mE/~

Hydrogen atom

V (r) = − e2

4πε0

− ~2

[− e2

4πε0

l(l + 1)

]u = Eu

− ~2

4πε0E

r− ~2

l(l + 1)

− ~2

2πε0~2~2

r− ~2

l(l + 1)

√−2mE/~

Hydrogen atom

V (r) = − e2

4πε0

− ~2

[− e2

4πε0

l(l + 1)

]u = Eu

− ~2

4πε0E

r− ~2

l(l + 1)

− ~2

2πε0~2~2

r− ~2

l(l + 1)

√−2mE/~

Hydrogen atom

V (r) = − e2

4πε0

− ~2

[− e2

4πε0

l(l + 1)

]u = Eu

− ~2

4πε0E

r− ~2

l(l + 1)

− ~2

2πε0~2~2

r− ~2

l(l + 1)

√−2mE/~

Hydrogen atom (cont.)

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

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

dρ2≈ u

u(ρ) = Ae−ρ +��Beρ

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

if we substitute

ρ ≡ κr , ρ0 ≡me2

2πε0~2κ

dρ2≈ u

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

if we substitute

ρ ≡ κr , ρ0 ≡me2

2πε0~2κ

dρ2≈ u

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

if we substitute

ρ ≡ κr , ρ0 ≡me2

2πε0~2κ

dρ2≈ u

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

if we substitute

ρ ≡ κr , ρ0 ≡me2

2πε0~2κ

dρ2≈ u

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

if we substitute

ρ ≡ κr , ρ0 ≡me2

2πε0~2κ

dρ2≈ u

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

if we substitute

ρ ≡ κr , ρ0 ≡me2

2πε0~2κ

dρ2≈ u

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

if we substitute

ρ ≡ κr , ρ0 ≡me2

2πε0~2κ

dρ2≈ u

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

if we substitute

ρ ≡ κr , ρ0 ≡me2

2πε0~2κ

dρ2≈ u

u(ρ) = Ae−ρ + Beρ

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

if we substitute

ρ ≡ κr , ρ0 ≡me2

2πε0~2κ

dρ2≈ u

κ2d2u

[1− me2

2πε0~2κ1

l(l + 1)

(κr)2

[1− ρ0

l(l + 1)

if we substitute

ρ ≡ κr , ρ0 ≡me2

2πε0~2κ

dρ2≈ u

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

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

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

this has a solution

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

ρ(l+2)

u(ρ) ∼ Cρl+1

this has a solution

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

ρ(l+2)

u(ρ) ∼ Cρl+1

this has a solution

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

ρ(l+2)

u(ρ) ∼ Cρl+1

this has a solution

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

ρ(l+2)

u(ρ) ∼ Cρl+1

this has a solution

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

ρ(l+2)

u(ρ) ∼ Cρl+1

this has a solution

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

ρ(l+2)

u(ρ) ∼ Cρl+1

this has a solution

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

ρ(l+2)

u(ρ) ∼ Cρl+1

this has a solution

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

ρ(l+2)

u(ρ) ∼ Cρl+1

this has a solution

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

ρ(l+2)

u(ρ) ∼ Cρl+1

this has a solution

l(l + 1)

u(ρ) = Cρl+1 +D

dρ= (l + 1)Cρl − l

ρ(l+1)

dρ2= l(l + 1)Cρl−1 + l(l + 1)

ρ(l+2)

u(ρ) ∼ Cρl+1

Asymptotic behavior

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 2 4 6 8 10

Asymptotic behavior

0 2 4 6 8 10

Asymptotic behavior

0 2 4 6 8 10

Asymptotic behavior

0 2 4 6 8 10

The wavy part

The polynomial can be determined by substituting this solution into theoriginal Schrodinger equation for u(ρ)

(l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρdv

= ρle−ρ[

(l + 1− ρ)v + ρdv

lρl−1e−ρ[

(l + 1− ρ)v + ρdv

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

= ρle−ρ{

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

(l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρdv

= ρle−ρ[

(l + 1− ρ)v + ρdv

lρl−1e−ρ[

(l + 1− ρ)v + ρdv

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

= ρle−ρ{

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

dρ= (l + 1)ρle−ρv

− ρl+1e−ρv + ρl+1e−ρdv

= ρle−ρ[

(l + 1− ρ)v + ρdv

lρl−1e−ρ[

(l + 1− ρ)v + ρdv

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

= ρle−ρ{

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

dρ= (l + 1)ρle−ρv − ρl+1e−ρv

+ ρl+1e−ρdv

= ρle−ρ[

(l + 1− ρ)v + ρdv

lρl−1e−ρ[

(l + 1− ρ)v + ρdv

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

= ρle−ρ{

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

dρ= (l + 1)ρle−ρv − ρl+1e−ρv + ρl+1e−ρ

= ρle−ρ[

(l + 1− ρ)v + ρdv

lρl−1e−ρ[

(l + 1− ρ)v + ρdv

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

= ρle−ρ{

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

= ρle−ρ[

(l + 1− ρ)v + ρdv

lρl−1e−ρ[

(l + 1− ρ)v + ρdv

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

= ρle−ρ{

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

= ρle−ρ[

(l + 1− ρ)v + ρdv

lρl−1e−ρ[

(l + 1− ρ)v + ρdv

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

]= ρle−ρ

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

= ρle−ρ[

(l + 1− ρ)v + ρdv

dρ2= lρl−1e−ρ

[(l + 1− ρ)v + ρ

− ρle−ρ[

(l + 1− ρ)v + ρdv

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

]= ρle−ρ

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

= ρle−ρ[

(l + 1− ρ)v + ρdv

[(l + 1− ρ)v + ρ

]− ρle−ρ

[(l + 1− ρ)v + ρ

+ ρle−ρ[−v + (l + 1− ρ)

dρ+ ρ

]= ρle−ρ

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

= ρle−ρ[

(l + 1− ρ)v + ρdv

[(l + 1− ρ)v + ρ

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

= ρle−ρ{

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

= ρle−ρ[

(l + 1− ρ)v + ρdv

[(l + 1− ρ)v + ρ

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

]= ρle−ρ

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

= ρle−ρ[

(l + 1− ρ)v + ρdv

[(l + 1− ρ)v + ρ

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

]= ρle−ρ

{ρd2v

+ 2(l + 1− ρ)dv

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

= ρle−ρ[

(l + 1− ρ)v + ρdv

[(l + 1− ρ)v + ρ

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

]= ρle−ρ

{ρd2v

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The wavy part

= ρle−ρ[

(l + 1− ρ)v + ρdv

[(l + 1− ρ)v + ρ

]− ρle−ρ

[(l + 1− ρ)v + ρ

]+ ρle−ρ

[−v + (l + 1− ρ)

dρ+ ρ

]= ρle−ρ

{ρd2v

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

The Schrodinger equation for v(ρ)

Substituting into the Schrodinger equation for u(ρ)

0 =d2u

dρ2−[

1− ρ0ρ

+l(l + 1)

= ρle−ρ{ρd2v

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

}− ρle−ρ

[ρ− ρ0 +

l(l + 1)

= ρd2v

dρ2+ 2(l + 1− ρ)

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

0 =d2u

dρ2−[

1− ρ0ρ

+l(l + 1)

= ρle−ρ{ρd2v

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

}− ρle−ρ

[ρ− ρ0 +

l(l + 1)

= ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v(ρ) =∞∑j=0

0 =d2u

dρ2−[

1− ρ0ρ

+l(l + 1)

= ρle−ρ{ρd2v

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

}− ρle−ρ

[ρ− ρ0 +

l(l + 1)

= ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v(ρ) =∞∑j=0

0 =d2u

dρ2−[

1− ρ0ρ

+l(l + 1)

= ρle−ρ{ρd2v

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

}− ρle−ρ

[ρ− ρ0 +

l(l + 1)

= ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v(ρ) =∞∑j=0

0 =d2u

dρ2−[

1− ρ0ρ

+l(l + 1)

= ρle−ρ{ρd2v

dρ2+ 2(l + 1− ρ)

[ρ− 2(l + 1) +

l(l + 1)

}− ρle−ρ

[ρ− ρ0 +

l(l + 1)

= ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v(ρ) =∞∑j=0

The recursion relation for v(ρ)

0 = ρd2v

dρ2+ 2(l + 1− ρ)

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

dρ=∞∑j=0

jcjρj−1

=∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

assume a polynomial solution

and take derivatives

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1

=∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1

=∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1

=∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1

=∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1 =

∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1 =

∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1 =

∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1 =

∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1 =

∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1 =

∞∑j=0

(j + 1)cj+1ρj

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

0 = ρd2v

dρ2+ 2(l + 1− ρ)

dρ+ [ρ0 − 2(l + 1)]v

v =∞∑j=0

dρ=∞∑j=0

jcjρj−1 =

∞∑j=0

(j + 1)cj+1ρj

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

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

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)

looking at the asymptotic behavior for large j

cj+1 '2j

j(j + 1)cj

j + 1cj −→ cj =

v(ρ) = c0

∞∑j=0

= c0e2ρ −→ u(ρ) = c0ρ

l+1eρ

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

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

cj+1 '2j

j(j + 1)cj

j + 1cj −→ cj =

v(ρ) = c0

∞∑j=0

= c0e2ρ −→ u(ρ) = c0ρ

l+1eρ

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

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

cj+1 '2j

j(j + 1)cj

j + 1cj −→ cj =

v(ρ) = c0

∞∑j=0

= c0e2ρ −→ u(ρ) = c0ρ

l+1eρ

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

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

cj+1 '2j

j(j + 1)cj

j + 1cj −→ cj =

v(ρ) = c0

∞∑j=0

= c0e2ρ −→ u(ρ) = c0ρ

l+1eρ

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

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

cj+1 '2j

j(j + 1)cj

j + 1cj −→ cj =

v(ρ) = c0

∞∑j=0

= c0e2ρ −→ u(ρ) = c0ρ

l+1eρ

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

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

cj+1 '2j

j(j + 1)cj

j + 1cj −→ cj =

v(ρ) = c0

∞∑j=0

= c0e2ρ −→ u(ρ) = c0ρ

l+1eρ

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

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

cj+1 '2j

j(j + 1)cj =

j + 1cj

−→ cj =2j

v(ρ) = c0

∞∑j=0

= c0e2ρ −→ u(ρ) = c0ρ

l+1eρ

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

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

cj+1 '2j

j(j + 1)cj =

j + 1cj −→ cj =

v(ρ) = c0

∞∑j=0

= c0e2ρ −→ u(ρ) = c0ρ

l+1eρ

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

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

cj+1 '2j

j(j + 1)cj =

j + 1cj −→ cj =

v(ρ) = c0

∞∑j=0

= c0e2ρ −→ u(ρ) = c0ρ

l+1eρ

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

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

cj+1 '2j

j(j + 1)cj =

j + 1cj −→ cj =

v(ρ) = c0

∞∑j=0

j!ρj = c0e

−→ u(ρ) = c0ρl+1eρ

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

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

cj+1 '2j

j(j + 1)cj =

j + 1cj −→ cj =

v(ρ) = c0

∞∑j=0

j!ρj = c0e

2ρ −→ u(ρ) = c0ρl+1eρ

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)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

2m= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

2m= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

2m= − me4

8π2ε20~2ρ20

En = −

4πε0

= −13.6eV

n2, n = 1, 2, 3, . . .

Limiting the sum

cj+1 =

[2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

0 = 2(jmax + l + 1)− ρ0

n ≡ jmax + l + 1

ρ0 = 2n

E = −~2κ2

2m= − me4

8π2ε20~2ρ20

En = −

4πε0

n2= −13.6eV

n2, n = 1, 2, 3, . . .

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

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1

m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1

m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1

m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ= 2n

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ= 2n

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ= 2n

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ= 2n

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ= 2n

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ= 2n

4πε0~2

a =4πε0~2

= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ= 2n

4πε0~2

a =4πε0~2

me2= 0.529× 10−10m

n = jmax + l + 1 → l = n − 1− jmax

n = 1, 2, · · · l = 0, 1, · · · , n − 1 m = 0,±1, · · · ,±l

solving for κ

d(n) =n−1∑l=0

(2l + 1) = n2

ρ0 =me2

2πε0~2κ= 2n

4πε0~2

a =4πε0~2

me2= 0.529× 10−10m

Hydrogen wave functions

It is now possible to assemble the entire hydrogen atom wave function

ψnlm(r , θ, φ) = Rnl(r)Yml (θ, φ)

)=∞∑j=0

)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

)l+1e−r/nav(r/na)

cj+1 =

{2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

v(ρ) = L2l+1n−l−1(2ρ)

Lpq−p(x) ≡ (−1)p(

Lq(x) = ex(

)q (e−xxq

)=∞∑j=0

Rnl(r) =1

)l+1e−r/nav(r/na)

cj+1 =

{2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

v(ρ) = L2l+1n−l−1(2ρ)

Lpq−p(x) ≡ (−1)p(

Lq(x) = ex(

)q (e−xxq

)=∞∑j=0

Rnl(r) =1

)l+1e−r/nav(r/na)

cj+1 =

{2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

v(ρ) = L2l+1n−l−1(2ρ)

Lpq−p(x) ≡ (−1)p(

Lq(x) = ex(

)q (e−xxq

)=∞∑j=0

but the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials

Rnl(r) =1

)l+1e−r/nav(r/na)

cj+1 =

{2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

v(ρ) = L2l+1n−l−1(2ρ)

Lpq−p(x) ≡ (−1)p(

Lq(x) = ex(

)q (e−xxq

)=∞∑j=0

but the v(r/na) functions are justa set of functions known as the as-sociated Laguerre polynomials

Rnl(r) =1

)l+1e−r/nav(r/na)

cj+1 =

{2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

v(ρ) = L2l+1n−l−1(2ρ)

Lpq−p(x) ≡ (−1)p(

Lq(x) = ex(

)q (e−xxq

)=∞∑j=0

Rnl(r) =1

)l+1e−r/nav(r/na)

cj+1 =

{2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

v(ρ) = L2l+1n−l−1(2ρ)

Lpq−p(x) ≡ (−1)p(

Lq(x) = ex(

)q (e−xxq

)=∞∑j=0

Rnl(r) =1

)l+1e−r/nav(r/na)

cj+1 =

{2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

v(ρ) = L2l+1n−l−1(2ρ)

Lpq−p(x) ≡ (−1)p(

Lq(x) = ex(

)q (e−xxq

)=∞∑j=0

Rnl(r) =1

)l+1e−r/nav(r/na)

cj+1 =

{2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

v(ρ) = L2l+1n−l−1(2ρ)

Lpq−p(x) ≡ (−1)p(

Lq(x) = ex(

)q (e−xxq

)=∞∑j=0

Rnl(r) =1

)l+1e−r/nav(r/na)

cj+1 =

{2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

v(ρ) = L2l+1n−l−1(2ρ)

Lpq−p(x) ≡ (−1)p(

Lq(x) = ex(

)q (e−xxq

)=∞∑j=0

Rnl(r) =1

)l+1e−r/nav(r/na)

cj+1 =

{2(j + l + 1)− ρ0

(j + 1)(j + 2l + 2)

v(ρ) = L2l+1n−l−1(2ρ)

Lpq−p(x) ≡ (−1)p(

Lq(x) = ex(

)q (e−xxq

)C. Segre (IIT) PHYS 405 - Fall 2018 October 09, 2018 17 / 42

Laguerre polynomials

Lq(x) = ex(

)q (e−xxq

L0 = 1

L1 = −x + 1

L2 = x2 − 4x + 2

associated Laguerre polynomials

Lpq−p(x) ≡ (−1)p(

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

Lq(x) = ex(

)q (e−xxq

)L0 = 1

L1 = −x + 1

L2 = x2 − 4x + 2

Lpq−p(x) ≡ (−1)p(

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

Lq(x) = ex(

)q (e−xxq

)L0 = 1

L1 = −x + 1

L2 = x2 − 4x + 2

Lpq−p(x) ≡ (−1)p(

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

Lq(x) = ex(

)q (e−xxq

)L0 = 1

L1 = −x + 1

L2 = x2 − 4x + 2

Lpq−p(x) ≡ (−1)p(

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

Lq(x) = ex(

)q (e−xxq

)L0 = 1

L1 = −x + 1

L2 = x2 − 4x + 2

Lpq−p(x) ≡ (−1)p(

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

Lq(x) = ex(

)q (e−xxq

)L0 = 1

L1 = −x + 1

L2 = x2 − 4x + 2

Lpq−p(x) ≡ (−1)p(

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

Lq(x) = ex(

)q (e−xxq

)L0 = 1

L1 = −x + 1

L2 = x2 − 4x + 2

Lpq−p(x) ≡ (−1)p(

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

Lq(x) = ex(

)q (e−xxq

)L0 = 1

L1 = −x + 1

L2 = x2 − 4x + 2

Lpq−p(x) ≡ (−1)p(

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

Lq(x) = ex(

)q (e−xxq

)L0 = 1

L1 = −x + 1

L2 = x2 − 4x + 2

Lpq−p(x) ≡ (−1)p(

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

Complete hydrogen solution

ψnlm =

)3 (n − l − 1)!

2n[(n + l)!]3e−r/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 =

)3 (0)!

2[(1)!]3e−r/a

)0 [L10(2r/a)

]Y 00 (θ, φ)

a3e−r/a

1√4π

=1√πa3

e−r/a

ψnlm =

)3 (n − l − 1)!

2n[(n + l)!]3e−r/na

)l [L2l+1n−l−1(2r/na)

]Yml (θ, φ)

ψ100 =

)3 (0)!

2[(1)!]3e−r/a

)0 [L10(2r/a)

]Y 00 (θ, φ)

a3e−r/a

1√4π

=1√πa3

e−r/a

ψnlm =

)3 (n − l − 1)!

2n[(n + l)!]3e−r/na

)l [L2l+1n−l−1(2r/na)

]Yml (θ, φ)

ψ100 =

)3 (0)!

2[(1)!]3e−r/a

)0 [L10(2r/a)

]Y 00 (θ, φ)

a3e−r/a

1√4π

=1√πa3

e−r/a

ψnlm =

)3 (n − l − 1)!

2n[(n + l)!]3e−r/na

)l [L2l+1n−l−1(2r/na)

]Yml (θ, φ)

ψ100 =

)3 (0)!

2[(1)!]3e−r/a

)0 [L10(2r/a)

]Y 00 (θ, φ)

a3e−r/a

1√4π

=1√πa3

e−r/a

ψnlm =

)3 (n − l − 1)!

2n[(n + l)!]3e−r/na

)l [L2l+1n−l−1(2r/na)

]Yml (θ, φ)

ψ100 =

)3 (0)!

2[(1)!]3e−r/a

)0 [L10(2r/a)

]Y 00 (θ, φ)

a3e−r/a

1√4π

=1√πa3

e−r/a

Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October...

Documents

Transcript of Today’s Outline - October 09, 2018segre/phys405/18F/lecture_14.pdf · Today’s Outline - October...