today’s outline - october 03, 2019csrri.iit.edu/~segre/phys405/19f/lecture_13.pdf · today’s...

Today’s Outline - October 03, 2019

• The 3D Schrodinger equation

• Separation of variables

• The angular solution

• Radial solutions

• Example 4.1 - Particle in a bubble

No lecture on Tuesday, October 8, and Thursday, October 10: Videos willbe prerecorded on Friday, October 4 in room 204 SB at 10:00 and 11:25and posted on blackboard

Homework Assignment #06:Chapter 3: 16,18,26,27,33,38due Tuesday, October 8, 2019, in my mailbox

C. Segre (IIT) PHYS 405 - Fall 2019 October 03, 2019 1 / 26

3D Schrodinger equation

The three-dimensionalSchrodinger equation,written in Cartesiancoordinates is

pn =~i

where the term in paren-theses is simply the Lapla-cian, ∇2

with volume elementd3~r = dx dy dz , we have

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

Ψn(~r , t) = ψn(~r)e−iEnt/~

pn =~i

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

pn =~i

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

pn =~i

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

pn =~i

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

pn =~i

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

pn =~i

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

pn =~i

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

pn =~i

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

pn =~i

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

pn =~i

if V = V (~r) only

i~∂Ψ

2mΨ + VΨ

2m(p2x + p2y + p2z )Ψ + VΨ

= − ~2

∂x2+

∂y2+

)Ψ + VΨ

i~∂Ψ

∂t= − ~2

2m∇2Ψ + VΨ

∫|Ψ(~r , t)|2 d3~r = 1

Spherical coordinates

Many common 3-dimensional potentials in physics are central and onlydepend on the distance from the origin. Thus spherical coordinates aremost useful.

z(r,θ,φ)

Using the radial distance r , the polar an-gle θ, and the azimuthal angle φ

, the in-finitesmal volume element can be written

dV = dr

rdθr sin θdφ

and the Laplacian then becomes

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

z(r,θ,φ)

dV = dr

rdθr sin θdφ

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

z(r,θ,φ)

dV = dr

rdθr sin θdφ

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

z(r,θ,φ)

Using the radial distance r , the polar an-gle θ, and the azimuthal angle φ, the in-finitesmal volume element can be written

dV = dr

rdθr sin θdφ

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

z(r,θ,φ)

dV = dr

rdθr sin θdφ

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

z(r,θ,φ)

dV = drrdθ

r sin θdφ

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

z(r,θ,φ)

dV = drrdθr sin θdφ

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

z(r,θ,φ)

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

z(r,θ,φ)

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

z(r,θ,φ)

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

z(r,θ,φ)

∇2 =1

(r2∂

r2sin θ

(sin θ

r2sin2 θ

∂φ2

)C. Segre (IIT) PHYS 405 - Fall 2019 October 03, 2019 3 / 26

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

the time-independent Schrodinger equation is thus

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

This is difficult to solve unless we can find separable solutions:

ψ(r , θ, φ) = R(r)Y (θ, φ) which gives

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

+ V (r)RY = E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

)the time-independent Schrodinger equation is thus

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

+ V (r)RY = E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

+ V (r)RY = E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

+ V (r)RY = E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

This is difficult to solve unless we can find separable solutions:ψ(r , θ, φ) = R(r)Y (θ, φ) which gives

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

+ V (r)RY = E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

+ V (r)RY = E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

+ V (r)RY = E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

+ V (r)RY = E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

+ V (r)RY = E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

+ V (r)RY = E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+ V (r)RY

= E RY

∇2 =1

(r2∂

r2 sin θ

(sin θ

r2 sin2 θ

∂φ2

− ~2

(r2∂ψ

r2 sin θ

(sin θ

r2 sin2 θ

(∂2ψ

∂φ2

)]+V (r)ψ = Eψ

− ~2

(r2∂RY

r2 sin θ

(sin θ

r2 sin2 θ

(∂2RY

∂φ2

)]+V (r)RY = ERY

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+ V (r)RY = E RY

Separation of variables

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

+ V (r) = E

multiplying by −2mr2/~2 and bringing the E over[

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

finally, we regroup the radial and angular parts

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

)]+ V (r)

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

)]+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

)]+ V (r) = E

multiplying by −2mr2/~2 and bringing the E over

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

)]+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

)]+ V (r) = E

multiplying by −2mr2/~2 and bringing the E over[1

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

)]+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

)]+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

− 2mr2

~2[V (r)− E ]

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

)]+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

)]− 2mr2

~2[V (r)− E ] = 0

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

)]+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

)]− 2mr2

~2[V (r)− E ] = 0

finally, we regroup the radial

and angular parts

− ~2

(r2∂R

r2 sin θ

(sin θ

r2 sin2 θ

(∂2Y

∂φ2

)]+V (r)RY = E RY

dividing by RY

− ~2

Rr2∂

(r2∂R

Yr2 sin θ

(sin θ

Yr2 sin2 θ

(∂2Y

∂φ2

)]+ V (r) = E

(r2∂R

Y sin θ

(sin θ

Y sin2 θ

(∂2Y

∂φ2

)]− 2mr2

~2[V (r)− E ] = 0

finally, we regroup the radial and angular partsC. Segre (IIT) PHYS 405 - Fall 2019 October 03, 2019 5 / 26

The radial and angular equations{1

(r2∂R

)− 2mr2

~2[V (r)− E ]

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

Since the two parts depend on different variables, the only way they cancancel is to each be equal to a separation constant, which we will take tobe l(l + 1). This gives two independent equations:{

)− 2mr2

~2[V (r)− E ]

}= l(l + 1)

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

We will first solve the angular equation

(r2∂R

)− 2mr2

~2[V (r)− E ]

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

Since the two parts depend on different variables, the only way they cancancel is to each be equal to a separation constant, which we will take tobe l(l + 1). This gives two independent equations:

)− 2mr2

~2[V (r)− E ]

}= l(l + 1)

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

(r2∂R

)− 2mr2

~2[V (r)− E ]

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)− 2mr2

~2[V (r)− E ]

}= l(l + 1)

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

(r2∂R

)− 2mr2

~2[V (r)− E ]

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)− 2mr2

~2[V (r)− E ]

}= l(l + 1)

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

(r2∂R

)− 2mr2

~2[V (r)− E ]

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)− 2mr2

~2[V (r)− E ]

}= l(l + 1)

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

The angular solution

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

begin by multiplying by Y sin2 θ

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

This clearly is separable in the two angular variables by assumingY (θ, φ) = Θ(θ)Φ(φ){

[sin θ

(sin θ

+ ΘΦl(l + 1) sin2 θ

final separation can be achieved by dividing by Θ(θ)Φ(φ)

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

[sin θ

(sin θ

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

[sin θ

(sin θ

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

[sin θ

(sin θ

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

[sin θ

(sin θ

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

[sin θ

(sin θ

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

[sin θ

(sin θ

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

This clearly is separable in the two angular variables by assumingY (θ, φ) = Θ(θ)Φ(φ)

[sin θ

(sin θ

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

[sin θ

(sin θ

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

[sin θ

(sin θ

)]+ ΘΦl(l + 1) sin2 θ

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

[sin θ

(sin θ

)]+ ΘΦl(l + 1) sin2 θ

sin θ

(sin θ

sin2 θ

(∂2Y

∂φ2

)}= −l(l + 1)

sin θ∂

(sin θ

(∂2Y

∂φ2

)= −l(l + 1) sin2 θY

rearranging

sin θ∂

(sin θ

)+ l(l + 1) sin2 θY +

(∂2Y

∂φ2

[sin θ

(sin θ

)]+ ΘΦl(l + 1) sin2 θ

final separation can be achieved by dividing by Θ(θ)Φ(φ)C. Segre (IIT) PHYS 405 - Fall 2019 October 03, 2019 7 / 26

Second angular separation constant{1

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

The separation constant is taken to be m2

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

−m2 =1

)A look ahead:

We now have identified 3 separation constants of the 3DSchrodinger equation, each of which corresponds to a quantum numberand operator which commutes with the others.

Separation constant Operator Quantum number

l(l+1) L2

m2 Lz m

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

−m2 =1

)A look ahead:

l(l+1) L2

m2 Lz m

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

−m2 =1

)A look ahead:

l(l+1) L2

m2 Lz m

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

−m2 =1

A look ahead:

l(l+1) L2

m2 Lz m

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

−m2 =1

)A look ahead:

l(l+1) L2

m2 Lz m

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

−m2 =1

)A look ahead: We now have identified 3 separation constants of the 3DSchrodinger equation, each of which corresponds to a quantum numberand operator which commutes with the others.

l(l+1) L2

m2 Lz m

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

−m2 =1

l(l+1) L2

m2 Lz m

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

−m2 =1

l(l+1) L2

m2 Lz m

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

−m2 =1

l(l+1) L2

m2 Lz m

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

[sin θ

(sin θ

)]+ l(l + 1) sin2 θ

−m2 =1

l(l+1) L2

m2 Lz m

Solution for Φ(φ)

Rearranging the equation for Φgives a straightforward differen-tial equation to solve

if we insist on continuity,Φ(φ) = Φ(φ + 2π) we can putconditions on the value of m

dφ2= −m2Φ

Φ(φ) = Ae±imφ

Ae±imφ = Ae±im(φ+2π) = Ae±imφe±2πim

m = 0,±1,±2, . . .

m is called the “magnetic” quantum number and must be an integer. Wewill have to impose more restrictions later.

The integration constant A can be absorbed into the constant determinedfor the solution of the Θ equation, and we have taken care of the negativesolutions by letting m be negative, so the final solution is

Φ(φ) = e imφ, m = 0,±1,±2,±3, . . .

Solution for Φ(φ)

dφ2= −m2Φ

Φ(φ) = Ae±imφ

m = 0,±1,±2, . . .

Φ(φ) = e imφ, m = 0,±1,±2,±3, . . .

Solution for Φ(φ)

dφ2= −m2Φ

Φ(φ) = Ae±imφ

m = 0,±1,±2, . . .

Φ(φ) = e imφ, m = 0,±1,±2,±3, . . .

Solution for Φ(φ)

dφ2= −m2Φ

Φ(φ) = Ae±imφ

m = 0,±1,±2, . . .

Φ(φ) = e imφ, m = 0,±1,±2,±3, . . .

Solution for Φ(φ)

dφ2= −m2Φ

Φ(φ) = Ae±imφ

Ae±imφ = Ae±im(φ+2π)

= Ae±imφe±2πim

m = 0,±1,±2, . . .

Φ(φ) = e imφ, m = 0,±1,±2,±3, . . .

Solution for Φ(φ)

dφ2= −m2Φ

Φ(φ) = Ae±imφ

m = 0,±1,±2, . . .

Φ(φ) = e imφ, m = 0,±1,±2,±3, . . .

Solution for Φ(φ)

dφ2= −m2Φ

Φ(φ) = Ae±imφ

m = 0,±1,±2, . . .

Φ(φ) = e imφ, m = 0,±1,±2,±3, . . .

Solution for Φ(φ)

dφ2= −m2Φ

Φ(φ) = Ae±imφ

m = 0,±1,±2, . . .

Φ(φ) = e imφ, m = 0,±1,±2,±3, . . .

Solution for Φ(φ)

dφ2= −m2Φ

Φ(φ) = Ae±imφ

m = 0,±1,±2, . . .

Φ(φ) = e imφ, m = 0,±1,±2,±3, . . .

Solution for Φ(φ)

dφ2= −m2Φ

Φ(φ) = Ae±imφ

m = 0,±1,±2, . . .

Φ(φ) = e imφ, m = 0,±1,±2,±3, . . .

Solution for Θ(θ)

The equation for Θ(θ), after rearranging, is

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

This has a more com-plex solution. Lookingat the first term

sin θd

(sin θ

)= sin2 θ

dθ2+ sin θ cos θ

What must be the form of the function Θ(θ)?

A polynomial in cos θ andsin θ will satisfy this differential equation.

The solution is an associated Legendre function, Pml :

Θ(θ) = APml (cos θ)

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Solution for Θ(θ)

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

sin θd

(sin θ

)= sin2 θ

dθ2+ sin θ cos θ

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Solution for Θ(θ)

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

sin θd

(sin θ

)= sin2 θ

dθ2+ sin θ cos θ

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Solution for Θ(θ)

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

sin θd

(sin θ

= sin2 θd2Θ

dθ2+ sin θ cos θ

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Solution for Θ(θ)

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

sin θd

(sin θ

)= sin2 θ

+ sin θ cos θdΘ

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Solution for Θ(θ)

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

sin θd

(sin θ

)= sin2 θ

dθ2+ sin θ cos θ

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Solution for Θ(θ)

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

sin θd

(sin θ

)= sin2 θ

dθ2+ sin θ cos θ

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Solution for Θ(θ)

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

sin θd

(sin θ

)= sin2 θ

dθ2+ sin θ cos θ

What must be the form of the function Θ(θ)? A polynomial in cos θ andsin θ will satisfy this differential equation.

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Solution for Θ(θ)

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

sin θd

(sin θ

)= sin2 θ

dθ2+ sin θ cos θ

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Solution for Θ(θ)

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

sin θd

(sin θ

)= sin2 θ

dθ2+ sin θ cos θ

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Solution for Θ(θ)

sin θd

(sin θ

)+[l(l + 1) sin2 θ −m2

]Θ = 0

sin θd

(sin θ

)= sin2 θ

dθ2+ sin θ cos θ

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

Legendre polynomials

Pl (x) is the l th Legendre Polyno-mial which is defined by the Ro-drigues formula

Pl (x) ≡ 1

(x2 − 1)l

P0 = 1

P1 = x

2(3x2 − 1)

2(5x3 − 3x)

8(35x4 − 30x2 + 3)

8(63x5 − 70x3 + 15x)

-1 0 +1

Pl (x) ≡ 1

(x2 − 1)l

P0 = 1

P1 = x

2(3x2 − 1)

2(5x3 − 3x)

8(35x4 − 30x2 + 3)

8(63x5 − 70x3 + 15x)

-1 0 +1

Pl (x) ≡ 1

(x2 − 1)l

P0 = 1

P1 = x

2(3x2 − 1)

2(5x3 − 3x)

8(35x4 − 30x2 + 3)

8(63x5 − 70x3 + 15x)

-1 0 +1

Pl (x) ≡ 1

(x2 − 1)l

P0 = 1

P1 = x

2(3x2 − 1)

2(5x3 − 3x)

8(35x4 − 30x2 + 3)

8(63x5 − 70x3 + 15x)

-1 0 +1

Pl (x) ≡ 1

(x2 − 1)l

P0 = 1

P1 = x

2(3x2 − 1)

2(5x3 − 3x)

8(35x4 − 30x2 + 3)

8(63x5 − 70x3 + 15x)

-1 0 +1

Pl (x) ≡ 1

(x2 − 1)l

P0 = 1

P1 = x

2(3x2 − 1)

2(5x3 − 3x)

8(35x4 − 30x2 + 3)

8(63x5 − 70x3 + 15x)

-1 0 +1

Pl (x) ≡ 1

(x2 − 1)l

P0 = 1

P1 = x

2(3x2 − 1)

2(5x3 − 3x)

8(35x4 − 30x2 + 3)

8(63x5 − 70x3 + 15x)

-1 0 +1

Pl (x) ≡ 1

(x2 − 1)l

P0 = 1

P1 = x

2(3x2 − 1)

2(5x3 − 3x)

8(35x4 − 30x2 + 3)

8(63x5 − 70x3 + 15x)

-1 0 +1

Pl (x) ≡ 1

(x2 − 1)l

P0 = 1

P1 = x

2(3x2 − 1)

2(5x3 − 3x)

8(35x4 − 30x2 + 3)

8(63x5 − 70x3 + 15x)

-1 0 +1

Associated Legendre functions

Unlike the Legendre polynomials, the Associated Legendre functions arenot necessarily polynomials

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

for the specific valueof l = 2, we startwith the Legendrepolynomial

2(3x2 − 1)

imposing the condi-tion |m| ≤ l

(3x2 − 1

(cos2 θ − 1

P12 = −

(1− x2

)1/2· 3x

= −3 cos θ sin θ

(1− x2

= 3 sin2 θ

P32 ≡ 0

if x = cos θ, the Pml become polynomials

Thus, the meaningful values of m are integers which range from −l to +land the full solution for both quantum numbers is:

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

P12 = −

(1− x2

)1/2· 3x

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

P12 = −

(1− x2

)1/2· 3x

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

P12 = −

(1− x2

)1/2· 3x

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

)P12 = −

(1− x2

)1/2· 3x

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

P12 = −

(1− x2

)1/2· 3x

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

P12 = −

(1− x2

)1/2· 3x

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

P12 = −

(1− x2

)1/2· 3x

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

P12 = −

(1− x2

)1/2· 3x

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

P12 = −

(1− x2

)1/2· 3x

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

)P12 = −

(1− x2

)1/2· 3x

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

)P12 = −

(1− x2

)1/2· 3x = −3 cos θ sin θ

(1− x2

= 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

)P12 = −

(1− x2

)1/2· 3x = −3 cos θ sin θ

(1− x2

)· 3 = 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

)P12 = −

(1− x2

)1/2· 3x = −3 cos θ sin θ

(1− x2

)· 3 = 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

2(3x2 − 1)

(3x2 − 1

(cos2 θ − 1

)P12 = −

(1− x2

)1/2· 3x = −3 cos θ sin θ

(1− x2

)· 3 = 3 sin2 θ

P32 ≡ 0

l = 0, 1, 2, . . . ; m = −l ,−(l − 1), . . . ,−1, 0, 1, . . . (l − 1), l

Angular wave functions

The full angular solutions, Y are

Y (θ, φ) = APml (cos θ)e imφ, Pm

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P12 = −3 sin θ cos θ

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P23 = 15 sin2 θ cos θ

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

Y (θ, φ) = APml (cos θ)e imφ

, Pml (x) ≡ (−1)m(1− x2)|m|/2

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

l (x) ≡ (−1)m(1− x2)|m|/2(

)|m|Pl (x)

P00 = 1

P01 = cos θ

P11 = − sin θ

P02 = 0.5(3 cos2 θ − 1)

P22 = 3 sin2 θ

P03 = 0.5(5 cos3 θ − 3 cos θ)

P13 = −1.5 sin θ(5 cos2 θ − 1)

P33 = −15 sin θ(1− cos2 θ)

-1 -0.5 0 0.5 1

3D angular wavefunctions – l = 1, 0

|Y 00 |

|Y 00 | |Y−11 | |Y 0

1 | |Y+11 |

Re{Y 00 }

Re{Y 00 } Im{Y−11 } Im{Y 0

1 } Im{Y+11 }

|Y 00 |

|Y 00 | |Y−11 | |Y 0

1 | |Y+11 |

Im{Y 00 }

Re{Y 00 } Im{Y−11 } Im{Y 0

1 } Im{Y+11 }

|Y 00 |

|Y−11 | |Y 01 | |Y+1

Im{Y 00 }

Re{Y 00 }

Re{Y−11 } Re{Y 01 } Re{Y+1

|Y 00 |

|Y−11 | |Y 01 | |Y+1

Im{Y 00 }

Re{Y 00 }

Im{Y−11 } Im{Y 01 } Im{Y+1

3D angular wavefunctions – l = 2

|Y−22 | |Y−12 | |Y 02 | |Y+1

2 | |Y+22 |

Re{Y−22 } Re{Y−12 } Re{Y 02 } Re{Y+1

2 } Re{Y+22 }

|Y−22 | |Y−12 | |Y 02 | |Y+1

2 | |Y+22 |

Im{Y−22 } Im{Y−12 } Im{Y 02 } Im{Y+1

2 } Im{Y+22 }

|Y−33 | |Y−23 | |Y−13 |

|Y 03 |

Re{Y−33 } Re{Y−23 } Re{Y−13 }

Re{Y 03 }

|Y−33 | |Y−23 | |Y−13 |

|Y 03 |

Im{Y−33 } Im{Y−23 } Im{Y−13 }

Re{Y 03 }

Im{Y 03 }

|Y 03 |

|Y+13 | |Y+2

3 | |Y+33 |

Re{Y 03 }

Re{Y=13 } Re{Y+2

3 } Re{Y+33 }

|Y 03 |

|Y+13 | |Y+2

3 | |Y+33 |

Im{Y 03 }

Re{Y 03 }

Im{Y+13 } Im{Y+2

3 } Im{Y+33 }

Spherical harmonics

When normalized, these angular functions are called spherical harmonics

Y ml (θ, φ) = ε

√(2l + 1)(l − |m|)!

4π(l + |m|)!e imφPm

l (cos θ), ε =

{(−1)m, m ≥ 0

1, m ≤ 0

∫ 2π

∫ π

l (θ, φ)]∗[Y m′l ′ (θ, φ)] sin θ dθ dφ = δll ′δmm′

where l is called the azimuthal quantum number and m is called themagnetic quantum number.

Now there is only the radial equation to be solved, but this depends on thespecifics of the potential.

Spherical harmonics

Y ml (θ, φ) = ε

√(2l + 1)(l − |m|)!

l (cos θ), ε =

{(−1)m, m ≥ 0

1, m ≤ 0

∫ 2π

∫ π

Spherical harmonics

Y ml (θ, φ) = ε

√(2l + 1)(l − |m|)!

l (cos θ), ε =

{(−1)m, m ≥ 0

1, m ≤ 0

∫ 2π

∫ π

Spherical harmonics

Y ml (θ, φ) = ε

√(2l + 1)(l − |m|)!

l (cos θ), ε =

{(−1)m, m ≥ 0

1, m ≤ 0

∫ 2π

∫ π

Spherical harmonics

Y ml (θ, φ) = ε

√(2l + 1)(l − |m|)!

l (cos θ), ε =

{(−1)m, m ≥ 0

1, m ≤ 0

∫ 2π

∫ π

Radial solution

Returning to the radial equation, where the potential is included

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

A change of variables to u(r) ≡ rR(r) simplifies the equation

dr− u

dr2− du

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Radial solution

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

dr− u

dr2− du

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Radial solution

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

dr− u

dr2− du

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Radial solution

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

dr− u

dr2− du

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Radial solution

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

dr− u

dr2− du

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Radial solution

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

dr− u

dr2− du

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Radial solution

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

dr− u

dr2− du

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Radial solution

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

dr− u

dr2− du

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Radial solution

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

dr− u

dr2− du

= rd2u

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Radial solution

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

dr− u

dr2− du

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Radial solution

)− 2mr2

~2[V (r)− E ]R = l(l + 1)R

dr− u

dr2− du

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

Equation for u(r)

We can now obtain a differential equation for u(r)

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

dr2− 2mr2

~2[V (r)− E ] u = l(l + 1)u

dr2−[

~2V (r)u + l(l + 1)u

]= −2mr2

− ~2

[V (r) +

l(l + 1)

]u = Eu

This looks like a time-independent Schrodinger equation with an effectivepotential Veff (r) whose solution is normalized as

∫∞0 |u(r)|2 dr = 1

Equation for u(r)

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

dr2− 2mr2

~2[V (r)− E ] u = l(l + 1)u

dr2−[

~2V (r)u + l(l + 1)u

]= −2mr2

− ~2

[V (r) +

l(l + 1)

]u = Eu

∫∞0 |u(r)|2 dr = 1

Equation for u(r)

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

dr2− 2mr2

~2[V (r)− E ] u = l(l + 1)u

dr2−[

~2V (r)u + l(l + 1)u

]= −2mr2

− ~2

[V (r) +

l(l + 1)

]u = Eu

∫∞0 |u(r)|2 dr = 1

Equation for u(r)

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

dr2− 2mr2

~2[V (r)− E ] u = l(l + 1)u

dr2−[

~2V (r)u + l(l + 1)u

]= −2mr2

− ~2

[V (r) +

l(l + 1)

]u = Eu

∫∞0 |u(r)|2 dr = 1

Equation for u(r)

dr2− 2mr2

~2[V (r)− E ]

r= l(l + 1)

dr2− 2mr2

~2[V (r)− E ] u = l(l + 1)u

dr2−[

~2V (r)u + l(l + 1)u

]= −2mr2

− ~2

[V (r) +

l(l + 1)

]u = Eu

∫∞0 |u(r)|2 dr = 1

Example 4.1

Consider the potential for the infi-nite spherical well

inside the well the wave equation is

dividing by ~2/2m and rearranging

substituting k =√

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− 2mE

[l(l + 1)

r2− k2

We must solve this for each value of l separately. The l = 0 case is thesimplest

Example 4.1

substituting k =√

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− 2mE

[l(l + 1)

r2− k2

Example 4.1

substituting k =√

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− 2mE

[l(l + 1)

r2− k2

Example 4.1

substituting k =√

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− 2mE

[l(l + 1)

r2− k2

Example 4.1

substituting k =√

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− 2mE

[l(l + 1)

r2− k2

Example 4.1

substituting k =√

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− 2mE

[l(l + 1)

r2− k2

Example 4.1

substituting k =√

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− 2mE

[l(l + 1)

r2− k2

Example 4.1

substituting k =√

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− 2mE

[l(l + 1)

r2− k2

Example 4.1

substituting k =√

V (r) =

{0, r ≤ a

∞, r ≥ a

Eu = − ~2

l(l + 1)

[l(l + 1)

r2− 2mE

[l(l + 1)

r2− k2

Example 4.1 (cont.)

For l = 0, we need to impose theboundary condition u(a) = 0.

this has the usual solution

since R(r) = u/r , the cos(kr) solu-tion would be infinite at the originand thus B = 0

applying the boundary condition

thus, ka = nπ where n is an integerand the energies for the l = 0 caseare

Normalizing (A =√

2/a) andadding the angular part of the wavefunction Y 0

0 = 1/√

dr2= −k2u

u(r) = A sin(kr) +��B cos(kr)

= A sin(kr)

u(a) = A sin(ka) = 0

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

u(r) = A sin(kr) + B cos(kr)

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

u(r) = A sin(kr) + B cos(kr)

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Example 4.1 (cont.)

Normalizing (A =√

0 = 1/√

dr2= −k2u

= A sin(kr)

En0 =n2π2~2

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

ψn00 =1√2πa

sin(nπr/a)

Solutions for l = 0

ψn00 =1√2πa

sin(nπr/a)

Solutions for l = 0

ψn00 =1√2πa

sin(nπr/a)

Solutions for l = 0

ψn00 =1√2πa

sin(nπr/a)

Solutions for l = 0

ψn00 =1√2πa

sin(nπr/a)

Solutions for l = 0

ψn00 =1√2πa

sin(nπr/a)

Solutions for l = 0

ψn00 =1√2πa

sin(nπr/a)

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

l (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

l (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

l (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

l (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

l (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

l (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

l (θ, φ)

today’s outline - october 03, 2019csrri.iit.edu/~segre/phys405/19f/lecture_13.pdf · today’s...

Documents