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Lecture 16 of a quantum mechanics course, focusing on working the angles, spherical harmonics, and angular momentum. Topics include the end of one-dimensional systems, skipping chapters on path integrals and uncertainty, three-dimensional systems, spherical geometry, and the time-dependent schrödinger equation. Emphasis is placed on bound vs scattering energies.
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end of 1-D, SHO via Dirac [section 7.4]
skip Chapter 8 on Path Integrals...(don’t read)
skip Chapter 9 on Uncertainty bits...(read)
3-D Systems [section 10.2] (might be back to 10)
skip Chapter 11 on Symmetries...(might also be back)
Spherical geometry...
`,m
Harmonics [section 7.5]
2 , L
x , L
y , L
z (^) , L
±
/ Rotation ops [section 7.2,7.3,7.4,7.5]
1-D problems: work in position or momentum space..
SHO: ladder operators, along with energy eigenstates
ˆa ±
mω
i ˆp
mω
ˆx )
ˆx
=
mω
ˆa
ˆ
a − )
and
ˆp
=
i √
mω 2
ˆa
−
ˆa − )
Given the ground state soln (and its energy):
ψ
0 ( x ) =
mω
ℏ ) 1 / 4 e ( −
(^) mω 2 ℏ
x 2 )
and
0
=
ω
Easy to generalise
time-dependent
1-D S.E. to 3-D:
i ℏ
∂t∂
(^) | Ψ(
r , t
) 〉
=
r , t
) 〉
just modify the classical Hamiltonian operator:
x, y, z
m
p x 2
p y 2
p z 2 )
x, y, z
Wherein replace:
p x
→ −
i ℏ
∂ ∂x
(^) ,
p y
→ −
i ℏ
∂ ∂y
(^) ,
p z
→ −
i ℏ
∂z∂
(^).
i ℏ
∂t ∂
2
m
2 Ψ +
‘Laplacian’
2
≡
∂ 2
∂ 2 x + ∂ 2
∂ 2 y + ∂ 2
∂ 2 z (^).
canonical commutation relations
infinite 3-D square well (separation of variables)
Time-indep S.E.:
ℏ 2
2 m (^) ∇
2 ψ
V ψ
Eψ
Atoms etc... intuitive to work in spherical not cartesian
where, for example, the potential is only
r ) .
z
=
r (^) cos
(^) θ
x
r (^) cos
(^) φ
sin
(^) θ
y
=
r
sin
(^) φ
(^) sin
(^) θ
r = √ x 2 + y 2 + z 2
cos
(^) θ
r z
tan
(^) φ
x y
Applying chain rule...
∂x
∂x ∂r
∂r
∂x ∂θ
∂θ
∂x ∂φ
∂φ
gets us to the Laplacian in Spherical coordinates:
2
=
r 1 2
∂r
r 2
∂ ∂r
r 2 sin
(^) θ
∂θ
sin
(^) θ
∂θ
r
2
sin
2
θ
(
2
2 φ
)
r ) +
θ, φ
So introduce an (arbitrary) separation constant
r ) =
, and
θ, φ
(^1) Consider the angular equation first (radial later!)
sin
θ
∂θ
sin
θ
∂θ ∂Y
sin
2 θ ( ∂ 2 Y
2 φ
)]
sin
(^) θ
∂θ
sin
θ
∂θ ∂Y
∂ 2 φ ) = − (
sin
2 θ
Try further separation of variables
θ, φ
θ )Φ(
φ ) :
sin
(^) θ
∂θ
sin
θ ∂ Θ
∂θ
2 θ ]
2 Φ
2 φ
So introduce another separation constant
m
2 ...
Another separation constant
m
2
means two angular eqns:
sin
(^) θ
∂θ
sin
θ ∂ Θ
∂θ
2 θ
=
m
2
2 Φ
∂ 2 φ = − m 2
The second eqn has solns
φ ) =
exp(
imφ
Boundary condition on aziumuthal:
φ
π ) = Φ(
φ )
e imφ
e 2 πim
e imφ
thus
exp(
πim
so we have
m
The coefficient
we absorb over into the polar soln...
Legendre polynomial
` ( x )
has degree
with parity (eveness or oddness) depending on
assoc Legendre funks
` (^) m
( x ) = (
x 2 ) | m |
2
d dx
| m | P ` ( x ) :
Note
for
m
then
` (^) m
( x ) = 0
(so
possible
m
For
(^0 )
= 1
. For
(^1 )
= cos
(^) θ, P
(^1 )
= sin
(^) θ
(^2 ) (^) ( x ) =
2 ( x ) =
x 2 −
(^2 ) (^) ( x ) =
√ 1 − x 2 d
dx
2 ( x ) = 3
x √ 1 − x 2
(^2 ) (^) ( x ) = (
− x 2 ) ( d
dx
) 2 P 2 ( x
x 2 )
(^2 )
=
2 1 (^) (3 cos
2 θ
−
(^2 )
= 3 sin
(^) θ
(^) cos
(^) θ, P
(^2 )
= 3 sin
2 θ
Finally... we need to normalise these angular functions,
ψ
( r, θ, φ
R ( r ) Y (
θ, φ
R ( r ) Y (
θ, φ
r )Θ(
θ )Φ(
φ ) :
∫ | ψ | 2 d 3 r = ∫ | ψ (
r, θ, φ
2
r 2 sin
(^) θ drdθdφ
= ∫ | R | 2 r 2
dr
θ, φ
2 sin
(^) θ dθdφ
require both radial and angular integrals normalised:
∞
0 | R | 2 r 2
dr
and
2 π
0
π
0
θ, φ
2 sin
(^) θ dθdφ
normalised angular functions are spherical harmonics:
m l
θ, φ
π
m
( ` + | m |
e imφ
l (^) m
(cos
(^) θ
)
azimuthal quantum number,
m
magnetic quantum
separation of variables gave
ψ
n,`,m
` (^) ( r, θ, φ
n ` m
` 〉
spherical harmonics
m `
`
θ, φ
are the eigenfunctions of
2 Y
m `
`
2
sin
(^) θ
∂θ
sin
(^) θ
∂θ
sin
(^) θ
∂φ
m
`
= ℏ 2 ( `m
`
`
z (^) Y
m `
` = [ − i ℏ ∂
∂φ
m `
= ℏ m Y m `
`
ie. good quantum numbers
n ` m
` 〉 = E n |
n ` m
` 〉 ,
Shankar 12.2: operator formalism, rotate in
x
−
y
plane:
φ 0 k~ )] = lim
n →∞
ℏ i
N φ
L z ) N = e
iφ
0 L z (^) / ℏ
in position-space:
e − φ 0 ∂/∂φ
ψ
( ρ, φ
ψ
( ρ, φ
φ
0 )
Following Shankar 12.5: Assume angular basis states...
2 | αβ
α | αβ
and
z (^) | αβ
β | αβ
Define raising/lowering operators:
L ± = L x ±
iL
y
they commute as
z (^) , L
± ] =
±
and
2 , L
± ] = 0
z (^) ( L
| αβ
L + L z + ℏ L + ) |
αβ
L + β + ℏ L + ) |
αβ
β + ℏ ) L + |
αβ
similarly
αβ
L
2 | αβ
αL
| αβ
but there are bounds on
α − β 2 ≥ 0
since
αβ
2 −
z 2 (^) | αβ
αβ
x 2
y 2 | αβ
| αβ
max
and
−
| αβ
min