Download Lecture 29: Motion in a Central Potential Phy851 Fall 2009

Lecture 29:
Motion in a Central Potential
Phy851 Fall 2009
Side Remarks
• Counting quantum numbers:
– 3N quantum numbers to specify a basis
state for N particles in 3-dimensions
– It will go up to 5N when we include spin
– When does it work:
• All of the standard basis choices
– Position eigenstates, Momentum eigenstates,
angular momentum eigenstates, …
• Any basis formed from energy eigenstates
of an analytically solvable system:
– Harmonic oscillator states, hydrogen orbitals,
...
– These problems are solvable due to a high
degree of symmetry
• Any basis formed from eigenstates of an
exactly solvable system plus a weak
symmetry breaking perturbation
– We can watch the levels evolve as we
increase the perturbation strength, and
therefore keep track of the quantum
numbers
– When it does not work
• Strongly interacting systems with minimal
symmetry
– These are problems that you could only
solve numerically, they won’t be
encountered in class or in textbooks
`And’ versus ‘or’ paradigm
I = I (C ) ⊗ I ( R )
(R)
(R)
I ( R ) = I bound
+ I continuum
• Hilbert subspace Hierarchy:
H
H (C ) , H ( R )
(R)
(R)
H bound
, H continuum
– To specify the state of the full system, we
(R)
must specify a state in H
AND a
(C )
state in H
– To specify the state of the relative motion
(R)
H
we may specify a state entirely
in
bound
(R)
OR a state entirely in H continuum OR a
state partially in both
Review of Separation of Variables and
Angular Momentum
H = H CM + H r
E = ECM
(C )
⊗ Er
(R)
E = ECM + Er
• The center-of-mass motion is that of a free
particle
• We thus only need to determine to state of
relative motion:
Pr2
L2
Hr =
+
+ V ( R)
2
2 µ 2 µR
• As long as V depends only r and not on
θ, φ
then simultaneous eigenstates of Hr , L2 and
Lz exist:
[ L2 , R] = 0
[ L2 , Pr ] = 0
[ Lz , R ] = 0
[ Lz , Pr ] = 0
€
Simultaneous Eigenstates of Energy
and Angular Momentum
E
E > Ec :
The spectrum of
Hr typically has
bound-states
and continuum
states
Continuum
r
V(r)
E < E c:
Bound States
• Bound state basis:
n, l, m = n, l
{ n,l,m } :
(r )
H n, l, m = En n, l, m
€
Lz n, l, m = hm n, l, m
• Continuum basis:
k,l,m = k,l
(r )
H k , l, m = E ( k ) k , l, m
L2 k , l, m = h 2 l(l + 1) k , l, m
€
Lz k , l, m = hm k , l, m
l, m
The energy levels
and radial
wavefunctions can
be found via the
series solution
method
L2 n, l, m = h 2 l(l + 1) n, l, m
{ k,l,m } :
⊗
(Ω)
⊗
l,m
(Ω)
These are the
‘scattering partial
waves’, we will
study them next
semester
Derivation of Radial Wave Equation
• We start from the energy eigenvalue
equation:
En n, l, m = H r n, l, m
• Hit with {|rlm〉} basis state from left:
R r , l, m = r r , l, m
En r , l, m n, l, m = r , l, m H r n, l, m
• Define the radial wavefunction:
ψ n ,l (r ) = r , l, m n, l, m
r , θ , φ n, l, m = ψ n ,l (r )Ylm (θ , φ )
Pr2
1
L2
Enψ n ,l (r ) = r , l, m
n, l, m +
r , l, m 2 n, l, m
2µ
2µ
R
+ r , l, m V ( R) n, l, m
 h 2  ∂ 2 2 ∂  h 2 l(l + 1)


 2 +
 +
Enψ n ,l (r ) =  −
+ V (r ) ψ n ,l (r )
2
r ∂r 
2µ r

 2 µ  ∂r
Comes from Laplacian in
Spherical Coordinates
Solving the Radial Wave eq.
 h 2  ∂ 2 2 ∂  h 2 l(l + 1)


 2 +
 +
En ,lψ n ,l (r ) =  −
+ V (r ) ψ n ,l (r )
2
r ∂r 
2 µr

 2 µ  ∂r
1
• Let: ψ n ,l ( r ) = Rn ,l ( r )
r
1
1
ψ n′ ,l (r ) = Rn′ ,l (r ) − 2 Rn ,l (r )
r
r
ψ n′′,l (r ) =
1
2
2
′
′
′
Rn ,l (r ) − 2 Rn ,l (r ) + 3 Rn ,l (r )
r
r
r
2
1
2
2
ψ n′′,l (r ) + ψ n′ ,l (r ) = Rn′′,l (r ) − 2 Rn′ ,l (r ) + 3 Rn ,l (r )
r
r
r
r
2
2
+ 2 Rn′ ,l (r ) − 3 Rn ,l (r )
r
r
1
= Rn′′,l (r )
r
 h 2 ∂ 2 h 2 l(l + 1)

En ,l Rn ,l (r ) =  −
+
+ V (r )  Rn ,l (r )
2
2
2µ r
 2 µ ∂r

Effective 1-D motion
 h 2 ∂ 2 h 2 l(l + 1)

 Rn ,l (r )
En ,l Rn ,l (r ) = −
+
+
V
(
r
)
2
2
2 µr
 2 µ ∂r

• The wavefunction, Rn,l(r), is that of a particle
of mass µ in one dimension, subject to the
effective potential:
h 2 l(l + 1)
Veff (r ) =
+ V (r )
2
2µ r
You should have
seen this before in
classical mechanics
• For l ≠0, angular momentum creates as a
repulsive effective potential
• Example:
V (r ) = −
a
r
h 2 l(l + 1) a
Veff (r ) =
−
2
2µ r
r
E
V l (r )
Veff(r)
V(r)
r
Example 1 : Spherically symmetric
Harmonic Oscillator
1
• Let:
V (r ) = µω 2 r 2
2
 h 2 ∂ 2 h 2 l(l + 1) 1
2 2
En ,l Rn ,l (r ) =  −
+
+ µω r  Rn ,l (r )
2
2
2 µr
2
 2 µ ∂r

• Switch to dimensionless variables:
r = λρ
Rn ,l (λρ ) = u (ρ )
En = hωε n
λ=
h
µω
 h 2 ∂ 2 h 2 l(l + 1) 1
2 2 2
hωε nu ( ρ ) =  −
+
+ µω λ ρ u ( ρ )
2
2
2 2
2 µλ ρ
2
 2 µλ ∂ρ

• Dropping common factors gives:
 l(l + 1) 1 2

1
0 = − u ′′ + 
+ ρ − ε n u
2
2
2
 2ρ

How to solve the differential equation
 l(l + 1)

2
u ′′ +  −
− ρ + 2ε n u = 0
2
ρ


• From ‘Handbook of Mathematical Functions’,
p. 781:
• Authors: Abramowitz and Stegun
• No copyright, full text free online
– If:

1 − 4α 2 
2
 y = 0
y′′ +  4nr + 2α + 2 − x +
2
4x 

– Then solution is:
y ( x) = Ne
−
x2
2
x
α+
1
2
L(nαr ) ( x 2 )
Generalized
Laguerre Polynomial
• Physicists never solve differential equations
by hand
• Let:
1 − 4α 2
= −l(l + 1)
4
1
α 2 = l2 + l +
4
α =l+
1
2
4nr + 2α + 2 = 2ε n
ε n = 2nr + α + 1
3
ε n = 2nr + l +
2
Full Solution to spherical harmonic
oscillator
ε n = 2nr + l +
3
2
nr = 0,1,2,3, K
• nr is the number of nodes in the radial
wavefunction
– Note that 2nr+l is always an integer
• We can define the principle quantum
number:
n = 0,1,2, K
n = 2nr + l
3
εn = n +
2
– Solution is then:
u ( ρ ) = Ne
ρ2
−
2
ρ
l +1
1
(l+ )
2
n −l
2
L
(ρ 2 )
• In original Units we have:
3

En = hω  n + 
2

n!
ψ n ,l , m ( r , θ , φ ) =
e
Γ(n + l + 3 / 2 )
−
r2
2 λ2
r l ( l +1/ 2 )  r 2  m
Ln −l  2 Yl (θ , φ )
l +1
λ
λ 
2
Normalization constant also
from Abramowitz and Stegun
Degeneracy of nth level
n = 2nr + l
nr = 0,1,2,3, K

 l = 0,1,2,3, K
• Case I: n is even:
– l must then be even also:
– l max= n :
k max
l = 2k
n
=
2
– Degeneracy factor:
dl =
n/2
l
∑ 1 = 2l + 1
m=−l
n/2
n/2
n 
d n = ∑ d k = ∑ (4k + 1) = 4∑ k + + 1
2 
k =0
k =0
k =1
N
1
k = ? N (N + 1)
1 nn  n 
∑
dn = 4
 + 1 +  + 1
2
k =1
2 22  2 
n 
= (n + 1) + 1
N
2 
1 2
N+1
d n = (n + 3n + 2 )
2
Degeneracy Continued…
• Case II: n is odd:
– l is odd:
– l ≤ n:
l = 2k + 1
k max =
n −1
2
d k = (2l + 1) = 2(2k + 1)+ 1 = 4k + 3
dn =
( n −1) / 2
( n −1) / 2
( n −1) / 2
k =0
k =0
k =1
∑ dk =
 n −1 
(
)
4
k
+
3
=
4
k
+
3
∑
∑  2 + 1


1 n −1  n −1   n −1 
=4
+ 1 + 3
+ 1

2 2  2
  2

1

= (n + 2 ) n + 
2

1 2
d n = n + 3n + 2
2
(
)
• Result is same for odd or even n!
Summary
• For Spherically Symmetric Harmonic
Oscillator, we have:
1
V (r ) = mω 2 r 2
2
3

En = hω  n +  n = 0,1,2, K
2

1 2
d n = (n + 3n + 2 )
2
n!
ψ n ,l , m ( r , θ , φ ) =
e
Γ(n + l + 3 / 2 )
−
r2
2 λ2
r l ( l +1/ 2 )  r 2  m
Ln −l  2 Yl (θ , φ )
l +1
λ
λ 
2
n = 2nr + l
h
λ=
mω
nr , l = 0,1,2, K , ∞
n

n
=
0
,
1
,
2
,
K
,

n fixed →  r
2
l = 0,1,2, K , n
n
dn
0
1
nr=0, l=0, m=0
1
3
nr=0, l=1, m= -1,0,1
2
6
nr=1, l=0, m=0; nr=0, l=2, m= -2,-1,0,1,2
3
10
nr=1, l=1, m= -1,0,1;
nr=0, l=3, m= -3,-2,-1,0,1,2,3
…
…
…