5.73 Lecture #21
21 - 1
3D-Central Force Problems I
Read: C-TDL, pages 643-660 for next lecture.
All 2-Body, 3-D problems can be reduced to
* a 2-D angular part that is exactly and universally soluble
* a 1-D radial part that is system-specific and soluble by 1-D techniques in
which you are now expert
Next 3 lectures:
Correspondence Principle
→ all matrix elements
Commutation Rules



Roadmap
1. define radial momentum pr = r-1(q ⋅ p - ih)
r
r
r
2. define orbital angular momentum L = q × p
[
]


also L × L = ihL and L i ,L j = ih ∑ ε ijk L k


k
general definition of angular
momentum and of “vector
operators”
3. separate p2 into radial and angular terms: p2 = p2r + r −2 L2
4. find Complete Set of Commuting Observables (CSCO) useful for blockdiagonalizing H
H,L2 = [H, L i ] = L2 ,L i = 0
H,L2 ,L i
CSCO
[
]
[
]
L, M L universal basis set
p2
h 2 l(l + 1)
5. separate radial H l (r) = r + V(r) +
2µ
part of H:
2µr 2
1-D Schröd Eq.
effective radial
potential
V l (r)
6. ALL Matrix Elements of Angular Momentum Components Derived from
Commutation Rules.
7. Spherical Tensor Classification of all operators.
⇓
8. Wigner-Eckart Theorem → all angular matrix elements of all operators.
I hate differential operators. Replace them using exclusively simple Commutation
Rule based Operator Algebra.
revised October 21, 2002 @ 10:15 AM
5.73 Lecture #21
21 - 2
Lots of derivations based on classical VECTOR ANALYSIS — much will be set aside
as NONLECTURE
Central Force Problems: 2 bodies where interaction force is along the vector
1
•
r
r1
CM
r
•
q
r
q1
r
q cm
12
r
r2
• 2
r
r
q1 − q 2
r
r
r
q 2 = q1 + q12
r
r
r
q12 = q 2 − q1
= î ( x 2 − x1 ) + ĵ( y 2 − y1 ) + k̂ (z 2 − z1 )
[
]
r
2
2
2 1/2
r ≡ q12 = ( x 2 − x1 ) + ( y 2 − y1 ) + (z 2 − z1 )
r
q2
origin •
also C.M. Coordinate system
r
r r
r1 = q1 − q cm
r
r
v
r2 = q 2 − q cm
[ r1
[ r2
]
r = m1 M ]
r = m2 M
H = H translation + H center of mass
motion of fictitious
free translation particle of mass
of C of M of
system of mass µ = m1m2
M = m1 + m2
m1 + m2
in coordinate system
with origin at C of M (CTDL page 713)
2
Ptrans
=
+ V
2( m1 + m2 ) constant
LAB
√
H
translation
BODY
1 2
√
H
P + V( r)
=
C. M .
2µ cm free{
rotation
(no θ ,φ
dependence)
free translation of
system with respect to
lab (not interesting)
motion of particle of
mass µ with respect
to origin at c. of m.
2
GOAL IS TO SIMPLIFY Pcm
because that is only place where θ,φ degrees of freedom appear.
revised October 21, 2002 @ 10:15 AM
5.73 Lecture #21
21 - 3
r
1. Define Radial Component of P cm
Correspondence Principle
* classical mechanics
* Cartesian Coordinates

* symmetrize to avoid failure to satisfy Commutation Rules
** verify that all three derived operators, p, pr and L
• are Hermitian
• satisfy [q,p]=ih
Purpose of this lecture is to walk you through the standard vector analysis and
Quantum Mechanics Correspondence Principle procedures
r
q ≡ îx + ĵy + k̂z
r
p ≡ îp x + ĵp y + k̂p z
[
]
1/2
= [q ⋅ q ]1/2 =|q|
r
r
find radial (i.e. along q) part of p
r ≡ x2 + y2 + z2
r
r
project p onto q
q ⋅ p = q p cos θ
r
q
θ
r
p
q⋅p
cos(q, p) =
{ qp
θ
radial component of p is
r
r
obtained by projecting p onto q
p r = p cos θ = p
q⋅p q⋅p
=
r
qp
r r
so from standard vector analysis we get p r = r −1q ⋅ p
revised October 21, 2002 @ 10:15 AM
5.73 Lecture #21
21 - 4
This is a trial form for pr, but it is necessary, according to Correspondence
Principle, to symmetrize it.
pr =
[
1 −1
r (q ⋅ p + p ⋅ q ) + (q ⋅ p + p ⋅ q )r −1
4
]
arrange terms in all possible orders!
NONLECTURE (except for Eq. (4))
SIMPLIFY ABOVE Definition to pr = r −1(q ⋅ p − ih)
(r is not a vector)
r r
[q, p] is a vector commutator — be careful
r r
[q, p] = [x,px ] + [y,py ] + [z,pz ] = 3ih
r r
∴ p ⋅ q = q ⋅ p − [q, p]
pr =
[
r r
r r
1 −1
r (2q ⋅ p − [q, p]) + (2q ⋅ p − [q, p])r −1
4
]
(1)


1

=  r −1 4q ⋅ p − r −1 2q ⋅ p + 2q ⋅ pr −1 − 6ihr −1 
3
4 14442444

add and subtract 2r −1q⋅p
3
1
= r −1q ⋅ p − ihr −1 + q ⋅ p,r −1
2
2
[
]
[
(2)
(3)
−1
LEMMA: need more general Commutation Rule for which q ⋅ p,r
is a special case
]
0
r
r r
1st simplify: [ f(r),q ⋅ p] = q ⋅ [ f(r), p] + [ f(r), q ] ⋅ p
revised October 21, 2002 @ 10:15 AM
5.73 Lecture #21
21 - 5
but, from 1-D, we could have shown
[f (x), p]φ = f (x) hi ∂∂x φ − hi ∂∂x (f (x)φ)
=
h
[f (x)φ′ − f ′φ − fφ′] = ihf ′(x)φ
i
∂f
[f (x), p] = ih ∂x
for 1 - D
Thus, in 3-D, the chain rule gives
√∂f ∂r √∂f ∂r √ ∂f ∂r 
r
r
p
h
f
(
),
=
i
[
]  i ∂r ∂x + j ∂r ∂y + k ∂r ∂z 


evaluate these first
[
]
[
1/2
∂r
∂ 2
=
x + y 2 + z2
= x x2 + y2 + z2
∂x ∂x
∂r ∂r
etc. for
&
∂y ∂z
]
−1/2
= x /r
r
r
∂f q
∂f √ x √ y √ z 
Thus [f (r ), p] = ih  i + j + k  = ih
∂r r
r
r
∂r  r
r r
∂f
∂f  x 2 + y 2 + z2 
⋅
r
q
p
q
r
p
h
= ih r
(
),
(
),
=
⋅
=
f
f
i
[
] [
] ∂r 

r
∂r


[f(r),q ⋅ p] = ih
∂f
r
∂r
this is a scalar, not a
vector, equation
(4)
But we wanted to evaluate the commutation rule for f(r) = r–1
∂  1
r , q ⋅ p = ih   r = −ihr −1
∂r  r 
[
−1
]
(5)
plug this result into (3)
pr = r −1q ⋅ p −
RESUME
HERE
(
3
1
ihr −1 + ihr −1
2
2
)
pr = r −1(q ⋅ p − ih)
(6)
This is the compact but non-symmetric result we got starting
with a carefully symmetrized starting point – as required by
Correspondence Principle.
revised October 21, 2002 @ 10:15 AM
5.73 Lecture #21
*
21 - 6
This result is identical to result obtained from standard vector
analysis IN THE LIMIT OF h → 0.
Still must do 2 things:
show [r,pr] = ih
show pr is Hermitian
[r, pr ] = [r, r −1(q ⋅ p − ih)]
0
0
[
]
= r −1[r, q ⋅ p] − r −1[r, ih] + r, r −1 (q ⋅ p − ih)
= r −1[r, q ⋅ p]
Use Eq. (4)
[r, q ⋅ p] = ihr
*
*
∴ [r, p r ] = ih
we do not need to confirm that pr is Hermitian because it was constructed
from a symmetrized form which is guaranteed to be Hermitian.
Correspondence Principle!
2. Verify that Classical Definition of Angular Momentum is Appropriate for
QM.
î ĵ k̂
r r r
L=q×p= x y z
px p y pz
(7)
We will see that this definition of an angular momentum is acceptable as far
as the correspondence principle is concerned, but it is not sufficiently general.
NONLECTURE
r
What about symmetrizing L?
L x = yp z − zp y = p z y − p y z
r r
= −( p × q )x
PRODUCTS OF
NON-CONJUGATE
QUANTITIES
∴ p × q = –L
revised October 21, 2002 @ 10:15 AM
5.73 Lecture #21
21 - 7
q×p+p×q = 0
r
q × p − p × q = 2L
symmetrization is impossible!
antisymmetrization is unnecessary!
r
But is L Hermitian as defined?
(q × p)† ≠ p† × q† !
BE CAREFUL:
go back to definition of vector cross product
L x = ypz − zpy
L†x = p†z y† − p†y z† = pz y − py z = ypz − zpy = L x
(p,q are Hermitian)
r
∴ L is Hermitian as defined.
RESUME
3A. Separate p2 into radial and angular terms.
GOAL:
vector analysis
p2 = pr2 + r −2 L2
r r r
p = p|| + p⊥
r
(|| and ⊥ with respect to q)
p⊥
r
p
r
q
Classically
p||
r
L 
678

r
r
r
p = r −2 q(q ⋅ p) − q × (q × p)

r r
component
in
q,p
component
 r
r
|| to q
(8)
(9)
plane which is ⊥ to q
(is the sign correct?)
r–2 is needed in both terms to remain dimensionally correct.
revised October 21, 2002 @ 10:15 AM
5.73 Lecture #21
21 - 8
talk through this vector identity
1st term (p|| ):
q ⋅ p = q p cos θ
r
r
q / q = unit vector along q
2nd term (p⊥ ):
q × p points ⊥ up out of paper
thumb
thumb
}
q
finger
palm
× q{
× p is in plane of paper in opposite direction of p ,
⊥
finger
hence minus sign.
Is it necessary to symmetrize Eq. (9)?
NONLECTURE
Examine Eq. (9) for QM consistency
x component
[(
) (
p x = r −2 x xp x + yp y + zp z − yL z − zL y
but
px = r
(
)
)]
yL z − zL y = y xp y − yp x + z( xp z − zp x )
−2
[(x
2
2
2
)
0
0
]
+ y + z p x + ( xy − yx )p y + ( xz − zx)p z = p x
similarly for py , p z
r
Symmetrize? No, because 2 parts of p
are already shown to be Hermitian.
RESUME
revised October 21, 2002 @ 10:15 AM
5.73 Lecture #21
21 - 9
3B. Evaluate p⋅p
r
p2 = pr –2 [q(q ⋅ p) − q × (q × p)]
(10)
[goal is p2 = pr2 + r−2L2 ]
r
commute p through r-2 to be able to take advantage of classical vector triple product
NONLECTURE
r −2
[p,r
] = −ihî ∂∂x r−2 + ĵ ∂∂y r−2 + k̂ ∂∂z r−2 
r
= 2ihr −4q
∂f 

Recall [f(x),p x ] = ih ∂ x 
∂ −2
∂r
1 2x
because
r = −2r −3
= −2r −3  
= −2x r 4


∂x
∂x
2 r
(
)
r
r
p2 = r −2 (p + 2ihr −2 q )[q(q ⋅ p) − q × (q × p)]
r
r
r
thus pr −2 = r −2 p + 2ihr −2 q
(11)
(12)
get 4 terms
p2 = r −2 (p ⋅ q )(q ⋅ p) − r −2 p ⋅ [q × (q × p)] + r −2 (2ih)r −2 (q ⋅ q )(q ⋅ p) − r −2 (2ih)r −2 q ⋅ [q × (q × p)]
I
II
III
IV
rpr + ih
I = r −2 (q ⋅ p − 3 ih)(q ⋅ p) −2
−1
r (q ⋅ p − ih)(q ⋅ p) = r pr (q ⋅ p)
III = r −2 ( 2 ih)(q ⋅ p)

( )
II = −r −2 (p × q) ⋅ (q × p) = −r −2 ±L2 = r −2 L2
0
IV = −r 4 ( 2 ih)(q × q) ⋅ (q × p)
p2 = r −1pr (rpr + ih) + r −2 L2 = r −1 [rpr − ih]pr + r −1pr ih + r −2 L2 = p2r + r −2 L2
(13)
rpr-[r,pr]
revised October 21, 2002 @ 10:15 AM
5.73 Lecture #21
21 - 10
RESUME
This p2 = pr2 + r −2 L2 equation
is a very useful and simple form for p2 – separated into additive radial and
angular terms! If H can be separated into additive terms, then the eigenvectors
can be factored.
SUMMARY
pr = r −1(q ⋅ p − ih)
radial momentum
p2 = pr2 + r −2 L2
separation of radial and angular terms

p2  L2
H= r +
+
V(r)

2µ  2µr 2

eventually
h 2 l(l + 1)
+ V(r)
V l (r) =
2µr 2
→ CSCO
Next: properties of L i , L2 
revised October 21, 2002 @ 10:15 AM