5.73 Lecture #22
22 - 1
3D-Central Force Problems II
[x,p] = ih → vector commutation rules: generalize from 1-D to 3-D
conjugate position and momentum components in Cartesian
coordinates
Last time:
Correspondence Principle Recipe
Cartesian and vector analysis
Symmetrize (make it Hermitian)
classical in h → 0 limit
Derived key results:
[f (x), px ] = ih ∂∂xf
[
]
f (r ) , q ⋅ p = ih
*pr = r −1 (q ⋅ p − ih)
∂f
df ∂r
r based on
and r = x 2 + y2 + z2
∂r
dr ∂x
[
]
1/2
(came from symmetrization in Cartesian coordinates)
*p2 = p2r + r −2 L2
operator algebra gave simple separation of variables
*L = q × p
not necessary (or possible) to symmetrize
 L2

V
*H =
+
+
(
r
)

2 µ  2 µr2

Vl (r ) radial effective potential
p2r
TODAY
We do not yet know anything about L2 nd Li.
[purpose is mostly to practice [,] and angular momentum algebra]
Obtain angular Momentum Commutation Rules → Block diagonalize H
εijk Levi-Civita Antisymmetric Tensor
useful in derivations, vector commutators, and remembering stuff.
Next Lecture: Begin derivation of all angular momentum matrix elements
starting from Commutation Rule definitions of angular momentum.
revised 21 October, 2002 @ 10:19 AM
5.73 Lecture #22
22 - 2
GOALS
1.
2.
3.
4.
5.
[Li , f (r )] = 0
[Li , pr ] = 0
any scalar function of scalar r.
difficult - need εijk
[L , p ] = 0
[L , L ] = 0 ( but [L , L ] ≠ 0!)
i
2
r
2
i
i
C.S.C.O.
2
j
H , L2 , Li → block diagonalize H
These 1-4 are chosen to show that all terms in H commute with L2 and Li
1.
[Lz , f(r)] = [xpy − ypx , f(r)] = x[py , f ] + [x, f ]py − y[px , f ] − [y, f ]px
[ x, f ] = 0,
r
[y, f ] = 0 because [q, f(r)] = 0 î + 0 ĵ + 0 k̂
∂f ∂r
∂f x
= ih
∂r ∂x
∂r r
[Lz,f (r )] = −ih ∂∂fr x yr − y xr  = 0
recall [f (r ), px ] = ih
2.
[Lz , pr ] = [Lz , r −1 (q ⋅ p −0 ih)] = [Lz , r −1q ⋅ p]
= [L z , r −1 ]q ⋅ p + r −1 [L z , q ⋅ p]
r
r
[Lz , q ⋅ p] = q ⋅ [Lz , p] + [Lz , q] ⋅ p two vector commutators on RHS
r
Note that q is not f(r )!
need to define special notational trick to evaluate these DIFFICULT
COMMUTATORS
revised 21 October, 2002 @ 10:19 AM
5.73 Lecture #22
22 - 3
ε ijk
Levi-Civita Symbol
adjacent interchange
ε xyz = ε yzx = ε zxy = +1
ε yxz = ε zyx = ε xzy = −1
2 repeated indices
ε xxy = etc.= 0
cyclic order
[
]
I claim L i , p j = ih ∑ ε ijk pk . This will become the definition of a “vector operator”
k
with respect to L.
Nonlecture: Verify claim for 1 of 3 × 3 = 9 possible cases
let i = x, j = y
[Lx ,py ] = [ypz − zpy ,py ] = [ypz ,py ] + 0 0
= [y,py ]pz + y[pz ,py ]
= ihpz
Now check this using εijk
[L ,p ] = ih∑ ε
x
y
xyk
[
0
0
pk = ih ε xyx px + ε xyy py + ε xyzpz
k
= ihpz .
]
OK
All other 8 cases go similarly
Other important Commutation Rules
[Li ,p j ] = ih∑k ε ijkpk 

[Li ,q j ] = ih∑k ε ijkq k 
[Li ,L j ] = ih∑k ε ijk Lk
general definition of
a “vector” operator
general definition of an
“angular momentum.” Works
even for spin where q × p
definition is inapplicable
All angular momentum matrix elements will be derived from these commutation
rules.
FOR THE READER: VERIFY ONE
COMPONENT OF EACH OF THE
THREE ABOVE COMMUTATORS
revised 21 October, 2002 @ 10:19 AM
5.73 Lecture #22
22 - 4
[Li ,L j ] = ih∑k ε ijk Lk is identical to
L × L = ihL
r r
expect 0! because vector cross product A × B = | A || B | sin θeˆ AB
(
)
 î ĵ k̂ 

 



L × L =  L x L y Lz  = î  L y Lz − Lz L y  + ĵ Lz L x − L x Lz  + k̂  L x L y − L y L x 

 


L L L 





 x y z
[
]
= ih îL x + ĵL y + k̂Lz = ihL
This vector cross product definition of L is more general than q × p because there
is no way to define spin in q × p form but S × S = ihS is quite meaningful.
Can one generalize that, if L × L = ihL (instead of 0), and the [Li, Lj] and [Li, pj]
commutation rules have similar forms, that L × p = ihp? NO! Check for yourself!
2.
Continued.
[L ,p ] = r q ⋅ [L ,pr ] + r [L ,qr ] ⋅ p
r
L
p
,
[ ] = ih∑ (ˆi ε + ˆjε + kˆ ε )p
−1
z
−1
r
z
i
ixk
sum of 3 terms
k
vector commutators
z
iyk
izk
k
r
q ⋅ [L i , p] = ih ∑ xε ixk + yε iyk + zε izk pk
k
(
= ih ∑ ε ijk q jpk
)
only one of these terms
is nonzero
(1)
j, k
revised 21 October, 2002 @ 10:19 AM
5.73 Lecture #22
22 - 5
r r
and the other term [L i , q ] ⋅ p
[L ,qr ] = ih∑
i
k
[ˆi ε
[L ,qr ] ⋅ p = ih∑ [ε
i
ixk
ixk
]
+ ˆj ε iyk + kˆ ε izk qk
qk px + ε iyk qk py + ε izk qk pz
k
= ih ∑ ε ijk qk p j = ih ∑ ε ikjq jpk
j, k
]
(k↔
(k↔j labels permuted)
k, j
= −ih ∑ ε ijk q jpk
k, j
(2)
switch order of j and k
putting Eqs. (1) and (2) together
[
]
q ⋅ [L i ,p] + [L i ,q ] ⋅ p = ih ∑ ε ijk q jp k − ε ijk q jp k = 0!
j,k
Elegance and power of εijk notation!
We have shown that:
* [Li,pr] = 0 for all i
* easy now to show [Li,pr2] = 0
[
] j( [
] j[
] ]
] [
Finally L i ,L2 = ∑ L i ,L2j = ∑ L j L i ,L j + L i ,L j L j
 
 
 
= ∑ L j ih Σ ε ijk L k + ih Σ ε ijk L k L j 
 k
  k
 
j 
same trick: permute j ↔ k indices in second term
εijk = −εikj


− ih Σ ε ijk L j L k
 k

=0
[
]
[
] [
]


But be careful: L i ,L2j = L j L i ,L j + L i ,L j L j = ih L j Σ ε ijk L k + Σ ε ijk L k L j


k
k
because this is a sum only over k, so can’t combine and cancel terms.
revised 21 October, 2002 @ 10:19 AM
5.73 Lecture #22
22 - 6
for i = x, j = y
[L ,L ] = ih[L L
x
2
y
y z
]
+ Lz L y ≠ 0!
so we have shown
[L2 ,Li ] = 0
[L2 ,f(r)] = 0
[Li ,f(r)] = 0
[L2 ,pr ] = 0
[Li ,pr ] = 0
∴ L2, Li, H all commute — Complete Set of Mutually Commuting Operators
eigenfunction of L2 with
eigenvalue h 2 L( L + 1)
So what does this tell us about L H L ′ = ?
BLOCK DIAGONALIZATION OF H!
ψ = χ ( r ) L2 , L z = nLM L
Basis functions
radial
special
angular
universal
eigenfunctions of Lz
eigenfunctions of L2
which radial eigenfunction?
Next time I will show, starting from
[L ,L ] = ih Σ ε
i
j
k
ijk
Lk
, that
*
L2 nLM L = h 2L (L + 1) nLM L
L = 0,1,…
*
L z nLM L = hM L nLM L
M L = −L, −L + 1,… + L
also derive all L x and L y matrix elements in nLM L basis set.
revised 21 October, 2002 @ 10:19 AM