Lecture 26: Angular Momentum II Phy851 Fall 2009

Lecture 26:
Angular Momentum II
Phy851 Fall 2009
The Angular Momentum Operator
• The angular momentum operator is defined
as:
r r r
L = R× P
• It is a vector operator:
r
r
r
r
L = Lx e x + L y e y + Lz e z
• According to the definition of the crossproduct, the components are given by:
Lx = YPz − ZPy
Ly = ZPx − XPz
Lz = XPy − YPx
Commutation Relations
• The commutation relations are given by:
[ Lx , Ly ] = ihLz
[ Ly , Lz ] = ihLx
[ Lz , Lx ] = ihLy
• These are not definitions, they are just a
consequence of [X,P]=ih
• Any three operators which obey these
relations are considered as ‘generalized
angular momentum operators’
• Compact notation:
[ L j , Lk ] = ihε jkl Ll
Summation over l is implied
‘Levi Cevita tensor’
ε jkl
0 if any two indices
are the same
1 cyclic permutations
of x,y,z (or 1,2,3)
-1 cyclic permutations
of z,y,x (or 3,2,1)
Simultaneous Eigenstates
• In the HW will see that:
[ L2 , Lx ] = 0
2
[ L , Ly ] = 0
[ L2 , Lz ] = 0
– Where:
L2 = L2x + L2y + L2z
• This means that simultaneous eigenstates of
L2 and Lz exist
– Let:
L2 a, b = a a, b
Lz a , b = b a , b
Why L2 and Lz and
not Lx or Ly?
– We want to find the allowed values of a and b.
Algebraic solution to angular
momentum eigenvalue problem
• In a analogy to what we did for the Harmonic
Oscillator, we now define raising and lowering
operators:
L+ = Lx + iL y
L− = Lx − iL y
Analogy is not exact:
1
(X − iP )
A† =
2
1
(X + iP )
A=
2
• Lets consider the action of L+ first:
recall our approach
for SHO:
†
(
)
(
A A A† ε = (ε + 1) A† ε
)
Lz ( L+ a,b ) = Lz Lx a,b + iLz Ly a,b
[ Lz , Lx ] = ihLy
€
[ Lz , Ly ] = −ihLx
→ Lz Lx = ihLy + Lx Lz
→ Lz Ly = −ihLx + Ly Lz
Lz ( L+ a,b ) = ihLy a,b + Lx Lz a,b + hLx a,b + iLy Lz a,b
= ( Lx + iL y ) Lz a, b + h ( Lx + iL y ) a, b
Lz (L+ a, b )= (b + h ) L+ a, b
Ladder Argument
Lz (L+ a, b )= (b + h ) L+ a, b
•
Conclusion: if |a,b〉 is an eigenstate of Lz with
eigenvalue b, then there is also an eigenstate
|a,b+h〉 with eigenvalue b+h
•
Thus we can say:
L+ a, b = C+ ab a, b + h
•
Similarly, we can readily show that:
Lz (L− a, b )= (b − h ) L− a, b
–
So that we must have also:
L− a, b = C− ab a, b − h
•
This allows us to say:
1. If |a,b〉 exists, then |a,b+h〉 exists or C+ab=0
2. If |a,b〉 exists, then |a,b-h〉 exists or C-ab=0
Establishing Some Lower Bounds
• Starting from:
Lx = YPz − ZPy
• It follows that
L†x = Pz†Y † − Z † Py†
= PzY − ZPy
= YPz − Py Z
= Lx
– Which makes sense because Lx is an observable
• Therefore, we have:
a, b L2x a, b = a, b L†x Lx a, b
= Lx a , b
• Result:
a, b L2x a, b ≥ 0
a, b L2y a, b ≥ 0
2
The Trick
• We note that:
L2 − L2z = L2x + L2y
(
)
a, b L2 − L2z a, b = a, b L2x a, b + a, b L2y a, b
• This means that:
(
)
a, b L2 − L2z a, b ≥ 0
• By definition, we have:
(L
2
)
(
)
− L2z a, b = a − b 2 a, b
• Combining the two gives us:
a − b2 ≥ 0
a≥b
a≥0
2
− a ≤b≤ a
The conclusion
a≥0
•
− a ≤b≤ a
Conclusion:
1. There is a bmin for every allowed a
2. There is a bmax for every allowed a
− a ≤ bmin (a )
•
We are just saying
that b is bounded
from above and
below
bmax (a ) ≤ a
Thus we must have:
L+ a, bmax (a ) = 0
–
–
So that no states with higher b can exist
Which requires:
C+ abmax ( a ) = 0
–
Similarly,
C− abmin ( a ) = 0
Just a Little Further to Go…
C+ abmax ( a ) = 0
C− abmin ( a ) = 0
• If we have:
L+ a, bmax (a ) = 0
• Then we must have:
L− L+ a, bmax (a ) = 0
• But:
L− L+ = (Lx − iL y )(Lx + iL y )
= L2x + iLx Ly − iL y Lx + L2y
= L2 − L2z − hLz
= L2x + L2y + i[ Lx , Ly ]
Starting to see the light at the end of the
tunnel…
L− L+ = L2 − L2z − hLz
L− L+ a, bmax (a ) = 0
• This means:
2
2
L
−
L
(
z − hLz ) a,bmax = 0
2
a
−
b
( max − hbmax ) a,bmax = 0
€
€
2
max
a −b
2
max
a=b
− hbmax = 0
(a ) − hbmax (a )
Not quite done yet…
L− L+ a, bmax (a ) = 0
2
a = bmax
(a ) − hbmax (a )
• Similarly, we can show that:
L+ L− a, bmin (a ) = 0
• Which together with:
L+ L− = ( Lx + iLy )( Lx − iLy )
= L2x − iLx Ly + iLy Lx + L2y
= L2x + L2y − i[Lx ,Ly ]
= L2 − L2z + hLz
• Gives us:
€
2
a = bmin
(a ) + hbmin (a )
€
Keep moving forward…
a = bmax (a )(bmax (a ) + h )
a = bmin (a )(bmin (a ) − h )
• From a=a we get:
2
2
bmax
(a ) + hbmax (a ) = bmin
(a ) − hbmin (a )
2
2
bmin
(a ) − hbmin (a ) − bmax
(a ) − hbmax (a ) = 0
• The solution is:
bmin
1
2
= h ± h 2 + 4bmax
+ 4hbmax
2
(
This is a perfect square
bmin
+
1
= (h ± (h + 2bmax ))
2
bmin = h + bmax
bmin < bmax
-
bmin = −bmax
)
Ladder Termination Requirement
• MAIN POINT SO FAR:
– For a given a, bmin(a) and bmax(a) are the only b
values from which the lowering/raising
operators do not create new states.
• So we can build a ladder by starting at bmin
and acting with L+ to create a new state at
bmin+h, etc…
bmin
bmin+h
bmin+mh
• The only way the ladder will terminate is if
we have
bmax = bmin + Nh
– For some integer N
• With bmin=-bmax, this leads to:
bmax
2bmax = Nh
N
= h
2
• This means that the maximum z-component
of angular momentum must always be a halfinteger or whole integer times h
Quantization of angular momentum
• Lets replace bmax with the symbol l  l=N/2.
• The allowed eigenvalues of Lz are then:
b = mh; m = −l, − l + 1, − l + 2, ..., + l
• From the condition:
a = bmax (a )(bmax (a ) + h )
• We see that the corresponding eigenvalue of
L2 is:
a = hl(hl + h )
a = h 2 l(l + 1)
• Conclusion:
– We can relabel the states as:
a , b → l, m
– Then we have:
L2 l, m = h 2 l(l + 1) l, m
L z l , m = hm l , m
The FINAL SLIDE
• Now we have:
(L+ )† = L−
L+ a,b
L+ a, b
2
2
= C+ab
2
= a, b L− L+ a, b
= a − b 2 − hb
€
C+ ab = a − b 2 − hb
– Likewise:
C− ab = a − b 2 + hb
a = h 2l(l + 1)
b = hm
L± l,m = h l(l + 1) − m(m ± 1) l,m ± 1
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