41
Chapter 4 Angular Momentum .
Chapter 4
Angular Momentum
.
A central force is any force that is directed along the radius, i.e., it is in the form
rF ˆF
Now since V
F
For any central force problem
rVVor
From this definition we conclude that any central force doesn't do any torque
about the origin
The Angular momentum must be conserved
ˆ
sin
1ˆ1ˆˆ
V
r
V
rr
r
VrF
0
V0and
V
Quantum mechanically, for the angular momentum to be conserved, it must
commute with the Hamiltonian, i.e.,
The Hamiltonian for a central force problem is now written as
)1(2
22
rVH
Let us prove this relation by expressing the angular momentum as an operator
The operator that represents the angular momentum is
irprL
Let us, first, derive some properties concerning the angular momentum
0, HL
xLx ,(i) xzpyp yz , 0, xLx
(ii) xx pL , xyz pzpyp , 0, xx pL
(iii) yLx , yzpyp yz , ypz y , ziyLx ,
(iv) yx pL , yyz pzpyp , zy ppy, zyx pipL ,
(v) yx LL , zxx xpzpL , zxxx pLxpzL ,, xypi yxpi zyx LiLL ,
(v) xLL ,2 xzyx LLLL ,222 zxzxzzyxyxyy LLLLLLLLLLLL ,,,,
zyyzyzzy LLiLLiLLiLLi 0,2 xLL
To generalize, if i,j and k are in cyclic order, we can show that
)2(, kji xixL
)3(, kji pipL
)4(, kji LiLL
)5(0,2 jLL
Applying the Heisenberg's Uncertainty principle to the relation of Eq.(4) we
conclude that
)6(21
kji LLL
Angular Momentum and Rigid Rotations
Consider the spatial displacement along the x-axis caused by the unitary
operator Ux such that
)()()( xx axxxU
In general if the displacement is arbitrary we can write, with the vector a is
infinitesimal
)(!2
)()()(2
22
xdx
dax
dx
daxaxBut x
xx
)()(2
1)(2
22
xexdx
da
dx
daxU xxax
xx
)()( xexUor xxi pa
x
)()( rerUorpa
r
i
par
i
eU
operatorunitaryThe is causing a displacement by a vector a.
Let us find the unitary operator that causes a
rotation by some angle.
Consider the rotation of the coordinates
about the z-axis by an angle , as shown.
We can write
,cossin,sincos yxyyxx
But for infinitesimal we have
1cos,sin
zzxyyyxx ,,
Since the direction of the angle is along the axis of rotation so we can write
,0,0
Knowing that 0,, xyr
Eq. can be combined as
rrr
In general, suppose that f(r) is an arbitrary differentiable function in space. If the
vector r is displaced by a to a new point r+a, a new function F(r) is obtained
such that
rar fF arr fF
Rotation about an axis defined by the unit vector n and the rotation angle (a) shows the displacement a(r) of the point whose position vector is r. (b) illustrates the active rotation of a function or state f(r) about an axis n perpendicular to the plane of the figure: f(r) → F(r) = f(r - a).
Let a be a rotation by an angle about an axis
ra
)7(rarr ffFf
with the direction of is along the axis of rotation.
Using the relation ACBBACCBACBA
.
rfrrf
)8(fLi
f
rfpri
rfrrfrFf
If a is infinitesimal rarr ffF
L is the generator of infinitesimal rotations.
Now integrating Eq.(8) we get
0
Ldi
f
dfF
f
L
i
rf
rF
)(
)(ln
Li
erf
rF
)(
)(
)9()()()( rfUrferF R
Li
)10(withL
i
R eU
Is the unitary rotational operator that is causing a rotation by an angle .
UR rotates a state into a new state such that
rr RU
Multiplying the last equation from the left by the adjoint of UR
rrr RRR UUU††
It also takes an arbitrary observable A onto a generally different observable A'.
For any operator A, we define a rotationary transformed operator A' such that
rrr†
RRR AUUAUA
)11(†RRAUUA
For infinitesimal it easy to proof that
)12(,† ALiAUAUA
That is, if an operator commute with L it will be invariant under rotation.
Now let f=x in Eq.(7) and the axis of rotation is taken to be the z-axis, so we
write
xaxax
Similarly we can show that
xayay y 0and zazaz
Combining the last three equation we can write
)13(rr
ya
zyx
kji
rφa x00
ˆˆˆ
yax x
Which is exactly like Eq.
We can generalize to say that any vector operator V is rotated under
infinitesimal rotation as
)14(VV
Substituting Eq.(14) into Eq.(12) we get
VVLV-VV
,
i
)15(ˆˆ VLV, nin
Let us derive some commutation relations.
Letting V=r and in Eq.(15) then we obtain zn ˆˆ
rr, z ziL ˆ
From which we re-derive Eq.(2) which is
kji xixL ,
A scalar operator S is an operator whose expectation value is invariant under
rotation and which therefore transforms according to the rule
0 SSS
Let the scalar operator S be represented by the scalar product between two
vector operators, i.e., BA S
BABA S
Using Eq.(14), the above equation becomes
0 BABA
S
It follows, from Eq.(12), that a scalar operator commute with L, i.e,
0L,S
If S is replaced by the potential energy then we expect
)16(0, rVL
For central force problem the potential energy is invariant under rotation
(rotational symmetry).
Angular Momentum and Kinetic Energy
prprL
2
mlilmkjijk prprL 2
Where the Levi-Cevita symbol (third-rank tensor) is defined as
equal are indices moreor when two
)npermutatioodd(213,321,132
)npermutatioeven(312,231,123
0
1
1
ijk
ijk
ijk
Using the identity kljmkmjlilmijk
Using the relation
3
1
3
1j kkjijki BABA
mlkjkljmkmjl prprL 2jkkjkjkj prprprprL 2
jkkjjk iprrp But jkkjjk irppr and
priprpripprrpiprrprpr kkjjkjkkjjkjkj
22
jkkjkjjkkj irpprprpr and prirpprprirppr kkjj
Knowing that
iprrp 3
pripriprprpr jkkj
32 pripr
2
2
The expression of L2 becomes
priprprpriprprihprL
2222222 2
rr
irrr
rirrpr
ˆ
sin
1ˆ1ˆˆBut
rr
rr
rrprL 22222
)17(22222
rr
rprL
)18(222
22
2
2
22
rr
rrr
LpT
But L commute with any radial derivative
)19(0, TL
From Eq.(16) and Eq.(19)
0, HL L is constant for any central force problem.
rr
r
r
rr
rr
rprL
222222
Reduction of the Central Force Problem
0,Since 2 HL
H and L2 have a common set of eigen functions
)20(22 L
)21( ErVTH
Substituting for T from Eq.(18) into Eq.(21) we get
)22(22
22
2
2
2
ErV
rr
rrr
L
Now substituting Eq.(20) into Eq.(22) we obtain
)23(22
22
2
2
2
ErV
rr
rrr
)24(,,,Letting YrRr
Eq.(23) now reads
)25(22 2
22
2
2
rERrRrVrdr
dr
dr
d
r
)26(Lettingr
rUrR
Then Eq.(25) becomes
)27(22 2
2
2
22
rEUrUrVrdr
Ud
With the condition that U(0)=0 for to be finite at r=0
Eq.(27) is similar to the 1-D Schrödinger equation with an additive term 2
2
2 r
which can be considered as a centrifugal potential.
We will return to the solution of Eq.(27) later.
The Eigenvalue Problem of L2
Now we want to solve Eq.(20), that is
)28(Now
ri
prL
ˆsin
1ˆ1ˆwith
rrr
r
Using the relations
cossinrx
sinsinry
cosrz
)20(22 L
and the relations
kjir ˆcosˆsinsinˆcossinˆ
kji ˆsinˆsincosˆcoscosˆ
Eq.(28) can now be expressed in spherical coordinates as
)29(ˆsin
1ˆ
iL
ji ˆcosˆsinˆ
It is clear from Eq.(23) that L must commute with any function of r and with
any derivative of r.
sin
11
00
ˆˆˆ
rrr
r
r
ri
L
To express L2 we have first to express each component of L. Now
)30(
xy
yx
iypxpL xyz
yyry
r
yBut
Using the relations
222 zyxr
r
z1cos
x
y1tanand
r
y
y
r
22 zr
yzr
y
22
andyx
x
y
)31(2222
yx
x
zr
yzr
rr
y
y
xxrx
r
x
)32(2222
yx
y
zr
xzr
rr
x
x
Substituting Eqs.(31 & 32) into Eq.(30)
22
2
2222
2
22 yx
y
zr
yxzr
rr
yx
yx
x
zr
xyzr
rr
xy
iLz
)33(22
2
22
2
iyx
y
yx
x
iLz
In the same way it straight forward to prove that
)34(cotcossin
iLx
)35(cotsincos
iLy
From Eqs.(33-35) we also get
)36(sin
1sin
sin
12
2
222222
zyx LLLL
Now returning to Eq.(20) and using Eq.(24) we get
rRYrRYL ,, 22
But, as it is clear from Eq.(36), L2 doesn't operate on R(r)
)37(,, 22 YYL
Substituting for L2 from Eq.(36) into Eq.(37)
)38(,,sin
1sin
sin
12
2
2
YY
Eq.(38) becomes ,Letting Y
2
2
2sinsin
sin d
d
d
d
d
d
Multiplying by
2sin
22
2
sin1
sinsin
d
d
d
d
d
d
)39(1
sinsinsin
2
22
d
d
d
d
d
d
Since each side of Eq.(39) depends on different variable each side must
equal to some constant (m2)
)40(02
2
2
m
d
d
)41(0
sinsin
sin
12
2
2
m
d
d
d
d
The Eigenvalue of Lz
The solution of Ed.(40) is
)42( ime
But since 2
2imim ee 12 ime
,2,1,0 m
If one compare eq.(40) with Eq.(33) he concludes that
222 mLz
)43( mLz
This means that mħ is the Eigenvalue of the operator Lz .
The Eigenvalue of L2
Rewriting Eq.(41) with the substitution cosu
)44(0
11
2
22
uu
u
m
d
du
du
d
Eq.(44) is called the associated Legendr's equation. For m=0 This equation
reduces to the well-known Legendr's differential equation.
)45(01 2
u
d
du
du
d
Applying the series solution technique to Eq.(45) we have
0k
kkuau Substituting back into equation (43) we get
02110000
2
k
kk
k
kk
k
kk
k
kk aukauakkuakk
Equating the coefficient of uk to zero, we get
02112 2 kk akkkakk
)46(12
12 kk a
kk
kka
Again we can generate either an even or odd series by setting 00 01 aora
Now it is clear from Eq.(46) that 2
2
k
k
a
a
kk
k
Now let us examine the series ln(1+x)
0
111ln
k
kk
k
xx
22
k
k
a
a
k
k
both the even & the odd series violate the boundary conditions
uuu as1lnlikebehaves
Which diverge at and0or1 u
0 u
u
02 la from Eq.(46) we get 1 ll
Substituting for in Eq.(20) we get
)47(1 22 llL
This means that l(l+1)ħ2 is the Eigenvalue of the operator L2 .
The associated Legender's functions are related to the Legender's functions by
the relation
)48(1 22 uPdu
duuP lm
mm
l
m
)49(1!2
1with 2 l
l
l
ll udu
d
luP
)50(21,0
2 21
l
ll txPtxttxg
To solve this dilemma we have to terminate the series after a finite number of
terms, say l,
)52(!
!
12
21
1ll
ml
ml
ml
ml
lduuPuP
)53(121 11 ulPuuPluPl lll
mnforxdx
d mn
n
0thatNoting
From Eqs.( 48 & 49) we conclude that for the wave functions not to be zero we
must have
lmllmlml 2
)51(12
21
1llll
lduxPuP
The Spherical Harmonics:
The solution to Eq.(38) which is normalized over the entire solid angle are
called the spherical harmonics and are given for m0 as
)54(cos1!
!
4
12,
m
limmm
l Peml
mllY
)55(,1,with m
lmm
l YY
)56(sin,,2
0 0mmll
ml
ml ddYY
The total eigenfunction for any central force problem takes the form
,mlnlnlm YrRΨ
Then the probability of finding the particle in a given volume element is
dddrrYrR mlnl sin, 22
Such that 1sin,
0 0
2
0
22
dddrrYrR mlnl
But since, from Eq.(56) 1sin,
0
22
0
dddrY ml
Then we expect 1
0
22
drrrRnl
So we define 2
,mlY as the angular probability density.
Which is associated with the probability of finding the particle within a solid
angle d about the origin.
and we define 22rrRnl as the radial probability density.
Which is associated with the probability of finding the particle in a spherical
shell of thickness dr a distance r from the origin.
s-shell (Sharp) p-shell (Principal) d-shell (Diffuse)
The Ladder Method of Angular Momentum :
Our problem is to find the eigenvalues of L2 and Lz using Dirac notations. Let
the eigenket of the two operators to be denoted by such that ml,
)57(,, 22 mlmlL l
)58(,, mlmlL mz
Where l and m are to be determined. It is know that
2222zyx LLLL
22222zzyx LLLLL mlLmlmlLml z ,,,, 2
)59(2ml
Let us now introduce the non-Hermitian operators
)60(yx iLLL
)61(yx iLLL
Now it is straight forward to show that
)62(2, zLLL
)63(0, 2 LL
)64(0, 2 LL
Also we have
zyxyxxyyxyxyx LLLLLLLiLLiLLiLLLL 2222
)65(22zz LLLLL
zyxxyyxyxyxyx LLLLLLLiLLiLLiLLLL 2222
)66(22zz LLLLL
Also we have
)67(,, LLLiLiLLLL xyzyxz
)68(,, LLLiLiLLLL xyzyxz
Now operating on Eq.(58) by L+
)69(,, mlLmlLL mz
From Eq.(67) we have LLLLL zz Eq.(69) becomes
mlLmlLLL mz ,,
)66(,1, mlLmlLL mz
Similarly by operating on Eq.(54) by L- we can show that
)67(,1, mlLmlLL mz
From Eqs.(66 & 67) we can conclude that the operators L+ and L- are,
respectively, the raising and the lowering operators.
By successive operations with L+ and L- on Eq.(58) we can show that m are
integrally spaced. Now from Eq.(59) we have
)68(maxmax m
For m to be integrally spaced and satisfy Eq.(65) , max is either an integer or a
half integer.
Thus we set max =l where l here is an integer or a half integer according to the
value of l. Eq.(68) now reads
)69(,,,,2,1,0,23
21 mlormlll m
As the maximum value of m is l, and using the fact that L+ is a raising operator
)70(0, llL
Now letting m=l in Eq.(57)
)71(,, 22 llllL l
Substituting for L2 from Eq.(62)
llllLLLL lzz ,, 22
Using Eqs.(58 & 69) llllll l ,, 2222
)72(1 lll
Let us now find the eigenvalues of L+ and L- . We have from Eq.(66)
)73(1,, mlCmlL
Similarly, it easy to show that
222 ,, CmlLLLml zz Using Eqs.(57 & 58)
222 ,1, Cmlmmllml 11 mmllC
)74(1,11, mlmmllmlL
Multiplying by its C.C
)75(1,11, mlmmllmlL
