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Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Angular momentum
Reminder: orbital momentumAngular momentum operator : orbital momentum and spin
In this lectures we show how properties of angular momentumcan be derived from properties of angular momentum operators.
-angular momentum operators and matrix representation-spin, spin ½ states.-magnetic moment and Pauli equation.
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Reminder, orbital momentum
�
L z
� � ��
i
�� �
� � � m � � ��
L 2 Y lm
�
,
� � 2 Y lm
�
,
�
�� l l
� 1 m= -l, -l +1 ..., 0, 1 , l-1, l
� � � ei m
� � m � integer�
L 2 | l,m> � � 2 l l
� 1 | l,m>
�
L z | l,m> � �
m | l,m>
[
�
L 2 ,
�
L j ] � 0 [
�
L j ,
�
L k ] � �
jki i
� �
L i
We have derived the form of eigenvalues of orbital momentum by solving the eigenvalue equations and imposing “reasonable condition”. We obtained that l=integerand m= integer and changes from -l to l.
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Angular momentumEvery operator which fulfills commutation rules of orbital momentumoperator will be called “angular momentum operator”. We will see whatwe can learn about eigenstates from commutation rules alone, we willsee that there are only two possible physically meaningful types of operatorslike that- orbital momentum (L) and spin (S).
[
J 2 ,
J j ] ! 0 j ! x , y , z [
J j ,
J k ] ! "
jki i
#
J i
J 2 !
J x2 $
J y2 $
J z2 [
J x ,
J y ] ! i #
J z[
J z ,
J x ] ! i #
J y[
J y ,
J z ] ! i #
J x
%
J 2 | j,m> & ' 2 j j
(1 | j,m>
)
J z | j,m> * +
m | j,m>
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
(m) Rising and lowering operators J +
, J x
-
i J y and J -
. J x
/ i J y
0 J -
1 J +
2J - | j,m> 2 3 <jm| J + J - | j,m> 4 <jm| J 2 5 J z
2 6 7
J z | jm> 8 9 2 [ j j
: 1 ; m m < 1 ]
m= -j, -j +1 ..., 0, 1 , j-1, j j=0, 1/2, 1,3/2, 2, 5/2....
J 2 J - | j,m> = J - J 2 |jm> > ?
j j
@ 1 J - |j,m>
Eigenstate with the same j
J z J - | j,m> A J - J z
B C
J - |jm> D Em F 1 J - |j,m> Eigenstate with m-1
J x
G
i J y J x
H i J y
I J x2 J
J y2 K
i [ J y , J x ] L J 2 M J z2 N O
J z
J x
P i J y J x
Q
i J y
R J x2 S
J y2 T i [ J y , J x ] U J 2 V J z
2 W XJ z
J + | j,m> 2 Y <jm| J - J + | j,m> Z <jm| J 2 [ J z2 \ ]
J z | jm> ^ _ 2 [ j j
` 1 a m m
b 1 ]thus:
[ j j
c 1 d m m e 1 ] f 0 [ j jg 1 h m m
i 1 ] j 0
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Matrix representation for J=1
J +
k J x
l
i J y and J -
m J x
n i J y
o J x
p 12
J +
q
J - ; J y
r i2
J +s J -
J - | j,m> t u
[ j j
v 1 w m m x 1 ] | j,m-1>
J + | j,m> y z
[ j j
{ 1 | m m
} 1 ] | j,m+1>
<1,n| J x | 1,m> ~ 12
�
[ 2 � m m
� 1 ]
�
n , m
� 1� 1
2�
[ 2 � m m � 1 ]
�
n , m � 1
<1,n| J y | 1,m> � i2
�
[ 2 � m m
� 1 ]�
n , m
� 1 � i2
�
[ 2 � m m � 1 ]
�
n , m � 1
J x
� � 12
0 1 01 0 10 1 0
J y� � 1
2
0 � i 0i 0 � i0 i 0
J z
� � 1 0 00 0 00 0 � 1
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Matrix representation for S=1/2
J +
� J x
i J y and J -
¡ J x
¢ i J y
£ J x
¤ 12
J +
¥
J - ; J y
¦ i2
J +§ J -
J - | j,m> ¨ ©
[ j j
ª 1 « m m ¬ 1 ] | j,m-1>
J + | j,m> ®
[ j j
¯ 1 ° m m
± 1 ] | j,m+1>
<1/2,n| S x | 1,m> ² 12
³
[ 34
´ m m
µ 1 ]
¶
n , m
· 1¸ 1
2
¹[ 3
4
º m m » 1 ]
¼
n , m ½ 1
<1,n| S y | 1,m> ¾ i2
¿
[ 34
À m m
Á 1 ]Â
n , m
à 1 Ä i2
Å
[ 34
Æ m m Ç 1 ]
È
n , m É 1
Sx
ÊË
20 11 0
SyÌ
Í2
0 Î ii 0
Sz
ÏÐ
21 00 Ñ 1
m= +1/2, -1/2 .:S=J
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Spin ½ projectionsImagine unit vector in the direction {x,y,z} ={ } sin
Ò
cos
Ó
,sin
Ô
sin
Õ
,cosÖ × Ø
n
As with usual vectors we project spin vector on a unit vector by a scalar product
Ù
S n
ÚÛÝÜ
S Þ ß
n à á
S x
â sin
ã
cos
äå æ
S y
ç sin
è
sin
éê ë
S z
ì cosí
î
Sn
ïð
2cos
ñ
sin
ò
e
ó i ô
sin
õ
ei
ö ÷ cos
ø
The eigenvectors of this matrix will have well defined spin projection along n
ù
Sn
cos
ú
2
sin
û
2ei
üý
þ
2
cos
ÿ
2
sin
�2
ei
�
�Sn
� sin
�
2ei
�
cos
�
2
� �
2
sin
�
2ei
�
cos
2
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Spin projections
Now remember that eigenvectors of S_z, with well defined spin projection alongz are
�
Sz10
� �
210
�
Sz01
� � �
201
cos
�
2
sin
�
2ei
� � cos
�
210
� sin
�
2ei
� 01
Thus the probability to find out eigenvector of S_z- measure projection of hbar/2 in thespin directed over n is cos2
�2
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Magnetic Moment
Current (charge) going in a “loop” has magnetic moment. Energy of magnetic momentin magnetic field is ( classical) is � �! " �
B
Classically, magnetic moment of a loop of current is proportional to I*A = current*areaIf we consider a particle with charge q going around in a loop then
I # q2 $ r %
v; A & ' r 2 ; I ( A ) qvr
2* qL
2m
+!,- q
.
L2m
;
/!01 gq
2
J2m
The mystery explained only by Dirac equation is that for an elementary spin=1/2particle giro-magnetic factor g=2. While for orbital momentum we have g=1.Magnetic moment is a tool to study structure of particles. Neutron whichhas no charge has a magnetic moment 354
N
67 3.83 e
8
S2mN
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Pauli equation
i
9 ::
t
;
+<
-
= > ? 22m
@ 2 AV
Br C q
2m
D
L E F
B G gS q
2m
H
B I J
S
K
+L
-
Unlike orbital momentum, spin is not related to position variable ( space), thusspin and space variables are totally independent
[
M
S ,
N
r ] O 0 we can write a wave function of spin=1/2 particle inthe following way :
P Q
r , sz
R S
+
T
r 10
UV
-
W
r 01
X Y+
Zr[
-\
r
amplitudes to find the particle with appropriatevalues of s_z ( 1/2, -1/2)
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Spin precession
i
] ddt
^
+_
-
` gS e
2m
a
B b c
S
d
+e
-q=-e
If particle is not supposed to move in space- change its position variablesPauli equation reduces to the one above.
i
f ddt
g
+h
-
i gS e
2m
j
B k lS
m
+n
-
o gS e
p
2mBz
12
1 00 q 1
r+s
-
Lets assume B field is along z
i
t ddt
uwv .
x y
gS e
z
4 mB-
{w| .
ddt
}�~ .
� �
igS e
4 mB-
��� .
���
i � ��� .
�w� .
� A � . exp �
i � t
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Spin precession, cont
�
t � 0, sx
� � �
2 � 12
10
� 12
01
� 12
11
Lets assume that the state is an eigenstate of S_x at t=0, with projection+ h/2
In this case:
�
t � 12
exp � i � texp i � t
We can calculate expectation values of s_x and s_y operators to see iftheyoscillate with time
<
�
t | Sx |
�
t > � 12
exp i t , exp ¡ i ¢ t
£
20 11 0
12
exp ¤ i ¥ texp i ¦ t
. § 12
¨
cos2 © t
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Spin precession
<
ª
t | Sy |
«
t > ¬ 12
exp i t , exp ® i ¯ t
°
20 ± ii 0
12
exp ² i ³ texp i ´ t
. µ 12
¶
sin 2 · t
While the z component ( expectation value of S_z) is zero to start with and does not change (why ?). Spin precesses with the frequency
2 ¸¹ gS e
2 mB-
º eBm
Setting in electron mass and magnetic field 1tesla we get the frequency of 180 billion rotations per second. (in muon experiment we are settinghere at the department spin precession will be used to measure parity violationin the muon decay)
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Addition of angular momenta
In most of life problems we have many particles, and each of this particlescan be have both orbital angular momentum (L) and spin (S). From classical mechanics we know that it is the total angular momentum of thesystem which is conserved
Quantum mechanically there are the following questions to be asked1) What are the allowed states of the total angular momentum? Is total angular momentum still angular momentum- does it have all the properties of an angular momentum operator?2) If so: what is the relation between the states of the total angular momentum and the states of individual angular momenta J_i.
i
»
J i ,
¼
J i
½ ¾
L i
¿ À
S i
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Addition of J_1 and J_2Lets imagine we have two particles, with angular momenta Á
J 1 ,
Â
J 2 ,
Ã
J 1,2
Ä Å
L 1,2
Æ Ç
S 1,2
We do not yet know how they are composed of spin and orbital momentum, lets assume now that J_1 and J_2 are simply angular momentaand that quantum mechanically they have all the properties of angular momentum operators, thus both for J_1 and J_2 we have
[
È
J 1,22 ,
É
J 1,2 j ] Ê 0 j Ë x , y , z
Ì
J 1,22 Í Î
J 1,2 x2 Ï Ð
J 1,2 y2 Ñ Ò
J 1,2 z2 [
Ó
J x ,
Ô
J y ] Õ i Ö ×
J z etc...Ø
J 1,22 | j,m> 1,2
Ù Ú 2 j j
Û 1 |j,m> 1,2Ü
ÝJ 1,2 z | j,m> 1,2
Þ ß
m | j,m> 1,2
m= -j, -j +1 ..., 0, 1 , j-1, j
1) Question: Is an angular momentum operator ? àâá
Jã äâå
J 1
æ çâè
J 2
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Properties of
éëê
J
ì íëî
J 1
ï ðëñ
J 2
The bet is, that J is also an angular momentum operator. Lets checkan example commuting rule :
[
ò
J x ,
ó
J y ] ô ? i
õ ö
J z
÷
J x
ø ù
J 1 x
ú û
J 2 x ,
ü
J z
ý þ
J 1 z
ÿ �
J 2 z ,
�
J y
� �J 1 y
� �
J 2 y
[
�
J 1 x
� �
J 2 x ,
J 2 y
�
J 1 y ] �
J 2 x ,
�
J 2 y
� �
J 1 x ,
�
J 1 y� i
� �
J 2 z
�
i
�
J 1 z
� i
� �
J z
Since because the particles 1,2 have“ nothing to do” with each other and their respective angular momentumoperators operate on states of different particles.
�
J 1 x ,
�
J 2 y
� 0,
�
J 2 x ,
�
J 1 x
� 0
Thus we must have :
J 2 | j,m> ! " 2 j j
# 1 |j,m>
$
J z | j,m> % &
m | j,m>
m= -j, -j +1 ..., 0, 1 , j-1, j 2) The question now is : what are j,min relation to j_1, j_2, m_1, m_2 ??
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Combined J state of both particles
'
J 12 | j1, m1 > ( ) 2 j1 j1
* 1 | j1, m1 >
+
J 22 | j2, m2 > , - 2 j2 j2
. 1 | j2, m2 >
/
J 1z | j1, m1 > 0 1
m1 | j1, m1 >
2
J 2z | j2, m2 > 3 4m2 | j2, m2 >
Here index (1) and (2) refer to particles. Now, for the total angular momentum we have :
5
J 1
6 7
J 22 | j,m> 8 9 2 j j
: 1 |j,m>;
J 1z< =
J 2z | j,m> > ?
m | j,m>
j2, m2, j1, m1 and j , mWe will find the relation between
| j2, m2 > | j1, m1 >Lets denote the state in which one particle has j_1,m_1 and other j_2, m_2by . Lets check what are the values of J, J_z
@
J 1z
A B
J 2z | j2, m2 > | j1, m1 > C D
m1 | j2, m2 > | j1, m1 > E F
m2 | j2, m2 > | j1, m1 > G H
m1
I
m2 | j2, m2 > | j1, m1 >
J m K m1
L
m2
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
“ m” and “j” of combined state
Thus “z” projection of the combined momentum is just the sum of z projections for individual particles. What about j ?
m M m1
N
m 2
m_1= -j_1, -j_1 +1 ..., 0, 1 , j_1-1, j_1
m_2= -j_2, -j_2 +1 ..., 0, 1 , j_2-1, j_2
Since m_max = j_1+j_2 and m_min= -j_1 -j_2 we must have
jmax
O j1
P
j2
| j2, m2 > | j1, m1 >
Note that a corresponding number of m states is2(j_1+j_2)+1
Note that we have (2 j_1 +1)(2j_2+1) of such states.Thus much more
Thus there must be other values of j possible. What are they ?
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Finding j values, rising and lowering operatorsJ +
Q J x
R
i J y and J -
S J x
T i J y
U J -
V J +
W
J x
X
i J y J x
Y i J y
Z J x2 [
J y2 \
i [ J y , J x ] ] J 2 ^ J z2 _ `
J z
J x
a i J y J x
b
i J y
c J x2 d
J y2 e i [ J y , J x ] f J 2 g J z
2 h iJ z
J - | j,m> j k
[ j j
l 1 m m m n 1 ] | j,m-1>
J + | j,m> o p
[ j j
q 1 r m m
s 1 ] | j,m+1>
Lets make use of this operators to try to find values of j :
t
J 2 | j2, m2 > | j1, m1 > u J z2 v w
J zx
J - J + | j2, m2 > | j1, m1 > ==
y 2 m1
z
m22 { | 2 m1
}
m2 | j2, m2 > | j1, m1 > ~
J 1 � .
�
J 2 � . J 1 � .
�
J 2 � . | j2, m2 > | j1, m1 >
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Are eigenstates of J1,J1Z, J2, J2Z eigenstates of J,Jz ?
�
J 2 � J z2 � � 2 J z
�
J - J + =J z
2 � �
J z
�
J 1 � .
�
J 2 � . J 1 � .
�
J 2 � . =J z
2 � �
J z
�
J 12 � J 1z
2 � �
J 1z
�
J 22 � J 2z
2 � �
J 2z
�
J 1 � . J 2 � .
J 2 ¡ . J 1 ¢ . = J z
2 £ J 2z2 ¤ J 1z
2 ¥
J 12 ¦
J 22 §
J 1 ¨ . J 2 © .ª
J 2 « . J 1 ¬ .
J 2 |j1,m1>|j2,m2> m1
®
m22 ¯ m1
2 ° m22 ±
j1 j1² 1 ³
j2 j2
´ 1 |j1,m1>|j2,m2> +j1 j1
µ 1 ¶ m1 m1
· 1 j2 j2
¸ 1 ¹ m2 m2
º 1 |j1,m1-1>|j2,m2+1> +j1 j1
» 1 ¼ m1 m1
½ 1 j2 j2
¾ 1 ¿ m2 m2
À 1 |j1,m1+1>|j2,m2-1>
It is easier to use this form of J^2
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
continuedWe see that in general is not an eigenstate of combinedJ^2 (even if it is an eigenstate of combined J_z) unless :
| j2, m2 > | j1, m1 >
Á
j1 j1
 1 à m 1 m 1
Ä 1
Å
j 2 j 2
Æ 1 Ç m 2 m 2
È 1 É 0 and
Ê
j 2 j 2
Ë 1 Ì m 2 m 2
Í 1
Î
j1 j1
Ï 1 Ð m 1 m 1
Ñ 1 Ò 0
This happens in two cases :
m 1
Ó j 1 or m 2
ÔÕ j 2 and m 2
Ö j 2 or m 1×Ø j1
It should be clear that two cases:
correspond to j=j_1+j_2. Lets check it:
m 1
Ù j 1 and m 2
Ú j 2
m 1ÛÜ j 1 and m 2
ÝÞ j 2
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Combined states
THUS :
two cases:
correspond to
m 1
ß à j 1 and m 2
á â j 2
j ã j1
ä j 2
J 2 |j1,m1>|j2,m2> å j1
æ
j22 ç j1
2 è j22 é
j1 j1
ê 1 ë
j2 j2
ì 1 |j1,m1>|j2,m2> == j1
í
j22 î
j1
ï
j2 |j1,m1>|j2,m2> ð j1
ñ
j2
ò 1 j1
ó
j2 |j1,m1>|j2,m2>
What are about other states of j which must exist? What is the minimal value of j ?
We can figure that out by checking norms :
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Minimal value of j:
J + | j,m> 2 ô <jm| J - J + | j,m> õ <jm| J 2 ö J z2 ÷ ø
J z | jm> ù ú 2 [ j j
û 1 ü m m
ý 1 ] =
þ 2 [ j j
ÿ 1 � m1
�
m 2 m1
�
m 2
� 1 ]
Lets assume that j1>j2 and insert m1=j1 and m2=-j2. Then
J + | j,m> 2 � � 2 [ j j
� 1 � j1
� j2 j1
j2
1 ]
� j
�
j1
j2
If j1<j2 we insert m1=-j1 and m2=j2 and considering both of these cases we conclude:
� j
�
j1
� j2
We also conclude that since m1 and m2 can change by 1, also combinedj value changes in steps of 1.
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Glebsch-Gordan coefficients
We thus have :
| j � j1
�
j2 , m � j1
�
j2 > � | j2 , m2
� j2 > | j1 , m1
� j1 >| j � j1
�
j2 , m �� j1
� j2 > � | j2 , m2
�� j2 > | j1 , m1
! j1 >
But in general:
| j , m " m1
#
m2 > $
m1 , m % m1
C m1 , m2j | j2 , m2 > | j1 , m1 >
Glebsch -Gordan coefficientsj1
&
j2
'
j
(
j1
) j2Lets check the number of states in the right hand side and left hand side:
LH *
j 1
+ j 2
j 1
,
j 2
2j
- 1 . j1
/
j2
0 1 1 j12 j2
3
j1
4
j2
5 1 j1
6
j2
7 j1
8 j2
9 1 j1
: j2 =
2= 2j2
; 1 <
2j1
=
j1
>
j22 ? j1
@ j22 A 2j2
B 1 C
2j1
D
4j E 2j1
F 1 2j2
G 1 H RH
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
Glebsch Gordan coefficients
How to find all the coefficients ? General formula is not available, butone can find them constructing the |j,m> states. For example, itis easy to build all states with j=j1+j2 and all values of m by applyingm-lowering operator J- :
J I . | j J j1
K
j2 , m L j1
M
j2 > N j j
O 1 P m m Q 1 |j,m-1>
J R . | j2 , m2
S j2 > | j1 , m1
T j1 > U J 1 V . W J 2 X . | j2 , m2
Y j2 > | j1 , m1
Z j1 > = j1 j1
[ 1 \ m1 m1
] 1 | j1 , m1
^ 1 > | j2, m2 > + j2 j2
_ 1 ` m2 m2
a 1 | j1 , m1 > | j2, m2
b 1 > =2j1 | j1 , j1
c 1 > | j2, j2 > d + 2 j2 | j1 , j1 > | j2, j2
e 1 >
| j1
f
j2, j1
g
j2
h 1 > i 2j1
2 j1
j
j2
| j1 , j1
k 1 > | j2, j2 > l 2 j2
2 j1
m
j2
| j1 , j1 > | j2, j2
n 1 >
| j1
o
j2, j1
p
j2
q 1 > r j1
j1
sj2
| j1 , j1
t 1 > | j2, j2 > u j2
j1
v
j2
| j1 , j1 > | j2, j2
w 1 >
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
J1=1/2 and j2=1/2
Let's try the last formula for two spins j1=1/2 and j2=1/2 . We expect:j=j1+j2=1 and j=j1-j2 =0. We should have the following four states:
j=1 m=1,0,-1 and j=0, m=0 From what we know already we can construct two states :|1,1> = |1/2,1/2> |1/2,1/2> and|1,-1> = |1/2,-1/2>|1/2,-1/2>.Using the formula:
| j1
x
j2, j1
y
j2
z 1 > { j1
j1
|
j2
| j1 , j1
} 1 > | j2, j2 > ~ j2
j1
�
j2
| j1 , j1 > | j2, j2
� 1 >
We get :| 1,0 > � 1
�
21
�
2 � 1
�
2| 1
�
2 , � 1
�
2 > | 1�
2,1
�2 > � 1
�
21
�
2 � 1
�
2| 1
�
2 ,1
�
2 > | 1
�
2, � 1
�
2 >
| 1,0 > � 12
| 1
�
2 , � 1
�
2 > | 1�
2,1�
2 > � 12
| 1
�
2 ,1
�
2 > | 1
�
2, � 1
2 >
What is the j=0,m=0 state ? It must be orthogonal, normalized and have Jz=0
Anna Lipniacka www.ift.uib.no/~lipniack/Angular momentum
j1=1/2,j2=1/2
The only possibility left for j=1,m=0 state is :
| 0,0 > ¡ 12
| 1
¢
2 , £ 1
¤
2 > | 1
¥
2,1
¦
2 > § 12
| 1
¨
2 ,1
©
2 > | 1
ª
2, « 1
¬
2 >
| 1,0 > 12
| 1
®
2 , ¯ 1
°
2 > | 1
±
2,1
²
2 > ³ 12
| 1´
2 ,1
µ
2 > | 1
¶
2, · 1
¸
2 >
|1,1> = |1/2,1/2> |1/2,1/2> |1,-1> = |1/2,-1/2>|1/2,-1/2>
How to check if j=0 is true ? We can use ¹
J 2 º J z2 » J 2z
2 ¼ J 1z2 ½
J 12 ¾
J 22 ¿
J 1 À . J 2 Á .
Â
J 2 Ã . J 1 Ä .
Check it as a part of homeworks ...This notes will continue...