angular momentum - universitetet i...

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.


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.


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>


(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


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


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

Ã 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


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


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


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


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)


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


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


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)


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


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


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 ??


“ 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 ?


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 >


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


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

Â 1 Ã 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


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 :


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.


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


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 >


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


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...

angular momentum - universitetet i...

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