Angular Momentum Matrix Elements. * all [, ]=0 Commutation Rules needed to block diagonalize

Angular Momentum Matrix Elements

LAST TIME: * all [, ]=0 Commutation Rules needed to block diagonalize

* ε_ijk Levi-Civita antisymmetric tensor — useful properties

* Commutation Rule DEFINITIONS of Angular Momentum and

“Vector” Operators

TODAY: Obtain all angular momentum matrix elements from the commutation rule definition of an angular momentum, without ever looking at a differential operator or a wavefuncton. Possibilities for phase inconsistencies. [Similar derivation for angular parts of matrix elements of all spherical tensor operators, .]

H p L

r r

= µ +

µ + ( )



 



r V

2 2

2 2 2 in nLM_L basis set

L L L

L V V

i j ijk k

k

i j ijk k

k

i i ,

,

[ ]

⁼

[ ]

⁼

∑

∑

h

h ε ε

1. Define Components of Angular Momentum using a Commutation Rule.

2. Define eigenbasis for J² and J_z |λµ〉

3. show λ ≥ µ²

4 raising and lowering operators (like a^†, a and x p

~± i~)

J±|λµ〉 gives eigenfunction of J^z belonging to µ ± _h eigenvalue and eigenfunction of J² belonging to λ eigenvalue

5. Must be at least one µMAX and one µMIN such that J–(J+|λµ^MAX〉) = 0

J+(J–|λµ^MIN〉) = 0 This leads to µmax =

h n h n n

2 ² 2 2 1

 

 =  +

 

,λ .

6. Obtain all matrix elements of Jx, Jy, J±, but there remains a phase ambiguity

7. Standard phase choice: “Condon and Shortley”

T_q^{( )k}

Classification of operators: universality of matrix elements.

1. Commutation Rule

This is a general definition of angular momentum (call it J, L, S, anything!).

Each angular momentum generates a state space.

2. eigenfunctions of J² and J_z exist

but what are the values of λ,µ?

J² and J_z are Hermitian, therefore λ,µ are real

3. find upper and lower bounds for µ in terms of λ : λ ≥ µ²

completeness J J

2 λ λ λ

λ λ

µ = µ

µ = µ µ

z

λµ

(

^J2−^J2z

)

λ^{µ = − µ}λ ²

but J J J J

J J J J

J J

2 2 2 2

2 2 2 2

2 2 2

= + +

− = +

− µ = µ + µ

x y z

z x y

x y

λ λ λ

λ λ λ λ λ

λ λ λ λ

− µ =

λ

[ µ ′ ′ µ ′ ′ µ µ + µ ′ ′ µ ′ ′ µ µ ]

′ ′µ

∑

2 ,

J J

J J

x x

y y

(Hermitian operators. Guaranteed by symmetrization.)

Hermitian:

Thus and

and

_{MAX MIN}

′ ′ µ µ = µ ′ ′ µ

− µ =  µ ′ ′ µ + µ ′ ′ µ





 ≥

− µ ≥ ≥ µ ≥

µ ≤ µ ≥ −

′ ′µ

∑

λ λ λ λ

λ λ λ λ λ

λ λ

λ λ

λ

J J

J J

x x

x y

*

,

,

/ /

2 2 2

2 2

1 2 1 2

0

0 0

J J_{i j ijk k}J

k

i

[ ]

, ^{= ∑h ε}

Want to show that this is ≥ 0.

4. Raising/Lowering Operators

J

_±

≡ J

^x

± ⁱ J

^y

( not Hermitian: J

₊^†

= J

₋

)

J J J J J J

J J J J

J

J J J J J

z z x z y

y x x y

z z

i

i i i i

,

_±

, ,

±

± ± ±

[ ] ⁼ [ ] ^± [ ]

= ± − ( ) ^{= ±} [ ^± ]

= ±

= ±

h h h

h

h

J J J J J

J J

J

z ± ± z ±

± ±

±

( µ ) ⁼ ( ^µ ) ^{± µ}

= µ µ ± µ

= µ ± ( ) ( ^µ )

λ λ λ

λ λ

λ

h h h

J J

J J

±

±

( ^λ µ ) is an eigenfunction of belonging to eigenvalue ^{µ ±} . Thus “raises” or “lowers” eigenvalue in steps of .

z z

h h

just like a,a

^†

( )

right multiply by λµ

Similar exercise for to get effect of on eigenvalue of e already know that

belongs to same eigenvalue of as has no effect on .

J J J J

J J J J J J J J

J J J J J

J J

J

2 2

2 2 2 2

2 2

2

0 0

,

, , , ,

± ±

±

± ± ±

±

±

[ ]

[ ] ⁼ [ ] ^± [ ] ^{= ( [ ] = )}

( µ ) ⁼ ( ^µ ) ^{= ( µ )}

( µ ) ^µ

x

i

y

W

i

λ λ λ λ

λ λ

λ

* upper and lower bounds on µ are ±λ^1/2

* J_± raises/lowers µ by steps of _h

* Since

The only nonzero matrix elements of J_i in the |λµ〉 basis set are those where

∆µ = 0, ±h and ∆ λ = 0. As for derivation of Harmonic Oscillator matrix

elements, we are not assured that all µ differ in steps of _h. Divide basis states into sets related by integer steps of _h in µ.

5. For each set, there are ^{µMIN and µMAX:λ≥ µ2}

λ = µ

MAX

+ µ

2

h

MAX

J_x J J J_y J J

=

(

++ −

)

⁼ i

(

+⁻ −

)

1 2

1

and 2 ,

Thus, for each set

_MAX

MIN

J J

+

−

µ =

µ =

λ λ

0 0

Similarly for

_MIN

MIN

µ

µ =

J J

+ −

λ ⁰

0 = µ = (

²

−

²

− ) ^µ

= ( − µ − µ ) ^µ

J J

− +

λ J J J λ

λ λ

MAX MAX

MAX 2

MAX MAX

z

h

z

h

but

but thus

J J J J J J J J J J J J

J J J J

J J J

J J J

J J J J

J J J J

− +

− +

= ( − ) ( ⁺ ) ^{= + + −}

= + + [ ]

= + + ( )

= + −

+ = −

= − −

x y x y x y x y y x

x y x y

x y z

x y z

x y z

z

i i i i

i i i

2 2

2 2

2 2

2 2

2 2 2 2

2 2

,

,

h h

hh J

_z

Thus µ

_MAX

= −µ

_MIN

OR µ

_MAX

= µ

_MIN

− h

λ = µ

MIN

− µ

2

h

MIN

subtract 2 equations for 0 =

0 =

MAX 2

MIN 2

MAX MIN

MAX MIN MAX MIN

λ

µ − µ + µ ( + µ )

µ + µ

( ) ( ^{µ − µ +} )

h

h

Thus for each set of λµ µ , goes from µ

_MAX

to µ

_MIN

in steps of

h

µ

MAX

= µ

MIN

+ n

_h

= µ –

MAX

+ n

_h

µ

MAX

= n 2 ^h

Thus µ is either integer or half integer or both!

Now, to specify allowed values of λ:

λ = µ + µ =  

 + 

 

 =  +

 



MAX MAX

2

2

2

2 2 2 2 1

h

n

h h

n

h h

n n

let n 2 ≡ j

µ =

µ = −

= ( + )

MAX MIN

h h h

j j λ

²

j j 1

!

j either integer or half integer or both

J J

_{+ −}

= J

²

− J

²z

+ h J

z

(impossible)

Thus there will at worst be only two non-communicating sets of λµ〉 because if µ were both integer and 1/2-integer, each would form a set of µ-values, the members of which would be separated in steps of h.

now factor

this equation

rename our basis states

6. J_x, J_y, J_± matrix elements

recall page 23-3, but in new notation

normalization factor

arbitrary phase factor from taking square root

J J

2 2

1

jm j j jm

jm m jm

z

=

(

+

)

= h h

1= ±1 ± =1

( ) ( )

=

=

± ± ± ±

± ±

±

jm jm N jm N jm

N N

J J

J J

†

† *

† _m !

1= N_±² jmJ J_{m ±} jm

J J J J J J J J J J

J J J J J J

J J J

m ± =

( ) (

^±

)

^{= + ±}

[ ]

= − ±

( )

^{= −}

= −

(

±

)

x y x y x y x y

z z z z

z z

i i i

i i m

h mh

h

2 2

2 2 2 2

2

,

1 1 1

1 1 1

2 2 2

1 2

= [ ( + ) ⁻ ( ^{( ± )} ) ]

= [ ( + ) ^{− ( ± )} ]

±

±

− − ±

N j j m m

N j j m m e

ⁱ

h h

h

/ δ

J

_±

^jm =

^h

[ ^{j j} ( + ¹ ) ^{− m m ( ± 1 )} ]

^{1 2/}

^{jm ± 1 e}

^−iδ±

jm± =1 N± ±J jm

(

J raises / lowers by ± m 1

)

valid for all operators that satisfy

OK to define an basis set for any angular momentum operator defined as above.

A A_{i j} A

k ijk k a

i am

[

,

]

^{= Σh ε}

(to be determined below)

use this to evaluate matrix elements of J_mJ_±

Usual phase choice is for all

the “Condon and Shortley” phase choice sometimes – so be careful

δ

δ

±

±

=

= ± π

( )

0

2

j m , :

′ ′ = − { [ ( + ) ^{− ( + )} ]

− [ ( + ) ^{− ( − )} ] }

′ ′ +

′ −

j m jm i j j m m

j j m m

y j j m m

m m

J ^h

2 1 1

1 1

1

1 2

1

1 2

δ δ δ

/

/ std. phase choice: δ_± = 0

( or _h δ δ

jj_′ m m_{′ ±}¹

[ j j ( + ¹ ) ⁻ m m ^{( ) ′} ]

^{1 2/}

)

two sign surprises

^{j m} ′ ′ ^J

^±

^jm = ^{h δ δ}

^{j j′ m m′ ±1}

[ ^{j j} ( + ¹ ) ^{− m m ( ± 1 )} ]

^{1 2/}

since J_x = ¹₂

(

J⁺+J⁻

)

′ ′ = { [ ( + ) ^{− ( + )} ]

+ [ ( + ) ^{− ( − )} ] }

′ ′ +

′ −

j m jm j j m m

j j m m

x j j m m

m m

J ^h

2 1 1

1 1

1

1 2

1

1 2

δ δ δ

/

/

J_y J J

= ₂¹i

(

⁺− ⁻

)

This phase choice leaves all matrix elements of and real and positive, but those of imaginary

if use this gives real and imaginary

J J J

J

J J J

2

2

,

, ,

x y

y x

±

± = + π ±

[

^δ

]

remember matrix elements of x and p in harmonic oscillator basis set?

Summary

′ ′ = ( ) +

=

± = ( ) [ ( ) ^{+ − ( ± )} ]

+ = ( + ) ⁺ ( ⁻ )

= ( ) −

′ ′

+ − + −

+

j m jm j j

jm jm k m

jm jm i ij j j m m

i j i j

i i ij

j j m m

x y

J J J

J J J J J J

J

2 2

1 2

1

1 2 1 1

1 2

1 2 1

2

δ δ

h

h

h

r

r m

√

√ √

√ √ √ √

√ √

/

++ ¹ ₂ ^{J √}

⁻

( ) ⁱ + ^{ij √}