Angular momentum: definition and commutation• classically, this is a central concept (no pun intended)!• if the (conservative) force is central L is conserved• the quantum mechanical implications are profound indeed• if the situation is that one body ‘orbits’ around another one, do the usual reduction to the equivalent ‘one-body’ problem, with the reduced mass replacing m in all formulae

• study of the commutation relations is most revealing• all coordinates commute, and all momentum components commute

R its toon acts where

tion,representaposition in the ˆˆ:ˆ:

everything

i

rprLprL

x

yy

xiLz

xx

ziLy

zz

yiL zyx ˆˆˆ

zy

ypxzpxLx

pppxpxipx

zyx

zyzyx

and for thingsame

commute operators theseall because 0ˆˆ,ˆˆˆ,ˆˆ,ˆ

and for results analogous with 0ˆ,ˆˆ,ˆˆ,ˆ

Now for some non-zero commmutators

CABCBACBA ˆ,ˆˆˆˆ,ˆˆˆ,ˆ theoremuseful a

yiLzxiLyziLxcyclicity

zizipxxpxx

pxzpzxpxxpzxLx

xzy

zz

xxzxy

ˆˆ,ˆˆˆ,ˆˆˆ,ˆby

ˆ00ˆ0ˆ,ˆˆˆˆ,ˆ

ˆ,ˆˆˆˆ,ˆˆˆ,ˆˆˆ,ˆˆ,ˆ

yxzxzyzyx

zzzxzx

xxxxzxxxyx

piLppiLppiLpcyclicity

pipippxpxp

ppzpzppxppzpLp

ˆˆ,ˆˆˆ,ˆˆˆ,ˆby

ˆ0ˆ00ˆ,ˆˆˆˆ,ˆ

ˆ,ˆˆˆˆ,ˆˆˆ,ˆˆˆ,ˆˆ,ˆ

The key non-zero commmutators• most interesting of all is angular momentum with itself• one can simply move around operators that commute, preserving the order otherwise

DBCADCBABDACBDCADCBAnd ˆ,ˆˆˆˆˆ,ˆˆˆˆ,ˆˆˆˆˆ,ˆˆˆ,ˆˆ theoremuseful 2 a

yxzxzyzyx

zxyyx

yz

xz

zyxyzzxz

zxyzyx

LiLLLiLLLiLLcyclicity

Lipypxipxipyi

ppzx

pzpy

pxpzpzpzpxpypzpy

pxpzpzpyLL

ˆˆ,ˆˆˆ,ˆˆˆ,ˆby

ˆˆˆˆˆˆˆˆˆ

}00ˆˆ,ˆˆ0{}0000{

}0000{}0ˆˆ,ˆˆ00{

ˆˆ,ˆˆˆˆ,ˆˆˆˆ,ˆˆˆˆ,ˆˆ

ˆˆˆˆ,ˆˆˆˆˆ,ˆ

Expressing components of L in sphericals I

• this is in a succinct and ‘hybrid’ notation

x

y

z

s

z

yxzyxr

xxrx

r

x

arctanarctan:arctan

whererule,chain by the

22222

22

222

2

22

2

2

2

2222

222

1

1

2

2

1

1

2

222

s

y

sr

xz

rr

x

x

y

yx

x

zs

x

sz

z

rr

x

x

y

yxz

x

rzyx

x

xx

y

z

yx

flipsign a and swappingjust 22

yxs

x

sr

yz

rr

y

y

Expressing components of L in sphericals II

• we can now construct the angular momentum operators

• for z, things proceed a bit differently and it is a lot simpler!

2222

2

2

22

2222

01

1

2

222

r

s

rr

z

z

s

sz

z

rr

z

z

yx

rzyx

z

zz

yx

222

2

2

222ˆ

s

xz

s

yi

s

xz

sr

yz

r

ys

rr

zy

r

yzi

s

x

sr

yz

rr

yz

r

s

rr

zyi

yz

zyiLx

coscotsin

sin

coscossin

sin

sinsinˆ22

i

r

rr

r

riLx

Expressing components of L in sphericals III, and an astounding eigenresult

• Sweet!! The functions are eigenfunctions of the Lz operator,

with eigenvalue mħ!!• therefore, the spherical harmonics are eigenfunctions as well with that same eigenvalue

cotsincosˆ

ˆ222

iL

r

s

rr

zx

s

y

sr

xz

rr

xziL

y

y

imemLiL

s

y

s

x

sr

xyz

sr

xyz

rr

xy

r

xyi

s

y

sr

xz

rr

xy

s

x

sr

yz

rr

yxiL

zz

z

2

1)( since )()(ˆˆ

ˆ

2

2

2

2

22

2222

Working out the square of L: another astounding eigenresult

• compare this with Legendre’s equation: exactly the same thing

)()1()(sin

1sin

sin

12

2

2

ll PllP

• evidently, one can simultaneously determine both the magnitude of the angular momentum, and its z component• claim: [L2, Lz] = 0 proof: you do it on paper!• claim: [L2, L] = 0 L2 commutes with any component of L• we’ve already established that the components don’t commute

2

2

222222

sin

1sin

sin

1ˆˆˆˆˆˆ

zyx LLLL LL

),(),(ˆ and ),()1(),(ˆ

)()1()(ˆ)()1()(ˆ

22

2222

m

lm

lzm

lm

l

ml

mlll

YmYLYllYL

PllPLPllPL

Two new angular momentum operators, built from Lx

and Ly, and a bizarre identity

• interesting commutation relations

• since [L2, L] = 0 [L2, L ± ]

• a fantastic and bizarre operator identity:

operators lowering and raising be will theseand ˆˆ:ˆyx LiLL

LLLixLiiLiLLiLLLL xyyyzxzzˆˆˆˆˆˆ,ˆˆ,ˆˆ,ˆ

zyxzyxyxyx

xyyxyxyxyx

LLLLiiLLLLiLL

LLLLiLLLiLLiLLL

ˆˆˆˆˆˆˆ,ˆˆˆ

ˆˆˆˆˆˆˆˆˆˆˆˆ

222222

22

zzzz

zyxzyx

LLLLLLLLLL

LLLLLLLL

ˆˆˆˆˆˆˆˆˆˆ

ˆˆˆˆˆˆˆ:ˆ sincebut

2222

22222222

zzyyxyy

zzyxxxx

LiiLiLLiLLLL

LLiiLLiLLLL

ˆ0ˆˆ,ˆˆ,ˆˆ,ˆ

ˆˆ0ˆ,ˆˆ,ˆˆ,ˆ

Learning about raising and lowering

• we see how the raising and lowering takes place: is unchanged, while is raised or lowered by one unit of angular momentum• there is therefore a ‘ladder’ of states for a given • but once the z component of Lz gets as big as (or nearly as big as)

L itself the process must stop: there must be a’top’ state Ytop

• second, test it with Lz :

eigenvaluewith ˆ of eigenstatean also is ˆ

ˆˆˆˆˆˆˆˆˆˆ

z

zzz

LYL

YLYLYLYLLLYLLYLL

• now to ask: suppose some function Y is an eigenfunction of both L2 (with eigenvalue ) & Lz (with eigenvalue )

• what is the effect on Y of L ± ? first, test it with L2:

toptoptoptop :ˆ where0ˆ YYLYL z

eigenvalue with ˆ of eigenstatean also is ˆ

ˆˆˆˆˆˆ

2

22

LYL

YLYLYLLYLL

Trying to raise the top, or lower the floor

• now lower the states one by one with the lowering operator• same each time, but is knocked down by ħ• not yet clear what the multiplicative factor might be…• finally we arrive at the ‘unlowerable’ bottom state Ybot

botbotbotbot :ˆ where0ˆ YYLYL z

? know wesince mean, thisdoeswhat

topbot

botbottoptop

• we therefore havetoptoptoptop :ˆ where0ˆ YYLYL z

toptoptoptoptop2toptop

toptop2

toptop2

top2

0

ˆˆˆˆˆˆˆˆˆ

YYY

YLYLYLLYLLLLYL zzzz

botbotbotbotbot2botbot

botbot2

botbot2

bot2

0

ˆˆˆˆˆˆˆˆˆ

YYY

YLYLYLLYLLLLYL zzzz

Sorting our the relationship between top and bot

• in principle n may be odd or even• it may be shown (Griffiths problem 4.18)

• two must differ by some integer nћ 0 wherebottop nn

22top

2botbotbot

bot2bot

2bot

2bot

botbotbotbot

that have weso and

2)1(22

)1(12

1

nn

n

n

nnnn

nnn

nn

• the spherical harmonics are not eigenfunctions of the raising and lowering operators

11 111ˆ m

lm

lm

l YmlmlYmmllYL