So as an exercise in using this notation let’s look at

)()( yx

)()( ygx

ygx

yxg

The indices indicate very specific matrix or vector components/elements. These are not matrices themselves, but just numbers, which we can reorder as

we wish. We still have to respect the summations over repeated indices!

yxg??

(g) = gAnd remember we just showed

xyyx

i.e.

yxyx )()( All dot products are INVARIANT

under Lorentz transformations.

yxg

even for ROTATIONS as an example, considerrotations about the z-axis

331212

212100

)sincos)(sincos(

)sincos)(sincos()()(

yxyyxx

yyxxyxyx

332111221222

222122121100

sincossincossincossincossincossincos yx

yxyxyxyxyxyxyxyxyx

33221100 yxxxxxyx yx

The relativistic transformations:

)(cEpp xx

)( xpcE

cE )( xcpEE

suggest a 4-vector);( p

cEp that also

transforms by

pp )(

so pp should be an invariant!

22

2p

cE

Ec

In the particle’s rest frame:px = ? E = ? pp = ?0 mc2 m2c2

In the “lab” frame:

)(cEpp xx

)( xpcE

cE

)0( mc = mv

= = mc

2222222)()( cmcmpp

so

2222 )1( cm

22cm

Limitations of Schrödinger’s Equation

1-particle equation ),()(),(

2),( 2

1

22

txxVtxxm

txt

i

),,(),()()(

),,(2

),,(2

),,(

212121

2122

2

2

2

2121

2

2

2

21

txxxxVxVxV

txxxm

txxxm

txxt

i

2-particle equation:

mutual interaction

But in many high energy reactions

the number of particles is not conserved!

np+e++e

n+p n+p+3

e+ p e+ p + 6 + 3

i.e. a class of differential eq's

to which Schrodinger's

equations all belong!

Sturm-Liouville Equations

then notice we have automatically

a class of differential equations that include:

Legendre's equationthe associated Legendre equationBessel's equationthe quantum mechanical harmonic oscillator

whose solutions satisfy: 3** )( drmnmn for different

eigenfunctions, n0

If we adopt the following as a definition of the "inner product"

3* dr compare this directly to the vector "dot product"

nnnuv*

*

eigenvalues are REALand

different eigenfunctionsare "orthogonal"

Recall: any linear combination of simple solutions to a differential equation is also a solution,

and, from previous slide:

nmnm mn

Thus the set of all possible eigenfunctions (basic solutions) provide an "orthonormal" basis set and any general solution to the differential equation becomes expressible as

n

nna where 3* dra nnn

any general solution will be a function in the "space" of all possible solutions (the solution set) sometimes called a Hilbert Space (as opposed to the 3-dimensional space of geometric points.

What does it mean to have a matrix representation of an operator? of Schrödinger’s equation?

nnn EH

nEnH nwhere n represents all distinguishing quantum numbers

(e.g. n, m, ℓ, s, …)

nmnn EnmEnHm

Hmn

HHnEnH

mEHm

n

m

since†

E1 0 0 0 0 . . .H = 0 E2 0 0 0 . . . 0 0 E3 0 0 . . .

0 0 0 E4 0 . . .:

.100 : ·

010 : ·

001 : ·, , ,

...with the “basis set”:

This is not general at all (different electrons, different atoms require different matrices)Awkward because it provides no finite-dimensional representation

That’s why its desirable to abstract the formalism

azreaz /

23

1001

aZrea

Zra

Z 2/23

200 21

21

iaZr ereaZ

sin

81 2/

25

121

cos2

1 2/25

210aZrre

aZ

iaZr ereaZ sin

81 2/

25

211

aZrea

rZaZr

aZ 3/

2

2223

300 21827381

1

Hydrogen Wave Functions 10000 ::

01000 ::

00100 ::

00010 ::

00001 :: 0

0000

Angular Momentum|lmsms…>

l = 0, 1, 2, 3, ...Lz|lm> = mh|lm> for m = l, l+1, … l1, lL2|lm> = l(l+1)h2|lm>

Sz|lm> = msh|sms> for ms = s, s+1, … s1, sS2|lm> = s(s+1)h2|sms>

Of course |nℓm> is dimensional again

But the sub-space of angular momentum(described by just a subset of the quantum numbers)

doesn’t suffer this complication.

can measure all the spatial (x,y,z) components (and thus L itself) of vmrL

not even possible in principal !

rixyx

irL

,,

ix

yy

xiLzSo, for

example

azimuthalangle inpolar

coordinates

Angular Momentum nlml…

Lz lm(,)R(r) = mħ lm(,)R(r)for m = l, l+1, … l1, l

L2lm(,)R(r)= l(l+1)ħ2lm(,)R(r) l = 0, 1, 2, 3, ...

Measuring Lx alters Ly (the operators change the quantum states).The best you can hope to do is measure:

States ARE simultaneously eigenfunctions of BOTH of THESE operators!We can UNAMBIGUOULSY label states with BOTH quantum numbers

ℓ = 2mℓ = 2, 1, 0, 1, 2

L2 = 2(3) = 6|L| = 6 = 2.4495

ℓ = 1mℓ = 1, 0, 1

L2 = 1(2) = 2|L| = 2 = 1.4142

2

1

0

1

0

Note the always odd number of possible orientations:

A “degeneracy” in otherwise identical states!

Spectra of the alkali metals

(here Sodium)all show

lots of doublets

1924: Pauli suggested electrons posses some new, previously un-recognized & non-classical 2-valued property

Perhaps our working definition of angular momentum was too literal…too classical

perhaps the operator relations

yzxxz

xyzzy

zxyyx

LiLLLL

LiLLLL

LiLLLL

may be the more fundamental definition

Such “Commutation Rules”are recognized by mathematicians as

the “defining algebra” of a non-abelian

(non-commuting) group[ Group Theory; Matrix Theory ]

Reserving L to represent orbital angular momentum, introducing the more generic operator J to represent any or all angular momentum

yzxxz

xyzzy

zxyyx

JiJJJJ

JiJJJJ

JiJJJJ

study this as an algebraic group

Uhlenbeck & Goudsmit find actually J=0, ½, 1, 3/2, 2, … are all allowed!

ms = ± 12

spin “up”spin “down”

s = ħ = 0.866 ħ 3 2

sz = ħ 12

| n l m > | > = nlm12

12

10( )

“spinor”

the most general state is a linear expansion in this 2-dimensional basis set

1 0 0 1( ) = + ( ) ( )

with 2 + 2 = 1

spin : 12p, n, e, , , e , , , u, d, c, s, t, b

leptons quarks

the fundamental constituents of all matter!

SPINORBITAL ANGULAR

MOMENTUMfundamental property

of an individual componentrelative motionbetween objects

Earth: orbital angular momentum: rmv plus “spin” angular momentum: I in fact ALSO “spin” angular momentum: Isunsun

but particle spin especially that of truly fundamental particlesof no determinable size (electrons, quarks)

or even mass (neutrinos, photons)

must be an “intrinsic” property of the particle itself

Total Angular Momentumnlmlsmsj… l = 0, 1, 2, 3, ...

Lz|lm> = mħ|lm> for m = l, l+1, … l1, lL2|lm> = l(l+1)ħ2|lm>

Sz|lm> = msħ|sms> for ms = s, s+1, … s1, sS2|lm> = s(s+1)ħ2|sms>

In any coupling between L and S it is the TOTAL J = L + s that is conserved.

ExampleJ/ particle: 2 (spin-1/2) quarks bound in a ground (orbital angular momentum=0) stateExamplespin-1/2 electron in an l=2 orbital. Total J ?

Either3/2 or 5/2possible

BOSONS FERMIONS

spin 1 spin ½ e,p, n,

Nuclei (combinations of p,n) can haveJ = 1/2, 1, 3/2, 2, 5/2, …

BOSONS FERMIONS

spin 0 spin ½

spin 1 spin 3/2

spin 2 spin 5/2 : :

“psuedo-scalar” mesons

quarks and leptonse,, u, d, c, s, t, b,

Force mediators“vector”bosons: ,W,Z“vector” mesonsJ

Baryon “octet”p, n,

Baryon “decupltet”

Combining any pair of individual states |j1m1> and |j2m2> forms the final “product state”

|j1m1>|j2m2>

What final state angular momenta are possible?What is the probability of any single one of them?

Involves “measuring” or calculating OVERLAPS (ADMIXTURE contributions)

|j1m1>|j2m2> = j j1 j2;m m1 m2 | j m >

j=| j1j2 |

j1j2

Clebsch-Gordon coefficients

or forming the DECOMPOSITION into a new basis set of eigenvectors.

Matrix Representationfor a selected j

J2|jm> = j(j+1)h2| j m >Jz|jm> = m h| j m > for m = j, j+1, … j1, jJ±|jm> = j(j +1)m(m±1) h | j, m1 >

The raising/lowering operators through which we identifythe 2j+1 degenerate energy states sharing the same j.

J+ = Jx + iJy

J = Jx iJy

2Jx = J+ + J Jx = (J+ + J )/2

Jy = i(J J+)/2

adding

2iJy = J+ J

subtracting

The most common representation of angular

momentum diagonalizes the Jz operator:

<jn| Jz |jm> = mmn

1 0 00 0 00 0 -1

Jz =(j=1)

2 0 0 0 00 1 0 0 00 0 0 0 00 0 0 -1 00 0 0 0 -2

Jz =(j=2)

J | 1 1 > =

J±|jm> = j(j +1)m(m±1) h | j, m1 >

J | 1 0 > =

J | 1 -1 > =

J | 1 0 > =

J | 1 -1 > =

J | 1 1 > =

| 1 0 > 2| 1 -1 > 2

0

| 1 0 > 2| 1 1 > 2

0

J =

J =

< 1 0 | 0 0 0 0 00 02

2

0 00 0 0 0 0

22

< 1 -1 |

< 1 0 |

< 1 1 |

0210

210210210

020202

020

21 xJ

020

202020

020202

020

21

iii

i

ii

iiJ y

100010001

zJ

For J=1 states a matrix representation of the angular momentum operators

Which you can show conform to the COMMUTATOR relationship

you demonstrated in quantum mechanics for the differential operators

of angular momentum

[Jx, Jy] = iJz

Jx Jy Jy Jx =

100000001

0000

0

0000

0

22

22

22

22i

ii

ii

ii

ii

= iJz

100010001

1 zJJ

2/1002/12/1 z

JJ

2/300002/100002/100002/3

2/3 zJJ

2000001000000000001000002

2 zJJ

x

y

zz′R(1,2,3) =

11

11cossin0sincos0

001

1

y′1

=x′

x

y

zz′R(1,2,3) =

11

11cossin0sincos0

001

1

y′1

=x′

2

2

2 x′′

z′′

=y′′

22

22

cos0sin010

sin0cos

x

y

zz′R(1,2,3) =

11

11cossin0sincos0

001

1

y′1

=x′

2

2

2 x′′

z′′

=y′′

22

22

cos0sin010

sin0cos

3

y′′′

z′′′ =

x′′′

3

3

1000cossin0sincos

33

33

R(1,2,3) =

11

11cossin0sincos0

001

22

22

cos0sin010

sin0cos

1000cossin0sincos

33

33

These operators DO NOT COMMUTE!

about x-axis about

y′-axis about z′′-axis1st

2nd3rd

Recall: the “generators” of rotations are angular momentum operators and they don’t commute!

but as nn

Infinitesimal rotations DO commute!!

10010

01

1000101

2

2

3

3

1000101

10010

01

3

3

2

2

1001

1

2

3

23

1001

1

2

3

23