SU(2) Combining SPIN or ISOSPIN ½ objects gives new statesdescribed by the DIRECT PRODUCT REPRESENTATION

built from two 2-dim irreducible representations: one 2(½)+1 and another 2(½)+1 yielding a 4-dim space.

isospinspace

+½

½

=

=

which we noted reduces to 2 2 = 1 3

the isospin 0singlet state

( 12

ispin=1triplet

SU(2)- Spin added a new variable to the parameter space defining all state functions

- it introduced a degeneracy to the states already identified; each eigenstate became associated with a 2+1 multiplet of additional states

- the new eigenvalues were integers, restricted to a range (- to + ) and separated in integral steps

- only one of its 3 operators, J3, was diagonal, giving distinct eigenvalues. The remaining operators, J1 and J2, actually mixed states.

- however, a pair of ladder operators could constructed: J+= J1 + iJ2 and J= J1 - iJ2

which stepped between eigenstates of a given multiplet.

n

-1/2 +1/2 -1 0 1

010

100

000

6

000

001

010

1

000

00

00

2 i

i

001

000

100

4

00

000

00

5

i

i

00

00

000

7

i

i

000

010

001

3

3200

0310

0031

8

The SU(3) Generators are Gi = ½i

just like the Gi = ½i are for SU(2)

The ½ distinguishesUNITARY from ORTHOGONAL

operators.

i appear in the

SU(2) subspacesin block diagonal

form.3’s diagonal entries

are just the eigenvaluesof the isospin projection.

8 is ALSO diagonal! It’s eigenvalues must represent a NEW QUANTUM number!

Notice, like hypercharge (a linear combination of conserved quantities),8 is a linear combinations of 2 diagonal matrices: 2 SU(2) subspaces.

In exactly the same way you found the complete multipletsrepresenting angular momentum/spin, we can define

T± G1± iG2

U± G6± iG7

V± G4± iG5

The remaining matrices MIX states.

000

001

010

1

000

00

00

2 i

i

000

010

001

3

T±, T3 are isospin operators

By slightly redefining our variables we can associate the eigenvalues of

8 with HYPERCHARGE.

831

832 )()( GY

3200

0310

0031

3

18

T3|t3, y = t3|t3, y t3I3

Y |t3, y = y|t3, y

VVY

UUY

TY

VVT

UUT

TTT

,

,

0,

,

,

21

3

21

,3

3

The COMMUTATION RELATIONS establish the stepping properties of these ladder (raising/lowering) operators

T3(T+|t3, y)

I3

Y V+

V

T T

U

U

Y (T+|t3, y) = T+(Y |t3, y) = y(T+|t3, y)

= (t3+1)(T+|t3, y)= T+(T3+ 1)|t3, y= (T+T3+ T+)|t3, y

000

00

00

2

12

1

21

3

3200

0310

0031

3

18

To be applicable to quantum mechanics, the lowest dimensional representation of SU(3)

– the set of 3-dimensional matrices – must act on, and their eigenvalues describe,

a set of real physical states,with quantum numbers:

T3 = Y =

ISO-SPIN I3 +1/2 0 -1/2HYPERCHARGEBARON NUMBERSTRANGENESSCHARGE

+1/3 -2/3 +1/3+1/3 +1/3 +1/3

Q = I3 + ½Y Y B+S

0 -1 0-1/3-1/3+2/3

Since U is an assumed symmetry (mixing states within a multiplet but remaining invariant to strong interactions)

consider: * e-iG**

which will also satisfy all the same equations as .

Since Gi = (Gi)* are obviously also traceless and hermitian (and satisfy the same algebraic Commutation Relations as the Gi)

we have a completely equivalent alternate set of generators for these SU(3) transformations

~

Mathematically we say we have a dual representation

(or an adjunct basis set).

Physically we interpret this as another possible set of fundamental states

carrying the minimum quanta of isospin and hypercharge,

though now the eigenvalues (the diagonal elements of 3 and 8)

change signs.

+1

+1

1

1

down up

strange

+1

+1

1

1

u d

s

The anti-particle quarks!

I3

Y

As m strictly adds when combining | j1 m1 > | j2 m2 >

the quantum numbers t3 , y must as well.+1

+1

1

1

down up

strange

I3

Y

This quark multiplet simply plots the points representing the 3 possible quark states.

+1

+1

1

1

dn up

s

I3

Y up/up

A 2-quark state of up pairs would have a total t3=+1 and y=+2/3

+1

+1

1

1

d u

s

I3

Y uuud

usA ud state have a total

t3=0 and y=+2/3

+1

+1

1

1

d u

s

I3

Y

The 33=9 possible quark-quark states form 33= 36 mulitplets

+1

+1

1

1

+1

+1

1

1

Unfortunately the 36 mulitplets of quark-quark combinations include fractionally charged states,

which do not seem to correspond to any known particles.

But by adding a 3rd quark (to the qq states we’ve built so far):

+1

+1

1

1

+1

+1

1

1

Which we can directly compare to the known spin-1/2 baryons

to these 63=18 qqq states which can be separated into

63= 810 mulitplets

+1

+1

1

1

+1

+1

1

1

n p

0 +

0

the spin ½+

baryon octet the spin 3/2

+

baryon decuplet

*

* *0 *+

Notice if you add a quark to an antiquark (or antiquark to a quark):

+1

+1

1

1

33=9 new states are defined, separable into 33= 18 mulitplets to be compared directly to the known 9 spin-0 and 9 spin-1 mesons

+1

+1

1

1

+1

+1

1

1

K0

0 +

0

K K

the spin 0

meson octet the spin 1

meson nonet

+1

+1

1

1

0 +

0, 0

K+ K*0

K* K*

K*+

1974 Accelerators 1st breached the NEXT energy thresholdand began creating NEW, HIGHER MASS particles!

Endowed with another quantum number:

requiring one more quark that carried it.

As we will see later, by this time theorists had alreadyextended u-d isospin symmetry into an anticipated s-c symmetry

both subsets of larger SU(4) multiplets

c

sd u

p n + + 0

+ 0

LDirac=iħcmc2

Look at the FREE PARTICLE Dirac Lagrangian

because ei and in all pairings this added phase cancels!

Dirac matrices Dirac spinors(Iso-vectors, hypercharge)

This is just an SU(1) transformation, sometimescalled a “GLOBAL GAUGE TRANSFORMATION.”

Which is OBVIOUSLY invariant under the transformation

ei (a simple phase change)

What if we GENERALIZE this? Introduce more flexibility to the transformation? Extend to:

but still enforce UNITARITY?ei(x)

LOCAL GAUGE TRANSFORMATION

Is the Lagrangian still invariant?

(ei(x)) =

LDirac=iħcmc2

So:

L'Dirac = ħc((x))

iħcei(x)( )ei(x)mc2

i((x)) + ei(x)()

L'Dirac =

ħc((x))iħc( )mc2

LDirac

For convenience (and to make subsequent steps obvious) define:

(x) (x)ħc q

L'Dirac = q

()LDirac

then this is re-written as

recognize this????

cqe /

the current of the charge carrying particle described by as it appears in our current-field interaction term

L=[iħcmc2qA

L'Dirac = q

()LDirac

If we are going to demand the complete Lagrangian be invariant under even such a LOCAL gauge transformation,

AAand that defines its transformation

under the same local gauge transformation

i.e., we must assume the full LagrangianHAS TO include a current-field interaction:

something that can ABSORB (account for) that extra term,it forces us to ADD to the “free” Dirac Lagrangian

L=[iħcmc2qA

•We introduced the same interaction term 3 weeks back following electrodynamic arguments (Jackson)

A ) that “couples” to If we chose to allow gauge invariance, it forces to introduce a vector field (here that means

The search for a “new” conserved quantum number shows that for an SU(1)-invariant Lagrangian, the free Dirac Lagrangain is “INCOMPLETE.”

A' = A + is exactly (check your notes!) the rule for GAUGE TRANSFORMATIONS already introduced in e&m!

•the transformation rule

•the form of the current density is correctly reproduced

The FULL Lagrangian also needs a term describing the free particles of the GAUGE FIELD (the photon we demand the electron interact with).

Of course NOW we want the Lagrangian term that recreates that!

Furthermore we now demand that now be in a form that is both Lorentz and SU(3) invariant!

titis

rki eCeCedk

trA 213

3

2)2(),(

We’ve already introduced the Klein-Gordon equation for a massless particle, the result, the solution

A = 0 was the photon field, A

We will find it convenient to express this term in terms of the ANTI-SYMMETRIC electromagnetic field tensor

More ELECTRODYNAMICS: The Electromagnetic Field Tensor

• E, B do not form 4-vectors

• E, B are expressible in terms of and A

but A=(V,A) and J=(c, vx, vy, vz) do!

the energy of em-fields is expressed in terms of E2, B2

• F = AA transforms as a Lorentz tensor!

xV

txA

AAF

)(

011001 = Ex since tAVE /

yxA

xyA

AAF

)()(

122112= Bz since AB

In general

xA

t

xAxxx AAF

0000

= Ex

0000 xt

xAx

Axx FAA

yxxy FF

00 xx FF

= Bz

etcFF xx 00 =

Actually thedefinition youfirst learned:

0

0

0

0

xyz

xzy

yzx

zyx

BBE

BBE

BBE

EEE

Fik = Fki =

While vectors, like J transform as “tensors” simply transform as

JJ

FF

zyxxxx

, , , ; 00

zyxxxx

g

,,,;00

x' = x or

x = x'

Under Lorentz transformations

xx )(

xx 1

dxxd

xddx 1

xxx

x

x

So, simply by the chain rule:

xx

1

and similarly:

xddx

dxxd

1

4 E

JBct

Ec

41

cEEEc

zzyyxx 4 xc

xyzzy JEBB 40 (also xyzyzxzxy)

both can be re-written with

04000

000 JFFFFc

zz

yy

xx

xc

xzxz

yxy

xxx

JFFFF 400

(with the same for xyz)

All 4 statements can be summarized in

JF

c)(4 zyx ,,,0

The remaining 2 Maxwell Equations: 0 B

01

tB

cE

are summarized by

0ijkkijjki

FFF ijk = xyz, xz0, z0x, 0xy

Where here I have used the “covariant form”

0

0

0

0

xyz

xzy

yzx

zyx

BBE

BBE

BBE

EEE

F

= g g F =