8:00am 9:0010:0011:0012:00pm 1:00 2:00 3:00 4:00 5:00 6:00 7:00

Monday Tuesday Wednesday Thursday Friday

Apr 25???

Make up schedule

April 11catch a1:00pmflight

April 27leaveearly

morningby car

April 11114 Ferguson

Apr 24???

Apr 24???

The Nuclear pp cycle producing energy in the sun

6 protons 4He + 6+ 2e + 2p 26.7 MeVBegins with the reaction

eedpp

0.26 MeV neutrinos

500 trillion solar neutrinos every second!

ALL NEUTRINOS ARE LEFT-HANDED

ALL ANTI-NEUTRINOS ARE RIGHT-HANDED

Helicity = ms/s = 1

Helicity = ms/s = 1

Dirac Equation (spin-½ particles)

( p m 0

j 0 j

j 0

p • ( ) = ( ) 0 0

0 p • p • 0

where p • pxpypz0 11 0

0 -ii 0

1 00 -1

pz pxipy

px+ipy pz

( 0 p0 • p m

Our “Plane wave” solutions (for FREE Dirac particles)

r,t) = a exp[i/h(Et-p • r)]u(E,p) a e(i/h)xpu(E,p)

which gave

( p mu = ( )( )E/cmc p•uA

p•E/cmc uB

from which we note:

uA = ( p • uB uB = ( p • uA cmc

cmc

Dirac Equation (spin-½ particles)Ec

multiply from left by (-i1 recall i0123

-i31 = -i1)223 = +i23

= +i23)( )( ) = +ii1)( ) = 1

p • )I )= im3

E

c

since =

since (i)

0 1-1 0

0 1-1 0

-1 0 0 -1

so px 1 px 1I

px1py 2pz 3 = m

-i30 = +i0123= 5

-i32 = 2-i33 = 3

p • )I )= im3

E

c

This gives an equation that looks MORE complicated! How can this form be useful?

For a ~massless particle (like the or any a relativistic Dirac particle E >> moc2)

E=|p|c as mo0 (or at least mo<<E)

p|p • )I )=

Which then gives:

or:

p • I )=

^

What do you think this looks like?

p • I

^ is a HELICITY OPERATOR!

I = 2

00

2

2

In Problem Set #5 we saw that if the z-axis was chosen to be the direction of a particle’s momentum

2122

2

1 , , ,

0

(0

(

vvu

c

mcE

c

mcE

u

were all well-definedeigenspinors

of Sz

i.e. p • I )u(p)= u(p)

^ “helicity states”

p • I )=

^

p • I )

^5 “measures”the helicity of

So

2122

2

1 , , ,

0

(0

(

vvu

c

mcE

c

mcE

u

Looking specifically at

5u(p) = =

01

10 uA

uB

uB

uA

B

A

upmcE

c

upmcE

c

)(

)(

2

2

)()(

0

0)(

2

2pu

mcE

pcmcE

pc

For massless Dirac particles (or in the relativistic limit)

5u(p)=

)(

)(0

0)(pu

pE

c

pE

c

p • I)u(p)

^

We’ll find a useful definition in the “left-handed spinor”

uL(p)= u(p)(1 5)

2

Think:“Helicity=1”

In general NOT an exact helicity state (if not massless!)

Since 5u(p) = ±u(p) for massless or relativistic Dirac particles

)()1( 521 pu 0 if u(p) carries helicity +1

u(p) if u(p) carries helicity 1if neither it still measures how close this state is to being pure left-handed

separates out the “helicity 1 component”

Think of it as a “projection operator” that picks out the helicity 1 component of u(p)

Similarly, since for ANTI-particles: 5 v(p) = (p· I)v(p)

again for m 0

we also define: vL(p)= v(p)(1 5)

2

with corresponding “RIGHT-HANDED” spinors:

uR(p) = u(p)(1 5)

2 vR(p)= v(p)(1 5)

2

and adjoint spinors like0†

LLuu 0

2)51(0

2

51 )( †† uu

since

5†= 5

2)51(0 †u

since 5 = - 5

2)51( u

Chiral Spinors Particles

uL = ½(1 5)u

uR = ½(1+ 5)u

uL = u ½(1 5)

uR = u ½(1 5)

Anti-particles

vL = ½(1 5)v

vR = ½(1 5)v

vL = v ½(1 5)

vR = v ½(1 5)

Note: uL+ uR = ( )u + ( )u =1 5

21 5

2u

and also: ( ) ( ) u =1 5

21 5

21 2 5 + 5)2

4( ) u

2 2 5

4= ( ) u 1 5

2= ( ) u

Chiral Spinors Particles

uL = ½(1 5)u

uR = ½(1+ 5)u

uL = u ½(1 5)

uR = u ½(1 5)

Anti-particles

vL = ½(1 5)v

vR = ½(1 5)v

vL = v ½(1 5)

vR = v ½(1 5)

note also: ( ) ( ) u =1 5

21 5

2

1 2 5 + 5)2

4

( ) u

2 2 5

4= ( ) u 1 5

2= ( ) u

while: ( ) ( ) u =1 5

21 5

2

1 5)2

4

( ) u = 0 Truly PROJECTION OPERATORS!

Why do we always speak of beta decay as a process “governed by the WEAK FORCE”?

What do DECAYS have to do with FORCE?

Where’s the FORCE FIELD? What IS the FORCE FIELD?

What VECTOR PARTICLE is exchanged?

ne-

p

e

_What’s been “seen” We’ve identified complicated 4-branch vertices, but only for

the mediating BOSONS…Not the FERMIONS!

+ +

e

e

_

semi-leptonic decays

leptonic decay

p

e+

n

e

_We’ve also “seen” the inverse of some of these processes:

ee

The semi-leptonic decays (with participating hadrons) must Internally involve the transmutation of individual quarks:

ddu

ud

ee

_

u

??

u

d_

+

??

duu e+

ud

d

??e

_

Protons, quarks, pions and muons are all electrically chargedso do participate in:

e

e p

p

Can we use QED as a prototype by comparing epn

e

enpe

or to eppe

???

e

en

p

lepton baryon

Charge-carrying currentsimply a charged vector

boson exchange!(we’ve already seen gluons carry color)

d e

u e

Then

might explain -decay!

+2/3

1/3 1

quark-flavor

coupling

ℓ-ℓ

coupling

e

e

u d _

To explain+ decay:

requires a +1charge carrierWhat about decays?

and

e

e

_

explains decays

but coupling only to theleft-handed particle statescoupling strength modulatedby left-handed components

e e

W

d e

W

W u e

u e

e

W

d

We’ve seen the observed weak interactions: e+e+np + e+e

p +e n + e

could all be explained in termsof the interaction picture of vector boson exchanges if we imagine a the existence of a W

W

We’ve identified two fundamental vertices to describe the observed “weak” interactions.

e W

e

or u W

d

Quark couplingFlips isospin!

Changes mass!Changes electric charge!

Lepton couplingChanges electric charge!

Changes mass!

Some new “weak charge” that couples

to an energy/momentum carrying W ±

e

e

Continuing the analogy to

qJleptonA

qJleptonA

e e

In general for a Quantum Mechanical charge carrier, the expression for “current” is of the form

)(~

)ˆ(

perator

opp

m

e

im

e

o†

††

but these newest currents would have to allow

eOe coupling to a “weak-field” W

Which must carry electric charge (why?)but not couple to it (why?)

If this interaction reflects a symmetry, how many weak fields must there be?

U(1)

SU(2)

SU(3)

one field (the photon)

YANG MILLS:

(gluons)

3 fields

COLOR: 8 fields

U(1) is clearly inadequate

U(2) would mean 3 weak fields we know we need W+, W

Could there be a neutral W0 ?

But YANG-MILLS assumes we have “ISO”DOUBLET states!

Left-handed weak iso-doublets (in a new weak “iso”-space)

ud

e

e

Right-handed weak iso-singlets

uR

dR

eR

L

L

NOT part of a doublet…NOT linked by the weak force to neutrinos

NOTE: there is NO (e )R

We’ve discovered we do have:

Left-handed weak iso-doublets (in a new weak “iso”-space)

u +½d ½

e +½e½

Right-handed weak iso-singlets

uR 0 dR 0eR 0

L

L

With ISO-SPIN we identified a complimentary “hypercharge”

representing another quantum valuethat could be simultaneously diagonalized

with ISO-SPIN operators. Wegeneralize that concept into a NEW

HYPERCHARGE in this “weak” space.

YL = 2Q – 2I3weak

Left-handed I3weak

uL +½dL ½

(e)L +½eL½

Right-handed

uR 0dR 0eR 0

YR = 2(Q) – 2(0) = 2Q

YL = 2(-1) – 2(-1/2)YL = 2(0) – 2(+1/2)

YL = 2(-1/3) – 2(-1/2)YL = 2(2/3) – 2(1/2)

= 1= 1

= 1/3= 1/3

Not all weak participants have ELECTRIC CHARGE • Its NOT electric charge providing the coupling

All weak participants (by definition) carry weak iso-spin

u +½ d ½

e +½e½

u +½ YL = 1/3 d ½ YL = 1/3

e +½ YL = 1e½ YL = 1

L

L

e W

e

u W

d

Butinteractions

are only well-defined by the theory if the fermion legs to a vertex have

equal coupling strengths

L

L

YL = 2Q – 2I3weak

With DOUBLET STATES and an associated “charge” defined

we can attempt a Yang-Mills gauge-field model

to explain the weak force

but with somewarnings...

e W 0

e

The Yang-Mills theory requires introducing a 3rd field:

Could this be the photon?How do we distinguish thisprocess from exchange?

Maybe the noted U(1) symmetry is part of a much larger symmetry:

U(1) SU(2) ?

UEM(1) UY(1) ×SUL(2)

U(1) U(1) ×SU(2)

Straight from U(1) and the SU(2) Yang-Mills extension, consider:

iiWTigBig

21 2 YD

charge-like couplingto a photon-like field

some new Yang-Mills coupling

Ti=i/2 for left-handed doublets

= 0 for right-handed singlets

This looks like it could be U(1)

with = q and B A Yg1

2This all means we now work from a BIG comprehensive Lagrangian

fiif

fWTigBigi

)

2(

21 Y

summed over all possible fermions f to include terms foru, d, c, s, t, b, e, , , e, ,

fiif

fWTigBigi

)

2(

21 Y

which contains, for example:

L

Lei

iLe e

WigBigieLL

)

22()(

21

0 †† Y

R

RR

eBigie )2

()(1

0

Y†

L

Li

iL

d

uWigBigidu

LL)

22()(

21

0

†† Y

RR

RuBigiu )

2()(

1

0

Y

†

RR

RdBigid )

2()(

1

0

Y

†

plus similar terms for , , c, s, , , t, b,

Straight from U(1) and the SU(2) Yang-Mills extension, consider:

iiWTigBig

21 2 YD

charge-like couplingto a photon-like field

some new Yang-Mills coupling

Ti=i/2 for left-handed doublets

= 0 for right-handed singlets

This looks like it could be U(1)

with = q and B A Yg1

2This all means we now work from a BIG comprehensive Lagrangian

fiif

fWTigBigi

)

2(

21 Y

summed over all possible fermions f to include terms foru, d, c, s, t, b, e, , , e, ,

fiif

fWTigBigi

)

2(

21 Y

which contains, for example:

L

Lei

iLe e

WigBigieLL

)

22()(

21

0 †† Y

R

RR

eBigie )2

()(1

0

Y†

L

Li

iL

d

uWigBigidu

LL)

22()(

21

0

†† Y

RR

RuBigiu )

2()(

1

0

Y

†

RR

RdBigid )

2()(

1

0

Y

†

plus similar terms for , , c, s, , , t, b,