8:00am 9:00 10:00 11:00 12:00pm 1:00 2:00 3:00 4:00 5:00 6:00 7:00
DESCRIPTION
Make up schedule. Monday Tuesday Wednesday Thursday Friday. 8:00am 9:00 10:00 11:00 12:00pm 1:00 2:00 3:00 4:00 5:00 6:00 7:00. Apr 24???. Apr 25???. April 11 114 Ferguson. April 27 leave early morning by car. April 11 catch a 1:00pm flight. - PowerPoint PPT PresentationTRANSCRIPT
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,