for spin ½ s = ·p=h helicity “ handedness ” for a moving particle state, its lab frame...
TRANSCRIPT
For spin½S =
·p=H
Helicity “handedness” For a moving particle state, its lab frame velocity defines an obvious
direction for quantizationms
sfraction of spin “aligned” in this direction
|Sz|
|S|=
mSħ
s(s+1)ħ=
mS
s(s+1)
spin
s
spin
sv v
though12
^
Notice individual spin-½ particles haveHELICITY +1 (ms = +½) RIGHT-HANDEDHELICITY 1 (ms = ½) LEFT-HANDED
spin
sv
However:
HELICITY +1 (ms = +½) RIGHT-HANDEDHELICITY 1 (ms = ½) LEFT-HANDED
not “aligned”just mostly so
57735.03
1
)2/3)(2/1(
2/1
)1(
ss
ms
But helicity (say of an electron) is not some LORENTZ-INVARIANT quantity! Its value depends upon the frame of reference: Imagine a right-handed electron traveling to the right when observed in a frame itself moving right with a speed > v.
It will be left-handed!
So HELICITY must NOT appear in the Lagrangian for any QED or QCD process (well, it hasn’t yet, anyway!).
HELICITY is NOT like some QUANTUM NUMBER.It is NOT unambiguously defined.
But what about a massless particle (like the or…the neutrino?)
m < 5.1 eV << me = 0.511003 MeV e
m < 160 keV
m < 24 MeV
Recall for a massless particle: v = c
Which means it is impossible (by any change of reference frame)to reverse the handedness of a massless particle.
HELICITY is an INVARIANT
a fundamental, FIXED property of a neutrino or photon.
Experimentally what is generally measured is a ratio comparing the number of a particles in a beam, or from a source,
that are parallel or anti-parallel to the beams direction.
Helicity =
NN
NN
Longitudinal polarization turns out to be hard to measure; Transverse polarization is much easier to detect.
There are several schemes for rotating the polarization of massive particles.
decay source
aluminum
analyzer
light element(metallic)reflector
++++++
+ + ++
++
e
to analyzer
Electro-static bending magnetic bending precesses spin
Coulomb scatteringdoesn’t alter
spin direction!
EE E
BB
B
Crossed magnetic/electric fields: E B selects the velocity v= cEB
but the spin precesses about the B-filed direction
Can be built/designed to rotate the spin by a pre-calculated amount(say 90O)
Following any scheme for rotating spins, beams of particles can beSpin analyzed by punching through a thin foil of some heavy element!
Head-on view of approaching nuclei
+ +
mc
eh
, oppositely aligned!
electron passing nuclei on the right
+ +
pr, oppositely aligned!
“orbital” angular momentum of nuclei
( up!)
0|||| BB
positive!
The interaction makes the potential energy increase with r BU
Sees B of approaching nuclei UP
BU
positive
negative
The interaction makes the potential energy increase with r
Ur
Fr
So gives a positive (repulsive) force
which knocks electrons to the RIGHT!
electron passing nuclei on the left
+ +
pr
“orbital” angular momentum of nuclei
( down!)
0|||| BB
negative!
Sees B of approaching nuclei DOWN
Ur
Fr
So gives an attractive forceknocks these electrons to the RIGHT again!
When positivemore electronsscatter LEFTthan RIGHT
When negativemore RIGHT than LEFT
EXPERIMENTALLYThe weak decay products , e
H = + for e, vc
H = for e, vc
predominantlyright-handed
predominantlyleft-handed
Until 1960s
assumed, like s neutrinos come in both helicities: +1 and -1…created in ~equal numbers (half polarized +1, half 1)
1961 1st observed PION DECAYS at REST(where , come out back-to-back)
_
_
spin-0
spins ,
(each spin-½)oppositely
aligned!Were these half +1, half -1?
No! Always polarized RIGHT-HANDED! So these must be also!
++
Each ALWAYS left-handed!
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!