For spin  ½ S =   ·p=H Helicity “ handedness ” For a moving particle state, its lab frame velocity defines an obvious direction for quantization mssmss fraction of spin “aligned” in this direction |S z | |S| = m S ħ  s(s+1)ħ = m S  s(s+1) spin s s v v though 1212    ^ ^ Notice individual spin-½ particles have HELICITY +1 (m s = +½) RIGHT-HANDED HELICITY  1 (m s =  ½) LEFT-HANDED

spin s v However: HELICITY +1 (m s = +½) RIGHT-HANDED HELICITY  1 (m s =  ½) LEFT-HANDED not “aligned” just mostly so 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 << m e = 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 = 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e to analyzer Electro-static bendingmagnetic bending precesses spin Coulomb scattering doesn’t alter spin direction!

E E E B B B Crossed magnetic/electric fields: E  B selects the velocity v= c EBEB but the spin precesses about the B-filed direction Can be built/designed to rotate the spin by a pre-calculated amount (say 90 O ) Following any scheme for rotating spins, beams of particles can be Spin analyzed by punching through a thin foil of some heavy element!

Head-on view of approaching nuclei ++   ,  oppositely aligned !

electron passing nuclei on the right ++   ,  oppositely aligned ! “orbital” angular momentum of nuclei (  up!) positive! The  interaction makes the potential energy increase with r Sees B of approaching nuclei UP 

positive negative The interaction makes the potential energy increase with r  So gives a positive (repulsive) force which knocks electrons to the RIGHT!

electron passing nuclei on the left ++   “orbital” angular momentum of nuclei (  down!) negative! Sees B of approaching nuclei DOWN  So gives an attractive force knocks these electrons to the RIGHT again!

When positive more electrons scatter LEFT than RIGHT When negative more RIGHT than LEFT EXPERIMENTALLY The weak decay products , e H = + for e ,   vcvc H =  for e ,   vcvc predominantly right-handed predominantly left-handed

Until 1960s assumed, like  s neutrinos come in both helicities: +1 and -1 …created in ~equal numbers (half polarized +1, half  1) st 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 = m s /s =  1 Helicity = m s /s =  1

Dirac Equation (spin-½ particles) (   p   m    0  j  0  j  j 0 p ( ) = ( ) 0    0 0 p   p  0 where p  p x   p y   p z  i i p z p x  ip y p x +ip y  p z (  0 p 0   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)x   p  u (E,p) which gave (   p   m  u = ( )( ) E/c  mc  p  u A p  E/c  mc u B from which we note: u A = ( p   u B u B = ( p   u A c  mc  c  mc 

Dirac Equation (spin-½ particles) EcEc multiply from left by (-i  1      recall    i  0  1  2  3 -i      3  1 = - i  1 ) 2  2  3 = + i  2  3 = + i  2  3 ) ( )( ) = + i  i  1 ) ( ) =  1      p  )I )  =  i m      3   EcEc since    =     since (  i )   so  p x  1   p x  1 I    p x  1  p y  2  p z  3 = m  -i      3  0 = +i  0  1  2  3 =  5 -i      3  2 =  2  -i      3  3 =  3 

    p  )I )  =  i m      3   EcEc 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 >> m o c 2 ) E=|p|c as m o  0 (or at least m o <<E)   p|    p  )I )  =   Which then gives: or:      p  I )  =   ^ What do you think this looks like? p  I  ^ is a HELICITY OPERATOR!  I =   0 0 

In Problem Set #5 we saw that if the z-axis was chosen to be the direction of a particle’s momentum were all well-defined eigenspinors of S z i.e.   p  I ) u(p)  =  u(p)  ^ “helicity states”      p  I )  =   ^       p  I )   ^  5 “measures” the helicity of  So

Looking specifically at  5 u(p) = = uAuBuAuB uBuAuBuA For massless Dirac particles (or in the relativistic limit)  5u(p)= 5u(p)=  p  I ) u(p)  ^

We’ll find a useful definition inthe “left-handed spinor” u L (p)= u(p) (1  5 ) 2 Think: “Helicity=  1” In general NOT an exact helicity state (if not massless!) Since  5 u(p) = ±u(p) for massless or relativistic Dirac particles 0 if u(p) carries helicity +1 u(p) if u(p) carries helicity  1 if 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: v L (p)= v(p) (1  5 ) 2 with corresponding “RIGHT-HANDED” spinors: u R (p) = u(p) (1  5 ) 2 v R (p)= v(p) (1  5 ) 2 and adjoint spinors like since  5 † =  5 since  5   = -    5

Chiral Spinors Particles u L = ½(1  5 ) u u R = ½(1+  5 ) u u L = u ½(1  5 ) u R = u ½(1  5 ) Anti-particles v L = ½(1  5 ) v v R = ½(1  5 ) v v L = v ½(1  5 ) v R = v ½(1  5 ) Note: u L + u R = ( ) u + ( ) u = 1     5 2 u and also: ( ) ( ) u = 1      2  5 +   5 ) 2 4 ( ) u 2  2  5 4 = ( ) u 1   5 2 = ( ) u

Chiral Spinors Particles u L = ½(1  5 ) u u R = ½(1+  5 ) u u L = u ½(1  5 ) u R = u ½(1  5 ) Anti-particles v L = ½(1  5 ) v v R = ½(1  5 ) v v L = v ½(1  5 ) v R = v ½(1  5 ) note also: ( ) ( ) u = 1      2  5 +   5 ) 2 4 ( ) u 2  2  5 4 = ( ) u 1   5 2 = ( ) u while: ( ) ( ) u = 1       5 ) 2 4 ( ) u = 0 Truly PROJECTION OPERATORS!