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Wave functions of Baryons.
Baryon Magnetic Moments
Baryon masses.
Need to explain Parity and Charge
Conjugation
Hadrons Magnetic moments
 mq related to the intrinsic spin S of the quark.
• m = (q/mc)S
q
q h
m iz   m i S z  
Sz  

mc
mc 2
• and therefore for each spin-up quark:
2 eh
mu  

3 2mu c
1 eh
md   

3 2md c
1 eh
ms   

3 2 ms c
Spin down just changes the sign
Hadron Magnetic moments
• Need a particles which are long-lived and have
some intrinsic spin. Proton!
pspin up
1 2u u d   u u d   u u d 



3 2  permutations

Total Magnetic Moment should equal the vector sum of the
magnetic moments of the constituent quarks.



pspin up m1  m 2  m 3 z pspin up
 pspin up m u Suz  m u Suz  m d S dz pspin up
Reminder: The order of the spin arrows designates which quark has that spin.
Hadron Magnetic moments
pspin up
1 2u u d   u u d   u u d 



3 2  permutations

Doing the calculation for the first term:
2
 

 2 

 u  u  d   m1  m 2  m 3 z u  u  d  
3 2 
2
h
 2 m u  m d 
9
2
So we expect mproton to be:
1
h
m proton  (4m u  m d )
3
2
Hadron Masses
• Seems Simple enough
– Just add up the masses of the quarks
• Mp = Mu + Md = 2*Mu = 620 MeV/c2
• Experimentally  Mp = 139 MeV/c2
– What????
p+ is |u, d-bar>.
This is a particle made up of two like-sign charged quarks.
Why doesn’t it fly apart?
Hadron Masses
Hyperfine splitting in hydrogen atom:
Caused by the spin of the electron interacting with the spin of the proton
Ehf 
8p p e 2
3c
2
 Aem 
 0  
Se  S p
me m p
2
Se  S p
me m p
 ~ 6.0  10 6 eV
S1  S 2
M (meson)  m1  m2  As 
m1m2
Hadron Masses
Masses are more equal, Force is much more powerful.
Fit to some meson masses and find
As = 160*(4pmu/h)2 MeV/c2
S1•S2 Meson
3
 h
p
4
1
 h
r
4
3
 h
K
4
1
 h
K*
4
2
2
2
2
Calculated
140
780
484
896
Observed
138
776
496
892
Hadron Masses
Amazingly we can take the meson mass formula as the lead for
estimating baryon masses:
 S1  S2 S3  S2 S1  S3 
M (baryon)  m1  m2  m3  As  



m3m2
m1m3 
 m1m2
Fit to some baryon masses and find
As’ = 50*(4pmu/h)2 MeV/c2
Caution: There are tricks you need in order to calculate those spin
dot products. Example: if all masses are equal (proton, neutron):
J 2  S1  S 2  S3   S1  S 2  S3  2S1  S 2  S3  S 2  S1  S3 
2
h2

2
9



j
j

1


4 
2
2
2
Again see Griffiths, page 182.
More Conserved Stuff
• We need to cover some more conserved quantum
numbers and explain some notation before moving
on.
• Parity and Charge Conjugation:
– Parity Y(x,y,z)Y(-x,-y,-z) not reflection in a mirror!
– Define the parity operator ‘P’ such that:
• P | Y(x,y,z)> = | Y(-x,-y,-z)>
• |> is an eigenstate of P if P|> = p|>
• P2|> = p2|> = |> so p = 1
• Parity is a simple group. Two elements only.
Eigenstates of Parity
• Suppose we have a force that only acts radially
between two particles.
– Then the wave function Y = (r)qqbar
• P | q>  | q> = -P| qbar>
• Parity is a Multiplicative quantum number, not additive.
– Given q1 and q2
– J = S 1 + S2
– P = P1*P2
Eigenstates of Parity
• For once, Baryons are easy!
– For Mesons with no ang. Momtenum
• P|Yb>|Ybbar> = -1 |Yb>|Ybbar>
– DEFINE:
P |Yb> 1
P |Ybbar> -1
– So in general, for baryons with orbital angular
momentum between the quarks:
• P |Yb> = (-1)l |Yb>
• Unfortunately, because baryon number is conserved
Eigenstates of Parity
 (r) can be separated into the angular part
Ylm(,) and a purely radial part so:
 (r) = (r) Ylm(,) space-part of wave function
– P Ylm(,) = (-1)l Ylm(,)
– And P| Y > = (-1)l pq pqbar| Y > = (-1)l(1)(-1)| Y >
– P| Y > = (-1)l+1 | Y >
• For MESONS only (since pq=1, pqbar=-1)
Charge Conjugation
• C is an operator which turns all particles into
antiparticles:
– C |q> = |q-bar>
• changes sign of charge, baryon #, flavour quan. Num.
• Leaves momentum, spin, position, Energy unchanged.
• Most particles are NOT eigenstates of C
– C |Y>  a|Y>
– eg.

(where a = number)
C ud   u d
  ud
Charge Conjugation
• Neutral Mesons are eigenstates of C

qr1 

q r2 
|Y>=|Y(space)>|Y(spin)>|q,qbar>
If we apply C to the diagram on the left we change
nothing but the ‘particleness’.
This doesn’t effect |q,qbar>
but has the same effect on |Y(space)> as if we’d used
the parity operator.
C |Y(space)> = (-1)l+1 |Y(space)>
Charge Conjugation
• Neutral Mesons are eigenstates of C

qr1 
|Y>=|Y(space)>|Y(spin)>|q,qbar>
If we apply C|Y(spin)> what do we get?
Lets try this on a S=1 or 0 meson |ms> = |0>

q r2 
C 0  C q q  q q 
 q q   q q 
2
2
  1
C |Y(spin)> = (-1)s+1 |Y(spin)> so
C |Y> = (-1)l+s |Y> neutral mesons only
s 1
0
Conserved by Strong force:
• Isospin, Quark Flavor
–(I, I3, U, D, S, C, B, and T)
• Parity
• Charge Conjugation
• Electric Charge
• Energy/momentum
• Angular Momentum / Spin