Transcript Particle Physics Option

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Spin Algebra for a spin operator ‘J’:
J , J   i
i
j
ijk
J k
‘Isospin operator ‘I’ follows this same algebra
Isospin is also additive. Two particles with Isospin Ia and Ib will
give a total Isospin I = Ia + Ib
By defining I+ = I1 + iI2 and I- = I1 - iI2 we could ‘Raise’ and ‘Lower’ the
third component of isospin:
I-|i,m> = [i(i+1)-m(m-1)]1/2|i,m-1>
I+|i,m> = [i(i+1)-m(m+1)]1/2|i,m+1>
NOTICE: I+|1/2,-1/2>
= I+|d> = |u> (or -|d-bar>)
All part of what we called SU(2)
• Concept Developed Before the Quark Model
• Only works because M(up)  M(down)
• Useful concept in strong interactions only
• Often encountered in Nuclear physics
• From SU(2), there is one key quantum number I3
Up quark  Isospin = 1/2; I3 = 1/2
Anti-up quark  I = 1/2; I3 = -1/2
Down quark
 I = 1/2; I3 = -1/2
Anti-down quark  I = 1/2; I3 = 1/2
Graphical Method of finding all the
possible combinations:
1). Take the Number of possible states each
particle can have and multiply them.
This is the total number you must have
in the end. A spin 1/2 particle can have
2 states, IF we are combining
two particles:
2 X 2 = 4 total in the end.
I3
1
1/2
0
-1/2
2) Plot the particles as a function of the
I3 quantum numbers.
-1
Graphical Method of finding all the
possible combinations:
Group A
I3
Group B
I3
Triplet
Singlet
Sum
I3
1
1
1
1/2
1/2
1/2
0
0
-1/2
-1/2
-1/2
-1
-1
-1
0
Graphical Method of finding all the
possible combinations:
We have just combined two fundamental representations of
spin 1/2, which is the doublet, into a higher dimensional
representation consisting of a group of 3 (triplet) and another
object, the singlet.
What did we just do as far as the spins are concerned?
Quantum states: Triplet I = |I, I3>
|1,1> = |1/2,1/2>1 |1/2,1/2>2
|1,0> = 1/2 (|1/2,1/2>1 |1/2,-1/2>2 + |1/2,-1/2>1 |1/2,1/2>2 )
|1,-1> = |1/2,-1/2>1 |1/2,-1/2>2
Singlet
|0,0> = 1/  2 (|1/2,1/2>1 |1/2,-1/2>2 - |1/2,-1/2>1 |1/2,1/2>2)
Reminder: u = |1/2,1/2>
u-bar or d = |1/2,-1/2>
d 
1 1
,
2 2
Quantum states: Triplet I = 1 |I, I3>
|1,1> = |1/2,1/2>1 |1/2,1/2>2 = -|ud>
p  
 0
p 
p  
 
Singlet
  
 0  |1,0> = 1/2(|1/2,1/2>1 |1/2,-1/2>2 + |1/2,-1/2>1 |1/2,1/2>2 )
= 1/2(|uu> - |dd>)
 
  
  |1,-1>= |1/2,-1/2> |1/2,-1/2> = |ud>
1
2
|0,0>=1/2(|1/2,1/2>1 |1/2,-1/2>2 - |1/2,-1/2>1 |1/2,1/2>2
=1/2(|uu> + |dd>)
Must choose either quark-antiquark states, or q-q states. We look for
triplets with similar masses. MESONs fit the bill!
p+,p0,p- and +, 0, - (q-qbar pairs). 0, 0, and 0 are singlets.
WARNING: Ask about |1,0> minus sign or read Burcham & Jobes pgs. 361 and 718
But quarks are also in groups of 3
so we’d like to see that structure
too:
I3
3/2
I3
I3
1
a
1
1
1/2
1/2
0
0
s
1/2
0
-1/2
-1/2
-1/2
-1
-1
-1
-3/2
Isospins of a few baryon and meson states:
3 3
  ,
2 2

3 1
  ,
2 2
3 1
  ,
2 2
1 1
p  ,
2 2
n 
1 1
,
2 2
   1,1
0 
 0  1,0
1 1
,
2 2
1 1
  ,
2 2

   1,1
0
 
3 3
,
2 2
p 
1 1
,
2 2
1 1
n  ,
2 2
p   1,1
p  1,0
0
p  1,1

1 1
K  ,
2 2

K0 
1 1
,
2 2
1 1
K  ,
2 2
0
 0  0   0  0,0
K 
1 1
,
2 2