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Spin Algebra for a spin operator ‘J’: ‘Isospin operator ‘I’ follows this same algebra Isospin is also additive. Two particles with Isospin I a and I b.

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Presentation on theme: "Spin Algebra for a spin operator ‘J’: ‘Isospin operator ‘I’ follows this same algebra Isospin is also additive. Two particles with Isospin I a and I b."— Presentation transcript:

1 Spin Algebra for a spin operator ‘J’: ‘Isospin operator ‘I’ follows this same algebra Isospin is also additive. Two particles with Isospin I a and I b will give a total Isospin I = I a + I b Last Time: By defining I + = I 1 + iI 2 and I - = I 1 - iI 2 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)

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 I 3 Up quark  Isospin = 1/2; I 3 = 1/2 Anti-up quark  I = 1/2; I 3 = -1/2 Down quark  I = 1/2; I 3 = -1/2 Anti-down quark  I = 1/2; I 3 = 1/2

3 I3I3 0 1/2 -1/2 1 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. 2) Plot the particles as a function of the I 3 quantum numbers.

4 I3I3 0 1/2 -1/2 1 Graphical Method of finding all the possible combinations: I3I3 0 1/2 -1/2 1 I3I3 0 1/2 -1/2 1 Triplet Singlet Group A Group B Sum

5 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, I 3 > |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 )

6 Quantum states: Triplet I = 1 |I, I 3 > |1,1> = |1/2,1/2> 1 |1/2,1/2> 2 = -|ud> |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 |1/2,-1/2> 2 = |ud> 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 =1/2(|uu> + |dd>) Reminder: u = |1/2,1/2> u-bar or d = |1/2,-1/2> Must choose either quark-antiquark states, or q-q states. We look for triplets with similar masses. MESONs fit the bill!  +,  0,  - 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

7 But quarks are also in groups of 3 so we’d like to see that structure too: I3I3 0 1/2 -1/2 1 I3I3 0 1/2 -1/2 1 I3I3 0 1/2 -1/2 1 3/2 -3/2 s a

8 Isospins of a few baryon and meson states:


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