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Isospin and SU(2) Symmetry

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Isospin and SU(2) Symmetry Isospin and SU(2) Symmetry Javier M. G. Duarte∗ Massachusetts Institute of Technology, MA 02142 (Dated: April 28, 2009) We describe isospin, an abstract property of elementary particles, which reveals a deep symmetry of the strong force. Particles are assigned a total isospin quantum number, I and an isospin projec- tion in one direction, I3, in analogy with angular momentum. Exploiting the fact that the strong force is invariant under any rotation in `isospin space,' that is, the strong force treats all particles with the same total isospin I equally, we may derive relations between the cross sections for various interactions mediated by the strong force. Specifically, we calculate the relative cross sections for Nπ scattering, ∆ ! Nπ decays, and Kp scattering. We discuss the underlying symmetry group SU(2) and its representations as well as how this symmetry is `broken.' Finally, with the introduction of a new quantum number, known as strangeness, and selection rules, we utilize isospin to calculate weak decay rates. I. INTRODUCTION AND MOTIVATION n p S=0 When faced with a new problem, a guiding principle of quantum mechanics and physics in general is to ex- ploit its symmetry. Symmetry is a powerful tool, which 0 has allowed physicists to learn about systems, which are − S=−1 otherwise too complicated to investigate. At its most fun- damental level, a symmetry is simply the invariance of a system after an operation has been applied. Symmetry − 0 can come in a variety of forms, including space-time sym- S=−2 metries, such as rotational invariance, and internal sym- metries, which describe the relations between elementary constituents of matter. But how can we as physicists Q=−1 Q=0 Q=1 exploit symmetry for problem solving? Emmy Noether provided the critical link. In rough terms, Noether's the- orem states that for each symmetry of a system (or equiv- FIG. 1: The baryon octet, comprised of the eight lightest alently the Lagrangian specifying that system), there is baryons. Particles along the same horizontal line have the an associated conserved quantity [1]. This gave physicists same strangeness S, while those along the same dotted slanted an excellent strategy: discover a symmetry of the prob- lines have the same charge Q. lem, derive its associated conserved quantity, and use it to drastically simplify calculations. One question that begged for a simple explanation in not discriminate on the basis of the direction of isospin. the early nineteenth century was: Why do the proton In other words, any rotation of a particle's isospin vector and the neutron have such similar masses but differ in will not affect the way it couples to the strong force. The electric charge? Heisenberg suggested in 1932 that the strong force only cares about the magnitude of I. The proton and neutron could be thought of as different states relationship between isospin and other quantum numbers of the same particle: `spin up' and `spin down' nucleon. was worked out empirically later. This was the beginning of isospin (originally isotopic spin In the 1960s, particle physics was comprised of a grab- and sometimes isobaric spin). By analogy with angular bag of particles and properties with no rhyme or rea- momentum, the proton and the neutron can be assigned son until Murray Gell-Mann proposed his Eightfold Way a vector quantity known as isospin. If we take I · I the for classifying the eight lightest baryons. A baryon is total isospin vector squared and I3 the isospin projection an elementary particle containing three quarks{protons in one direction to be our complete set of commuting and neutrons are both baryons. He found that they observables, an isospin state is denoted by jII3i. The could be arranged in terms of certain quantum numbers: proton and the neutron are charge and strangeness, as seen in figure 1. He general- 1 1 1 1 ized Heisenberg's notion by saying that each row in his p = j 2 2 i; n = j 2 − 2 i (1) diagram would be completely degenerate if the strong Heisenberg proposed that the strong interaction does force was the only force mediating interactions between quarks, was involved. Heisenberg's postulate and Gell-Mann's isospin assign- ment scheme allow us to perform calculations. Nonethe- ∗Electronic address: [email protected] less, isospin is not an absolute conservation law. Only 2 momentum. Since we are only able to specify the magni- tude and a single component of each summand, the mag- nitude of the sum ranges from jI(1) − I(2)j to jI(1) + I(2)j. Clebsch-Gordan coefficients describe the possible states, in the bassis of total isospin, and their probabilities, as seen in figure 2. A derivation of the Clebsch-Gordan co- efficients is carried out in Appendix C of reference [2]. In particular, consider the state of a neutral pion and a proton. We may use the Clebsch-Gordan table to de- compose the two single-particle isospin states into total isospin states. q π0 + p : j10ij 1 1 i = 2 j 3 1 i − p1 j 1 1 i (4) 2 2 3 2 2 3 2 2 1 1 1 FIG. 2: The Clebsch-Gordan coefficients for 2 × 2 and 1 × 2 . Note that for each cell of the table there is an implied square III. RELATIVE DECAY RATES AND CROSS root so that −1=2 should be read as −p1=2. Taken from [3] SECTIONS An important application of isospin conservation in the the strong force is invariant under rotations in abstract strong interaction is the calculation of relative decay rates isospin space. However, the weak force respects isospin and cross sections (the probability for a decay or a scatter via selection rules that allow for the derivation of weak to occur, respectively). An essential concept is that of a decay rates. matrix element M for a particular interaction, which is related to its cross section. The matrix element connecting the initial and final states, and is given by II. ASSIGNMENT AND ADDITION OF i f ISOSPIN Mif = h f jAif j ii; (5) The assignment of isospin proceeds as follows: a mul- where Aif is an isospin operator (not to be confused tiplet of 2I + 1 particles is given a total isospin I and with the Hamiltonian) which only depends on the to- each member is given a projection isospin I , which takes 3 tal isospin, so that A = A1=2 for initial and final states on the values −I; −I + 1; :::; +I in order of increasing of I = 1=2 and A = A3=2 for states of I = 3=2. Further, charge. We can skip some of the difficulty of assign- conservation of isospin demands that A = 0 for initial ment by using the empirically-determined Gell-Mann{ and final states of varying isospin. Knowing this, we de- Nishijima formula: fine an amplitude for each total isospin I, 1 I = Q − (A + S) (2) MI = h (I)jAI j (I)i (6) 3 2 Regardless of the type of interaction (decay or scatter), where Q is charge in units of elementary charge, A is the probability is always proportional to the absolute baryon number and S is strangeness1. Equation 2 implies square of the matrix element [2]. that the lines of constant I3 in figure 1 are horizontal. As a simple exercise, the pions form a triplet 2 2 σ / jMif j = jh f jAif j iij ; (7) (π−; π0; π+) with A = S = 0 so their isospin assignments can be inferred from equation 2. The idea is that since the strong force treats all total isospin I particles the same, we need only specify the parameter for each total isospin in order to complete − 0 + MI π = j1 − 1i; π = j10i; π = j11i (3) the calculation of relative cross sections. In order to add two isospin vectors, we invoke the for- malism from the quantum mechanical addition of angular A. Nπ ! Nπ Scattering One of the simplest calculations is nucleon-pion scat- 1 These are quantum numbers from particle physics: A is the num- tering. To proceed, we first write down the possible chan- ber of baryons (or three times the number of quarks) and S is nels through which the interaction may occur. There are the number of `strange' particles. six elastic processes (same particles come out that went 3 in) and four charge-exchange processes. (a)π+ + p ! π+ + p (b)π0 + p ! π0 + p (c)π− + p ! π− + p (d)π+ + n ! π+ + n (e)π0 + n ! π0 + n (f)π− + n ! π− + n (8) (g)π+ + n ! π0 + p (h)π0 + p ! π+ + n (i)π0 + n ! π− + p (j)π+ + p ! π0 + n 1 Since the pion has I = 1 and the nucleon has I = 2 , 1 3 the total isospin of the composite system must be 2 or 2 . We use figure 2 to compute the isospin decompositions: π+ + p : j11ij 1 1 i = j 3 3 i 2 2 q 2 2 π0 + p : j10ij 1 1 i = 2 j 3 1 i − p1 j 1 1 i 2 2 3 2 2 3 2 2 q π+ + n : j11ij 1 − 1 i = p1 j 3 1 i + 2 j 1 1 i 2 2 3 2 2 3 2 2 q π0 + n : j10ij 1 − 1 i = 2 j 3 − 1 i + p1 j 1 − 1 i 2 2 3 2 2 3 2 2 q π− + p : j1 − 1ij 1 1 i = p1 j 3 − 1 i + 2 j 1 − 1 i 2 2 3 2 2 3 2 2 − 1 1 3 3 π + n : j1 − 1ij 2 − 2 i = j 2 − 2 i (9) As prescribed by equation 5, we take the inner product of the initial and final state to find the coefficient in front of the scattering amplitude.
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