Problem Set # 4 SOLUTIONS

Wissink P40 Subatomic Physics I Fall 007 Problem Set # 4 SOLUTIONS 1. Gee! Parity is Tough! In lecture, we examined the operator that rotates a system by 180 about the -axis in isospin space. This operator, which flips the world in isospin space, is known as the charge symmetry operator, formally defined as ˆP CS Û isospin R (π) = exp( i π ˆτ /) (a) Show that the flavor wave functions of the π 0, η 0, ρ 0, and ω 0 mesons, as predicted by flavor SU(3), are all eigenstates of ˆP CS. Find the corresponding eigenvalues. (b) The mesons listed in part a are also eigenstates of the charge conjugation operator Ĉ. Systems that are eigenstates of both ˆP CS and Ĉ can be assigned a new quantum number, called G-parity, given by the product of the two eigenvalues. With this definition, show that the two decays both violate G-parity conservation. η 0 3 π 0 and ω 0 π + π (c) Experiment shows that G-parity is very nearly, but not exactly, conserved in strong interactions, and so both of the above decays have in fact been observed. These violations have been interpreted as evidence for isospin-mixing between the pseudoscalar mesons η 0 and π 0, and between the vector mesons ω 0 and ρ 0. Look up the measured partial decay widths for ω 0 π + π and for ρ 0 π + π, and use these to estimate the magnitude of the mixing between the two vector mesons. Note that these particles have almost the same mass, so differences in decay phase space are negligible. SOLUTION As discussed in class, the charge symmetry operator ˆP CS is an isospin flipper, and hence only operates in the u-d flavor subspace. In this SU() space, ˆPCS has the matrix form: 1

ˆP CS Û isospin R (π) = exp( i π ˆτ /) = iˆτ = [ 0 1 1 0 where the exponentiated operator was simplified by performing a series expansion and using properties of the τ matrices (identical to those of the Pauli σ matrices). Thus, ˆPCS operates on a quark state or an anti-quark state according to ˆP CS [ u d = [ d u and ˆPCS [ d ū = [ ū d where, as shown in lecture, the anti-quark state was chosen to keep the I 3 = + 1 particle on top and to preserve the form of the isospin operators. With these results, and recognizing that ˆτ has no effect on strange quarks or anti-quarks, we can easily work out the effect of the charge symmetry operator on various neutral mesons. Using the flavor wave functions worked out in class (where the two ket subscripts are the meson L and S), we find: ˆP CS π 0 = 1 ˆPCS uū d d 0,0 = 1 ( d)( d) uū 0,0 = π 0 ˆP CS η 0 = 1 ˆPCS uū + d d s s 0,0 = 1 ( d)( d) + uū s s 0,0 = + η 0 ˆP CS ρ 0 = 1 ˆPCS uū d d 0,1 = 1 ( d)( d) uū 0,1 = ρ 0 ˆP CS ω 0 = 1 ˆPCS uū + d d 0,1 = 1 ( d)( d) + uū 0,1 = + ω 0 Thus, we see that all four mesons are eigenstates of ˆP CS, with the π and ρ having eigenvalue 1, while the η and ω have +1. Not too surprisingly, the distinction is between the isotriplet (former) and isosinglet (latter) cases. We can summarize this concisely as ˆP CS q q I = ( 1) I q q I, where I is the isospin for the neutral meson. (b) We argued in lecture that if a meson is composed of a quark and anti-quark of the same flavor, configured in a state of orbital angular momentum L and total spin S, then by taking all particles to their anti-particles (charge conjugation), followed by particle interchange (to return the state to its original configuration), we arrive at a quantum number C = ( 1) L+S for that meson. The four mesons considered above all have L = 0, so C is just ±1 for S even (odd). The π 0 and η 0 mesons therefore have positive C-parity, while C = 1 for the ρ 0 and ω 0. If we introduce the concept of G-parity as the product of the ˆP CS and Ĉ operators, then the quantum number G = P CS C = ( 1) I+L+S, so G π = 1 G η = +1 G ρ = +1 G ω = 1 We now consider the meson decay η 0 3π 0. The eta and the pions all have J P = 0 (i.e., they are pseudo-scalars), so the decay can conserve total angular momentum and parity

as long as there is no orbital angular momentum (L = 0) in the final state. However, note that we have G = +1 in the initial state and G = ( 1) 3 = 1 in the final state decay violates G-parity. Nevertheless, this is one of the dominant decay modes for the η 0, accounting for 33% of all decays. This is because the decay to two pions if forbidden by CP conservation, which is a much stronger principle than G-parity or charge symmetry. For the decay ω 0 π + π, we have a vector meson (J P = 1 ) going to two pseudo-scalar mesons, so we must have L ππ = 1 to conserve total angular momentum. An odd L also conserves parity here, so it would seem the reaction is perfectly allowed. However, because we treat the two pions as identical bosons (when working in an isospin framework), their total wave function must be symmetric under particle interchange. L ππ = 1 implies an anti-symmetric spatial state, and S π = 0 yields a symmetric spin state the two pions must be in an anti-symmetric isospin configuration, with a wave function of the form π + π L=1 = 1 [ π + (1) π () π (1) π + () Like the neutral pion discussed in part a, the charged pions are also G-parity eigenstates, with negative eigenvalues: Ĝ π + = ˆP CS Ĉ u d = ( 1) L+S ˆPCS ūd = ( 1) 0 ( d)u = u d = π + Ĝ π = ˆP CS Ĉ dū = ( 1) L+S ˆPCS du = ( 1) 0 ū( d) = dū = π Putting this all together, we see that for the initial state G i = G ω = 1, while G f = (G π ) = +1, and so here also we find that G-parity is violated. (c) Our last task is to try to understand semi-quantitatively how this latter decay can proceed, despite violating G-parity. Our approach is to attribute the observed ω 0 π + π decay as resulting from a small I = 1 admixture in the predominantly I = 0 wave function for the ω. From our preceding discussion, we see that for the vector mesons, with L = 0 and S = 1, then G = ( 1) I+1, so isospin-mixing also brings in a component of the opposite G-parity. Thus, we can write the mixed neutral vector meson wave functions as ω real = α ω ideal + β ρ ideal ρ real = β ω ideal + α ρ ideal where the ideal mesons have the isospin and G-parities assigned above. The coefficients have been chosen to ensure that ρ ω real = 0, and we require that α + β = 1. With this ansatz, the ratio of observed partial decay widths for each meson to a two-pion final state will be equal to the ratio of the squared amplitudes for the I = 1, G = +1 component of these wave functions, i.e., Γ(ω π + π ) Γ(ρ π + π ) = β α = β 1 β = BRω ππ Γω tot BR ρ ππ Γρ tot = (0.0170)(8.49) (1.00)(149.4) = 9.7 10 4 where the branching ratios and total decay widths are from the PDG compilation. The result is that β 0.031, so there is a 3% amplitude of the wrong isospin for these mesons. 3

. The Flavor is Very Strange... (a) Following the example of what we did in class for the proton, write down quark-model SU() SU(3) spin-flavor wave functions for the Λ 0 and Σ 0 baryons. Note that these are both J P = 1 + particles, and each has a uds valence quark content, but they have total isospins I = 0 and I = 1, respectively. Assume the baryon has a positive spin projection m J = + 1 as you construct your wave functions. (b) Now use your wave functions to calculate the expectation value of the spin projection of the strange quark in each baryon in other words, determine how aligned the spin of s is with the spin of the baryon as a whole. More explicitly, evaluate the matrix elements where B = either Λ 0 or Σ 0. B, m B = +1/ Ŝs quark z B, m B = +1/ SOLUTION In this problem, we construct the spin-flavor wave functions for the two uds members of the J P = 1 + baryon octet. We know that to couple to a total spin of 1, the three quark spins must be in a configuration of mixed symmetry, either M S or M A ; and to keep the spin-flavor wave function fully symmetric, the flavor state must have a similar symmetry. We start with the Λ 0, which is isospin 0, and therefore must be flavor anti-symmetric under u d interchange. The flavor state must also be symmetric (anti-symmetric) under interchange of the first two quarks for the M S (M A ) configuration. Finally, we insist that the mixed-symmetry flavor states be orthogonal to each other and to the fully symmetric and fully anti-symmetric states, which are given by: uds S = 1 uds + dsu + sud + sdu + dus + usd } uds A = 1 uds + dsu + sud sdu dus usd } With these constraints, the flavor states are now fully defined, except for an overall (and irrelevant) phase. For I = 0, we have uds 0 M S = 1 [ 1 (sd + ds) u = 1 [ sdu + dsu usd sud 4 [ } 1 (us + su) d

uds 0 M A = 1 [ 1 (sd ds) u + = 1 1 [ sdu dsu + usd sud dus + uds [ 1 [ 1 } (us su) d (du ud) s We now form the complete spin-flavor wave function for a spin-up Λ 0 by building the two symmetric combinations Λ 0 = 1 [ MS uds 0 M S + MA uds 0 M A where the mixed-symmetry spin states were worked out in lecture. Putting all this together gives us Λ 0 = 1 1 ( )[ + sdu + dsu usd sud + 1 ( )[ } sdu dsu + usd sud dus uds After expanding this out and simplifying, we obtain the not-too-bad result: Λ 0 = 1 1 s d u s u d + s u d s d u + d s u u s d + u s d d s u + d u s u d s + u d s d u s } It is very useful to stare at this for a few minutes and check that 1. each term has the flavor content uds and the spin content ;. the overall normalization is correct; 3. it is fully symmetric under interchange of any two labels; 4. it is anti-symmetric under interchange of u and d. In addition, we note that the s-quark is spin-up in every single term, i.e., the spin of the Λ is (in this model) due entirely to the spin of the strange quark, while the u and d quarks always couple up to a spin of 0. This is not surprising: by design, the u and d couple up to an isospin of 0, which is isospin anti-symmetric, and so to keep the full spin-flavor wave function symmetric under u-d interchange, the state must be anti-symmetric in spin as well, so S ud = 0. We now repeat the same set of steps for the I = 1 Σ 0, which means that the flavor state must be symmetric under u d interchange. The state must also be symmetric (antisymmetric) under interchange of the first two quarks for the M S (M A ) configuration. As before, we again insist that the mixed-symmetry flavor states be orthogonal to each other 5

and to the fully symmetric and fully anti-symmetric states given earlier. These constraints define the flavor states to be: uds 1 M S = 1 [ 1 [ 1 [ 1 } (ds + sd) u + (us + su) d (du + ud) s = 1 1 [ dsu + sdu + usd + sud dus uds uds 1 M A = 1 [ 1 (ds sd) u + = 1 [ dsu sdu + usd sud [ } 1 (us su) d To write out the complete spin-flavor wave function for a spin-up Σ 0, we must again combine the M S flavor state above with the M S spin state, and the same for the M A flavor and spin states, just as we did for the Λ 0. Putting this all together gives us Σ 0 = 1 1 ( )[ + dsu + sdu + usd + sud dus uds + 1 ( )[ } dsu sdu + usd sud We expand these terms, cancel out as much as possible, and regroup to obtain the nottoo-pretty result: Σ 0 = 1 s d u +s u d s d u s u d s d u s u d + d s u +u s d d s u u s d d s u u s d + d u s +u d s d u s u d s +d u s u d s } Again, it is useful to check that 1. each term has the flavor content uds and the spin content ;. the overall normalization is correct; 3. it is fully symmetric under interchange of any two labels; 4. it is symmetric under interchange of u and d. Note that, in contrast to the wave function for the Λ 0, the u and d quarks must here couple to isospin I = 1, and so must also couple to S ud = 1, though sometimes (about 1/3 of the time) with S z = 0. This means that we would expect the s quark to be spin-down more often than spin-up, and thus on average to be anti-parallel to the spin of the Σ 0. We investigate this more quantitatively next.

(b) Using the above wave functions, we can now easily calculate the expectation value of the s-quark spin projection in each baryon. When expressed in these forms, we simply get a contribution of + 1 for each term containing an s, and 1 for each s. Doing the counting tells us: Λ 0 quark Ŝs z Λ 0 = 1 [ + 1 = + 1 1 and Σ 0 quark Ŝs z Σ 0 = 1 [ + 1 3 + 4 [ 1 3 = 1 Thus, our expectations from part a are indeed borne out, and we see that the Λ 0 is an excellent laboratory in which to study a polarized strange quark! 3. Magic Magnetic Moments Please do Exercises (a).0 and (b).1 on pages 5-57 in H & M. SOLUTION (a) In this problem, we investigate the radiative decay of the ω meson, i.e., the process ω π 0 + γ, using the simple quark model. Neither meson has any orbital angular momentum, so J = S in each case. Thus, to make the transition from a vector meson to a pseudoscalar meson, we must have S = 1, or a quark spin-flip in this process. To begin, we need the spin-flavor wave functions for the mesons involved. For the flavor state, we assume that (.54) in the text is correct, so ω = 1 ( uū + d d) Note that despite appearances, this state is anti-symmetric in flavor, as it should be since this is an I = 0 particle. For the spin configuration, we want to examine the m J = +1 state. This means m S = +1 also, so the q and q spin projections are both + 1. Altogether, we see: ω m=+1 = 1 uū + d d = 1 u ū + d d Unlike the previous problem, the ordering of the particles is not relevant here, because the quark and anti-quark are distinguishable (B = ±1/3). 7

Next, we work out the spin-flavor wave function for the neutral pion. This meson has isospin I = 1, so the flavor component will be symmetric, which means the spin (S = 0) must be anti-symmetric. Thus, π 0 = 1 uū d d = 1 u ū u ū d d +d d To make the transition from one meson state to the other via an electromagnetic process (emitting a photon) requires reversing the projection onto ẑ of either the q or q spin. To accomplish this, we apply a torque to the magnetic moment using the magnetic interaction µ B = µ σ B, with µ Qe/m. But we must also conserve total angular momentum in the process, which means (using the geometry shown in Fig..10) the photon must carry away one unit of h along the z-axis. This is, by convention, a right-circularly polarized photon, with the polarization vector ε R = 1 (1, i, 0) and so σ ε R = 1 [ σx + iσ y = σ = [ 0 0 1 0 As a quick sanity check, note that it is reasonable that the transition ω π 0 would involve the Pauli spin-lowering operator; there is no change in the flavor composition, but the meson wave functions make it clear we must have for exactly one of the two partons. Summing over these two possibilities (j = 1, ), the relevant matrix element is therefore π 0 µ i σ ε R ω(m J = +1) j = 1 u ū u ū d d d d µ i (σ ) i u ū +d d j = 1 u ū u ū d d d d µ u (u ū u ū ) + µ d (d d d d ) = 1 = µ u µ d µu ( 1 1) + µ d (1 + 1) } The above result differs from that in the text by an overall minus sign. This is not a problem, and simply reflects the fact that either of the meson wave functions could have been multiplied by ( 1) and would still satisfy all of our requirements. The transition probability, of course, is insensitive to this choice in phase. 8

(b) In this part of the problem, we examine possible radiative decays of other vector mesons, such as the φ and the ρ 0. For the φ, equation.54 in the text indicates that we have almost complete flavor mixing (compared to the SU(3) expectations), yielding a φ that is a pure s s flavor state. In this case, operating on the φ with ˆσ will have no effect, and the resulting state will have no overlap with the π 0 wave function we worked out above. Thus, the transition matrix element will vanish, and so we see that the quark model suggests that there should be no significant decay strength in the φ π 0 γ process. Turning now to the ρ 0, we see that this vector meson is isospin I = 1, and will have the same flavor configuration as the pseudoscalar π 0 (see Eq..47 in H & M, for example). Its spin-flavor wave function will therefore be identical to that of the ω 0, except for a relative minus sign between the u and d quark terms, i.e., ρ 0 m=+1 = 1 uū d d = 1 u ū d d If we stare at the algebra for a moment, we see that the effect of this sign change in evaluating the transition matrix element is only to reverse the sign of the µ d term. So, without redoing all the steps, we can see immediately that j π 0 µ i σ ε R ρ0 (m J = +1) = µ u + µ d To complete the problem, we need to calculate the ratio of the radiative decay rates for the ω 0 and the ρ 0. This rate is proportional to the square of the decay amplitude, and so we find Rate(ω 0 π 0 γ) = (µ u µ d ) Rate(ρ 0 π 0 γ) (µ u + µ d ) Treating the quarks as ideal (point-like) Dirac particles, we can express µ u and µ d in terms of the quark charges and masses, with µ f = Q f e/m f. If we assume perfect isospin symmetry, so m u = m d m, then µ u = e/3m and µ d = e/m. Using these values in the expression above yields Rate(ω 0 π 0 γ) Rate(ρ 0 π 0 γ) = [(1/3) ( 1/) [(1/3) + ( 1/) = [1/ [1/ = 9 In reality, we can check the Particle Data Group s booklet and find Rate(ω 0 π 0 γ) Rate(ρ 0 π 0 γ) = (0.087)(8.44 MeV) (7.9 10 4 )(149 MeV) =. Given possible differences in phase space, the approximate flavor content of each meson we assumed, and the fact that isospin is not a perfect symmetry, this isn t too bad! 9