PHY357S: Thursday, 12 February 1998

Answer 1

(a) (i) The binding energy of a hydrogen atom is

where Q_p is the proton charge, Q_e is the electron charge, and m_r is the reduced mass of the electron. (ii) For a proton-antiproton atom the binding energy is similarly

(b) The total angular momentum of an L=0 proton-antiproton atom is either J=0 or J=1, depending on whether the proton and antiproton spins or antiparallel or parallel. Pions have J^P=0^-, so if L_pp is the two-pion final state orbital angular momentum, then the parity of the final state is

A fermion has the opposite parity from their antifermion, so the L=0 proton-antiproton atom has parity

So L_pp=1 is required for any of the two boson decays, but p⁰p⁰ is not allowed because Bose-Einstein statistics prevents two identical spinless bosons from being in an antisymmetric state.

The C-parity of a particle-antiparticle system is obtained by looking at the wavefunction after turning the particle into an antiparticle and the antiparticle into a particle. If the spins of the particle and antiparticle are parallel (or zero), then charge conjugation and parity transformation are each equivalent to simply interchanging the two particles, so the C and P quantum numbers should be the same, i.e.

(More generally, from Lecture 12 we have C=(-1)^L+S for a fermion-antifermion pair.) In this case we have L_pp=1 and L=0, so the proton-antiproton spins must be aligned so S=1 in the initial state and

The final state pions are spinless, so again C and P should be the same

So a p⁺p^- final state is allowed.

C_final can also be worked out using C=(-1)^L+S applied to the quarks inside the p⁺p^- system. i.e. (p⁺)(p^-) is a (up anti-down) (anti-up down) system. If we mathematically regroup the quarks as (up anti-up) (anti-down down), we can see that C_final = C(up anti-up) C(anti-down down) = (-1)^Lu+Su(-1)^Ld+Sd = (-1)^Lu+Su+Ld+Sd. (Remember the C is a multiplicative quantum number.) L_u & S_u are just the relative angular momentum and to total spin of the (up anti-up) pair, and similarly L_d & S_dare for the (down anti-down) pair. Since we are grouping together quarks in different pions, we don't necessarily know what the values of L_u, S_u, L_d, & S_d are, but angular momentum conservation tells us that L_u+S_u+L_d+S_d=L=1, so we have

as before.

(Note: This is essentially a proof that C=(-1)^L+S is true for either bosons or fermions.)

Answer 2

(a) (a) If the deuteron did not have zero isospin, then one would expect it to have isospin partners. In particular, if it had isospin 1 (a possibility since it is made of two nucleons, each with I=¹/₂.), then one would expect pp and nn bound states as well. It is possible that the electrostatic repulsive force between the two protons might make the pp state unbound, but there is no obvious reason why an I=1 nn state should not exist if a I=1 pn state existed. (b) The total wave function is the product of the space, spin, and isospin parts:

The deuteron has spin 1, internal angular momentum mostly L=0, and parity -1 (see F&H Section 9.2). Since L = 0 y_space must be symmetric under interchange of the neutron and proton. Since spin and isospin are mathematically the same, if S=1 and I=1, then f_spin and j_isospin must have the same symmtry, so we would have Y_total symmetric which would violate the generalized Pauli principle. Since I=0 and I=1 are the two possible states we can form from a neutron and a proton, if I=1 is not allowed, the deuteron isospin must be I=0.

(b) This should be a strong interaction since all the particles are hadrons, so isospin should be conserved. All the obvious quantum numbers (energy, electric charge, baryon number, …) are conserved in this interaction. All the particles have zero isospin, so since isospin is conserved by the strong interaction, one can explain the absence of the reaction if the neutral pion does not have isospin 0.

Answer 3

(a) As discussed at the top of F&H page 247, particles and antiparticles interact differentil in matter and hence have different cross sections(e.g. F&H Figures 5.29, 6.32 & 14.20). This means that their absorption cross sections, s_abs, and absorption lengths (l_abs=1/ns_abs, n is the target density) will be different. So neutral kaons and antikaon beams will be attenuated differently.

(b) A K₂⁰ is the symmetric equal mixture of kaon and antikaon given by F&H Eqn. 9.81. Since kaons and antikaons are attentuated differently, there will no longer be equal amounts of kaon and antikaon after the K⁰₂ beam passes through a slab of matter. If there are not equal amounts of kaon and antikaon, the the beam cannot be a pure K₂ (or pure K₁) beam, but must be a mixture of K₂ and K₁.

(c) The purity of the K₂⁰ beam can be measured by measuring the rate of two-pion decays is the same before and after passing through the slab. (Note: I can't just say that the purity can be measured by showing that there are no two-pion decays. Since CP is violated, even a pure K⁰₂ beam may have two-pion decays, althought the fraction will be very small since CP violation is very small.)

Answer 4

(a) It does not occur through an hadronic decay because the mass difference (m(S⁰)-m(L⁰) = 1192.55-1115.63 = 76.92MeV) between the S⁰ and the L⁰ is less than a pion mass, and the pion is the lightest hadron.

(b) This is a ¹/₂⁺->¹/₂⁺ decay, so the photon angular momentum (see F&H Eqn. 10.79) is either 0 or 1, and the photon is carrying away even parity. j=0 is not possible, since the j=1 electric dipole term is the first term in the expansion (F&H Eqn. 10.59), and all other terms carry at least this much momentum. j=1 and p=+1 is an M1 transition (see F&H Eqn. 10.79) where the spin of the final state L⁰ is flipped compared to the initial S⁰.

The energy of the photon is 76.92 MeV. Extrapolating the M1 transition rate from F&H Fig. 10.10 into the crease of the book, we get a transition rate of about (1⁺¹_-0.5)10¹⁹s^-1, corresponding to a lifetime of about (10⁺¹⁰_-5)10^-20s. The observed lifetime of the S⁰ is (7±0.7)10^-20s, in good agreement with our estimate.