(a) The ground state deuteron 3S1 has positive parity. The electric dipole transition requires a change of parity between the initial and final states. Hence the parity of the free (n, p) system must have parity -1. Assume that the wave function of (n, p) can be written as \(\Psi\)(n, p) \(\sim\) \(\psi\)(rn, rp)x(n, p). For x = x13, after the nucleons are interchanged the wave function becomes \(\Psi\)(p, n) = (-1)l\(\Psi\)(n, p). For x = x01, after the nucleons are interchanged, the wave function becomes Q(p, n) = (-1)l+1\(\Psi\)(n, p).
A fermion system must be antisymmetric with respect to interchange of any two particles, which means that for the former case, l = 1, 3,. . . , and for the latter case, l = 0, 2, 4,. . . , and so the parities are -1(l = odd) and +1(l = even) respectively. Considering the requirement we see that only states with x = x13, i.e. spin triplet states, are possible. Further, S = 1, L = 1,3,. . . , and so J = 0, 1,2,. . . . As the deuterons are unpolarized, its spin wave function has the same probability of being x11, x10 or x1-1. Therefore, after the transition (n, p) can be represented by the product of a plane wave and the average spin wave function:

(b) The correlation of time and space is manifested in conservation of energy and conservation of momentum. In the center-of-mass coordinates, if the energy of the proton is measured to be Ep, the energy of the neutron is En = Ecm - Ep; if the momentum of the proton is p, the momentum of the neutron is -p. Let a be the spin function for "up" spin, and \(\beta\) be that for "down” spin. Then x11 = α(n)α(p), x1-1 = \(\beta\)(n)\(\beta\)(p), x10 = \(\frac1{\sqrt 2} [\alpha(n) \beta (p) + \alpha(p) + \beta(n)]\), and the spin wave function is

Thus, if the spin of p is detected to be up, we have

Hence the probability that the spin state of n is also up is
