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Forbidden transitions from triplet to singlet states in the deuteron?

Physics Asked on August 10, 2021

The dipolar moment for the deuteron is
$$vec{mu}_{d}=frac{mu_{N}}{2 h}left{left(g_{s}^{(p)}+g_{s}^{(n)}right) vec{S}+left(g_{s}^{(p)}-g_{s}^{(n)}right)left(vec{s}_{p}-vec{s}_{n}right)+vec{L}right}$$
we know could understad that
$$left(vec{s}_{p}-vec{s}_{n}right)=0$$
in the deuteron if we suppose that proton and neutron are too similar. But I find that people use to interpret that as forbbiden transition from triplet to singlet. Why transitions from triplet to singlet would be allowed if $left(vec{s}_{p}-vec{s}_{n}right)neq 0$?

One Answer

The tensor part of the nuclear force contains a term $-(vec{s_1}cdotvec{s_2})$. This means that the spin triplet state is at a lower energy than the spin singlet state; in fact, due to the deuteron's miniscule binding energy, the spin singlet state is unbound, making the spin triplet the only bound state.

Correct answer by probably_someone on August 10, 2021

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