# Why doesn't the deuterium nucleus have spin 0?

A deuterium nucleus is composed of a proton and a neutron. Both have spin 1/2 so I would expect the deuterium to have two possible spins: 1 for the triplet and 0 for the singlet. But apparently deuterium always has spin 1 and the spin 0 state doesn't exist. Why?

Deuterium is light enough that isospin is a good symmetry: it's fair to neglect the electric charge and consider only the strong interaction between the proton and neutron. In that case we should expect essentially the same excitation structure in the diproton, the dineutron, and the neutron-proton system.

The proton and neutron are both fermions, and a state containing two of them must be antisymmetric under exchange. If they are bound without any orbital angular momentum, the only way to make the state antisymmetric is for the two particles to occupy a spin singlet. So the ground state of the diproton or dineutron must have spin zero. Since the diproton and dineutron are both unstable, isospin symmetry tells us that spin-zero deuteron should also be unstable.

In the stable deuteron the state is made antisymmetric by the isospin part of the wavefunction: the deuteron is a (symmetric) spin triplet but an (antisymmetric) isospin singlet. (It's true but irrelevant that the deuteron wavefunction is only mostly $s$-wave; there's a small contribution of $d$-wave with two units of orbital angular momentum, but that doesn't change the symmetry arguments or the total nucleon spin.)

If you prefer, you can turn this argument around. If it's the case that the strong interaction is more important than electrical repulsion in light nuclei, and if you found a stable two-nucleon state with zero angular momentum, you would expect to find that two-nucleon state for all allowed values of the charge: the dineutron, the deuteron, and the diproton. We don't find any evidence for stable dineutrons or diprotons, and we also don't find any spinless bound deuterons.

• Related homework problem: physics.stackexchange.com/q/108830/44126 – rob Dec 30 '14 at 21:17
• 'Since the diproton and dineutron are both unstable, isospin symmetry tells us that spin-zero deuteron should also be unstable.' By this, do you mean that Isospin symmetry leads us to believe that since the I = 1 : $|pp$ and $|nn$ states are unstable, the $1/\sqrt(2) |pn + |np$ should also be unstable? Why is that so? – inya Apr 16 '18 at 23:42
• @inya There are other examples of mirror nuclei and mirror states in excited nuclei where the spin, parity, and isospin together are a better predictor of the total energy that the electric charge of the nucleus. That's the point of the end of this answer. – rob Apr 16 '18 at 23:53
• This is nice, but seems to kind of punt on the actual question of why the strong force has this characteristic of being more attractive in a S=1, T=0 state than in a S=0, T=1 state. For that you appeal to observation. Deuterium is a loosely bound system, and therefore dominated (Friar, PRC 30 (1984) 1084) by the long-range tail of the nuclear interaction, which is well described by the one-pion exchange potential (OPEP). OPEP has this property that the S=1, T=0 channel is attractive. See Naghdi, nucl-th/0702078, eq. 2.1, and for details Savage, tinyurl.com/y3rpmxws . – Ben Crowell May 26 at 19:21

I am bit confused by Rob's reasoning above, he does not talk about the parity of the strong wave function and the color degree of freedom and confinement which seems to me the key of the argument. Please let me know if there is any mistake about my reasoning below.

This is how I would argue. The wavefunction of the deuteron is the tensor product of four wavefunctions, each representing a particular degree of freedom:

$$\psi=\psi_{spatial}\psi_{spin}\psi_{flavour}\psi_{color}$$

We can assume that the strong interaction (QCD) is the only one relevant here and neglect all others, from the point of view of QCD a proton and a neutron are basically identical (and having have-integer spins are of course fermions) therefore we must require the total wave function $\psi$ to be antisymmetric under the exchange. Now:

Assume $l=0 \Rightarrow \psi_{spatial}$ has parity $+1$.

Since QCD does not distuinguish the flavour degree of freedom of the particles the flavour wave function is symmetric $\Rightarrow \psi_{flavour}$ has parity $+1$.

This is the key point, it looks like QCD only allows bound states that are singlets of color (confinement) singlets of color have parity $-1$, color is the name that was given to the strong charge and comes in three varieties: red, green, blue. So $\psi_{color}$ has parity $-1$.

As explained above we need $\psi$ to be antisymmetric, so we must require $\psi_{spin}$ to be symmetric. We know that two particles of spin $\frac{1}{2}$ can only couple to states of total spin $0$ or $1$ but of those two only the latter is symmetric. That is why the deuteron has spin 1.

• Nucleons, which are color singlets, don't have a color degree of freedom. That makes me doubt that the $-1$ needed for exchange antisymmetry could be hiding in the color part of the wavefunction. Where you write "flavor," I wrote "isospin"; the heavy flavors don't contribute. If you'd like me to elaborate further, please convert this to a new question and I'll answer it. – rob Jul 13 '17 at 22:30
• If I'm understanding correctly, then this seems to be making an argument about existence of spin singlet and triplet states. Both exist, it's just that the singlet state is lower in energy. Now it may happen, as it does in deuterium, that one of these states is not bound, but that's a different issue. – Ben Crowell May 26 at 19:10

Proton has a positive magnetic moment, whereas neutron has a negative one. As a matter of fact, there will be a small attractive contribution if their spins are parallel (and their magnetic moments antiparallel). Consequently, the lowest energy state of a proton-neutron system is the parallel spin state. The parity of deuteron is positive, thus orbital momentum should be either 0 or 2 and total spin is either $$l+s_p+s_n=0+1/2+1/2=1,$$ or $$l-s_p-s_n=2-1/2-1/2=1$$

Actually it is a mixture of both states, 96% with $l=0$ and 4% with $l=2$, as it comes out from the comparison of total magnetic moment of deuteron with the sum of proton and neutron magnetic moments. They differ considerably, that being the consequence of the contribution of the $l=2$ state.

• The magnetic interaction is negligible, as explained in rob's answer. – Ben Crowell Jul 11 '17 at 20:58

The reason is related to the strong interaction. The important parameters to describe the strong interaction between two particles are the scattering length (f0) and the effective range of the interaction (d0). The parameter f0 is related to the scattering cross section. The parameter d0 is related to the range of the potential. If a bound state can be formed, one needs an attractive potential and the maximum of the wave function inside the potential range. (check this paper "Measurements of momentum correlation and interaction parameters between antiprotons"). Put it in short, for a bound state to be formed, one needs an attractive potential and f0 ﹤ 0. For proton-proton, f0 = 7.78 fm, d0 = 2.78 fm; for n-n, f0 = 16.9 fm, d0 = 2.8 fm; and for pn singlet state, f0 = 23.75 fm, d0 = 2.75 fm. So you can't find the stable bound state for pp, nn, pn singlet in nature (All f0 > 0). And for pn triplet state, f0 = -5.4fm, d0 = 1.76fm (check this paper "Measurement of interaction between antiprotons"). So attractive potential and f0﹤0, you would have the pn triplet state bound state which is the deuteron.

• Please use MathJax for the equations and numbers, it makes things much more legible. Also, it'd be helpful if you could include links to the papers which you're citing or directing a reader to. Thanks! – user191954 May 31 '18 at 15:23
• Note also that while it's okay for you to cite your own work, it must be clear from the text of your answer that the work is yours, without the reader having to click through or do any sleuthing. – rob Jun 1 '18 at 17:03