Which of the two possible orientations of parallel spins is favoured by nucleon-nucleon interaction?

Question

The tensor part of the two-body nucleon-nucleon potential has the form $$S_{12}=V_T(r)[3(\vec{s}_1\cdot\hat{r})(\vec{s}_2\cdot\hat{r})-\vec{s}_1\cdot\vec{s}_2]$$ where ${\vec r}=r\hat{r}$ represents the vector joining the nucleon spins $\vec{s}_1, \vec{s}_2$ and $V_T(r)$ is a function of $r$. Pretending $\vec{s}_1$ and $\vec{s}_2$ as classical vectors, the second term of $S_{12}$ inside the bracket is minimum when $\vec{s}_1 ~||~ \vec{s}_2$. Keeping $\vec{s}_1 ~||~ \vec{s}_2$, there are two possible orientations of $\vec{s}_1$ and $\vec{s}_2$ w.r.t $\vec{r}$.

CASE-I: With $\vec{s}_1 ~||~ \vec{s}_2$, both the spins can be parallel to the vector joining them i.e. $\vec{s}_1|| \hat{r}$ and $\vec{s}_2|| \hat{r}$. In this case, we get, $$S_{12}=2V_T(r)s_1s_2.$$

CASE-II: With $\vec{s}_1 ~||~ \vec{s}_2$, both the spins can be perpendicular to the vector joining them i.e. $\vec{s}_1\perp \hat{r}$ and $\vec{s}_2\perp \hat{r}$. In this case, we get, $$S_{12}=-V_T(r)s_1s_2.$$

Do we know which of the two possibilities mentioned above is true?

Answer 1

As a commenter points out, this is a quantum-mechanical system, so there is no possibility for “perpendicular” nucleon spins. The aligned state is the spin triplet; the anti-aligned state is the spin singlet.

For a crude answer to this question, you could look at the mass-two system. The diproton and dineutron (which must be spin singlets in the $L=\text{even}$ orbital states, for antisymmetry under exchange) are unbound; the only bound state of the deuteron is a spin triplet. This is generally interpreted as part of the evidence for “isospin” symmetry: the two-nucleon system prefers energetically to make an isospin-zero, spin-one deuteron rather than the theoretically possible spin-zero deuteron.

Your specific reference to the tensor part of the two-nucleon potential is related to the existence of the $d$-wave admixture of the deuteron wavefunction, in a way that I understood as a PhD student but probably couldn’t explain in an answer today.

As a rule, even-even nuclei tend to have ground states with spin-parity $J^P = 0^+$, while odd-odd nuclei tend to have $J\neq 0$ ground states. (For that matter, there are only four stable odd-odd nuclei: deuterium, lithium-6, boron-10, and nitrogen-14.) A person who was groping their way towards nuclear structure might use a “cluster model” where an N-14 is a carbon-12 “core” bound to an $s$-wave deuteron; this reproduces nitrogen’s $J^P=1^+$.

But you don’t want to invest too much energy into this kind of cargo-cult nuclear structure; you rapidly have to make enough other assumptions that it’s better just to do it right. In the semi-empirical mass formula there is a term related to the “pairing energy” which ignores spin. In a serious nuclear potential, like the Argonne V18, there are absolutely spin terms; you might be able to answer your own question if you squint at the Argonne coefficients hard enough.