A deuterium nucleus is composed of a proton and a neutron. Both have spin $\tfrac12$ 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?
[Physics] Why doesn’t the deuterium nucleus have spin $0$
angular momentumisospin-symmetrynuclear-physicsquantum-spinstability
Related Solutions
Define $\mathbf{J}=\mathbf{S}_1 + \mathbf{S}_2$ where $\mathbf{J}$ by definition is the total angular momentum to two identical, say spin-half, angular momenta. Define $\sigma$ to be the operation that exchanges the two spins, i.e., it denotes the operation: $1\leftrightarrow 2$. Then, it is easy to see that $\sigma$ commutes with $\mathbf{J}$ as $\mathbf{S}_1 + \mathbf{S}_2=\mathbf{S}_2 + \mathbf{S}_1=\mathbf{J}$. (The argument can be extended to the case of the addition of $N$ identical spins where $\sigma$ gets replaced by any pairwise exchanges which generates the permutation group in $N$ elements.)
Coming back to the case of two spins, it follows that we can simultaneously diagonalise $J^2,J_z$ and $\sigma$. The action of $\sigma$ must necessarily take an eigenstate of $J^2,J_z$ to one with the same $J^2,J_z$ eigenvalues. But the addition of two angular momenta says, that the multiplicity of $J^2$ eigenvalues is one. This is not true when you add more than two spins. It follows that for each given value of $J^2$, the multiplet has a definite $\sigma$ eigenvalue. Since $\sigma^2=1$, its eigenvalues must be $\pm1$.
However, neither $S_{1z}$ or $S_{2z}$ commute with $\sigma$. Thus, we have $$ \sigma\ |\ell,m_1,\ell,m_2\rangle = |\ell,m_2,\ell,m_1\rangle\ , $$ in the notation where $\ell(\ell+1)$ is the $S_i^2$ eigenvalue and $(m_1,m_2)$ are the eigenvalues of $(S_{1z},S_{2z})$ respectively. So in general, the action of sigma is not diagonal, unless $m_1=m_2$. Coming to Prof. Balakrishnan's example. He considered two spin half-particles. The $(j=1,m=\pm1)$ states only arise when $m_1=m_2=\pm\tfrac12$ and from the above argument must be symmetric.However, the $(j=1,m=0)$ and $(j=0,m=0)$ states arises as linear combinations of $m_1=-m_2$ state which are not diagonal. However, since the $(j=1,m=0)$ state can be obtained by using the lowering operator $J_-$ (which commutes with $\sigma$), it will have the same $\sigma$ eigenvalue as the state it was constructed from, i.e., $(j=m=1)$, which is symmetric. The singlet state must be (i) orthogonal to the triplet $m=0$ state and must necessarily be an eigenstate of $\sigma$ (I can give the precise argument if you wish) (ii) antisymmetric.
For the sum of N identical spins, all one can say is that there exists a basis of multiplets organised/labelled by the representations of the permutation group (given by Young diagrams) in addition to the $J^2$ eigenvalue.
Since the post is asking about the shape of the deuteron, this answer is based around a picture, rather than a physical description.
Dominated by three components describing the interactions of the quark components of the neutron and proton, its shape is not spherical. Recent tests have shown no deviations in the predictions of standard nuclear physics.
The structure of the deuteron, the nucleus of the deuterium atom, is of prime importance to nuclear physicists. The deuteron is a bound state of one proton and one neutron, and it is the nucleus most often used in measurements of neutron structure. Studies of the deuteron have helped determine the role of non-nucleonic degrees of freedom in nuclei and the corrections from relativity. A recent series of Jefferson Lab measurements have focused on the role of quarks in the structure of the deuteron. At high-energy and high-momentum transfer, the deuteron is probed at a length scale smaller than the nucleon size and at an energy scale at which the physical picture simplifies — by considering quarks rather than numerous baryon resonances. Measurements of reaction cross sections confirm the approximate scaling behavior expected from the underlying quark structure, while polarization measurements show simple behavior that's in rough agreement with some quark-based calculations.
The deuteron has spin +1 ("triplet") and is thus a boson. The NMR frequency of deuterium is significantly different from common light hydrogen. Infrared spectroscopy also easily differentiates many deuterated compounds, due to the large difference in IR absorption frequency seen in the vibration of a chemical bond containing deuterium, versus light hydrogen. The two stable isotopes of hydrogen can also be distinguished by using mass spectrometry.
The triplet deuteron nucleon is barely bound at EB = 2.23 MeV, so all the higher energy states are not bound. The singlet deuteron is a virtual state, with a negative binding energy of ~60 keV. There is no such stable particle, but this virtual particle transiently exists during neutron-proton inelastic scattering, accounting for the unusually large neutron scattering cross-section of the proton.
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Best Answer
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.