# Relation between parity and symmetry for deuteron

I am currently reading the book Física Nuclear y de Partículas by Antonio Ferrer Soria. In this book the following claim is made (this is at translationn from the Spanish):

The wave function of deuteron is given by $$|d\rangle=|\phi(\mathbf{r})\rangle|\chi(s)\rangle|\psi(T)\rangle$$ where $$|\phi(\mathbf{r})\rangle$$ has information about angular momentum, $$\chi(s)\rangle$$ and $$|\psi(T)\rangle$$ are the spin and isospin corrections, respectively.

The wave function $$|d\rangle$$ is antisymmetric under exchange of nucleons since the nucleons are identical. The function $$|\phi(\mathbf{r})\rangle$$ has the symmetry of angular momentum $$\mathbf{L}$$ (as $$\mathbf{L}$$ is even we have $$|\phi(\mathbf{r})\rangle$$ is symmetric) and $$|\chi(s)\rangle$$ is symmetric because $$J_d=1$$. Hence $$|\psi(T)\rangle$$ must be antisymmetric.

The author seems to conclude symmtry properties from parity properties. For example, he says that $$|\phi(\mathbf{r})\rangle$$ is symmetric because $$\mathbf{L}$$ is even.

Question: What is the relation between symmetry and parity? Is it that even parity corresponds to symmetric functions?

If that's the case we know by experimentation that the parity of deuteron is even, but since $$|d\rangle$$ is antisymmetric, how is thas possible?

For an equal-mass two-body system, the symmetry under exchange is the same as the parity of the wavefunction. If you have structureless identical nucleons, you can always construct a coordinate system where their center of mass is at an origin, and then the transformation $$(x,y,z) \to (-x,-y,-z)$$ sends each nucleon to the other's location, no matter where they happen to be.

For a two-body system the angular momentum eigenstates are given by the spherical harmonics $$Y^{\ell m}$$, which have parity $$(-1)^\ell$$.

We know experimentally that the deuteron has one unit of angular momentum and even parity. Even parity tells us even $$L$$. It turns out to be mostly $$s$$-wave and about 4% $$d$$-wave, or mostly $$L=0$$ with a little $$L=2$$. We must have the symmetric spin triplet to get $$\vec J = \vec L + \vec S$$ to have unit magnitude; the symmetric spins are symmetric under exchange. The angular momentum part of the wavefunction has exchange symmetry $$\text{even}\times\text{even} = \text{even}$$.

The overall antisymmetry is required by the spin-statistics theorem. One way to interpret these results is that a $$J^P=1^+$$ deuteron, made of "approximately identical" nucleons, requires some other piece of its wavefunction to be antisymmetric under exchange. A creative theorist might call that extra piece "isospin," and examine its properties in other nuclei where the charge and mass differences between nucleons are negligible. I think that's historically what happened.

Contrast with molecular hydrogen, where the two nuclei are identical without the isospin degree of freedom. In that more-constrained system the states with symmetric spins can only have $$L=\text{odd}$$, while the states with antisymmetric spins can only have $$L=\text{even}$$. The ensembles with even and odd rotational states ("parahydrogen" and "orthohydrogen") act nearly like separate gas species, with slightly different densities and heat capacities. At the temperatures where hydrogen liquifies, the $$L=0$$ para ground state is strongly preferred over the $$L=1$$ ortho ground state, but the gas density is too low for much ortho-ortho to para-para scattering. When recently-warm hydrogen liquifies, the huge density increase dramatically increases the rate of ortho-to-para conversion, and so much heat is released that the most of the new liquid boils again. When pure parahydrogen liquifies, it stays liquid.

• Thank you so much for the answer, it clarifies some ideas I had. Can you please explain how is it posible that if the wave function for deuteron, being antisymmetric, has even partity? That's what troubles me the most. Commented May 25 at 23:52
• Parity is only the space-inversion transformation. Exchange symmetry applies to the entire wavefunction. For the deuteron, the spatial wavefunction is even, but the space, spin, and isospin parts have overall exchange symmetry $(+)(+)(-)=(-)$.
– rob
Commented May 26 at 0:27
• That's it! Thank you. Commented May 26 at 2:15

The answer could be found from the soln of Hydrogen atom problem, if you remember. Let's consider the state $$\psi_{100}\propto e^{-r/a}$$, which is represented by the $$10$$ color plot in the attached figure. This is clearly an even function, hence inversion of argument does not affect. But consider the wavefunction $$\psi_{210}\propto r e^{-r/a}$$, which is not symmetric under space inversion, see the graph $$21$$. So, the formula $$\mathcal{P}|\psi,l\rangle=(-1)^l|\psi,l\rangle$$ does make sense, or when angular momenta odd, parity is odd.

• The parity transformation in spherical coordinates is not $r\to -r$, but $(\phi,\cos\theta)\to(\phi+\frac{2\pi}2,1-\cos\theta)$.
– rob
Commented May 25 at 17:00