I'm having trouble understanding a specific part of the classic deuteron example when introducing Isospin. I have seen this exact example in multiple lectures and textbooks (recently "Nuclear Structure From A Simple Perspective" by R. Casten). The first part of building an antisymmetric wavefunction is pretty clear, but when later introducing the nuclear force to argue about bound and unbound states (or the other way around: unbound states $\Rightarrow$nuclear force) there is, to my understanding, a contradiction in a specific way to illustrate this example.

To recapitulate the argument: For the two nucleon system we have two aspects of interest for building a nuclear wavefunction: Spin and Isospin. Since the total wavefunction must be antisymmetric for fermions, we are restricted to the two combinations "antisymmetry in Isospin and symmetry in Spin" or "symmetry in Isospin and antisymmetry in Spin":

$$ \begin{matrix} |nn\rangle&\otimes&\frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\right) &\mathrm{dineutron}\\ |pp\rangle&\otimes&\frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\right) &\mathrm{diproton}\\ \frac{1}{\sqrt{2}}\left(|np\rangle + |pn\rangle\right) &\otimes&\frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle-|\downarrow\uparrow\rangle\right) &\mathrm{deuteron \:(unbound)}\\ \frac{1}{\sqrt{2}}\left(|np\rangle - |pn\rangle\right) &\otimes& \begin{cases} |\uparrow\uparrow\rangle\\ \frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\right)\\ |\downarrow\downarrow\rangle\\ \end{cases} & \mathrm{deuteron} \end{matrix} $$

First, all of these look viable, but when introducing the nuclear force, we see that half of these states are unbound. The illustration I am having trouble with looks like this (here "Kernphysik" by T. Mayer-Kuckuk, as more info is included): From "Kernphysik" by T. Mayer-Kuckuk

caption reads: "States of the two-nucleon-system. [...] Arrows denote spin direction."

The Isospin triplet has the same, raised energy (apart from the coulomb-part, which also makes the states distinguishable), while the isospin singlet has a lower energy. The reason for this is said to be the nuclear force, which favors parallel spins between nucleons.

The problem now is, that to me the $\frac{1}{\sqrt{2}}\left(|np\rangle - |pn\rangle\right) \otimes \frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\right)$ deuteron state also looks like it has antiparallel spins, and thus should also be raised if the explanation is right. But the picture and accompanying text (in Mayer-Kuckuk) clearly states that all $T=0, S=1$ states are lowered. In the text by R. Casten it says that the two up-arrows in the lowered state are symbolic for the both up and the both down case, completely leaving out the symmetric antiparallel case but explicitly excluding it from the lowered state, and still labeling the lower level with $T=0, S=1$ and the raised three with $T=1, S=0$.

I have talked to some people about this problem, and they were as confused as me, but there are three explanations on how to resolve this which come to mind:

  1. I misunderstood the specifics of the nuclear force: Attraction is stronger in the symmetric case rather than the parallel case. So the picture is correct.
  2. I misunderstood how parallel spins work and the $\frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\right)$ state somehow has parallel spins. So the picture would also be correct.
  3. The picture is incorrect. The $\frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\right)$ state is raised to either the same energy as the rest or somehow has its own energy.

I also think that this "problem" would be easily solvable by looking at the production rate of deuterium in proton-neutron scattering, but didn't find any useful data.

So, where did I go wrong in the explanation or what do I misunderstand? Where is the $\frac{1}{\sqrt{2}}\left(|\uparrow\downarrow\rangle+|\downarrow\uparrow\rangle\right)$ state?


1 Answer 1


Remember that spin-half is a two-state system, and that the spins always have some orientation in every coordinate system. You can get from $\newcommand\ket[1]{\left|{#1}\right>} \ket{\uparrow\uparrow}$ to $\ket{\downarrow\downarrow}$ without any physical transformation at all: just stand on your head and insist that everyone use your new coordinates.

Well, the symmetric combination $\left( \ket{\uparrow\downarrow}+\ket{\downarrow\uparrow} \right)/\sqrt2$ is what happens if you lie on your side instead of standing on your head. It is physically the same state as $\ket{\downarrow\downarrow}$, connected by a passive coordinate transformation. Meanwhile the antisymmetric spin singlet $\left( \ket{\uparrow\downarrow}-\ket{\downarrow\uparrow} \right)/\sqrt2$ retains its antisymmetry and its antiparallel pairing under all rotations.


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