2

If one interchanges two identical fermions in the wavefunction of a $N$-particle system, the total wavefunction changes by a sign i.e., the total wavefunction should be antisymmetric under the interchange of any two identical fermions. I have two very related doubts regarding the application of this principle.

DEUTERON

Deuteron is a bound state of a neutron and a proton. It is made of two non-identical fermions - the proton which is electrically charged and the electrically neutral neutron. Why should we insist that the total wavefunction (space $\times$ spin $\times$ isospin) be antisymmetric under the interchange of $n$ and $p$?

Note added in response to @naturallyinconsistent Well, one may argue that n and p are two different isospin states of the same particle "the nucleon". So with isospin symmetry, $n$ and $p$ may be considered identical (See the answer here by Lubos). But isospin symmetry is a good symmetry only if we can neglect the electric charge of the proton i.e. in a Universe governed only by strong interactions but not electromagnetism. In the real world, n and p are distinguished by their electromagnetic interactions.

BARYON

Similarly, consider a light baryon that is made of three different flavors of quarks $u,d,s$ which, as I understand, are non-identical fermions. Because the charge of $u$ is not equal to the charge of $d,s$. The mass of $u,d$ is approximately equal but not equal to that of $s$. Therefore, how can we claim that the total wavefunction (space $\times$ spin $\times$ isospin $\times$ color) will be antisymmetric under the interchange of any two quarks? It seems to me that the antisymmetry of the total wavefunction is at best an approximate requirement in the limit when mass differences can be neglected and differences in the electric charges can be neglected.

AccidentalFourierTransform
  • 56,647
Solidification
  • 12,736

1 Answers1

2

You are correct that there is no symmetry requirement for the exchange of non-identical fermions, but isospin is a useful approximate symmetry because it is essentially a short-cut for describing the spin-dependent effects of the strong interaction for protons and neutrons or up and down quarks. It turns out that when differences are small (e.g. the proton-neutron or $u-d$ mass difference), or irrelevant (e.g. the electric charge for strong interactions), then the wavefunctions that best describe strongly bound systems are those that are anti-symmetric in $p-n$ or $u-d$ exchange.

The important point is that the strong interaction is spin dependent, and it turns out that nucleons are more strongly bound when their spins are parallel than anti-parallel. The effects of the proton-neutron mass and charge differences are less than these spin-dependent effects. Since having the spins parallel is what we'd expect if two identical particles were antisymmetric in another way, inventing isospin is a useful way of explaining the deuteron without getting into the messy details of nuclear forces. (See "Why is nuclear force spin dependent?" and "Why doesn't the deuterium nucleus have spin 0?".)

The situation for baryons is similar. Assuming isospin symmetry helps explain the observed bound states, but ultimately this is a consequence of the small $u-d$ mass difference and the charge-independence and spin-dependence of QCD interactions (which are very hard to calculate).

David Bailey
  • 13,658