# Question about the wavefunctions of a system of non-identical fermions

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.

• where did you get the impression that the non-identical fermions are also anti-symmetrised? Dec 15, 2023 at 3:37
• @naturallyInconsistent I gave you two examples. The deuteron and the baryons. If you cared to read it, you'll understand that the question is precisely that. Why should the deuteron wavefunction be completely antisymmetric (when n and p are not identical)? Dec 15, 2023 at 3:51
• I'm precisely asking you who gave you that idea that those non-identical fermions are supposed to be skew-symmetrised? Dec 15, 2023 at 3:57
• Why is deuteron wavefunction completely antisymmetric? physics.stackexchange.com/q/4778/164488 Why are baryon wavefunctions completely antisymmetric? Also see students.iiserkol.ac.in/~mms15ms051/courses/PH4204/… Page 215, Example 2 Dec 15, 2023 at 4:09
• You should add these things to the question, because otherwise it is not easy to understand where you are coming from. What part of the answer given by Lubos do you not understand? While they are definitely not identical fermions on the fundamental level, the trick is to consider them as nucleons under isospin, in which case they are supposed to be identical under this scheme. Dec 15, 2023 at 4:18

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 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).