For a $\pi^+$ $\pi^-$ state, why does the wave function have to be symmetric? For example, in the isospin state $|I=1,I_3=0\rangle = \frac{1}{\sqrt{2}}(\pi^+ \pi^- -\pi^- \pi^+)$, the angular momentum has to be odd ($L=1,3...$) in order to keep wave function symmetric (since the isospin part is antisymmetric).

However, $\pi^+$ and $\pi^-$ are not identical bosons, so why does the wave function have to be symmetric?


1 Answer 1


In the isospin formalism you treat them as identical particles. If you have two electrons, one is spin up, and one is spin down, the electrons are still identical they just have different z-projections of their spin. This same logic applies to isospin. Instead of a spin up component and spin down component like the electron, you now have $\pi^+,\pi^0,\pi^-$ components of the pion particle, and these are projections in "isospin space". As there are three projections, it falls into the vector representation of the "isospin symmetry group", and the electric charge is the "isospin projection onto the third axis".

The isospin symmetry becomes an approximate symmetry when you include electromagnetic interactions, and the small mass differences in the quarks. However as long as all of these effects are small the approximate symmetry holds very well. The isospin breaking effects can be applied as corrections to the calculation.

  • $\begingroup$ Thanks very much, I think your comment is right. $\endgroup$ Jan 5, 2021 at 15:30

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