Under the exchange of particle, the state of Fermions and Bosons are anti symmetric and symmetric respectively. For example, if $\psi_1(x)$ and $\psi_2(x)$ are two one particle wavefunctions, two particle Fermionic and Bosonic wavefunctions are [not normalized] $$\psi_1(x_1)\psi_2(x_2)-\psi_1(x_2)\psi_2(x_1) \hspace{1cm} Fermionic$$ $$\psi_1(x_1)\psi_2(x_2)+\psi_1(x_2)\psi_2(x_1) \hspace{1cm} Bosonic$$

What will be the wavefunction of a two particle system consists of one Fermion and one Boson? Will it retain any symmetry property?

  • $\begingroup$ Short answer, there would be no symmetry. Longer answer, a system of more than one identical bosons and more than one identical fermions coupled would be modeled by a wavefunction that is symmetric with respect to the exchange of two bosonic degrees of freedom, and antisymmetric w.r.t. the exchange of two fermionic degrees of freedom, and has no symmetry w.r.t. the exchange of a bosonic and fermionic degree of freedom. $\endgroup$ – yuggib Nov 23 '17 at 7:57

As fermions and bosons can't be particles of the same species, they are not indistinguishable. (Anti-)Symmetrization only makes sense and is necessary for the exchanges of identical, indistinguishable particles. Thus, there'll be no symmetry with respect to the exchange of your boson and fermion.


It won't retain any symmetry. The symmetrization/antisymmetrization condition applies only to identical bosons/fermions. If you have one boson described by $\psi_1(r_1)$ and a fermion described by $\psi_2(r_2)$. Then the full wave function (in case of low energy interaction) is:


And swapping $r_1$ by $r_2$ gives you a completely different wavefunction for the system, it is nor symmetric nor antisymmetric.

The same argument works if they are not identical particles (bosons/fermions).

Edit: correction swapping coordinates changes substantially the problem.

  • $\begingroup$ While I get what you mean, saying "swapping $r_1$ and $r_2$ doesn't change anything" is not very well phrased. The entire point is that swapping $r_1$ and $r_2$, does change something, namely it exchanges the positions of the fermion and the boson, which is a measurable difference. $\endgroup$ – By Symmetry Nov 23 '17 at 10:54
  • $\begingroup$ @BySymmetry you're completely right, I will rephrase it. $\endgroup$ – Mauricio Nov 23 '17 at 11:15
  • $\begingroup$ @Mauricio, there can be many states which are neither symmetric nor anti symmetric under particle exchange. How have you chosen one? $\endgroup$ – Abu Saleh Musa Nov 24 '17 at 4:03
  • $\begingroup$ @AbuSalehMusa My fault! I should have pointed out that my solution only works in the case that each particle has a definite wavefunction and there is no correlation between the two. If not, then is a sum of the kind of states I wrote. The same happens when you symmetrize, you have to write a sum of the kind of states you wrote down. $\endgroup$ – Mauricio Nov 24 '17 at 8:52

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