If I have a system of say 2 identical bosons, and 2 identical fermions, what can I say about the combined wavefunction, in terms of exchange symmetries?

Intuitively, I would expect swapping any two of the identical particles to leave expectation values unchanged, so at least symmetric. But I can't justify to myself how to write the wavefunction. I've thought perhaps the product of a symmetric and an antisymmetric part?

How about if I was writing it in second quantisation?


1 Answer 1


Let the bosons have positions $x_1$ and $x_2$ and the fermions have positions $y_1$ and $y_2$. The wavefunction $\psi(x_1, x_2, y_1, y_2)$ satisfies $$\psi(x_1, x_2, y_1, y_2) = \psi(x_2, x_1, y_1, y_2) = - \psi(x_1, x_2, y_2, y_1).$$ However, there is no definite symmetry associated with swapping one of the $x$'s with one of the $y$'s.

This symmetry structure is analogous to a tensor with four indices, like the Riemann tensor $R_{\mu\nu\rho\sigma}$. Like our wavefunction, the Riemann tensor has definite symmetry associated with swapping the first two indices, and the last two indices, but there isn't any symmetry associated with swapping the first and third. We could decompose it into a symmetric and antisymmetric piece, but this isn't useful, because the $\mu\nu$ components play different physical roles from the $\rho \sigma$ components.

Similarly, for this four-argument wavefunction, you could write it as a sum of terms with definite symmetry everywhere. But this isn't useful because we can't mix the fermions and bosons up; they should be thought of as living in entirely different 'spaces'. It's like looking at, say, the Yang-Mills gauge field $F_{\mu\nu}^a(x)$ and trying to symmetrize $\mu$, $a$, and $x$.

These rules are easy to derive in second quantization. There, the wavefunction is defined as $$\psi(x_1, x_2, y_1, y_2) = \langle x_1, x_2, y_1, y_2 | \psi \rangle$$ where $a^\dagger_x$ creates a boson at $x$, $b^\dagger_x$ creates a fermion at $y$, and $$|x_1, x_2, y_1, y_2\rangle = a_{x_1}^\dagger a_{x_2}^\dagger b_{y_1}^\dagger b_{y_2}^\dagger |0 \rangle.$$ Using the fact that the $a^\dagger$'s commute and the $b^\dagger$'s anticommute gives the first equation.


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