# Why don't we have particles whose wavefunctions are symmetric wrt one exchange operator and anti-symmetric wrt other exchange operator?

Consider a system with $n$ identical particles. Let the wavefunction of the system be $\psi(r_1,\ldots, r_2)$. Let $P_{a,b}$ represent the exchange operator which exchanges particle $a$ with particle $b$. Similarly, $P_{c,d}$ represent the exchange operator which exchanges particle $c$ with particle $d$. Now, suppose $P_{a,b}(\psi(r_1,\ldots, r_2))= - \psi(r_1,\ldots, r_2)$ and $P_{c,d}(\psi(r_1,\ldots, r_2))= \psi(r_1,\ldots, r_2)$. Why don't we have particles with wavefunctions satisfying the above property?

• Um...take one state with the first property and one state with the second property. Write the combined state. You've got a state with the property you claim we never have. So I don't understand the question. Dec 27 '14 at 15:19
• @ACuriousMind The combined state will be still anti-symmetric wrt $P_{a,b}$ and symmetric wrt $P_{c,d}$. Dec 27 '14 at 15:23
• Isn't that what you want? Dec 27 '14 at 15:25
• You mean that particles $a,b,c,d$ are all of the same species? Then $a,b$ can't have exchange properties different from those of $c,d$ since this would make particles $c,d$ distinguishable from particles $a,b$. (It's also not consistent with the group structure on exchange operators, if we are also to be able to exchange $a,c$; $a,d$; $b,c$; and $b,d$.) Dec 27 '14 at 15:33
• Well, if the particles interact (e.g., they have electric charge), by treating the interaction as a perturbation to the free Hamiltonian, you find an exchange energy term. This term will be slightly different depending on if the exchange is symmetric or anti-symmetric. So if you move the pair $a,b$ far away from the pair $c,d$, so that the interaction between the pairs is negligible, you could tell which is which by measuring the energy of each pair. Dec 27 '14 at 15:44

It is not possible to have a state with four indistinguishable particles such that $P_{12} \psi = -\psi$ and $P_{34} \psi =\psi$, for an algebraic reason. Namely, the exchange operators have to form a representation of the permutation group $S_4$. It is rather well known that there are exactly two representations of $S_n$: the trivial representation where all exchange operators are $1$, and the parity representation where all single transpositions are represented by $-1$ and this is extended by the group law.
Thus either all $P_{nm} = 1$ or all $P_{nm} = -1$. Anything else is simply not consistent with the algebra of permutations.
A wavefunction is a normalised vector (or a ray) in the Hilbert space of vector states (I will assume finite degrees of freedom, so the C*-algebra of the system can be taken to be $B(H)$, with $H$ an $L^2(\mathbb R^n)$ space). Spin gives a superselection rule, and therefore there must be superselection sectors. It follows from the general theory that vector states from different sectors cannot be combined to yield another vector state, but just a statistical mixture. So a wavefunction is either bosonic or fermionic. To describe a state with mixed particles you need more general linear functionals which combine as convex combination of states from the superselection sectors.