Klein transformation

Suppose φ and χ are fields such that, if x and y are spacelike-separated points and i and j represent the spinor/tensor indices,

${\displaystyle [\varphi _{i}(x),\varphi _{j}(y)]=[\chi _{i}(x),\chi _{j}(y)]=\{\varphi _{i}(x),\chi _{j}(y)\}=0.}$

Also suppose χ is invariant under the Z₂ parity (nothing to do with spatial reflections!) mapping χ to −χ but leaving φ invariant. Obviously, free field theories always satisfy this property. Then, the Z₂ parity of the number of χ particles is well defined and is conserved in time (even though the number of χ particles itself depends on the choice of which splitting into a free Hamiltonian and an interacting Hamiltonian we make in the interaction picture, which doesn't even exist for interacting theories (the number is typically infinite)). Let's denote this parity by the operator K_χ which maps χ-even states to itself and χ-odd states into their negative. Then, K_χ is involutive, Hermitian and unitary.

Needless to say, the fields φ and χ above don't have the proper statistics relations for either a boson or a fermion. i.e. they are bosonic with respect to themselves but fermionic with respect to each other. But if you look at the statistical properties alone, we find it has exactly the same statistics as the Bose–Einstein statistics. Here's why:

Define two new fields φ' and χ' as follows:

${\displaystyle \varphi '=iK_{\chi }\varphi \,}$

and

${\displaystyle \chi '=K_{\chi }\chi .\,}$

This redefinition is invertible (because K_χ is). Now, the spacelike commutation relations become

${\displaystyle [\varphi '_{i}(x),\varphi '_{j}(y)]=[\chi '_{i}(x),\chi '_{j}(y)]=[\varphi '_{i}(x),\chi '_{j}(y)]=0.\,}$

Now, let's work with the example where

${\displaystyle \{\phi ^{i}(x),\phi ^{j}(y)\}=\{\chi ^{i}(x),\chi ^{j}(y)\}=[\phi ^{i}(x),\chi ^{j}(y)]=0}$

(spacelike-separated as usual).

Assume once again we have a Z₂ conserved parity operator K_χ acting upon χ alone.

Let

${\displaystyle \phi '=iK_{\chi }\phi \,}$

and

${\displaystyle \chi '=K_{\chi }\chi .\,}$

Then

${\displaystyle \{\phi '^{i}(x),\phi '^{j}(y)\}=\{\chi '^{i}(x),\chi '^{j}(y)\}=\{\phi '^{i}(x),\chi '^{j}(y)\}=0.}$

But what if we have more than two fields? In that case, we can keep on applying the Klein transformation to each pair of fields with the "wrong" commutation/anticommutation relations until we're done.

This explains the equivalence between parastatistics and the more familiar Bose–Einstein/Fermi–Dirac statistics.