Consider the fermionic operator $c_k, c^\dagger_k$, and where $k$ is discrete and unbounded. (Note: This situation frequently arises in bosonization.) Let the vacuum $|0\rangle$ be the state with all $k\leq 0$ filled and all $k>0$ empty.

There are two definitions of normal ordering:

  1. The normal ordering is obtained by substracting the vacuum expectation value.

  2. The normal ordering is obatained by putting $c_k$ right to $c_k^\dagger$ for $k > 0$, and putting $c_k^\dagger$ right to $c_k$ for $k \leq 0$.

My question is that for $c_{k+q}^\dagger c_k$ with $q>0$, the two definitions above seem to be different. According to the first definition, $:c_{k+q}^\dagger c_k:=c_{k+q}^\dagger c_k$ for all $k$ since $\langle0|c_{k+q}^\dagger c_k|0\rangle=0$. However, if we use the second definition, $:c_{k+q}^\dagger c_k:=c_k c_{k+q}^\dagger$ when $k\leq-q$. Since $c_k$ and $c_{k+q}$ anticommute, this expression is minus the first expression!

Then what is the correct definition of the normal ordering?

  • 1
    $\begingroup$ The normal ordering depends on what sort of vacuum you are concerned with. E.g. if there is BCS phase transition, $\langle0|c_{k+q}^\dagger c_k|0\rangle \neq 0$, and you have to resort to Bogoliubov-transformed operators to define vacuum and normal ordering. $\endgroup$
    – MadMax
    Commented Jul 3, 2019 at 15:09

1 Answer 1


Normal ordering for Fermionic operators is defined so that a minus sign is introduced when changing the order (otherwise the normal order of two creation operators that anticommute would be ill-defined — see, for instance, this wikipedia page), in contrast to the Bosonic case. Once this is taken into account, your discrepancy should disappear.


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