Apparent conflict between path integral formulation and canonical quantization for fermions In the path integral formulation of QED, $\psi$ and $\bar\psi$ are Grassmann fields. The anti-commutativity of each of the 4 components of $\psi$ (or $\bar\psi$) will force an expression like $(\bar\psi\psi)^5$ to be zero. So such an interaction term will drop out in the path integral.
However, this seems to be different in the canonical quantization context, where if we switch the order of $\psi$ and $\bar\psi$ we get an extra delta function that gives us infinity because all the operators are evaluated at the same spacetime point. So a $(\bar\psi\psi)^5$ term doesn't seem to drop out in the canonical quantization. 
Where am I going wrong?
 A: This is the whole zero-point/vacuum energy issue about in the context of canonical quantization. If you follow the 'normal ordered' approach, the 'extra delta function that gives us infinity' will drop out.
On the other hand, path integral formulation automatically takes care of normal ordering. In other words,  path integral formulation doesn't suffer from the zero-point energy issue.
An added note: A $(\bar{\psi}\psi)^5$ term demands at least 10 degrees of Grassmannian freedom. It vanishes since 4-component Dirac spinor only have $4*2 = 8$ degrees of Grassmannian freedom at a given spacetime point.  Such terms can potentially be non-zero if you expand the definition of $\psi$ to cover multiple fermions (e.g. electron + neutrino). 
A: Suppose you normal order the operator. Then it contains a string of 5 fermionic annihilation operators, all at the single point. But this must yield zero on all states, because of the Pauli exclusion principle and the fact that there are only 4 fermionic degrees of freedom at each point. So the operator is exactly zero.
