# What is my misunderstanding in Wick's theorem?

Trying to understand Wick's theorem, I took most of my knowledge from the corresponding Wikipedia article. The statement is that given the definition of normal ordering of operators $$A,B,C,\ldots$$ any product of those operators can be expressed as \begin{align} ABCDEF\ldots=\;&:ABCDEF\ldots: \\\\ &+\sum_{\mathrm{singles}} :A^\bullet B^\bullet CDEF\ldots: \\\\ &+\sum_{\mathrm{doubles}} :A^\bullet B^{\bullet\bullet} C^{\bullet\bullet}D^\bullet EF\ldots: \\\\ &+\ldots \end{align}

where $$:ABCDEF\ldots:$$ is the normal ordering of the operators and $$A^\bullet B^\bullet\equiv AB-:AB:$$ is their contraction.

Problem: Consider the quantum harmonic oscillator with ladder operators $$a,a^\dagger$$ and $$N=a^\dagger a$$ with commutation relations $$[a,a^\dagger]=1$$ and $$[N,a]=-a$$ and $$[N,a^\dagger]=a^\dagger$$. I am interested in changing the order of terms like $$N^na$$. Therefore I define the normal ordering as $$:Na:=aN$$ and hence $$N^\bullet a^\bullet=[N,a]=-a$$.

Now apply Wick's theorem to the term $$N^na$$ and we find \begin{align} N^na =& :N^na:+:N^{n-1}N^\bullet a^\bullet:+:N^{n-2}N^\bullet Na^\bullet:+\ldots \\\\ =&:N^na:+n:N^{n-1}N^\bullet a^\bullet: \\\\ =&aN^n-naN^{n-1} \end{align} where I made use of the fact that only contractions involving $$a$$ survive so that we only get terms with single contractions.

Apparently this is not the correct result, as a simple example with $$n=2$$ shows \begin{align} N^2a=&aN^2+[N^2,a] \\\\ &=aN^2+N[N,a]+[N,a]N\\\\ &=aN^2-Na-aN \\\\ &=aN^2-2aN-[N,a] \\\\ &=aN^2-2aN+a \end{align} where I used the fact that $$[A^2,B]=A[A,B]+[A,B]A$$. This result differs from the one obtained using Wick's theorem: $$N^2a=aN^2-2aN$$.

I am sure that the mistake lies in my understanding how to apply the theorem. Can someone point out what I did wrong?

The problem is that Wick's theorem assumes that the contraction $$N^{\bullet}a^{\bullet}$$ belongs to the center of the pertinent operator algebra, which is not true in OP's case.