Second Quantization: The Identity Operator does not Commute?

Let me take the simplest possible example. Consider the fermonic Fock-space $\Lambda^*(\mathbb{C}^n)$ built out of a finite-dimensional, oriented single-particle Hilbert space $\mathbb{C}^n$ with orientation $$[\psi_1,\cdots, \psi_n].$$ The identity operator on this Fock-space may be written as $$1= \sum_{j=1}^n c_j^\dagger c_j,$$ where $c_j,c_j^\dagger$ create and annihilate $\psi_j$ respectively. Therefore, $1$ should commute with every other operator. However, it doesn't. Take $c_i$, for example: $$[1,c_i]=\sum_{j=1}^n[c_i,c_j^\dagger c_j]=\sum_{j=1}^n(c_ic_j^\dagger c_j-c_j^\dagger c_jc_i)$$ $$~~~~~~~~~=\sum_{j=1}^n(\delta_{ij}c_j-c_j^\dagger c_ic_j-c_j^\dagger c_jc_i)=\sum_{j=1}^n\delta_{ij}c_j=c_i.$$ Does anyone else find this confusing?

The identity operator is not $\sum_j c^\dagger_j c_j = \sum_j |1_j\rangle\langle1_j|$ but rather $$\prod_j \big(|0_j\rangle\langle0_j| + |1_j\rangle\langle1_j| \big) = \prod_j\left(c^\dagger_j c_j + c_jc_j^\dagger\right) = \prod_j \left\{c_j^\dagger,c_j\right\}.$$Fortunately $\left\{c_j^\dagger,c_j\right\}=1$ for fermionic CAPs.
• It's worth mentioning that $N = \sum_j c_j^\dagger c_j$ is a special operator: the number operator, whose eigenvalues give the total number of fermions in the system. It therefore obviously does not commute with $c_j$, which changes the fermion number by one. – Mark Mitchison Dec 4 '15 at 22:17