I'm trying to prove by myself the canonical commutation relations (CAR): $$[a(\varphi),a(\psi)] = 0 \quad \mbox{and} \quad [a^{\dagger}(\varphi),a^{\dagger}(\psi)] = 0$$ on fermionic Fock spaces. Here $[A,B] := AB+BA$ and $\varphi, \psi$ are elements of a Hilbert space $\mathcal{H}$. The Fock space is given by $\mathcal{F}(\mathcal{H}) := \bigoplus_{n=0}^{\infty}A_{n}\mathcal{H}$ where $A_{n}$ is the antisymmetrization operator. I'm trying to address the first commutation relation.
Now, I know that: $$a(\varphi)A_{n}(\varphi_{1}\otimes \cdots \otimes \varphi_{n}) = \frac{1}{\sqrt{n}}\sum_{j=1}^{n}(-1)^{j+1}\langle \varphi, \varphi_{j}\rangle A_{n-1}(\varphi_{1}\otimes \cdots \otimes \hat{\varphi}_{j}\otimes \cdots \otimes \varphi_{n})$$ where $\hat{\varphi}_{j}$ means omit this entry.
Thus, my reasoning is the following: $$ a(\varphi)a(\psi)A_{n}(\varphi_{1}\otimes \cdots \otimes \varphi_{n}) = a(\varphi) \frac{1}{\sqrt{n}}\sum_{j=1}^{n}(-1)^{j+1}\langle \psi, \varphi_{j}\rangle A_{n-1}(\varphi_{1}\otimes \cdots \otimes \hat{\varphi}_{j}\otimes \cdots \otimes \varphi_{n})$$ Now, using the linearity of $a(\varphi)$ we get: \begin{align} a(\varphi)a(\psi)A_{n}(\varphi_{1}\otimes \cdots \otimes \varphi_{n}) &= a(\varphi) \frac{1}{\sqrt{n}}\sum_{j=1}^{n}(-1)^{j+1}\langle \psi, \varphi_{j}\rangle A_{n-1}(\varphi_{1}\otimes \cdots \otimes \hat{\varphi}_{j}\otimes \cdots \otimes \varphi_{n}) \\ &= \frac{1}{\sqrt{n}}\sum_{j=1}^{n}(-1)^{j+1}\langle \psi, \varphi_{j}\rangle a(\varphi) A_{n-1}(\varphi_{1}\otimes \cdots \otimes \hat{\varphi}_{j}\otimes \cdots \otimes \varphi_{n}) \\ &= \frac{1}{\sqrt{n}}\sum_{j=1}^{n}(-1)^{j+1}\langle \psi, \varphi_{j}\rangle \frac{1}{\sqrt{n-1}}\sum_{k=1}^{n-1}(-1)^{k+1}\langle \varphi, \varphi_{k}\rangle A_{n-1}(\varphi_{1}\otimes \cdots \otimes \hat{\varphi}_{j}\otimes \cdots \otimes \hat{\varphi}_{k}\otimes \cdots \otimes \varphi_{n}) \end{align} And we get the same expression but exchanging $\psi$ and $\varphi$ if we evaluate $a(\psi)a(\varphi)A_{n}(\varphi_{1}\otimes \cdots \otimes \varphi_{n})$. Now, the sum of these terms is supposed to be zero, which is not obvious to me but it probably is.
Is my reasoning correct? Is there an easier way to approach this?
There is a slight but significant error in your last term. The sum over $k$ should go from $1$ to $n$ excluding $j$, not from $1$ to $n-1$, but more importantly the sign should be $(-1)^{k+1}$ if $k<j$ and $(-1)^{k}$ if $k>j$. So:
$$a(\varphi)a(\psi) [A_n(\varphi_1\otimes .... \varphi_n)] = \frac1{\sqrt n}\frac1{\sqrt{n-1}}\left[ \begin{split} \sum_{j=2\\ k<j}^n (-1)^{j+k} \langle \psi,\varphi_j\rangle\langle\varphi,\varphi_k\rangle A_{n-2}(j,k)\\+ \sum_{j=1\\ k>j}^{n-1} (-1)^{j+k+1} \langle \psi,\varphi_j\rangle\langle\varphi,\varphi_k\rangle A_{n-2}(j,k) \end{split}\right]$$
If you look at $a(\psi)a(\varphi) A_n(\varphi_1\otimes...\otimes \varphi_n)$ the only thing that changes is that $\varphi$ is contracted with $\varphi_j$ and $\psi$ with $\varphi_k$ in the inner products. This can be corrected by substituting $j$ with $k$ and vise versa. If you do this the sum with $k<j$ swaps with the sum with $k>j$. But now the signs don't match, there is an extra factor $-1$. So $a(\varphi)a(\psi)=-a(\psi)a(\varphi)$.