Does this argument prove that all fermionic states have zero norm?

The following argument seems to show that all states created by a fermionic field have zero norm. This would surely cause problems in QFT, so I believe there must be an error somewhere, but I can’t find it.

Let $psi^alpha$ be a quantum field transforming in the $(frac{1}{2},0)$ irrep of $SL(2,mathbb{C})$. That is, under a boost $Lambda$, with unitary representative $U(Lambda)$, the field transforms as:

begin{equation} U(Lambda)^dagger psi^alpha(x) U(Lambda) = S(Lambda)^alpha_{spacespacebeta} psi^beta(Lambda^{-1}x) tag{1}end{equation}

with $S(Lambda)in SL(2,mathbb{C})$. The Hermitian conjugate field $psi^alpha(x)^dagger$ transforms via the complex conjugate of $S$:

begin{equation} U(Lambda)^dagger psi^alpha(x)^dagger U(Lambda) = (S(Lambda)^alpha_{spacespacebeta})^ast psi^beta(Lambda^{-1}x)^dagger tag{2}end{equation}

Define the array $T^{alphabeta} := langle Omega | psi^alpha(0) psi^beta(0)^dagger | Omega rangle $, where $|Omega rangle $ is the vacuum, which satisfies $U(Lambda)|Omega rangle = |Omega rangle$ for all $Lambda$. Inserting unitaries $U(Lambda)$ we find:

begin{align} T^{alphabeta} &=
langle Omega | underbrace{U(Lambda)^dagger psi^alpha(0) U(Lambda) }_{S(Lambda)^alpha_{spacespace rho} psi^rho(0)} space
underbrace{U(Lambda)^dagger psi^beta(0)^dagger U(Lambda)}_{(S(Lambda)^beta_{spacespace sigma})^ast psi^sigma(0)^dagger} |Omega rangle tag{3}
&= S(Lambda)^alpha_{spacespace rho} (S(Lambda)^beta_{spacespace sigma})^ast T^{rhosigma} tag{4}
&= left( S(Lambda) T S(Lambda)^dagger right)^{alpha beta}. tag{5}
end{align}
So for all $Sin SL(2,mathbb{C})$, the following matrix equation holds

$$ T = S T S^dagger. tag{6}$$

But this implies that $T=0$.$^ddagger$ Looking just at the diagonal entries of $T$, we see that $langle Omega | psi^alpha(0) psi^alpha(0)^dagger | Omega rangle = 0$ for each $alpha=1,2$ (no summation) and we conclude that $psi^alpha(0)|Omegarangle = 0$. Of course, this also implies that $psi^alpha(x)|Omegarangle = 0$ for all $x$.

$textbf{What has gone wrong?}$

^{$ddagger $ To see this, note that $SL(2,mathbb{C})$ contains $SU(2)$, so $UTU^dagger=T$ for all $Uin SU(2)$. So $T$ commutes with all of $SU(2)$, and therefore must commute with everything in the complex linear span of matrices in $SU(2)$. But the complex linear span of $SU(2)$ is simply all $2times 2$ matrices, so $T$ commutes with all $2times 2$ matrices, and so $Tpropto I_2$. Subbing back into (6) we find we must have $T=0$, since $S^dagger S neq I_2$ for some values of $Sin SL(2,mathbb{C})$.}