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

Question

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_{\space\space\beta} \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_{\space\space\beta})^\ast \psi^\beta(\Lambda^{-1}x)^\dagger \tag{2}\end{equation}

Define the array $T^{\alpha\beta} := \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^{\alpha\beta} &= \langle \Omega | \underbrace{U(\Lambda)^\dagger \psi^\alpha(0) U(\Lambda) }_{S(\Lambda)^\alpha_{\space\space \rho} \psi^\rho(0)} \space \underbrace{U(\Lambda)^\dagger \psi^\beta(0)^\dagger U(\Lambda)}_{(S(\Lambda)^\beta_{\space\space \sigma})^\ast \psi^\sigma(0)^\dagger} |\Omega \rangle \tag{3} \\ &= S(\Lambda)^\alpha_{\space\space \rho} (S(\Lambda)^\beta_{\space\space \sigma})^\ast T^{\rho\sigma} \tag{4}\\ &= \left( S(\Lambda) T S(\Lambda)^\dagger \right)^{\alpha \beta}. \tag{5} \end{align} So for all $S\in SL(2,\mathbb{C})$, 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 $\psi^\alpha(0)|\Omega\rangle = 0$. Of course, this also implies that $\psi^\alpha(x)|\Omega\rangle = 0$ 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 $U\in SU(2)$. 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 $2\times 2$ matrices, so $T$ commutes with all $2\times 2$ matrices, and so $T\propto I_2$. Subbing back into (6) we find we must have $T=0$, since $S^\dagger S \neq I_2$ for some values of $S\in SL(2,\mathbb{C})$.}

Answer 1

You're assuming T is finite.

But $\langle \Omega | \psi(x) \psi(x) | \Omega \rangle$ is not finite, even in free field theory.

Likewise, you're assuming that T transforms as an irrep, which it need not.

Answer 2

The equation (6) $$ T = S T S^\dagger $$ should be $$ T \rightarrow S T S^\dagger $$

If one enforces equation (6), which means $T$ is boost invariant, $T$ indeed has to be zero if $\psi$ belongs to $(\frac{1}{2},0)$ only.

However, if you allow mixture of $(\frac{1}{2},0)$ and $(0, \frac{1}{2})$ for $\psi$, then $T$ is not necessarily zero to be boost invariant. $T$ could be $\gamma_0$ or $ \gamma_1\gamma_2\gamma_3$.