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 $S\in SL(2,\mathbb{C})$.

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/174898/2451 , physics.stackexchange.com/q/174898/2451 and links therein. $\endgroup$
    – Qmechanic
    Feb 10 at 20:24
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    $\begingroup$ That is not at all what I was intending to claim. Apologies for my terrible phrasing. I'll edit the question later when I have time. $\endgroup$ Feb 10 at 20:52
  • $\begingroup$ Doesn't $S=$diag$(i,-i)$ break your example? $\endgroup$ Feb 10 at 21:52
  • $\begingroup$ @AccidentalFourierTransform because then $STS^\dagger=-T\neq T$. Here's the calculation in Wolfram Alpha: wolframalpha.com/input/?i=%7B%7Bi%2C0%7D%2C%7B0%2C-i%7D%7D*%7B%7B0%2C1%7D%2C%7B-1%2C0%7D%7D*%7B%7B-i%2C0%7D%2C%7B0%2Ci%7D%7D $\endgroup$ Feb 10 at 22:45
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    $\begingroup$ $T$ must satisfy $T=STS^\dagger$ for all $S\in SL(2,\mathbb{C})$. You proposed a $T$ which you claimed satisfied this. But I gave an example of an $S$ such that $T\neq STS^\dagger$ for your $T$. For the proof that, in fact, only $T=0$ can satisfy our requirement, see my new amended footnote in my question. $\endgroup$ Feb 10 at 23:00

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.

  • $\begingroup$ "You're assuming T is finite." Is this really all there is to it? In my mind, the proof that a spin-1 field has negative norm states ran similarly to what I wrote for spin-1/2. But perhaps some subtler argument is needed. $\endgroup$ Feb 10 at 19:08
  • $\begingroup$ "Likewise, you're assuming that T transforms as an irrep". Yes - my argument above applies only to the (1/2,0) irrep. Having said that, I believe the argument still runs through in the reducible (1/2,0)+(0,1/2) (Dirac spinor) case (I didn't post it since it would make the question too long). But if finiteness is really the issue (and can't be somehow regulated) then I agree that the proof fails. $\endgroup$ Feb 10 at 19:12
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    $\begingroup$ Your first two sentences are perfectly correct and they are the true answer to the question. (so +1). But the last sentence is false: nowhere in this post are reps assumed to be irreducible. For reducible reps, the matrix $S$ is block diagonal, but this plays no role in the (faulty) argument in the OP. (Well, irreducibility is used in the footnote, but this footnote is also wrong, but for different reasons). $\endgroup$ Feb 10 at 20:45
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    $\begingroup$ Given that my argument is doomed, how would one prove that spin-1 fields have negative norm states? I'd always thought the argument basically went the same as what I wrote. (If this question is too off-topic I can ask it separately, but any references/pointers would be appreciated) $\endgroup$ Feb 10 at 20:50
  • $\begingroup$ @JacobDrori It's better to ask such questions separately. You can link this question to connect them. But comments aren't guaranteed to last forever, so not good for archival. And comments don't bump questions on the front page, so the number of people who see the question is drastically reduced. $\endgroup$
    – user1504
    Feb 11 at 0:58

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$.

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    $\begingroup$ 6 is really is an equation. He's inserted $1=UU^\dagger$ in 3 places to generate it. $\endgroup$
    – user1504
    Feb 11 at 1:01
  • $\begingroup$ @user1504, $UU^\dagger \neq 1$ for boosts, albeit $UU^\dagger = 1$ for rotations. $\endgroup$
    – MadMax
    Feb 11 at 5:22
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    $\begingroup$ @MadMax That's wrong. While it is true that $\Lambda\Lambda^\dagger\neq1$ and $S(\Lambda)S(\Lambda)^\dagger\neq1$ for boots, we do have $U(\Lambda)U^\dagger(\Lambda)=1$ for all $\Lambda$. The finite-dimensional representation $S$ is non-unitary, but the infinite dimensional $U$ definitely is. $\endgroup$ Feb 11 at 12:13
  • $\begingroup$ @AccidentalFourierTransform, please give a precise definition of "infinite dimensional" $U(\Lambda)$ and prove $U(\Lambda)U^\dagger(\Lambda)=1$ for boosts. $\endgroup$
    – MadMax
    Feb 11 at 17:38
  • $\begingroup$ @MadMax If you don't know what "infinite-dimensional" means then you probably should not be trying to answer questions on QFT or representation theory. Anyway, I don't feel like wasting time today arguing with you, so I'll leave you with two wikipedia links, in case you honestly want to learn more: One basic idea of the Wightman axioms is that there is a Hilbert space upon which the Poincaré group acts unitarily, ... $\endgroup$ Feb 11 at 18:05

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