# Why must the gauge potential create negative norm states?

This question concerns the quantisation of the EM gauge potential $$A_\mu$$. When the Gupta-Bleuler formalism is introduced, it is usually stated that the creation/annihilation operators satisfy $$\langle 0|a_\mu (p)a_\nu^\dagger(p') |0\rangle \propto \delta(p-p') \eta_{\mu\nu}$$ due to Lorentz covariance. Since $$\eta_{\mu\nu}$$ is indefinite, this implies the existence of negative norm states.

But why exactly must $$\langle 0|a_\mu (p)a_\nu^\dagger(p') |0\rangle \propto \delta(p-p') \eta_{\mu\nu}$$ hold? Specifically, why the factor of $$\eta_{\mu\nu}$$? (I'm not worried about the $$\delta(p-p')$$ factor). Here's the way I've always thought about it (which I now believe is flawed):

Define the array $$T_{\mu\nu} = \langle 0|a_\mu (0)a_\nu^\dagger(0) |0\rangle$$. Then, since $$a_\mu$$ transforms as a 4-vector under a boost $$\Lambda$$ with unitary rep $$U(\Lambda)$$ -- that is, $$U(\Lambda)^\dagger a_\mu(p) U(\Lambda)=\Lambda_\mu^{\space\space\rho} A_\rho(\Lambda^{-1}p)$$ -- and the vacuum is invariant under $$U(\Lambda)$$, we find by inserting unitaries that:

\begin{align} T_{\mu\nu}&= \langle 0| U(\Lambda)^\dagger a_\mu(0) U(\Lambda) U(\Lambda)^\dagger a_\nu(0)^\dagger U(\Lambda) |0\rangle\\ &= \langle0| \Lambda_\mu^{\space\space\rho} a_\rho(0) \Lambda_\nu^{\space\space \sigma}a_\sigma^\dagger(0)|0\rangle\\ &= \Lambda_\mu^{\space\space\rho} \Lambda_\nu^{\space\space\sigma} T_{\rho\sigma} \end{align} In other words, $$T=\Lambda T \Lambda^T$$ for all boosts $$\Lambda$$, and I'm willing to believe that the only matrices satisfying this are scalar multiples of $$\eta_{\mu\nu}$$, concluding the proof.

The issue is that, as I've defined it, $$T_{\mu\nu}$$ is infinite: it contains a $$\delta(0)$$ factor. I'd always assumed that this infinity could somehow be dealt with rigorously, making the proof valid. But I recently posted this argument, which uses very similar reasoning to the above to reach the absurd conclusion that any spin-$$\frac{1}{2}$$ field must identically annihilate the vacuum. The flaw, I was told, is that my argument contains infinities. So now I'm worried about the above reasoning for $$A_\mu$$. Is there a way to make it rigorous? Why does the infinity cause a problem in the spin-$$\frac{1}{2}$$ case but not in the spin-1 case?

• exactly as in your previous question, the object $T$ does not exist. The fields $\phi,A,a,\psi,\dots$ are operator-valued distributions, states created by them do not have finite norm. The expression $a(0)a(0)$ is just meaningless: you cannot multiply distributions with coinciding singular support. The argument involving $T$ is just wrong, I'm not sure what else you want to know. – AccidentalFourierTransform Feb 11 at 12:15
• I think this is an XY-problem so it is hard to give useful feedback, but I imagine that what you want to do is to compute the norm of $|f\rangle:=\int f(\boldsymbol p)a^\dagger_\mu(\boldsymbol p)|0\rangle\ \mathrm d\boldsymbol p$, where $f$ is a smooth test function. Unlike $a^\dagger|0\rangle$, the ket $|f\rangle$ is a proper state, with well-defined norm. – AccidentalFourierTransform Feb 11 at 12:23
• I have never seen the negative norm states derived using the stress energy tensor (not to say that you can't do so). The derivation I am familiar with starts from the equal time commutation relation $[A_\mu(\vec{x}),\pi_\nu(\vec{y})]=i\eta_{\mu\nu}\delta^{(3)}(\vec{x}-\vec{y})$, where $A_\mu$ is the gauge field and $\pi_\nu$ is the conjugate momentum. The $\eta_{\mu\nu}$ appears here because you are quantizing the fields in a gauge where Lorentz invariance is manifest, so "what else" could you have there. This in turn implies commutators for the $a_\mu$ which leads to the negative norm states. – Andrew Feb 11 at 13:22
• David Tong's QFT lecture notes on QFT (chapter 6) are a good reference: damtp.cam.ac.uk/user/tong/qft.html – Andrew Feb 11 at 13:23
• "What else could you have there" is precisely the type of justification I'm trying to avoid. Also, the presence of the delta function in that equation is worrying, for the reason that @AccidentalFourierTransform gave – Jacob Drori Feb 11 at 13:26

## 1 Answer

It is kind of quantization requirement for "independent" $$a_{\mu}$$. Quantization by analogy with coordinate-momentum requirement. Find a review by Polubarinov on equations of QED.