In my class the teacher tried to show that Gupta-Bleuler doesn't work in non abelian gauge theories. But I didn't really understood what he did.

Here is what we did:

We have:

$$ \mathcal{L}=\mathcal{L_0}+\mathcal{L_{fix}} $$


$$ \mathcal{L_0}=-\frac{1}{4 I(R)}Tr(F^{\mu \nu}F_{\mu \nu})$$ $$ \mathcal{L_{fix}}=-\frac{1}{2 \xi}Tr((\partial_\mu A^\mu)^2)$$

After few manipulations we can show that:

$$ \mathcal{L_{fix}}=-A_a^\mu \partial_\mu \partial_\nu A_a^{\mu}$$

(in fact we have also a surface term that I will not show here as it is not important for the Lagrangian).

Then, we end up with the equations of motion:

$$0=D_\mu F^{\mu \nu} + \partial^{\nu}(\partial_\mu A^\mu) $$ $$ \Box(\partial_\mu A^\mu)=-ig \partial_\nu[A_\mu, F^{\mu \nu}].$$

Indeed to end up with the last equation we just applied $\partial_\nu$. I remind that: $D_\mu F^{\mu \nu}=\partial_\mu F^{\mu \nu}+ig[A_\mu, F^{\mu \nu}]$

My question is the following:

It is written that we can't decompose in $>0$ and $<0$ frequencies to impose Gupta-Bleuler condition: $\partial_\mu A^{(+)}(x)|\psi \rangle$ at each time.

I'm not sure to get why (it is probably very simple but I probably lack of some basis).

Is it because if I do an average on $|\psi \rangle$ (a physical state) the lhs of the equation will be $0$ as the $\Box$ and $\partial_\mu$ "commutes" with the ket. Then I will end up with something like:

$$0= -ig \partial_\nu\langle \psi |[A_\mu, F^{\mu \nu}]|\psi\rangle.$$

And this last term has no reason to be $0$ : we have a contradiction.

So in short term: As I'm not very confident with my understanding of all this I would like to check with you if my explanation is the right one or if I am totally wrong.

Extra question: If I was right, why couldn't we have on all physical states : $\partial_\nu\langle \psi |[A_\mu, F^{\mu \nu}]|\psi\rangle=0$? What would "break" if we also had this?


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