Relation between the trace anomaly and the energy-momentum tensor being off-shell

Let’s say we have a massless QED theory with a Lagrangian

begin{equation}
L=ibar{psi}not{D}psi-frac{1}{4}F_{munu}F^{munu}
end{equation}

The symmetric energy-momentum tensor is

begin{equation}
Theta^{munu}=frac{1}{2}bar{psi}Big{gamma^mu D^nu+gamma^nu D^muBig}psi-eta^{munu}ibar{psi}not{D}psi-F^{mulambda}F^nu_{ lambda}+frac{1}{4}eta^{mu nu}F^{sigmalambda}F_{sigmalambda}
end{equation}

The trace of this operator is

begin{equation}
Theta^mu_{ mu}=big(1-dbig)ibar{psi}not{D}psi+big(frac{d}{4}-1big)F^{sigmalambda}F_{sigmalambda}
end{equation}

The fermionic part is zero if we use Dirac’s equation, which can be stated as “The energy-momentum tensor is traceless on-shell“. On the other hand, if we are working on a $d=4$ spacetime, the photon part is inmediately traceless.

The last equation works if $Theta^mu _{ mu}$ is a quantum operator as well, since the equations of motion work for operators too. My question is: How is the path integral insertion $langleTheta^mu_{ mu}rangle$ not inmediately zero?

My preliminary answer is that when you actually calculate that using

begin{equation}
langleTheta^mu_{ mu}rangle=intmathcal{D}psimathcal{D}A_muTheta^mu_{ mu}e^{-S_E}
end{equation}

the operator inside the path integral is not on-shell, which means that it is not zero. However, if we were working only with the photon field in $d=4$ then there’s no way $langleTheta^mu_{ mu}rangle$ is not zero.

Is my reasoning correct?