Consider the following Lagrangian for fermions coupled to an external gauge field and chiral gauge field, $$ \mathcal{L} = -i\bar{\psi}\gamma^{\mu}\partial_{\mu} \psi - \bar{\psi}A_{\mu}\gamma^{\mu}\psi - \bar{\psi}A_{5,\mu}\gamma^{\mu}\gamma^5 \psi, $$ where I am using the Minkowski metric with signature $(-+++)$, $\bar{\psi} = \psi^{\dagger}\gamma^0$ and I assume $A_{\mu}$ and $A_{5,\mu}$ to be constant fields. The Hamiltonian corresponding to this Lagrangian reads
$$ \mathcal{H} = \psi^{\dagger}\big[i\gamma^0\gamma^i\partial_i + \gamma^0A_{\mu}\gamma^{\mu}+ \gamma^0A_{5,\mu}\gamma^{\mu}\gamma^5\big]. $$
I think I have derived that the following is the energy-momentum tensor of the theory $$ T^{\mu\nu} = -\frac{i}{4}\bar{\psi}\gamma^{\mu} \Big[\overset{\leftrightarrow}{\partial^{\nu}} - i A^{\nu} -iA_5^{\nu}\gamma^{5} \Big]\psi + (\mu\leftrightarrow \nu) - \eta^{\mu\nu}\mathcal{L}, $$ because I managed to show $\partial_{\mu}T^{\mu\nu}=0$. However, normally the $00$-th component should give $\pm\mathcal{H}$. However, evaluating the above expression at $\mu=\nu=0$ yields an extra contribution because the $A_0$ and $A_{5,0}$ do not drop out. Hence, I would like to modify the tensor $T^{\mu\nu}$ such that it does give me the Hamiltonian. I tried messing around with the sign in front of the (chiral) gauge field, but that messes up its conservation.
Does anyone know how to modify it such that I obtain the correct EM-tensor?
