Given a four component Dirac spinor $\psi$ (working classically) and the usual Dirac matrices $\gamma^{\mu}$, one can construct a Lagrangian
$$\mathcal{L} = \bar{\psi}(i\overset{\leftrightarrow}{{\not\partial}} - m)\psi = i\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi - i(\partial_\mu\bar{\psi})\gamma^{\mu}\psi - m\bar{\psi}\psi$$
Where I have used the convention (as I understand it - could be the problem)
$$\alpha \overset{\leftrightarrow}{\not \partial} \beta = \alpha\gamma^{\mu}(\partial_{\mu}\beta) - (\partial_{\mu}\alpha) \gamma^{\mu}\beta$$
The energy-momentum tensor has the regular form
$$\theta^{\mu\nu} = \dfrac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi_i)}\partial^{\nu}\phi_i - \eta^{\mu \nu}\mathcal{L}$$
Now, by writing the arrowed derivative explicitly as above and treating $\psi, \bar{\psi}$ as independent fields, this seems to indicate that
$$\theta^{\mu\nu} = (i\bar{\psi} \gamma^{\mu})\partial^{\nu}\psi - (i\gamma^{\mu}\psi)\partial^{\nu}\bar{\psi} - \eta^{\mu \nu}\mathcal{L} $$
However, my lecture notes seem to instead suggest that the second term of the RHS above should actually be
$$-i\bar{\psi}\gamma^{\mu} \overset{\leftarrow}{\partial^{\nu}}\psi = -i(\partial^{\nu}\bar{\psi})\gamma^{\mu}\psi$$
Now, unless $\gamma^{\mu}\psi$ and $\partial^{\nu}\bar{\psi}$ commute, I must be missing a subtle point/doing something stupid.
EDIT: If I were to guess I'd imagine that the product in the first term of the e-m tensor must be ordered to ensure that the object is a 4x4 matrix for consistency, whereas $\sim\gamma\psi\partial\bar{\psi}$ is clearly no such thing. Does that sound about right?
Any help is greatly appreciated!