I've recently learned of a method which can symmetrise the canonical energy-momentum tensor,
$$T_{\mu\nu} = \eta_{\mu\nu} \mathcal{L} – \sum_{a} \frac{\partial \mathcal{L}}{\partial (\partial^\mu \varphi_a)} \partial_\nu \varphi_a,$$
using the $(-,+,+,+)$ signature, where we add an additional total derivative term,
$$T_{\mu\nu} \rightarrow T_{\mu\nu} + \partial^\lambda \chi_{\lambda\mu\nu},$$
$$\chi_{\lambda\mu\nu} = K_{\lambda\mu\nu} + K_{\mu\nu\lambda} + K_{\nu\mu\lambda},\quad K_{\lambda\mu\nu} = -\frac{i}{2} \sum_{a,b} \frac{\partial \mathcal{L}}{\partial (\partial^\lambda \varphi_a) } (J_{\mu\nu})_{ab} \varphi_b,$$
where the $(J_{\mu\nu})_{ab}$ are the representations of the Lorentz algebra under which the fields $\varphi_a$ transform, given here.
Having confirmed this works for the electromagnetic Lagrangian, and understand that $\chi_{\lambda\mu\nu}$ is zero for scalars (which is fine as the original canonical SEM tensor is symmetric), I then attempted to try this on the Dirac Lagrangian,
$$\mathcal{L} = \bar\psi (i\displaystyle{\not}\partial – m) \psi,$$
But I've either made some mistakes, potentially in the handling of the Grassmann-valued fields, or just got a bit stuck at the end.
First, I found the canonical SEM tensor to be
$$T_{\mu\nu} = i \bar\psi \gamma_\mu \partial_\nu \psi,$$
Which is asymmetric in $\mu$ and $\nu$. Now, the correction term:
$$K_{\lambda\mu\nu} = -\frac{i}{2} \left( \frac{\partial \mathcal{L}}{\partial (\partial^\lambda \psi)} S_{\mu\nu} \psi + \frac{\partial \mathcal{L}}{\partial (\partial^\lambda \bar\psi)} S_{\mu\nu} \bar\psi \right),\quad S_{\mu\nu} = \frac{i}{4} \left[ \gamma_\mu, \gamma_\nu \right].$$
The $\frac{\partial \mathcal{L}}{\partial (\partial^\lambda \bar\psi)}$ factor is zero, so the second term disappears.
$$K_{\lambda\mu\nu} = \frac{1}{8} \frac{\partial \mathcal{L}}{\partial (\partial^\lambda \psi)} \left[ \gamma_\mu, \gamma_\nu \right] \psi,\quad \frac{\partial \mathcal{L}}{\partial (\partial^\lambda \psi)} = -i \bar\psi \gamma_\lambda$$
$$\Rightarrow \chi_{\lambda\mu\nu} = -\frac{i}{8} \bar\psi \left( \gamma_\lambda \left[ \gamma_\mu, \gamma_\nu \right] + \gamma_\mu \left[ \gamma_\nu, \gamma_\lambda \right] + \gamma_\nu \left[ \gamma_\mu, \gamma_\lambda \right] \right) \psi$$
$$= -\frac{i}{8} \bar\psi \left( \left\{ \gamma_\mu, \gamma_\nu \right\} \gamma_\lambda – \left\{ \gamma_\lambda, \gamma_\nu \right\} \gamma_\mu – \left[ \gamma_\mu, \gamma_\lambda \right] \gamma_\nu \right) \psi$$
$$= -\frac{i}{4} \bar\psi \left( -\eta_{\mu\nu}\gamma_\lambda + \eta_{\lambda\nu}\gamma_\mu + \eta_{\mu\lambda}\gamma_\nu + \gamma_\lambda \gamma_\mu \gamma_\nu \right) \psi.$$
Then we take the derivative:
$$\partial^\lambda \chi_{\lambda\mu\nu} = -\frac{i}{4} \bar\psi \left( -\eta_{\mu\nu}\displaystyle{\not}\partial + \gamma_\mu \partial_\nu + \gamma_\nu \partial_\mu + \displaystyle{\not}\partial \gamma_\mu \gamma_\nu – \overleftarrow{\displaystyle{\not}\partial} \eta_{\mu\nu} + \overleftarrow{\partial_\nu} \gamma_\mu + \overleftarrow{\partial_\mu} \gamma_\nu + \overleftarrow{\displaystyle{\not}\partial} \gamma_\mu \gamma_\nu \right) \psi$$
$$= -\frac{i}{4} \bar\psi \left( -\eta_{\mu\nu}\displaystyle{\not}\partial + \gamma_\mu \partial_\nu + \gamma_\nu \partial_\mu + \gamma_\mu \gamma_\nu \displaystyle{\not}\partial + \left[ \displaystyle{\not}\partial, \gamma_\mu \gamma_\nu \right] – \overleftarrow{\displaystyle{\not}\partial} \eta_{\mu\nu} + \overleftarrow{\partial_\nu} \gamma_\mu + \overleftarrow{\partial_\mu} \gamma_\nu + \overleftarrow{\displaystyle{\not}\partial} \gamma_\mu \gamma_\nu \right) \psi.$$
We can use the equations of motion to cancel some terms:
$$\displaystyle{\not}\partial \psi = -i m \psi,\quad \bar\psi \overleftarrow{\displaystyle{\not}\partial} = i m \bar\psi$$
$$\Rightarrow \partial^\lambda \chi_{\lambda\mu\nu} = -\frac{i}{4} \bar\psi \left( \gamma_\mu \partial_\nu + \gamma_\nu \partial_\mu + \left[ \displaystyle{\not}\partial, \gamma_\mu \gamma_\nu \right] + \overleftarrow{\partial_\nu} \gamma_\mu + \overleftarrow{\partial_\mu} \gamma_\nu \right) \psi,$$
But at this point I'm not quite sure how to simplify things any more.
Best Answer
I found the identity I needed as the fifth miscellaneous identity on the gamma matrices Wikipedia article:
$$\gamma_\sigma \gamma_\mu \gamma_\nu = -\eta_{\sigma\mu}\gamma_\nu - \eta_{\mu\nu}\gamma_\sigma + \eta_{\sigma\nu}\gamma_\mu - i \epsilon_{\rho\sigma\mu\nu} \gamma^\rho \gamma^5,$$
$$\left[ \gamma_\sigma, \gamma_\mu \gamma_\nu \right] = \gamma_\sigma \gamma_\mu \gamma_\nu - \gamma_\mu \gamma_\nu \gamma_\sigma = 2\eta_{\nu\sigma}\gamma_\mu - 2\eta_{\mu\sigma}\gamma_\nu,$$
$$\left[ \displaystyle{\not}\partial, \gamma_\mu \gamma_\nu \right] = 2\gamma_\mu \partial_\nu - 2\gamma_\nu \partial_\mu$$
$$\Rightarrow \partial^\lambda \chi_{\lambda\mu\nu} = -\frac{i}{4} \bar\psi \left( 3\gamma_\mu \partial_\nu - \gamma_\nu \partial_\mu + \overleftarrow{\partial_\nu} \gamma_\mu + \overleftarrow{\partial_\mu} \gamma_\nu \right) \psi.$$
Finally we can get the modified canonical SEM tensor:
$$T_{\mu\nu} + \partial^\lambda \chi_{\lambda\mu\nu} = i \bar\psi \gamma_\mu \partial_\nu \psi - \frac{i}{4} \bar\psi \left( 3\gamma_\mu \partial_\nu - \gamma_\nu \partial_\mu + \overleftarrow{\partial_\nu} \gamma_\mu + \overleftarrow{\partial_\mu} \gamma_\nu \right) \psi$$
$$= \frac{i}{4} \bar\psi \left( \gamma_\mu \partial_\nu + \gamma_\nu \partial_\mu - \overleftarrow{\partial_\nu} \gamma_\mu - \overleftarrow{\partial_\mu} \gamma_\nu \right) \psi.$$
This is symmetric in $\mu$ and $\nu$ as desired and expected.