Dirac field and stress-energy tensor density I read somewhere that stress-energy tensor density is a symmetric tensor. But if I take the Dirac Field tensor:
$$T^{\mu \nu}=i \psi^\dagger \gamma^0 \gamma^\mu  \partial^\nu \psi $$
How could I demonstrate this property?
 A: There is a lot of ambiguity in the definition of the stress-energy tensor. The stress-energy tensor is a conserved current, and like all conserved currents it is only defined up to a total divergence. I assume this $T_{\mu \nu}$ was calculated using the canonical prescription $ T^\mu_\nu=\frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^i)}\partial_\nu \phi^i-\mathcal{L}\delta^\mu_\nu $ (you seem to be missing the second piece, or you are dealing with a massless field). The canonical tensor is not symmetric for fields with spin. Essentially, the intrinsic angular momentum is also contributing to T. So you find a term $S^\lambda_{\mu \nu}$ satisfying $\partial_\lambda S^\lambda_{\mu \nu}\approx T_{[\mu \nu]}$ (S is antisymmetric in its first two indices, and thus has vanishing divergence) and add it to the canonical tensor. See this worked out in detail here
http://en.wikipedia.org/wiki/Belinfante%E2%80%93Rosenfeld_stress%E2%80%93energy_tensor
This procedure might seem a little random, but of course what you really should be doing is obtaining T from $T^{\mu \nu}=\frac{\delta S}{\delta g_{\mu \nu}}$ as in general relativity. This $T$ will always be symmetric, and is in fact the same as the Belinfante tensor. However, there is still ambiguity in this procedure. In order to obtain T this way, you have to "covariantize" the theory, promoting the metric to a dynamical field. This covariantization is ambiguous: you may couple the metric to the curvature non-minimally. These couplings vanish in the flat space limit, but can still affect the expression for T. But at least this expression will always be symmetric.
Hope this helps!
