quantum field theory - Is Hilbert Stress-energy tensor always same as Belinfante stress-energy tensor?

The canonical stress-energy (SE) tensor arises from Noether’s Theorem by employing the conserved currents associated with translational symmetries.

It's defined as $$T^{ab}=\frac{\partial \mathcal{L}}{\partial(\partial_a\phi)}\partial^b\phi+\mathcal{L}\eta^{ab}$$

However in general canonical SE tensor is not symmetric. In fact for a SE tensor $T_{ab}$, $T_{ab}+\partial^c\chi_{cab}$ is also a SE tensor for any $\chi_{cab}=-\chi_{acb}$.

So given a canonical SE tensor, we can always construct a symmetric SE tensor, called Belinfante SE tensor.

There is another way to define SE tensor in QFT in curved spacetime. That is Hilbert SE tensor which is defined as

$$T_{ab}=\frac{-2}{\sqrt{-g}}\frac{\delta(\mathcal{L}\sqrt{-g})}{\delta g^{ab}}$$

So Hilbert SE tensor is also a symmetric tensor.

My questions are

1. 
   

   Is Hilbert SE tensor always same as Belinfante SE tensor? If yes, how to prove.

   
   



2. 
   

   If the question (1)'s answer is yes, is Hilbert SE tensor or Belinfante SE tensor the unique symmetric SE tensor that you can construct?

   
   

Answer

1. 
   

   Well, firstly, we have to assume Lorentz covariance and general covariance of the theory. For non-relativistic theories all bets are off. Secondly, in case of fermions, one needs to generalize the Hilbert SEM tensor from variation wrt. a metric to a variation wrt. a vielbein, see e.g. my Phys.SE answer here. Then the generalized Hilbert SEM tensor is the canonical SEM plus the Belinfante-Rosenfeld improvement term. A proof is sketched in my Phys.SE answer here and links therein.

   
   


2. 
   
   

   No, a symmetric SEM tensor is not unique. It is in principle possible to add improvement terms that respects the symmetry and the conservations laws.