Okay, let’s start from the basics. Under infinitesimal scale transformation, a generic field \varphi ( x ) transforms according to \delta^{*} \varphi ( x ) \equiv \bar{\varphi} ( \bar{x} ) - \varphi ( x ) = \Delta \ \varphi ( x ) , where \Delta is the scaling dimension of the field. The canonical Noether current associated with this transformation (the dilatation current) is D_{c}^{\mu} = \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi )} \Delta \varphi + T^{\mu \nu} \ x_{\nu} , where T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi )} \partial^{\nu} \varphi - \eta^{\mu \nu} \mathcal{L} , is the canonical energy-momentum tensor. For a Lorentz scalar field, we have \Delta = 1 and \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \varphi )} = \partial^{\mu} \varphi . Therefore, the canonical dilatation current becomes D_{c}^{\sigma} = T^{\sigma \tau} \ x_{\tau} + \varphi \ \partial^{\sigma} \varphi , \ \ \ \ \ (1) and the associated scaling generator is D \equiv \int d^{3} x \ D_{c}^{0} ( x ) = \int d^{3} x \left( T^{0 \tau} \ x_{\tau} + \varphi \ \partial^{0} \varphi \right) . Now, let’s play with the second term in Eq(1). First, you can check that \varphi \partial^{\sigma} \varphi = \frac{1}{2} \ \partial^{\sigma} \varphi^{2} = - \frac{1}{6} \ \eta_{\rho \tau} \left( \eta^{\tau \sigma} \partial^{\rho} - \eta^{\tau \rho} \partial^{\sigma} \right) \varphi^{2} . Using the relation \eta_{\rho \tau} = \partial_{\rho} x_{\tau} , we find \varphi \ \partial^{\sigma} \varphi = - \frac{1}{6} \ ( \partial_{\rho} x_{\tau} ) \left( \eta^{\tau \sigma} \partial^{\rho} - \eta^{\tau \rho} \partial^{\sigma} \right) \varphi^{2} . Finally, with help of the identity ( \partial X ) Y = \partial ( X Y ) - X ( \partial Y ), we rewrite the above equation in the form \varphi \ \partial^{\sigma} \varphi = \frac{1}{6} \ x_{\tau} \partial_{\rho} \ ( \eta^{\tau \sigma} \partial^{\rho} \varphi^{2} - \eta^{\tau \rho} \partial^{\sigma} \varphi^{2} ) - \frac{1}{6} \ \partial_{\rho} ( x^{\sigma} \partial^{\rho} \varphi^{2} - x^{\rho} \partial^{\sigma}\varphi^{2} ) . \ \ (2) Now, substituting (2) in (1) we get D_{c}^{\sigma} + \partial_{\rho} X^{\sigma \rho} = \theta^{\sigma \tau} x_{\tau} , \ \ \ \ (3) where X^{\sigma \rho} = - X^{\rho \sigma} \equiv \frac{1}{6} \ ( x^{\sigma} \partial^{\rho} - x^{\rho} \partial^{\sigma} ) \varphi^{2} , and \theta^{\sigma \tau} = T^{\sigma \tau} + \frac{1}{6} \ ( \eta^{\sigma \tau} \partial^{2} - \partial^{\sigma} \partial^{\tau} ) \varphi^{2} . \ \ \ (4) Now, let us examine the left-hand-side of Eq(3) by writing D^{\sigma} \equiv D_{c}^{\sigma} + \partial_{\rho} X^{\sigma \rho} . The last term is the divergence of an anti-symmetric tensor (super-potential), so it does not contribute to the scaling current and charge. Indeed \partial_{\sigma} D^{\sigma} = \partial_{\sigma} D_{c}^{\sigma} and, since \int d^{3} x \ \partial_{\rho} X^{\rho 0} = \int d^{3} x \ \partial_{j} X^{j 0} = 0 , we find \int d^{3} x \ D^{0} ( x ) = \int d^{3} x \ D_{c}^{0} ( x ) = D . Thus, D^{\mu} is a legitimate dilatation current and, therefore, can be used instead of the canonical Noether current D_{c}^{\mu}. This implies that the tensor \theta^{\mu \nu} as given by eq(4) is a new legitimate energy-momentum tensor. Indeed it is: It is easy to check that the new tensor has the following properties \theta^{\mu \nu} = \theta^{\nu \mu} , \ \ \partial_{\mu} \theta^{\mu \nu} = \partial_{\mu} T^{\mu \nu} = 0 . And, more importantly, the added piece in (4) does not change the Poincare’ generators P^{\mu} = \int d^{3} x \ \theta^{0 \mu} = \int d^{3} x \ T^{0 \mu} , M^{\mu \nu} = \int d^{3} x \ ( x^{\mu} \theta^{0 \nu} - x^{\nu} \theta^{0 \mu} ) = \int d^{3} x \ ( x^{\mu} T^{0 \nu} - x^{\nu} T^{0 \mu} ) . So, given \theta^{\mu \nu} we can rewrite dilatation current in the following compact form D^{\sigma} = \theta^{\sigma \tau} x_{\tau} . So, \partial_{\mu} D^{\mu} = \theta^{\mu}_{\mu} . Thus, for scale-invariant scalar field theory, the new energy-momentum tensor is traceless \theta^{\mu}_{\mu} = 0. I believe you have realized now that your question is a trivial one as you can see that \int d^{3} x \ \theta^{0 \tau} \ x_{\tau} = \int d^{3} x \ ( T^{0 \tau} \ x_{\tau} + \varphi \ \partial^{0} \varphi ) = D .
Sam
and wait and see if anybody (other than myself) can sort out your confusion. As a hint you can check out this old thread
