In Carroll's Introduction to General Relativity: Spacetime and Geometry, I am going to express the components of the energy-momentum tensor of scalar field in three- vector notation, using the divergence, gradient, curl.
$$ T^{\mu\nu}_{\text{scalar}} = \eta^{\mu\lambda}\eta^{\nu\sigma}\partial_\lambda\varphi \partial_\sigma\varphi - \eta^{\mu\nu} \left(\frac{1}{2}\eta^{\lambda\sigma} \partial_\lambda\varphi \partial_\sigma\varphi + V(\varphi)\right) $$
I think that $\eta^{\mu\lambda}\eta^{\nu\sigma}\partial_\lambda\varphi \partial_\sigma\varphi$ can only be non-zero when $\mu = \lambda = i$ and $\nu = \sigma = j$, so that:
$$ \eta^{ii}\eta^{jj}\partial_i\varphi \partial_j\varphi = \partial_i\varphi\partial_j\varphi \quad(i,j=t,x,y,z) $$
$$ T^{ij}_{\text{scalar}} = \partial_i\varphi\partial_j\varphi - \eta_{ij}\left\{\frac{1}{2}\left[(\nabla \varphi)^2 - (\partial_t\varphi)^2\right] + V(\varphi)\right\}\quad (i,\ j = t,x,y,z) $$
But I found another solution here:
it says that:
$$ \eta^{\mu\lambda}\eta^{\nu\sigma}\partial_\lambda\varphi \partial_\sigma\varphi = (\eta^{\mu\lambda}\partial_\lambda\varphi) (\eta^{\nu\sigma}\partial_\sigma\varphi) = \begin{bmatrix} -\partial_t\varphi \\ \partial_x\varphi \\ \partial_y\varphi \\ \partial_z\varphi \end{bmatrix} \begin{bmatrix} -\partial_t\varphi & \partial_x\varphi & \partial_y\varphi & \partial_z\varphi \end{bmatrix} $$ which is a $4 \times 4$ matrix
and then
So that which is right? I am confused with the outer product, is it necessary?
