# Obtaining momentum operator $P^\mu$ from Lagrangian and energy-momentum tensor $T^{\mu\nu}$

I am pretty new to quantum field theory. Given the Lagrangian density, $$\mathcal{L} = \frac{1}{2} ( \partial_\mu \phi ) ( \partial^\mu \phi ) - \frac{1}{2} m^2 \phi^2$$ and its energy-momentum tensor $$T^{\mu \nu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi) } \partial^\nu \phi - g^{\mu \nu} \mathcal{L}$$ I would like to obtain $P^\mu$. I already know that $$P^j = \int d^3 x \pi_r(x) \frac{\partial \phi_r(x)}{\partial x_j}$$ and $$\pi_r(x) = \frac{\partial \mathcal{L}}{\partial \dot{\phi_r}} ,$$ but I do not really know how to calculate $P^\mu$. Anybody has some advice how to start?

• I'd put $P^\mu = T^{\mu0}$ – Phoenix87 Dec 28 '14 at 11:55
• @Phoenix87 - Rather $P^\nu = \int d^3 x T^{0\nu}$ – Prahar Dec 28 '14 at 18:43
• @Prahar agreed. – Phoenix87 Dec 28 '14 at 19:14

The momentum is given by $P^\mu = (E,\vec{P})$, and the operator acting as $[:P^\mu:,a^\dagger_k] = k^\mu a^\dagger_k$.
Energy is an eigenvalue of the Hamiltonian, which is given by $H = \int d^3x~T^{00}$ and $\vec{P}$ is given by your relation above, which from the definition of the stress energy tensor also equals $$P^j = \int d^3x~T^{0j}$$
You can combine these to give $$P^\mu = \int d^4x~T^{0\mu}$$