# About canonical momentum operator as generators of translation

As we know, translations in Minkowski space form a group, it may be represented by a unitary operator, satisfying:

$$U(a)U^\dagger(a) = 1 \tag{1}$$ $$U(a)U(b) = U(a+b) \tag{2}$$ $$U(0) = 1 \tag{3}$$

The U satisfying these is $$U(a) = e^{i a_\mu G^\mu} \tag{4}$$ where $G^\mu$'s are Hermitian operators(matrices), called the generators of the group.

In QFT, for a scalar field, we have $$U(a)^{-1}\phi(x)U(a) = \phi(x-a) \tag{5}$$ or $$[\phi(x), G^\mu] = i \partial^\mu \phi(x) \tag{6}$$

On the other hand, the canonical momentum operator is defined as $$P^\mu \equiv \int d^3 x T^{0\mu} = \int d^3 x \left(\frac{\partial \mathcal{L}}{\partial(\partial_0\phi)}\frac{\partial \phi}{\partial x_\mu} - g^{0\mu} \mathcal{L} \right) \tag{7}$$

In QFT textbooks, $P^{\mu}$'s are taken as generators of $U(a)$, $$U(a) = e^{i a_\mu P^\mu} \tag{8}$$

but no rigorous proofs. For example, the validity of $P^\mu$ as the generators may depend on the particular Lagrangian from which $P^\mu$ is constructed.

So, under what conditions, $P^\mu$ may be taken as generators of a translation group?

• From (6) you can see that the only possible solution for $G_\mu$ is $i\partial_\mu$ that is the definition of the momentum operator.
– Jon
Jan 29 '18 at 9:45
• possible duplicate: Conserved charges and generators. Jan 29 '18 at 15:27

Well, one normally assumes your Lagrangian density has a standard kinetic term, so it is bilinear in the canonical momenta $\pi=\partial_0 \phi$ and the space gradients of $\phi$. The canonical commutation relation, then, $$\left[\phi\left(\vec{x}\right), \pi\left(\vec{y}\right)\right] = i \delta\left(\vec{x} - \vec{y}\right),$$ in addition to the trivial ones, with vanishing right-hand side, yields $$[\vec{P},\phi(\vec{x})]= -i\nabla \phi (\vec{x}),$$ so that the Hadamard lemma reduces to $$e^{-i\vec{a}\cdot \vec{P}} \phi e^{i\vec{a}\cdot \vec{P}} = e^{-i\operatorname {ad} \vec{a}\cdot \vec{P} } \phi= e^{ -\operatorname {ad} \vec{a}\cdot \nabla } \phi (\vec{x})= \phi (\vec{x}-\vec{a}),$$ the standard Lagrange translation formula.