In a phi fourth theory, the Hamiltonian density is:
$$\mathcal{H}=\frac{1}{2}\pi^2+\frac{1}{2}(\nabla \phi)^2+\frac{1}{2}m^2\phi^2+\frac{\lambda}{4!}\phi^4$$
Now I impose the usual equal time canonical commutation relations for fields ($\hbar=1$)
$$[\phi(\vec{x}),\pi(\vec{y})]=i \delta^3(\vec{x}-\vec{y})$$
where
$$\pi=\frac{\partial \mathcal{L}}{\partial (\dot{\phi})} \equiv \dot{\phi}$$
Heisenberg equation of motion for the field is just the definition of the conjugate momentum
$$\frac{d}{dt}\phi(\vec{x},t)=\pi(\vec{x},t) $$
and for $\pi(\vec{x})$ I have to calculate the commutator (not writing time dependence)
$$[H,\pi(\vec{x},t)]=\int d^3x'\left[\frac{1}{2}\pi^2(\vec{x}')+\frac{1}{2}(\nabla \phi)^2(\vec{x}')+\frac{1}{2}m^2\phi^2(\vec{x}')+\frac{\lambda}{4!}\phi^4(\vec{x}'),\pi(\vec{x}) \right] $$
First term gives zero, third and fourth terms give $i\left(m^2\phi(\vec{x})+\frac{\lambda}{3!}\phi^3(\vec{x})\right)$
My question is, how can I calculate
$$\frac{1}{2}\int d^3x' [(\nabla \phi)^2(\vec{x}'),\pi(\vec{x})] $$
As an analogy with the integral of the commutator is the commutator of the integral, may I write $\nabla \phi^2=\nabla \phi \cdot \nabla \phi$ and integrate by parts? How can I show that that is true?
Best Answer
Here's a formal computation. First note that: $$[A^2,B]=AAB-ABA+ABA-BAA=A[A,B]+[A,B]A.$$
Also: $$[\frac{\partial}{\partial z'}\phi (\mathbf x'),\pi (\mathbf x)]=\frac{\partial}{\partial z'}[\phi (\mathbf x'),\pi (\mathbf x)]=\frac{\partial}{\partial z'}i\delta ^3 (\mathbf x ' -\mathbf x )=i\frac{\partial \delta}{\partial z'} (z'-z)\delta(x'-x)\delta (y'-y).$$
Recall that the derivative of a distribution $T$ is defined by: $$(T',f)=-(T,f').$$
So:
$$[(\frac{\partial}{\partial z'}\phi (\mathbf x'))^2,\pi (\mathbf x)]=2i\frac{\partial \delta}{\partial z'} (z'-z)\delta(x'-x)\delta (y'-y)\frac{\partial \phi}{\partial z'} (\mathbf x ')$$ and: $$\int \text {d} ^3 \mathbf x '[(\frac{\partial}{\partial z'}\phi (\mathbf x'))^2,\pi (\mathbf x)]=-2i\frac{\partial ^2\phi}{\partial z ^2} (\mathbf x ).$$