I'm having trouble understanding the step

$$\left[\pi (\vec{x},t),\int d^{3}y ~(\frac{1}{2} \pi (\vec{y},t)^{2}+\frac{1}{2}\phi (\vec{y},t)(-\nabla^{2} +m^{2})\phi (\vec{y},t)) \right]$$ $$ =\int d^{3}y ~(-i\delta^{(3)}(\vec{x}-\vec{y})(-\nabla^{2} +m^{2})\phi (\vec{y},t)) $$

I've tried using the relations $$[\phi (\vec{x},t), \pi (\vec{y},t)] = i\delta^{(3)}(\vec{x}-\vec{y})$$ and $$[A,BC] = [A,B]C + B[A,C], $$ but run into $$[\pi (\vec{x},t), (-\nabla^{2} +m^{2})\phi (\vec{y},t)] ,$$ which I don't know how to evaluate.

Any help would be appreciated.

  • 3
    $\begingroup$ Yep, and while you're at it, a more descriptive title would be helpful. $\endgroup$ – David Z Jul 9 '14 at 17:26
  • $\begingroup$ Seems to me like it follows directly from the equal-time commutation relations for $\phi$ and $\pi$ $\endgroup$ – ZachMcDargh Jul 9 '14 at 17:59
  • $\begingroup$ Have you tried going to Fourier space (that gets rid of these nasty derivatives ;))? $\endgroup$ – ACuriousMind Jul 9 '14 at 20:27
  • 2
    $\begingroup$ Nabla operator with a mass term which can be treated as an operator L acting only upon y coordinates not acting upon x coordinate. So it commute with any function of (x,t) i.e. you can pull out the operator out of commutator. Think of this as a partial derivatives acting upon the a function of (x,y,t). $\endgroup$ – user45765 Jul 10 '14 at 0:11

This is really straight forward, once you get used to the notation. (Don't you hate it when people say that?)

$$[\pi (\vec{x},t), (-\nabla^{2} +m^{2})\phi (\vec{y},t)] ,$$

Here you need to remember that $\nabla^2$ acts on the $\phi(\vec{y},t)$ only, so $\pi$ can pass right through this wave operator. Now when you evaluate the commutator you'll end up with something like $\phi (\vec{y},t)(-\nabla^{2} +m^{2})\delta^{(3)}(\vec x-\vec y)$, after which you use "self-adjointness" of $\nabla^2$ (really, integration by parts), to make the wave operator act on the first $\phi$. You might need to relabel variables afterwards.


$$i {{\partial}\over{\partial t}}\pi=[\pi,\int d^3x\tfrac{1}{2}\pi^2+\tfrac{1}{2}\phi()\phi]$$ $$=[\pi,\int d^3x\tfrac{1}{2}\phi()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\phi[\pi,()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\phi[\pi,()]\phi+\phi()[\pi,\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\{\phi\pi()\phi-\phi()\pi\phi+\phi()\pi\phi-\phi()\phi\pi\}$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\{\phi\pi()\phi-\phi()\phi\pi\}$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\{\pi\phi()\phi-\phi\pi()\phi\}$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+[\pi,\phi]()\phi$$ $$=\int d^3x[\pi,\phi]()\phi$$ $$=\int d^3x-[\phi,\pi]()\phi$$ $$=\int d^3x-i\delta()\phi$$ $$=-i()\phi$$ I'm not sure about the middle part, but I've used a few properties: (2.44), (2.20), (2.30), and of course the commutator identity. But I don't think that is the correct way of proofing this. (I'm struggling as well)


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