5
$\begingroup$

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.

$\endgroup$
  • 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
6
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

$$i {{\partial}\over{\partial t}}\pi=[\pi,\int d^3x\tfrac{1}{2}\pi^2+\tfrac{1}{2}\phi()\phi]$$ $$=[\pi,\int d^3x\tfrac{1}{2}\phi()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\phi[\pi,()\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\phi[\pi,()]\phi+\phi()[\pi,\phi]$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\{\phi\pi()\phi-\phi()\pi\phi+\phi()\pi\phi-\phi()\phi\pi\}$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\{\phi\pi()\phi-\phi()\phi\pi\}$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+\{\pi\phi()\phi-\phi\pi()\phi\}$$ $$=\tfrac{1}{2}\int d^3x[\pi,\phi]()\phi+[\pi,\phi]()\phi$$ $$=\int d^3x[\pi,\phi]()\phi$$ $$=\int d^3x-[\phi,\pi]()\phi$$ $$=\int d^3x-i\delta()\phi$$ $$=-i()\phi$$ 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)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.