Solving Perplexing Commutator for Simplification

[waht]

When simplifying this

\int d^3x' [\pi(x), \frac{1}{2}\pi^2(x') + \frac{1}{2} \phi(x')( -\nabla^2 + m^2)\phi(x')]

we know that

[\pi(x), \pi(x')] = 0

[\phi(x), \pi(x')] = -i\delta(x-x')

how does that simplify to

\int d^3x' \delta(x-x')( -\nabla^2 + m^2)\phi(x')

I know that

[\pi(x), \pi^2(x')] = 0

but not sure how does the laplacian gets factored out like that and one-half disappears?

 

[dextercioby]

Have you tried to see what part integration brings you ?

 

[waht]

By expanding the commutator we get,

<br /> \frac{1}{2} \int d^3x&#039; \pi(x)\phi(x&#039;)( -\nabla^2 + m^2)\phi(x&#039;) - \phi(x&#039;)( -\nabla^2 + m^2)\phi(x&#039;)\pi(x)<br />

the laplacian in the second term is sandwiched between two phis, integration by part doesn't seem to help to factor it out.

 

[lbrits]

I think you're better off using [A,BC] = [A,B]C + B[A,C] and then integration by parts. Also, it's a good time in your life to realize that a commutator behaves very much like a functional derivative (recall QM 101).