# Perplexing commutator

1. Apr 10, 2009

### 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?

Last edited: Apr 10, 2009
2. Apr 10, 2009

### dextercioby

Have you tried to see what part integration brings you ?

3. Apr 11, 2009

### waht

By expanding the commutator we get,

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

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

4. Apr 12, 2009

### 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).