- #1

waht

- 1,517

- 4

When simplifying this

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

we know that

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

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

how does that simplify to

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

I know that

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

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

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

we know that

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

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

how does that simplify to

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

I know that

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

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

Last edited: