Is Normal Ordering for Total Momentum Correct in QFT?

[moob301]

Homework Statement

I'm reading David Tong's QFT lecture notes. And he explains how to drive the free theory total momentum by quantum field operators from the classical field theory.
But I'm confusing on the normal ordering process a little bit.

Homework Equations

In his notation, the free field $$\phi(\vec{x})$$ and its momentum conjugate $$\pi(\vec{x})$$ are

$$\phi(\vec{x}) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\vec{p}}}}[a_{\vec{p}}e^{i\vec{p}\cdot\vec{x}}+{a^\dag}_{\vec{p}}e^{-i\vec{p}\cdot\vec{x}}]$$
and
$$\pi(\vec{x}) = \int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{\vec{p}}}{2}}[a_{\vec{p}}e^{i\vec{p}\cdot\vec{x}}-{a^\dag}_{\vec{p}}e^{-i\vec{p}\cdot\vec{x}}].$$

And he calculates the normal ordered total momentum from the classical one (the conserved momentum from the energy momentum tensor)
and says the normal ordered total momentum is :
$$\vec{P}=\int \frac{d^3p}{(2\pi)^3}\vec{p}{a^\dag}_{\vec{p}}a_{\vec{p}}.$$

The Attempt at a Solution

But the actual calculation goes like this:
$$\vec{P}=-\int d^3x \pi \nabla\phi$$
$$=\frac{1}{2}\int \frac{d^3p}{(2\pi)^3}\vec{p}(a_{\vec{p}}a_{-\vec{p}}+{a^\dag}_{\vec{p}}a_{\vec{p}}+a_{\vec{p}}{a^\dag}_{\vec{p}}+{a^\dag}_{\vec{p}}{a^\dag}_{-\vec{p}})$$
$$=\frac{1}{2}\int \frac{d^3p}{(2\pi)^3}\vec{p}(a_{\vec{p}}a_{-\vec{p}}+2{a^\dag}_{\vec{p}}a_{\vec{p}}+(2\pi)^3\delta(\vec{0})+{a^\dag}_{\vec{p}}{a^\dag}_{-\vec{p}})$$

So (my question is) what about the other terms? :
$$\frac{1}{2}\int \frac{d^3p}{(2\pi)^3}\vec{p}(a_{\vec{p}}a_{-\vec{p}}+{a^\dag}_{\vec{p}}{a^\dag}_{-\vec{p}})$$

Thanks in advance.

 

[gabbagabbahey]

HInt: What happens when you integrate an odd function over the reals?

 

[moob301]

HInt: What happens when you integrate an odd function over the reals?

I got it. Thank you very much.