QFT proof from Peskin and Schroeder

[Norman]

I am unsure if this is the proper forum for this, since it is not actually homework... but here goes anyway.

I am trying to Prove Peskin and Schroeder equation 2.33( the second equal sign)


$$P=-\int d^3 x \pi (x) \nabla \phi (x) = \int \frac{d^3 x}{(2 \pi)^3} p a^{\dagger}_p a_p$$

so far what I have done:
written the fields as the momentum space quantities, done the integral over the spatial coordinates to give me the delta function and integrated over the p' variables to give me this:

The last step forces p'=-p

$$\int \frac{d^3}{(2 \pi)^3} \frac{p}{2} (a^{\dagger}_{-p} a_{-p} + a^{\dagger}_{-p} a^{\dagger}_p - a_p a_{-p} - a_p a^{\dagger}_p )$$

I don't see how these operators cancel out to give :
$$() = 2a^{\dagger}_p a_p$$

Any help would be greatly appreciated... even just a hint would be very helpfull.
Thanks

 

[Dr Transport]

How does $$a_{-p}^{\dag}$$ relate to $$a_{p}$$? I can't remember, it was a long time ago when I did QFT. There has to be a commutation relation somewhere that will allow for combining terms.

 

[humanino]

I think you can cancel those 2 and 3 terms in which you only create or only destroy : they will give zero.
Then the first term, you change p to -p, and the last term you use commutation to order it right (create first, destroy after !) which switch sign. Maybe a problem with an overall constant at this last step.
Sorry, I am not rigorous. I do not have the book right here.

 

[Fredrik]

The integrals of $$p^i a(\vec p)a(-\vec p)$$ and $$p^i a^\dagger (\vec p) a^\dagger (-\vec p)$$ are both =0 because they are odd functions of $$\vec p$$.

Suppose that

$$f(-\vec p)=-f(\vec p)$$

Now look at this:

$$\int d^3p f(\vec p)=-\int_{\infty}^{-\infty}d^3q f(-\vec{q})=\int_{-\infty}^{\infty} d^3q f(-\vec q)=-\int_{-\infty}^{\infty} d^3q f(\vec q)=-\int d^3p f(\vec p)$$

1. Change variables, q=-p.
2. "Reverse" the integration interval.
3. Use f(-q)=-f(q).
4. Rename the integration variable to p.
5. Note that the equation is now in the form A=-A.

The same change of variables will help you deal with the first term. I think you should also check the signs in front of each term. They look wrong to me. I think they should be either -++- or +--+, but I could be wrong about that. I'm too lazy to check it carefully right now.