QFT proof from Peskin and Schroeder

Norman
Messages
895
Reaction score
4
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
 
Physics news on Phys.org
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.
 
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.
 
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.
 
Last edited:
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles. Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated... Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/ by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!

Similar threads

Replies
4
Views
2K
Replies
3
Views
2K
Replies
4
Views
2K
Replies
2
Views
1K
Replies
11
Views
2K
Back
Top