How to Express Charge Q Using Creation and Annihilation Operators in QFT?

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The discussion focuses on expressing the charge Q in quantum field theory using creation and annihilation operators. The user initially struggles with the normal ordering procedure and the integration over spatial variables. They derive an expression for Q involving double integrals over different momentum variables but are unsure how to simplify it. Guidance is provided on using delta functions to eliminate one of the momentum integrals, leading to a single integral. The user expresses gratitude and indicates they have resolved their confusion.
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Homework Statement


Express the charge Q in terms of the creation and annihilation operators.

Homework Equations


$$\phi_{(x)}=\int \dfrac {d^3 p} {(2\pi)^3} \dfrac {1} {2 \omega_p} (a_p e^{i x \cdot p} + b^{\dagger}_{p} e^{-i x \cdot p})$$
$$\pi_{(x)}=\dfrac {-i} {2}\int \dfrac {d^3 p} {(2\pi)^3} (b_p e^{i x \cdot p} - a^{\dagger}_{p} e^{-i x \cdot p})$$
$$Q=-i\int d^3 x(\pi\phi - \phi^* \pi^*)$$

The Attempt at a Solution



Hey guys, little bit stuck with the normal ordering procedure. So I've basically plugged in my expressions for pi and phi into the expression for Q and arrived at the following:
$$Q=-i \int d^3x \int \dfrac {d^3 p} {(2\pi)^3} \dfrac {-i} {2\omega_p} (b_p b^{\dagger}_p - a^{\dagger}_p a_p)$$
So I know that I shouldn't have the spatial integral, but I'm not sure how to get rid of it and I know I need to normal order the operators but I'm stuck there to :/ any guidance would be massively appreciated :)
 
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The ##p## in the integral for ##\pi## should be distinguished from the ##p## in the integral for ##\phi##. Use different symbols for these ##p##'s. Thus, the product ##\pi \phi## will be a double integral over the two different ##p##'s. The integration over ##x## is used to eliminate the exponentials and to introduce delta functions involving the two different ##p##'s. The delta functions allow you to integrate over one of the ##p##'s so that you end up with a single integral over the remaining ##p##.
 
Oh ok, thanks very much :) Looks like I have it now.
 

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