- #1

hellfire

Science Advisor

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## Main Question or Discussion Point

I am not familiar with calculations such as the following one, and I want to be sure I do the right steps before going on with Peskin & Schröder. I want to derive step by step the Hamiltonian of the real scalar field. I have no problem to arrive at the first part of (2.31) in page 21:

[tex]H = \int dx^3 \int \frac {d^3p d^3p^{\prime}}{(2 \pi)^6} e^{i (p + p \prime) x} \left( - \frac{\sqrt{\omega_p \omega_{p \prime}}}{4} (a_{p} - a^{\dagger}_{-p}) (a_{p \prime} - a^{\dagger}_{-p \prime}) + \frac{-p p^{\prime} + m^2}{4 \sqrt{\omega_p \omega_{p \prime} }} (a_{p} + a^{\dagger}_{-p}) (a_{p \prime} + a^{\dagger}_{-p \prime}) \right) [/tex]

This shall be equal to:

[tex]H = \int \frac{d^3p}{(2 \pi)^3} \omega_p \left(a^{\dagger}_{p} a_p + \frac{1}{2}[a_p, a^{\dagger}_{p}] \right)[/tex]

So, I figured out I can make use of:

[tex]\int dx^3 e^{i (p + p \prime) x} = (2 \pi)^3 \delta^{(3)} (p + p^{\prime})[/tex]

and

[tex] \int \delta(p + p^{\prime}) F(p^{\prime}) dp^{\prime} = F(- p) [/tex]

and I can set [itex]p\prime = - p[/itex] and [itex]\omega_p = \omega_p^{\prime}[/itex] considering that [itex]p^2 + m^2 = \omega^2_p[/itex], remove dp' and I arrive at:

[tex]H = \int \frac {d^3p}{(2 \pi)^3} \frac{\omega_p}{4}} \left(- a_p a_{-p} + a_p a^{\dagger}_{p} + a^{\dagger}_{-p} a_{-p} - a^{\dagger}_{-p} a^{\dagger}_{p} + a_p a_{-p} + a_p a^{\dagger}_{p} + a^{\dagger}_{-p} a_{-p} + a^{\dagger}_{-p} a^{\dagger}_{p} \right)[/tex]

which simplifies to

[tex]H = \int \frac {d^3p}{(2 \pi)^3} \frac{\omega_p}{2}} \left(a_p a^{\dagger}_{p} + a^{\dagger}_{-p} a_{-p} \right)[/tex]

This seams to lead to the required result if [itex]a_p = a_{-p}[/itex], but this is not the case, isn't it? So, I may have made some error. Comments are appreciated... (I hope I made no typo...)

[tex]H = \int dx^3 \int \frac {d^3p d^3p^{\prime}}{(2 \pi)^6} e^{i (p + p \prime) x} \left( - \frac{\sqrt{\omega_p \omega_{p \prime}}}{4} (a_{p} - a^{\dagger}_{-p}) (a_{p \prime} - a^{\dagger}_{-p \prime}) + \frac{-p p^{\prime} + m^2}{4 \sqrt{\omega_p \omega_{p \prime} }} (a_{p} + a^{\dagger}_{-p}) (a_{p \prime} + a^{\dagger}_{-p \prime}) \right) [/tex]

This shall be equal to:

[tex]H = \int \frac{d^3p}{(2 \pi)^3} \omega_p \left(a^{\dagger}_{p} a_p + \frac{1}{2}[a_p, a^{\dagger}_{p}] \right)[/tex]

So, I figured out I can make use of:

[tex]\int dx^3 e^{i (p + p \prime) x} = (2 \pi)^3 \delta^{(3)} (p + p^{\prime})[/tex]

and

[tex] \int \delta(p + p^{\prime}) F(p^{\prime}) dp^{\prime} = F(- p) [/tex]

and I can set [itex]p\prime = - p[/itex] and [itex]\omega_p = \omega_p^{\prime}[/itex] considering that [itex]p^2 + m^2 = \omega^2_p[/itex], remove dp' and I arrive at:

[tex]H = \int \frac {d^3p}{(2 \pi)^3} \frac{\omega_p}{4}} \left(- a_p a_{-p} + a_p a^{\dagger}_{p} + a^{\dagger}_{-p} a_{-p} - a^{\dagger}_{-p} a^{\dagger}_{p} + a_p a_{-p} + a_p a^{\dagger}_{p} + a^{\dagger}_{-p} a_{-p} + a^{\dagger}_{-p} a^{\dagger}_{p} \right)[/tex]

which simplifies to

[tex]H = \int \frac {d^3p}{(2 \pi)^3} \frac{\omega_p}{2}} \left(a_p a^{\dagger}_{p} + a^{\dagger}_{-p} a_{-p} \right)[/tex]

This seams to lead to the required result if [itex]a_p = a_{-p}[/itex], but this is not the case, isn't it? So, I may have made some error. Comments are appreciated... (I hope I made no typo...)

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