# Hamiltonian of the quantised Klein-Gordon theory

spaghetti3451
The Klein-Gordon field ##\phi(\vec{x})## and its conjugate momentum ##\pi(\vec{x})## is given, in the Schrodinger picture, by

##\phi(\vec{x})=\int \frac{d^{3}p}{(2\pi)^{3}} \frac{1}{\sqrt{2\omega_{\vec{p}}}}[a_{\vec{p}}e^{i\vec{p}\cdot{\vec{x}}}+a_{\vec{p}}^{\dagger}e^{-i\vec{p}\cdot{\vec{x}}}]##
##\pi(\vec{x})=\int \frac{d^{3}p}{(2\pi)^{3}} (-i)\sqrt{\frac{\omega_{\vec{p}}}{2}}[a_{\vec{p}}e^{i\vec{p}\cdot{\vec{x}}}-a_{\vec{p}}^{\dagger}e^{-i\vec{p}\cdot{\vec{x}}}]##

I would like to show that the Hamiltonian ##H## of the Klein-Gordon theory is given by

##H = \int \frac{d^{3}p}{(2\pi)^{3}}\omega_{\vec{p}}[a_{\vec{p}}^{\dagger}a_{\vec{p}}+\frac{1}{2}(2\pi)^{3}\delta^{(3)}(0)]##.

Here's my attempt:

##H=\frac{1}{2}\int d^{3}x [\pi^{2}+(\nabla\phi)^{2}+m^{2}\phi^{2}]##

##=\frac{1}{2}\int \frac{d^{3}x\ d^{3}p\ d^{3}q}{(2\pi)^{6}}\Big[ -\frac{\sqrt{\omega_{\vec{p}}\omega_{\vec{q}}}}{2} \Big( a_{\vec{p}}e^{i\vec{p}\cdot{\vec{x}}}-a_{\vec{p}}^{\dagger}e^{-i\vec{p}\cdot{\vec{x}}} \Big) \Big( a_{\vec{q}}e^{i\vec{q}\cdot{\vec{x}}}-a_{\vec{q}}^{\dagger}e^{-i\vec{q}\cdot{\vec{x}}} \Big)##
##+ \frac{1}{2\sqrt{\omega_{\vec{p}}\omega_{\vec{q}}}} \Big( i\vec{p}a_{\vec{p}}e^{i\vec{p}\cdot{\vec{x}}}-i\vec{p}a_{\vec{p}}^{\dagger}e^{-i\vec{p}\cdot{\vec{x}}} \Big)\cdot{\Big( i\vec{q}a_{\vec{q}}e^{i\vec{q}\cdot{\vec{x}}}-i\vec{q}a_{\vec{q}}^{\dagger}e^{-i\vec{q}\cdot{\vec{x}}} \Big)}+ \frac{m^{2}}{2\sqrt{\omega_{\vec{p}}\omega_{\vec{q}}}} \Big( a_{\vec{p}}e^{i\vec{p}\cdot{\vec{x}}}+a_{\vec{p}}^{\dagger}e^{-i\vec{p}\cdot{\vec{x}}} \Big)\Big(a_{\vec{q}}e^{i\vec{q}\cdot{\vec{x}}}+a_{\vec{q}}^{\dagger}e^{-i\vec{q}\cdot{\vec{x}}} \Big)\Big]##

##=\frac{1}{4}\int \frac{d^{3}p\ d^{3}q}{(2\pi)^{3}}\Big[-\sqrt{\omega_{\vec{p}}\omega_{\vec{q}}}\Big(a_{\vec{p}}a_{\vec{q}}\delta(\vec{p}+\vec{q})-a_{\vec{p}}^{\dagger}a_{\vec{q}}\delta(-\vec{p}+\vec{q})-a_{\vec{p}}a_{\vec{q}}^{\dagger}\delta(\vec{p}-\vec{q})+a_{\vec{p}}^{\dagger}a_{\vec{q}}^{\dagger}\delta(-\vec{p}-\vec{q})\Big)+\frac{1}{\sqrt{\omega_{\vec{p}}\omega_{\vec{q}}}}\Big(-\vec{p}\cdot{\vec{q}}a_{\vec{p}}a_{\vec{q}}\delta(\vec{p}+\vec{q})+\vec{p}\cdot{\vec{q}}a_{\vec{p}}^{\dagger}a_{\vec{q}}\delta(-\vec{p}+\vec{q})+\vec{p}\cdot{\vec{q}}a_{\vec{p}}a_{\vec{q}}^{\dagger}\delta(\vec{p}-\vec{q})-\vec{p}\cdot{\vec{q}}a_{\vec{p}}^{\dagger}a_{\vec{q}}^{\dagger}\delta(-\vec{p}-\vec{q})\Big)+\frac{m^{2}}{\sqrt{\omega_{\vec{p}}\omega_{\vec{q}}}}\Big(a_{\vec{p}}a_{\vec{q}}\delta(\vec{p}+\vec{q})+a_{\vec{p}}^{\dagger}a_{\vec{q}}\delta(-\vec{p}+\vec{q})+a_{\vec{p}}a_{\vec{q}}^{\dagger}\delta(\vec{p}-\vec{q})+a_{\vec{p}}^{\dagger}a_{\vec{q}}^{\dagger}\delta(-\vec{p}-\vec{q})\Big)\Big]##

##=\frac{1}{4}\int \frac{d^{3}p}{(2\pi)^{3}}\Big[- \omega_{\vec{p}} a_{\vec{p}} a_{-\vec{p}} +
\omega_{\vec{p}} a_{\vec{p}}^{\dagger} a_{\vec{p}} +
\omega_{\vec{p}} a_{\vec{p}} a_{\vec{p}}^{\dagger} - \omega_{\vec{p}} a_{\vec{p}}^{\dagger}
a_{-\vec{p}}^{\dagger} + \frac{1}{\omega_{\vec{p}}} \vec{p}^{2} a_{\vec{p}} a_{-\vec{p}} + \frac{1}{\omega_{\vec{p}}} \vec{p}^{2} a_{\vec{p}}^{\dagger} a_{\vec{p}} + \frac{1}{\omega_{\vec{p}}}
\vec{p}^{2} a_{\vec{p}} a_{\vec{p}}^{\dagger} + \frac{1}{\omega_{\vec{p}}} \vec{p}^{2}
a_{\vec{p}}^{\dagger} a_{-\vec{p}}^{\dagger} + \frac{m^{2}}{\omega_{\vec{p}}} a_{\vec{p}} a_{-\vec{p}} +
\frac{m^{2}}{\omega_{\vec{p}}} a_{\vec{p}}^{\dagger} a_{\vec{p}} + \frac{m^{2}}{\omega_{\vec{p}}} a_{\vec{p}} a_{\vec{p}}^{\dagger} + \frac{m^{2}}{\omega_{\vec{p}}} a_{\vec{p}}^{\dagger} a_{-\vec{p}}^{\dagger}\Big]##

##=\frac{1}{4}\int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\omega_{\vec{p}}}\Big[(-\omega_{\vec{p}}^{2}+\vec{p}^{2}+m^{2})a_{\vec{p}}a_{-\vec{p}}+(-\omega_{\vec{p}}^{2}+\vec{p}^{2}+m^{2})a_{\vec{p}}^{\dagger}a_{-\vec{p}}^{\dagger}+(\omega_{\vec{p}}^{2}+\vec{p}^{2}+m^{2})a_{\vec{p}}a_{\vec{p}}^{\dagger}+(\omega_{\vec{p}}^{2}+\vec{p}^{2}+m^{2})a_{\vec{p}}^{\dagger}a_{\vec{p}}\Big]##

##=\frac{1}{4}\int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\omega_{\vec{p}}}\Big[(-\omega_{\vec{p}}^{2}+\vec{p}^{2}+m^{2})(a_{\vec{p}}a_{-\vec{p}}+a_{\vec{p}}^{\dagger}a_{-\vec{p}}^{\dagger})+(\omega_{\vec{p}}^{2}+\vec{p}^{2}+m^{2})(a_{\vec{p}}a_{\vec{p}}^{\dagger}+a_{\vec{p}}^{\dagger}a_{\vec{p}})\Big]##

##=\frac{1}{2} \int \frac{d^{3}p}{(2\pi)^{3}}\omega_{\vec{p}}[a_{\vec{p}}a_{\vec{p}}^{\dagger}+a_{\vec{p}}^{\dagger}a_{\vec{p}}]##, where we used ##\omega_{\vec{p}}^{2}=\vec{p}^{2}+m^{2}## to eliminate the first term and simplify the second term

##=\frac{1}{2} \int \frac{d^{3}p}{(2\pi)^{3}}\omega_{\vec{p}}[[a_{\vec{p}},a_{\vec{p}}^{\dagger}]+a_{\vec{p}}^{\dagger}a_{\vec{p}}+a_{\vec{p}}^{\dagger}a_{\vec{p}}]##

##=\int \frac{d^{3}p}{(2\pi)^{3}}\omega_{\vec{p}}[a_{\vec{p}}^{\dagger}a_{\vec{p}}+\frac{1}{2}[a_{\vec{p}},a_{\vec{p}}^{\dagger}]]##

##=\int \frac{d^{3}p}{(2\pi)^{3}}\omega_{\vec{p}}[a_{\vec{p}}^{\dagger}a_{\vec{p}}+\frac{1}{2}(2\pi)^{3}\delta^{(3)}(0)]##

Is my working correct?

DuckAmuck

Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?

Gold Member
It's correct.