Hamiltonian of the quantised Klein-Gordon theory

  • #1
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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?
 

Answers and Replies

  • #2
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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?
 
  • #3
Demystifier
Science Advisor
Insights Author
Gold Member
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It's correct.

You should be more self-confident about your mathematical skills. :smile:
 
Last edited:
  • #4
1,344
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It's correct.

Thanks!

You should be more self-confident about your mathematical skills. :smile:

I am still very new to the kind of calculations which form the bread and butter of QFT and GR, and so I write such prosaic and detailed answers just to confirm with others that I don't have the occasional flawed conceptual understanding of how to do the calculations.
 

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