Total Momentum Operator for Free Scalar Field

  • Thread starter nicksauce
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  • #1
nicksauce
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Homework Statement


I want to show that
[tex]\mathbf{P} = -\int d^{3}x}\pi(x)\nabla\phi(x) = \int{\frac{d^{3}p}{(2\pi)^3}\mathbf{p}a_{p}^{\dagger}a_p[/tex]

for the KG field.


Homework Equations


[tex]\phi(x) = \int{\frac{d^{3}p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}(a_p + a_{-p}^{\dagger})e^{ipx}[/tex]

[tex]\pi(x) = -i\int{\frac{d^{3}p}{(2\pi)^3}\sqrt{\frac{\omega_k}{2}}(a_p - a_{-p}^{\dagger})e^{ipx}[/tex]


The Attempt at a Solution


I'm having trouble seeing why this is true. What happens to the [itex]a_pa_k[/tex] and [itex]a_p^{\dagger}a_k^{\dagger}[/itex]-like cross terms?
 

Answers and Replies

  • #2
gabbagabbahey
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I'd write [itex]\pi(x)[/itex] and [itex]\phi(x)[/itex] as

[tex]\pi(\textbf{x}) = -i\int{\frac{d^{3}\textbf{p}'}{(2\pi)^3}\sqrt{\frac{\omega_{p'}} {2}}(a_{\textbf{p}'} - a_{-\textbf{p}'}^{\dagger})e^{i\textbf{p}'\cdot\textbf{x}}[/tex]

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

Then just calculate [itex]\mathbf{\nabla}\phi[/itex], substitute everything in and do the integration over [itex]\textbf{x}[/itex] first. You should get something with a delta function like [itex]\delta^3(\textbf{p}+\textbf{p}')[/itex], which allows you to get rid of all the [tex]a_{\textbf{p}}a^{\dagger}_{-\textbf{p}'}[/itex]-like cross terms just by integrating over [itex]\textbf{p}'[/itex].
 

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