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I can't find this in similar questions, and I must be missing something very basilar since I can't find this in any textbook or online note: they just skip the passage.

So, from my course's notes, we have for example a complex scalar field:

$$ \phi(x) = \int \dfrac{d^3p}{(2\pi)^3} \dfrac{\sqrt{V}}{\sqrt{2E( \mathbf p)}} \left( a_{(+)} (\mathbf p ) e^{-ipx} + a_{(-)}^{\dagger } (\mathbf p ) e^{ipx} \right) $$

$$ \phi^{*}(x) = \int \dfrac{d^3p}{(2\pi)^3} \dfrac{\sqrt{V}}{\sqrt{2E( \mathbf p)}} \left( a_{(+)}^{\dagger } (\mathbf p ) e^{ipx} + a_{(-)} (\mathbf p ) e^{-ipx} \right) $$

and from the free $ S_0 = \int d^4x \left( \partial_\mu \phi^*(x) \partial^\mu \phi(x) - m^2 \phi^*(x) \phi(x) \right) $ with Noether's theorem for U(1) we get

$$ J^\mu(x) = i \left(\phi^*(x) \partial^\mu\phi(x) - \partial^\mu \phi^* (x) \phi(x) \right) $$

$$ Q = \int d^3x J^0(x) $$

QUESTION

So, how do I go from

$$ Q = i \int d^3x \dfrac{d^3p \ d^3q}{(2\pi)^6} \dfrac{V}{2\sqrt{E( \mathbf p)E( \mathbf q)}} \cdot \\ \cdot \left[ \left( a_{(+)}^{\dagger } (\mathbf p ) e^{ipx} + a_{(-)} (\mathbf p ) e^{-ipx} \right) \ iE(\mathbf q) \left( - a_{(+)} (\mathbf q ) e^{-iqx} + a_{(-)}^{\dagger } (\mathbf q ) e^{iqx} \right) + \\ - iE(\mathbf p) \left( a_{(+)}^{\dagger } (\mathbf p ) e^{ipx} - a_{(-)} (\mathbf p ) e^{-ipx} \right) \left( a_{(+)} (\mathbf q ) e^{-iqx} + a_{(-)}^{\dagger } (\mathbf q ) e^{iqx} \right) \right] $$

to?

$$ Q = \int d^3p \dfrac{V}{(2\pi)^3} \left( a_{(+)}^{\dagger } (\mathbf p ) a_{(+)}(\mathbf p ) - a_{(-)}^{\dagger } (\mathbf p ) a_{(-)}(\mathbf p ) \right) $$

At least, what mathematical formulas do I have to use?

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1 Answer 1

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It is pretty simple just use the following formula,

$$ \int d^3 x e^{i(p+q)x} = (2\pi)^3\delta(p+q)$$

and thus on integrating $d^3 q$ you will have $\sqrt{2E(p)2E(q)} = E(p)$ in the downstairs, and then it's pretty straightforward.

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  • $\begingroup$ Right, I was just dumb not searching properly for such formula... $\endgroup$ Commented Feb 15, 2015 at 19:31

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