In the srednicki notes he goes from(adsbygoogle = window.adsbygoogle || []).push({});

$$H = \int d^{3}x a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x) $$ to

$$H = \int d^{3}p\frac{1}{2m}P^{2}\tilde{a}^{\dagger}(p)\tilde{a}(p) $$

Where $$\tilde{a}(p) = \int \frac{d^{3}x}{(2\pi)^{\frac{3}{2}}}e^{-ipx}a(x)$$

Is this as simple as substituting or isn't there a commutator to get to

$$H = \int d^{3}p\frac{1}{2m}P^{2}\tilde{a}^{\dagger}(p)\tilde{a}(p)$$ ?

Do $$e^{ipx}$$ and $$P^{2}$$ commute? What about $$\nabla$$ and $$ a(x)$$?

This is really bugging me any help would be greatly appreciated!

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# A Canonical quantization of scalar fields

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