In the Srednicki notes (http://web.physics.ucsb.edu/~mark/ms-qft-DRAFT.pdf) page 36 he goes from

$$H = \int d^{3}x a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x) $$ to $$H = \int d^{3}p\frac{1}{2m}P^{2}\tilde{a}^{\dagger}(p)\tilde{a}(p) $$

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

I tried doing this by saying

$$H = \int d^{3}x \int \frac{d^{3}p}{(2\pi)^{3}}e^{-ipx} \tilde{a}^{\dagger}(p) \left(\frac{P^{2}}{2m}\right)e^{ipx}\tilde{a}(p) $$

But then I'm unsure how to proceed with commutators. Does $P^{2}$ commute with $e^{ipx}$? What about with $\tilde{a}(p)$?

Any help would be greatly appreciated.


Starting from $$ H = \int d^{3}x a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x) $$ and $${a}(x) = \int \frac{d^{3}p}{(2\pi)^{\frac{3}{2}}}e^{ipx}\tilde{a}(p)$$ (the second of which follows by inverting the expression above which defines the momentum space $a$ in terms of the position space $a$.

Plugging in for both operators, we have \begin{align*} H &= \int d^{3}x a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x)\\ &=\int d^{3}x \int \frac{d^{3}p'}{(2\pi)^{\frac{3}{2}}}e^{ip'x}\tilde{a}^{\dagger}(p') \left( \frac{- \nabla^{2}}{2m}\right) \int \frac{d^{3}p}{(2\pi)^{\frac{3}{2}}}e^{ipx}\tilde{a}(p)\\ &= \int d^{3}p'\int d^{3}p\tilde{a}^{\dagger}(p')\tilde{a}(p) \int \frac{d^{3}x}{(2\pi)^3} e^{ip'x}\left( \frac{- \nabla^{2}}{2m}\right)e^{ipx} \end{align*} after some rearranging. Then, taking the derivatives, this becomes \begin{align*} H &= \int d^{3}p'\int d^{3}p\tilde{a}^{\dagger}(p')\tilde{a}(p) \int \frac{d^{3}x}{(2\pi)^3} e^{ip'x}\left( \frac{p^2}{2m}\right)e^{ipx} \end{align*} which we can rearrange as \begin{align*} H &= \int d^{3}p'\int d^{3}p\frac{p^2}{2m}\tilde{a}^{\dagger}(p')\tilde{a}(p) \int \frac{d^{3}x}{(2\pi)^3}e^{ip'x}e^{ipx}. \end{align*} Recognizing the last integral as a representation of a delta function and then evaluating the integral over the primed momentum coordinates gives us our result: \begin{align*} H &= \int d^{3}p'\int d^{3}p\frac{p^2}{2m}\tilde{a}^{\dagger}(p')\tilde{a}(p) \delta^{(3)}(p-p')\\ &= \int d^{3}p\frac{p^2}{2m}\tilde{a}^{\dagger}(p)\tilde{a}(p). \end{align*}


Srednicki goes from $$ H = \int d^{3}x\ a^{\dagger}(x)\left( \frac{- \nabla^{2}}{2m}\right) a(x) $$ to $$ H = \int d^{3}p\ \frac{1}{2m}p^{2}\ \tilde{a}^{\dagger}(p)\tilde{a}(p) $$

where $p^2$ is just a number (an integration variable, not an operator). Therefore, it commutes with everything.

  • $\begingroup$ Is $e^{ipx}$ also just a number and so it also commutes with $\tilde{a}(p)$? $\endgroup$ – boson Mar 19 '16 at 20:03
  • $\begingroup$ @boson Yes. Here, everything commutes except for $a,a^\dagger$ (and $\nabla$) $\endgroup$ – AccidentalFourierTransform Mar 19 '16 at 20:04
  • $\begingroup$ But then what happens to the $\int \frac{d^{3}x}{(2\pi)^{3}}$? $\endgroup$ – boson Mar 19 '16 at 20:27
  • $\begingroup$ It's not true, though: $\nabla$ does not commute with the exponentials, and that matters: it's what brings down the two factors of $p$. $\endgroup$ – march Mar 19 '16 at 23:15
  • $\begingroup$ @march yes, of course. I thought that was self-evident and we didnt need to point it out. Read my last comment as "everything commutes modulo basic algebra" $\endgroup$ – AccidentalFourierTransform Mar 20 '16 at 10:44

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