Zee's QFT in a Nutshell pages 65-66. For a complex scalar QFT

$$ \varphi(\vec{x},t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}\left[a(\vec{k})\mathrm{e}^{-i(\omega_kt-\vec{k}\cdot\vec{x})} + b^\dagger(\vec{k})\mathrm{e}^{i(\omega_kt-\vec{k}\cdot\vec{x})}\right] \tag{17} $$

with creation/annihilation operators $a,a^\dagger,b,b^\dagger$ he defines the current $J$ and charge $Q$:

\begin{align} J_\mu & = i(\varphi^\dagger\partial_\mu\varphi - \partial_\mu\varphi^\dagger\varphi)\tag{18}\\ Q & = \int d^DxJ_0(x). \end{align}

I want to check that

\begin{align} \tag{1} Q = \int d^Dk[a^\dagger(\vec{k})a(\vec{k}) - b^\dagger(\vec{k})b(\vec{k})] \end{align}

That's what I've done so far (there are three integrals, the $dx$ one from the definition of $Q$, the $dk$ and $dl$ ones from the $\varphi$ and $\partial_\mu\varphi$ in the definition of $J$):

\begin{align} Q = & \int d^DxJ_0(x) = i\int d^Dx(\varphi^\dagger\partial_0 \varphi - \varphi\partial_0\varphi^\dagger)\\ % = \,& i\int d^Dx(\varphi^\dagger\pi - \varphi\pi^\dagger)\\ = & i(i\omega_k)\iiint d^Dxd^Dkd^Dl\frac{1}{(\sqrt{(2\pi)^D 2\omega_k})^2}\bigg[\\ & \left(a^\dagger(\vec{k})e^{ikx}+b(\vec{k})e^{-ikx}\right) \left(-a(\vec{l})e^{-ilx}+b^\dagger(\vec{l})e^{ilx}\right)-\\ & \left(a(\vec{k})e^{-ikx}+b^\dagger(\vec{k})e^{ikx}\right) \left(a^\dagger(\vec{l})e^{ilx}-b(\vec{l})e^{-ilx}\right) \bigg]\\ = & \iiint d^Dxd^Dkd^Dl\frac{1}{2(2\pi)^D}\bigg[\\ & a^\dagger(\vec{k})a(\vec{l})e^{ix(k-l)}+a(\vec{k})a^\dagger(\vec{l})e^{ix(l-k)}-b^\dagger(\vec{k})b(\vec{l})e^{ix(k-l)}-b(\vec{k})b^\dagger(\vec{l})e^{ix(l-k)} +\\ & \left(b^\dagger(\vec{k})a^\dagger(\vec{l})-a^\dagger(\vec{k})b^\dagger(\vec{l})\right)e^{ix(k+l)} + \left(b(\vec{k})a(\vec{l})-a(\vec{k})b(\vec{l})\right)e^{ix(k+l)} \bigg] \end{align}

IF I could set $k=l$, then this whole mess would clean up and I get the desired result: The last line should vanish because $[a, b]=[a^\dagger,b^\dagger]=0$. In the second-to-last line for $k=l$ we have $a^\dagger a+aa^\dagger = 2a^\dagger a + 1$ because of $[a,a^\dagger]=1$, and this would imply $(1)$.

But why can I set $k=l$? Maybe I can use $2\pi\delta(y)=\int dz e^{-iyz}$ somehow to get a $\delta(k-l)$ in the integral, but I'm too inexperienced in Fourier transforms of operators.

  • 1
    $\begingroup$ To be clear - space-time is $(D+1)$-dimensional? $\endgroup$ – Prahar Nov 15 '15 at 13:31
  • $\begingroup$ @Prahar yes, space-time is $(D+1)$-dimensional. To simplify the notation of the exponentials, I have used $xk = t\omega_k - \vec{x}\cdot\vec{k}$ with $\omega_k=\sqrt{k^2+m^2}$. $\endgroup$ – Bass Nov 15 '15 at 13:44
  • 2
    $\begingroup$ Why do you say "maybe" for your last attempt? That's exactly what you do - nothing else there depends on $x$, so you can just carry out the integral over it, and that gives you the delta function you need. $\endgroup$ – ACuriousMind Nov 15 '15 at 13:47
  • $\begingroup$ @ACuriousMind Arrrgh! Of course, I can just do the $dx$ integral first, then there are no operators in the integral! I desperately wanted to get rid of the $dl$ integral so I didn't care about the $dx$ one. ::sigh:: Thanks a lot! $\endgroup$ – Bass Nov 15 '15 at 13:56
  • $\begingroup$ @ACuriousMind what about the last line? There, I get a $\delta(k+l)$, so $l=-k$. If the c/a operators were symmetric in their argument, then the last line would vanish. $\endgroup$ – Bass Nov 15 '15 at 14:08

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