Introduction and Notation

Let $\phi(\vec{x})$ be the real Klein-Gordon (quantum) field, written as:

$$\phi(\vec{x})=\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}} $$

where $\omega_{p}=\sqrt{|\vec{p}|^2+m^2}$, $a_{\vec{p}}$, $a^{\dagger}_{-\vec{p}}$ the annihilation and creation operators, and let $\pi(\vec{x})$ the momentum density conjugate to $\phi(\vec{x})$, given by

$$\pi(\vec{x})=\int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}} $$

The question

The only non trivial equal-time commutator is

$$[\phi(\vec{x}),\pi(\vec{y})]=i\hbar\delta^{(3)}(\vec{x}-\vec{y}) $$

As the relation between $\phi(\vec{x})$ and the $a,a^{\dagger}$ is linear, and so is between $\pi(\vec{x})$ and them, I'm going to express the commutator obeyed by $a$ and $a^{\dagger}$. I'm failing to derive the inverse Fourier transform

$$ \int d^3x'\phi(\vec{x})e^{i\vec{p}'\cdot\vec{x}'}=\int\int d^3x'\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}e^{i\vec{p}'\cdot\vec{x}'}=\\=\int d^3x'e^{i\vec{p}'\cdot\vec{x}'} \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}=\\=(2\pi)^3\delta^{3}(\vec{p}')\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}$$

What I want is a Dirac delta $\delta^{3}(\vec{p}-\vec{p}')$, so where is my procedure failing? I know this is a math question, but given the physical context, it may fit here better.

Thanks for your time, any hint will be appreciated


EDIT 2 EDIT 1 Now as an answer

  • 3
    $\begingroup$ The argument of $\phi$ needs to be the integration variable, i.e. $x'$. $\endgroup$
    – Michael
    Commented Mar 16, 2013 at 13:24
  • $\begingroup$ Yep in the line right after "the inverse Fourier transform" $\endgroup$
    – twistor59
    Commented Mar 16, 2013 at 13:42
  • $\begingroup$ Ok so it goes as: multiply by $e^{i(p'x)}$ integrate with respect to $x$, and integrate with respect to $p$ with the help of $\delta(p-p')$? I'm going to edit the question to see if I did understood it right $\endgroup$
    – J L
    Commented Mar 16, 2013 at 13:42
  • 1
    $\begingroup$ Yes (assuming you're using $x$ rather than $x'$ as in the current version of the question) $\endgroup$
    – twistor59
    Commented Mar 16, 2013 at 13:48
  • 1
    $\begingroup$ Sorry I didn't see the prime! So your computation looks correct. w.r.to your final question: nothing else depends on x, so as far as the x integration is concerned, you treat everything else (inc. the p's) as constant, and you get the delta function. You could edit your question to its original form and post this computation as an answer. It's fine to post answers to your own questions. $\endgroup$
    – twistor59
    Commented Mar 16, 2013 at 17:00

1 Answer 1


For the field:

$$ \int d^3x\phi(\vec{x})e^{-i\vec{p}'\cdot\vec{x}}=\int\int d^3x\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}e^{-i\vec{p}'\cdot\vec{x}}=\\=\int d^3x e^{-i(\vec{p}'-{p})\cdot\vec{x}}\int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)=\\=\int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{p}}}\left(a_{\vec{p}}+a^{\dagger}_{-\vec{p}} \right)(2\pi)^3\delta(\vec{p}'-\vec{p})=\frac{1}{\sqrt{2\omega_{p}'}}\left(a_{\vec{p}'}+a^{\dagger}_{-\vec{p}'} \right)\equiv\tilde{\phi}(\vec{p}')$$

For the momentum:

$$ \int d^3x\pi(\vec{x})e^{-i\vec{p}'\cdot\vec{x}}=\int d^3x\int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)e^{i\vec{p}\cdot\vec{x}}e^{-i\vec{p}'\cdot\vec{x}}=\\=\int d^3x e^{-i(\vec{p}'-{p})\cdot\vec{x}}\int\frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)=\\=\int \frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{p}}{2}}\left(a_{\vec{p}}-a^{\dagger}_{-\vec{p}} \right)(2\pi)^3\delta(\vec{p}'-\vec{p})=(-i)\sqrt{\frac{\omega_{p'}}{2}}\left(a_{\vec{p}'}-a^{\dagger}_{-\vec{p}'} \right)\equiv\tilde{\pi}(\vec{p}')$$




Now we can use the fact that the CCR for the momentum and the field is almost the same as the CCR for the Fourier transforms:

$$[\phi(\vec{x}),\pi(\vec{y})]-i \delta^{(3)}(\vec{x}-\vec{y})=0$$

Multiplying by $e^{-i(\vec{p}\cdot\vec{x})}$ and $e^{-i(\vec{p}'\cdot\vec{y})}$ integrating with respect to $\vec{x}$ and $\vec{y}$

$$ \int d^3xd^3y\left( e^{-i(\vec{p}\cdot\vec{x})}\phi(\vec{x})e^{-i(\vec{p}'\cdot\vec{y})}\pi(\vec{y})-\pi(\vec{y})e^{-i(-\vec{p}'\cdot\vec{y})}\phi(\vec{x})e^{-i(\vec{p}\cdot\vec{x})}-i\delta^{(3)}(\vec{x}-\vec{y})\right)=0$$

We get the commutator in terms of the fields and momentum in the momentum basis:

$$[\tilde{\phi}(\vec{p}),\tilde{\pi}(\vec{p}')]=(2\pi)^3i\delta^{(3)}(\vec{p}-\vec{p}') $$

so that

$$\left[a_{\vec{p}},a^{\dagger}_{-\vec{p}'}\right]=\left[\frac{1}{2}\left(\sqrt{2w_{p}}\tilde{\phi}(\vec{p})+i\sqrt{\frac{2}{w_{p}}}\tilde{\pi}(\vec{p})\right),\frac{1}{2}\left(\sqrt{2w_{p'}}\tilde{\phi}(-\vec{p}')-i\sqrt{\frac{2}{w_{p'}}}\tilde{\pi}(-\vec{p}')\right)\right]=\frac{1}{2}\left(-i[\tilde{\phi}(\vec{p}),\tilde{\pi}(\vec{p}')]+i[\tilde{\pi}(-\vec{p}),\tilde{\phi}(-\vec{p}')] \right)=(2\pi)^3\delta({\vec{p}+\vec{p}'})$$

Final Remarks

1)I think there are some $\hbar$ missing and 2) I'm not sure about the symmetry between $\vec{p}$ and $-\vec{p}$ for the real KG field


I set $\hbar=1$, but the commutator still looks a little ugly with that $+$ sign


Given that the field is real may I say $\tilde{\phi}(\vec{p})=\tilde{\phi}^*(-\vec{p})=\tilde{\phi}(-\vec{p})$

  • 1
    $\begingroup$ In the very first integral, where you're transforming $\phi(x)$ to momentum space I think you should be integrating against $e^{-ip'x}$ (i.e. minus sign) and ditto when you're transforming $\pi(x)$ to momentum space $\endgroup$
    – twistor59
    Commented Mar 16, 2013 at 18:33
  • $\begingroup$ I was following Peskin's convention $\int d^4x e^{ikx}=(2\pi)^4\delta^{4}(k)$ $\endgroup$
    – J L
    Commented Mar 16, 2013 at 18:35
  • 2
    $\begingroup$ If you look at P&S p xxi, they note that in three-dimensional FTs, the signs in the exponents are reversed from the ones in their 4d equations, and 3d is what you're doing. $\endgroup$
    – twistor59
    Commented Mar 16, 2013 at 19:04
  • $\begingroup$ Thank you, sorry for that silly error. Tomorrow I will update my answer with that and the proof of the CCR in the momentum basis. $\endgroup$
    – J L
    Commented Mar 16, 2013 at 19:30

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