When we canonically quantize the scalar field in QFT, we use a Lorentz invariant integration measure given by $$\widetilde{dk} \equiv \frac{d^3k}{(2\pi)^3 2\omega(\textbf{k})}.$$

How can I show that it is Lorentz invariant?


closed as off-topic by akhmeteli, tpg2114, Abhimanyu Pallavi Sudhir, Waffle's Crazy Peanut, Qmechanic Nov 3 '13 at 18:52

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    $\begingroup$ Hi Ome. If you haven't already done so, please take a minute to read the definition of when to use the homework tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Nov 3 '13 at 13:37
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    $\begingroup$ I'd say that this question very much falls under the category homework. $\endgroup$ – Danu Nov 3 '13 at 14:30
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    $\begingroup$ Related: physics.stackexchange.com/q/53534 $\endgroup$ – joshphysics Nov 3 '13 at 18:23
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    $\begingroup$ This is certainly homework. Don't feel demeaned by your question being tagged as such, it really means that your question is just about problem-solving. You can improve your question, by the way, by showing your effort, and asking a conceptual question. Your question may then be reopened. If you do that, just ping me here, using "@DIMension10", and I'd vote to reopen. I'm not the downvoter, by the way. $\endgroup$ – Abhimanyu Pallavi Sudhir Nov 5 '13 at 14:48
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    $\begingroup$ +1 for this question and flagging the moderator, for the following reason: "This question belongs to the important conceptual (as well as technical indeed) questions and obstacles that usually hinders the understanding of QFT. It certainly is useful to a broad public!" $\endgroup$ – Noix07 Oct 12 '18 at 8:21

To show that this measure is Lorentz invariant you first need to explicitly write your integral as an integral over mass shell in 4D k-space. This could be done by inserting Dirac delta function $\delta[k^\mu k_\mu-m^2]$ and integrating over the whole 4D space.

Then you could apply the following transformations: \begin{align} \theta(k_0)\cdot\delta[k^\mu k_\mu-m^2] &= \theta(k_0)\cdot\delta[k_0^2-|\mathbf{k}|^2-m^2]\\ &=\theta(k_0)\cdot\delta\left[(k_0-\sqrt{|\mathbf{k}|^2+m^2})(k_0+\sqrt{|\mathbf{k}|^2+m^2})\right]\\ &=\frac{\delta\left[k_0-\sqrt{|\mathbf{k}|^2+m^2}\right]}{2\,k_0}, \end{align} where Heaviside function $\theta(k_0)$ is used to select only future part of the mass shell.

  • $\begingroup$ You nean $\theta(k_0) \delta(k^2 - m^2)$ since $\omega_{\vec{k}} = \sqrt{ |\vec{k}|^2 + m^2 } $. $\endgroup$ – nervxxx Nov 3 '13 at 15:59
  • $\begingroup$ @nervxxx: Thanks! I mistakenly thought we were talking about massless (or EM) field. Edited accordingly. $\endgroup$ – user23660 Nov 3 '13 at 16:23

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