# Lorentz Invariant Integration Measure [closed]

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?

• 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. – Qmechanic Nov 3 '13 at 13:37
• – joshphysics Nov 3 '13 at 18:23
• 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. – Abhimanyu Pallavi Sudhir Nov 5 '13 at 14:48
• +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!" – Noix07 Oct 12 '18 at 8:21
• Let me now clarify why this is of great conceptual importance and subtlety: it seems at first that one integrates on $\mathbb{R}^3$ the usual vector spaces. This does make sense but first, it is not a stable subspace of $\mathbb{R}^4$ under boosts and secondly the measure would not be invariant. (I suggest that the people who closed the question explicitly write down the image measure). The crucial point is that one integrates on a Lorentz invariant submanifold (the set of 4-vectors $k^{\mu}$ such that $k^{\mu} k_{\mu}=m^2$). So now we now see that we are dealing with integration of... – Noix07 Oct 12 '18 at 8:35

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.
• You nean $\theta(k_0) \delta(k^2 - m^2)$ since $\omega_{\vec{k}} = \sqrt{ |\vec{k}|^2 + m^2 }$. – nervxxx Nov 3 '13 at 15:59