quantum field theory - Is Lorentz invariant differential measure arbitrary?

In Srednicki, we chose a function $f(\mathbf k)$ to make $d^3\mathbf k/f(\mathbf k)$ Lorentz invariant. The way to do this is to first start from a 4 dimensional measure and multiply it by a Dirac delta with a Lorentz invariant parameter. The one the book chose is: $$ d^4k\delta(k^2+m^2)\theta(k^0), $$ where $\theta(x)$ is the unit step function to pick the positive energy. It argues that if we integrate it with $k^0$: $$ \int_{-\infty}^{+\infty}dk^0d^3k\delta(k^2+m^2)\theta(k^0)=\frac{d^3k}{\frac{\partial(k^2+m^2)}{\partial k^0}\Big\vert_{k^2+m^2=0}}=\frac{d^3k}{2\sqrt{m^2+\mathbf k^2}} \equiv (2\pi)^3\widetilde{dk}. $$

Now, if we replace $m^2$ with an arbitrary number, $\delta(k^2+m^2)$ will still be Lorentz invariant. Does this mean $\widetilde{dk}$ can choose an arbitrary $m$ and still be Lorentz invariant? Why do we choose $m^2$ here (apart from the requirement from dimensional analysis)?

Answer

Yes: any scalar $A$ makes $\delta(k^2+A)$ covariant. A different choice for $A$ changes the form of $\widetilde{\mathrm d k}$, which in turns changes the form of the creation/annihilation operators $a,a^\dagger$, the normalisation of 1-particle states and the form of the canonical commutation relations. We use $m^2$ because any other choice would make most expressions messy, and because we want momentum to be on-shell (that is, we want $k^2+m^2=0$, which is obviously enforced by the delta).

The truth is, that particular choice for $f(\boldsymbol k)$ is the natural one, because if you solve the KG equation in Fourier space for the four components of $k$ (instead of just $\boldsymbol k$) you'll see that this is the measure that naturally arises. The argument Srednicki gives for $f(\boldsymbol k)$ is a bit artificial IMHO: one could derive the same form for $\widetilde{\mathrm dk}$ in a more natural way, one that is manifestly covariant. See this post of mine for a different approach, where I derive the invariant measure from another perspective. Note that in that post I use a different convention for the metric ($k^2=m^2$ instead of $k^2=-m^2$) and I omit the factors $(2\pi)^3$.