special relativity - Lorentz transformation of classical Klein–Gordon field

I'm trying to see that the invariance of the Klein–Gordon field implies that the Fourier coefficients $a(\mathbf{k})$ transform like scalars: $a'(\Lambda\mathbf{k})=a(\mathbf{k}).$

Starting from the mode expansion of the field

$$\phi'(x)=\phi(\Lambda^{-1}x)= \int \frac{d^{3}k}{(2\pi)^{3}2E_{k}} \left( e^{-ik\cdot \Lambda^{-1}x}a(\mathbf{k}) +e^{ik\cdot \Lambda^{-1}x}b^{*}(\mathbf{k}) \right),$$

it's easy to see that it equals

$$\int \frac{d^{3}k}{(2\pi)^{3}2E_{k}} \left( e^{-i(\Lambda k)\cdot x}a(\mathbf{k}) +e^{i(\Lambda k)\cdot x}b^{*}(\mathbf{k}) \right).$$

when using the property $\Lambda^{-1}=\eta\Lambda^{T}\eta$. Then doing a change of variable $\tilde{k}=\Lambda k$ and considering orthochronous transformations so that the Jacobian is 1, then I get the wanted result ($a'(\Lambda\mathbf{k})=a(\mathbf{k})$) when comparing with the original mode expansion. However, this is not quite right as I would have to justify that $E_\tilde{k}=E_k$ but I can't see how.

Answer

The important insight is that it's actually the whole combination $$\frac{d^3 k}{2(2\pi)^3 E_\mathbf k}, \qquad E_\mathbf{k} = \sqrt{\mathbf k^2 + m^2}$$ that forms a Lorentz-invariant measure. To see this, note that if we define $k= (k^0, \mathbf k)$ and use the identity $$\delta(f(x)) = \sum_{\{x_i:f(x_i) = 0\}} \frac{\delta(x-x_i)}{|f'(x_i)|}$$ then we get $$\delta(k^2 - m^2)=\frac{\delta(k^0 - \sqrt{\mathbf k^2+m^2})}{2\sqrt{\mathbf k^2+m^2}} + \frac{\delta(k^0 + \sqrt{\mathbf k^2+m^2})}{2\sqrt{\mathbf k^2+m^2}}$$ so the original measure can be rewritten as $$\frac{d^3 k}{2(2\pi)^3 E_\mathbf k}=\frac{d^3k\,d k^0}{2(2\pi)^3 k^0}\delta(k^0 - \sqrt{\mathbf k^2+m^2}) = \frac{d^4k}{(2\pi)^3}\delta(k^2-m^2)\theta(k^0)$$ which is manifestly Lorentz invariant for proper, orthochronous Lorentz transformations. The rest of your manipulations go through unscathed, and you get the result you want!

Hope that helps!

Cheers!