# 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.

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!
In second step where you have put. $$\frac{d^3k}{2(2\pi)^3 E_k} = \frac{d^3k dk^0}{2(2\pi)^3 k^0}\delta(k^0-\sqrt{\mathbf{k}^2+m^2})$$ according to delta function property $$f(x)\delta(x-a) = f(a)\delta(x-a)$$