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.
 A: 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!
A: 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 $$f(x)\delta(x-a) = f(a)\delta(x-a)$$
Hence your derivation is not clear to me,
