Fourier expansions of Klein-Gordon field not Lorentz invariant?

Question

I’m working in Peskin and Schroeders book on QFT and noticed that they expanded a solution to the Klein Gordon equation in a manner that seems to me not to be be Lorentz invariant even though the solutions to the KGE should be scalars.

The Hamiltonian, being a function of $\phi$ and $\pi$, also becomes an operator. Our next task is to find the spectrum from the Hamiltonian. Since there is no obvious way to do this, let us seek guidance by writing the Klein-Gordan equation in Fourier space. If we expand the classical Klein-Gordan equation field as $$ \phi(\mathbf{x},t) = \int \frac{d^3p}{(2\pi)^3} e^{i \mathbf{p} \cdot \mathbf{x}} \phi(\mathbf{p},t) $$ (with $\phi^*(\mathbf{p}) = \phi(-\mathbf{p})$ so that $\phi(\mathbf{x})$ is real) ...

Am I correct that this combination is not Lorentz invariant? Or is there a reason that this does not matter at this stage?

Answer 1

A bit later they address the apparent issue with integrals over 3-momentum, which does not at first glance look Lorentz-invariant.

[I]n fact, the integral $$ \int \frac{d^3p}{(2\pi)^3} \frac{1}{2E_{\mathbf{p}}} = \left. \int \frac{d^4p}{(2\pi)^4} (2\pi) \delta(p^2 - m^2) \right|_{p^0>0} $$ is a Lorentz-invariant 3-momentum integral, in the sense that if $f(p)$ is Lorentz-invariant, so is $\int d^3p\ f(p) / (2E_{\mathbf{p}})$.

There's further discussion beyond that about what the right-hand side means with the $p^0>0$ restriction.