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!

Let's understand this statement in Hamiltonian formalism, where KG equation is equivalent to having the free scalar field hamiltonian and the Heisenberg equations of motion for the free fields.

Then $\phi(\vec{x},t) = \int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left( a_p e^{ipx} + a^\dagger_p e^{-ipx}\right)$ and the canonical conjugate $\pi(\vec{x},t) = \dot{\phi}(\vec{x},t)$, are the most general solution.

Now let's consider an interacting Hamiltonian $H = H_0 + \lambda V$, and DEFINE $$\Phi(\vec{x},t)\equiv e^{iHt}e^{-iH_0 t} \phi(\vec{x},t) e^{iH_0 t}e^{-iHt}$$
$$\Pi(\vec{x},t)\equiv e^{iHt}e^{-iH_0 t} \pi(\vec{x},t) e^{iH_0 t}e^{-iHt}$$
Then it is straight forward to show that $\Phi$ and $\Pi$ satisfy the canonical commutation relations, as well as the new interacting Heisenberg equations (notice that in the definition we use $H(\phi,\pi)$ and in the Heisenberg equations $H(\Phi,\Pi)$, we are allowed to do so because both are equal!). (Hint: to see that the new fields satisfy the full heisenberg equations notice that $\Phi(\vec{x},t) = e^{iHt}\phi(\vec{x},0)e^{-iHt}$) So in this sense they are the interacting fields, written in terms of the free fields.

Then we conclude that $$\Phi(\vec{x},t) = \int \frac{d^3 p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}}\left( \mathbb{a}(t)_p e^{ipx} + \mathbb{a}(t)^\dagger_p e^{-ipx}\right)$$
Where $$\mathbb{a}(t)_p\equiv e^{iHt}e^{-iH_0 t} a_p e^{iH_0 t}e^{-iHt}$$
and $\mathbb{a}_p(t)$ satisfies the required commutation relations as a consequence of its parents $\Phi$ and $\Pi$ doing so, or as can be verified directly using those of $a_p$.

Notice that this description is particularly useful for weakly coupled theories, since then $\mathbb{a}_p = a_p + \mathcal{O}(\lambda,a_p^2)$, then all our particle spectrum can be inferred from that of the free theory, unlike when this expansion is no longer valid, and the new creation operator can create states completely different in nature from what's contained in the free theory.

## Best Answer

Physically the creation of a particle with momentum $\mathbf{p}$ will be affected by the Lorentz group in the following manner: since $$ U(\Lambda)|\mathbf{p}\rangle = |\Lambda\mathbf{p}\rangle $$ $$ |\mathbf{p}\rangle=a^\dagger(\mathbf{p})|0\rangle $$ we get $$ U(\Lambda)a^\dagger(\mathbf{k})U^\dagger(\Lambda)=a^{\dagger}(\Lambda\mathbf{k}). $$ Indeed any transition amplitude gives: $$ \langle \Lambda\mathbf{p}| \Lambda \mathbf{q}\rangle=\langle0|a(\Lambda\mathbf{p})a^\dagger(\Lambda\mathbf{q})|0\rangle\\ \langle \Lambda\mathbf{p}| \Lambda \mathbf{q}\rangle= \langle \mathbf{p}|U^\dagger(\Lambda)U(\Lambda)|\mathbf{q}\rangle= \langle0|a(\mathbf{p})U^\dagger(\Lambda)U(\Lambda)a^\dagger(\mathbf{q})|0\rangle=\\ \langle0|U^\dagger(\Lambda) U(\Lambda) a(\mathbf{p})U^\dagger(\Lambda)U(\Lambda)a^\dagger(\mathbf{q})U^\dagger(\Lambda) U(\Lambda)|0\rangle=\langle0|U(\Lambda)a(\mathbf{p})U^\dagger(\Lambda)U(\Lambda)a^\dagger(\mathbf{q})U^\dagger(\Lambda)|0\rangle; $$ comparison yields the transformation formula for $a,a^\dagger$, where we have used the postulate: $U(\Lambda)|0\rangle = |0\rangle$. Then by taking the adjoint of the above:$$ U(\Lambda)a(\mathbf{k})U^\dagger(\Lambda)=a(\Lambda\mathbf{k}). $$ Now $$ U(\Lambda)\varphi(x)U^\dagger(\Lambda)= U(\Lambda)\int d\Omega_m\left(a(\mathbf{k})e^{-ik\cdot x}+a^\dagger(\mathbf{k})e^{+ik\cdot x}\right) U^\dagger(\Lambda)=\\ \int d\Omega_m\left(U(\Lambda)a(\mathbf{k})U^\dagger(\Lambda)e^{-ik\cdot x}+U(\Lambda)a^\dagger(\mathbf{k})U^\dagger(\Lambda)e^{+ik\cdot x}\right)=\\ \int d\Omega_m\left(a(\Lambda\mathbf{k})e^{-ik\cdot x}+a^\dagger(\Lambda\mathbf{k})e^{+ik\cdot x}\right) $$ changing variable and recalling $d\Omega_m$ is invariant under such change, which is in fact a boost, $\mathbf{k}=\Lambda^{-1}\mathbf{k}'$: $$ \int d\Omega'_m\left(a(\mathbf{k}')e^{-i(\Lambda^{-1}k')\cdot x}+a^\dagger(\mathbf{k}')e^{+i(\Lambda^{-1}k')\cdot x}\right)=\\ \int d\Omega'_m\left(a(\mathbf{k}')e^{-i(\Lambda^{-1}k')\cdot (\Lambda^{-1}x')}+a^\dagger(\mathbf{k}')e^{+i(\Lambda^{-1}k')\cdot (\Lambda^{-1}x')}\right) $$ where $x'=\Lambda x$. But the $\cdot$ product is invariant under $\Lambda$ so: $$ U(\Lambda)\varphi(x)U^\dagger(\Lambda) = \int d\Omega'_m\left(a(\mathbf{k}')e^{-ik\cdot x'}+a^\dagger(\mathbf{k}')e^{+ik'\cdot x'}\right)=\varphi(x'=\Lambda x). $$