Transformation of the scalar field under Lorentz Boost Assume a Lorentz transformation $\Lambda$ is to be implemented as the unitary operator $U(\Lambda)$ in the Hilbert space of quantum states of the Fock representation upon which the scalar Klein-Gordon field acts:
$$
\varphi(x)=\int\frac{d^3k}{\sqrt{2}k_0}\left(a(\mathbf{k})e^{-ik\cdot x}+a^\dagger(\mathbf{k})e^{+ik\cdot x}\right)\equiv \int d\Omega_m\left(a(\mathbf{k})e^{-ik\cdot x}+a^\dagger(\mathbf{k})e^{+ik\cdot x}\right) 
$$
where $d\Omega_m$ is the Lorentz-invariant measure element. 
How do the annihilation and creation operator transform?
How can I prove that
$$
U(\Lambda)\varphi(x)U^\dagger(\Lambda) = \varphi(\Lambda x)?
$$
 A: 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:
$$
U(\Lambda)\varphi(x)U^\dagger(\Lambda) = \int d\Omega'_m\left(a(\mathbf{k}')e^{-ik\cdot x'}+a^\dagger(\mathbf{k}')e^{+ik'\cdot x'}\right)=\varphi(x'=\Lambda x).
$$
