# [Physics] Transformation of the scalar field under Lorentz Boost

homework-and-exercisesklein-gordon-equationlorentz-symmetryquantum-field-theory

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
$$U(\Lambda)\varphi(x)U^\dagger(\Lambda) = \varphi(\Lambda x)?$$

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: $$U(\Lambda)\varphi(x)U^\dagger(\Lambda) = \int d\Omega'_m\left(a(\mathbf{k}')e^{-ik\cdot x'}+a^\dagger(\mathbf{k}')e^{+ik'\cdot x'}\right)=\varphi(x'=\Lambda x).$$