Transformation of the scalar field under Lorentz Boost

Question

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

Answer 1

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