Quantum field for Klein-Gordon Equation is defined as $$\phi(x)=\int\frac{d^3k}{(2\pi)^3\sqrt{2\omega_k}}(a_k e^{-ikx}+a^{\dagger}_ke^{ikx})$$ Please forgive me for my slopiness here regarding the vector k in creation and annihilation operator. Now my question is I don't see the R.H.S of above equation as manifestly Lorentz invariant as following:
- The 4-vector product in exponential is of course manifestly Lorentz invariant.
- But the issue is this term: $\int\frac{d^3k}{\sqrt{2\omega_k}}$ I know $\int\frac{d^3k}{2\omega_k}$ is manifestly Lorentz invariant because it can be written as $\int d^4k\hspace{3pt}\delta(k^2-m^2)\hspace{3pt}\theta(k^0)$. If I take $\phi(x)|0\rangle$ then it is manifestly Lorentz invariant because in my convention $a_k|0\rangle=\frac{1}{\sqrt{2\omega_k}}|k\rangle$.
So does this mean that $\phi(x)$ doesn't need to be manifestly Lorentz invariant cause if we're doing some observation ultimately we have to operate it on some state and that expression as seen above is manifestly Lorentz invariant.
