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What is the exact reason why the solution of the classical Klein-Gordon equation is written as a mode expansion with a Lorentz invariant measure and after that the coefficients are promoted to operators? Why do we want the measure to be Lorentz invariant? In particular, what do we want to be Lorentz invariant and what are the aforesaid Lorentz transformations acting on (4-coordinates,4-momenta,the scalar field itself,...)?

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    $\begingroup$ Wherever possible, working with Lorentz covariant (or invariant) objects is beneficial because it makes doing the same calculation in a different reference frame more straight forward. $\endgroup$
    – Charlie
    Commented Jan 30, 2021 at 0:27
  • $\begingroup$ Another thing to observe: simply carrying over the measure from NRQM: $d^3 \vec p$, does not allow you define unitary operators that correspond to all Lorentz transformations $\endgroup$
    – Nihar Karve
    Commented Feb 8, 2021 at 9:24

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