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For the free scalar field theory, the path integral has a differential operator term in the exponent,

$$ Z[J] = \int \mathcal{D}\phi \, \exp\left( i \left[ -\frac{1}{2} \int d^d x \, \phi(x) A \phi(x) + \int d^d x \, J(x) \phi(x) \right] \right). $$

When we solve the integral, we get the inverse ${A}^{-1}$ of the second-order differential operator ${A}$. This is interpreted as the quantum propagator, cf. e.g. this Phys.SE post. My question is, if this is interpreted as the quantum propagator, should we interpret the differential operator as the differential operator of the quantum Klein-Gordon equation for a free scalar particle? Instead of the usual interpretation as the differential operator of the classical Klein-Gordon field.

Qmechanic
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ConfusedPleb
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    How you "interpret" $A$ is rather up to you. It is just the usual KG differential opertor $-\nabla^2+m^2$ acting on functions in ${\mathbb R}^4$. – mike stone Sep 01 '23 at 15:51
  • So am I allowed to interpret it as the differential operator of the quantum Klein-Gordon equation? If so, that would make the term “second quantisation” actually valid. – ConfusedPleb Sep 03 '23 at 07:30

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