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YeaNah

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- TL;DR Summary
- Is the Lagrangian in the path integral for QFT actually classical? Since the inverse of the differential operator is taken to be the quantum propagator, shouldn't this mean the differential operator itself is a quantum equation?

Using free scalar field for simplicity.

Hi all, I have a question which is pretty simple, we have the path integral in QFT in the presence of a source term:

$$

Z[J] = \int \mathcal{D}\phi \, e^{i \int d^4x \left( \frac{1}{2} \phi(x) A \phi(x) + J(x) \phi(x) \right)}

$$

So far so good. Now this action is classical, right? Well when solve the integral in the presence of a classical source term we get:

$$ Z[J] = Z[0] e^{-\frac{i}{2} \int d^4x \, d^4y \, J(x) \mathcal{A}^{-1} J(y)} $$

where

$$\mathcal{A}^{-1} = D(x - y)$$

Here, ##\mathcal{A}^{-1} = D(x - y)## is commonly interpreted as a quantum propagator. But if this is so, then ##\mathcal{A}## must be the differential operator for the Klein-Gordon equation of a quantum scalar particle, not a classical Klein-Gordon field. Therefore, my question is, is the Lagrangian that goes into the path integral actually classical? Or is it in fact tied to single particle QM wavefunctions? Kind of like a sum over all possible histories the wavefunction can take?

Thanks.

$$\phi(x) = \int d^4y \, D(x - y) \mathcal{A} \phi(y).$$

Hi all, I have a question which is pretty simple, we have the path integral in QFT in the presence of a source term:

$$

Z[J] = \int \mathcal{D}\phi \, e^{i \int d^4x \left( \frac{1}{2} \phi(x) A \phi(x) + J(x) \phi(x) \right)}

$$

So far so good. Now this action is classical, right? Well when solve the integral in the presence of a classical source term we get:

$$ Z[J] = Z[0] e^{-\frac{i}{2} \int d^4x \, d^4y \, J(x) \mathcal{A}^{-1} J(y)} $$

where

$$\mathcal{A}^{-1} = D(x - y)$$

Here, ##\mathcal{A}^{-1} = D(x - y)## is commonly interpreted as a quantum propagator. But if this is so, then ##\mathcal{A}## must be the differential operator for the Klein-Gordon equation of a quantum scalar particle, not a classical Klein-Gordon field. Therefore, my question is, is the Lagrangian that goes into the path integral actually classical? Or is it in fact tied to single particle QM wavefunctions? Kind of like a sum over all possible histories the wavefunction can take?

Thanks.

$$\phi(x) = \int d^4y \, D(x - y) \mathcal{A} \phi(y).$$