Time ordered correlator from path integral: equation of motion?

Question

Consider a Lagrangian $L(\phi)$ for a field $\phi$ (assume it is a free real scalar for simplicity). Then the time ordered propagator can be expressed as a path integral $$ \langle\Omega|T\{ \phi(x) \phi(x') \}|\Omega\rangle = \int D\phi\ \phi(x) \phi(x') e^{i \int L} . $$ This is a standard result from Zee's textbook for example.

We know that the time ordered propagator should obey the equation of motion $$ ( -\Box_x^2 + m^2 )\langle\Omega|T\{ \phi(x) \phi(x') \}|\Omega\rangle = \delta(x - x') \ . $$ I would like to check that the path integral representation of the propagator satisfies the equation of motion.

However, applying $\Box_x$ to the path integral on the RHS of the above, does not seem to give me the right equation. In fact, since $( -\Box_x^2 + m^2 )\phi(x) =0$ it seems like $$ ( -\Box_x^2 + m^2 ) \int D\phi\ \phi(x) \phi(x') e^{i \int L} = 0 $$ This is obviously wrong, but I don't see why. Is it because of the end points on the integral?

Answer 1

Let $P(\Phi)$ be a polynomial in a set of field variables $\Phi= \{\phi_1(x), \phi_2(x),\ldots\}$, and consider a correlation function $$ \langle{P(\Phi)}\rangle= \frac 1 Z \int d[\Phi] P(\Phi)e^{-S[\Phi]}. $$ The functional integral over $\Phi$ should be unaffected by a linear shift in the integration variables $\phi_i(x) \to \phi_i(x) +\delta \phi_i(x)$. We therefore have
$$ 0= \frac 1 Z \int d[\Phi] \left\{\int \delta \phi_i(x) \left(\frac{\delta P}{\delta \phi_i(x)} - P(\Phi) \frac{\delta S}{\delta \phi_i(x)}\right)d^d x\right\} e^{-S[\Phi]}, $$ and hence $$ 0=\left\langle \int \delta \phi_i(x) \left(\frac{\delta P}{\delta \phi_i(x)} - P(\Phi) \frac{\delta S}{\delta \phi_i(x)}\right)d^d x\right\rangle $$ for any $\delta\phi_i(x)$. So, for each point $x$ and field $\phi_i$, we have $$ \left\langle \frac{\delta P}{\delta \phi_i(x)}\right\rangle= \left\langle P(\Phi) \frac{\delta S}{\delta \phi_i(x)}\right\rangle, $$ which is the quantum counterpart of the classical equation of motion$$ \frac{\delta S}{\delta \phi_i(x)}=0. $$

When the fields in $P= \phi(x_1)\phi(x_2) \cdots \phi(x_n)$ are bosonic $$ \frac{\delta P}{\delta \phi(x)}= \sum_{k=1}^n \delta(x-x_k)\phi(x_1) \phi(x_2)\cdots \widehat{\phi(x_k)} \cdots \phi(x_n), $$ where the hat denotes the omission of that factor.

In your case the functional derivative $\frac{\delta S}{\delta \phi_i(x)}$ is the KG equation and you get your equation by taking $P= \phi(x)$ to be a one point function.