In QM using Feynman path integral(FPI) we derive the propagator of free particle which comes out to $$(f(t))e^{iS_{cl}/\hbar}.$$

But in QFT the Feynman propagator is derived using the differential equation $$(\Box_x +m^2)G_F(x-y)=-i\delta^4(x-y).$$

We take the Fourier transform of the above equation and use a proper contour to define the propagator.

My question is: whether we can get $G_F(x-y)$ using FPI: $$G_F(x-y)=\int\mathcal{D}\phi \hspace{2pt} e^{iS/\hbar}$$

where $$S=\int\mathcal{L}\hspace{3pt}\mathrm{d}^4x$$ and $$\mathcal{L}=\frac{1}{2}\Big(\partial_\mu\phi\partial^\mu\phi-m^2\phi^2\Big)~?$$

  • 2
    $\begingroup$ The fact that the two point function is equal to the Green function is only true for Gaussian (free) theories. It's essentially the same as for a multivariate Gaussian. $\endgroup$
    – lcv
    Commented Mar 20, 2020 at 7:20

2 Answers 2


Yes, it can be easily derived.

First some notation. We denote by $\left< \Omega \right>$ (the quantum expectation value) the path integral

$$ \left< \Omega \right> = \int \mathcal{D} \phi \, e^{i S[\phi]} \Omega[\phi]. $$

Here $\Omega[\phi]$ is an arbitrary functional of $\phi$, though we shall not promise that all possible functionals give rise to well defined expectation values.

We will be manipulating abstract expressions – at this point the path integral is just a symbol that we have yet to define if we want to make all that is outlined below precise.

We have assumed in the formula above that the path integral measure $\mathcal{D} \phi$ is normalized such that $\left<1\right> = 1$.

Now consider an auxiliary functional $\Omega[\phi] = \phi(y)$ where $y$ is just an arbitrary point in spacetime. Consider the variation of the path integral under an infinitesimal shift $\phi(x) \rightarrow \phi(x) + \delta \phi(x)$:

$$ \left< \phi(y) \right> = \int \mathcal{D} \phi \, e^{i S[\phi]} \phi(y), $$ $$ \delta \left< \phi(y) \right> = \int \mathcal{D} \phi \, \left( \delta e^{i S[\phi]} \cdot \phi(y) + e^{i S[\phi]} \cdot \delta \phi(y) \right) = \left< i \delta S \cdot \phi(y) + \delta \phi(y) \right>. $$

Now plug the identity $$ \delta \phi(y) = \int d^4 x \, \delta^{(4)}(x-y) \, \delta \phi(x) $$ into that formula. You will get

$$ \delta \left< \phi(y) \right> = \int d^4 x \, \left< i \frac{\delta S}{\delta \phi(x)} \cdot \phi(y) + \delta^{(4)}(x-y) \right> \delta \phi(x). $$

Here we can move $\delta \phi$ in and out of the quantum expectation braces because it is fixed, unlike $\phi(x)$ which we integrate over.

Note that $\delta \left< \phi(y) \right> = 0$ by definition of the expectation bracket (it does not depend on shifts in the dummy integration variable). We have therefore the so-called quantum action principle:

$$ \left< \frac{\delta S}{\delta \phi(x)} \cdot \phi(y) \right> = - i \delta^{(4)} (x - y). $$

This is in contrast with the classical theory where $\delta S / \delta \phi$ vanishes. In fact, there should be a factor of $\hbar$ in the formula above if we give up on natural units, which convinces us that the formula above is the quantum generalization of the least action principle. The term proportional to the Dirac delta function is called the contact term.

Now it is time to plug the concrete expression for $S[\phi]$. In a class of free theories, $\delta S/\delta \phi$ is linear in $\phi$. For example, for the Klein-Gordon theory,

$$ S[\phi] = \int d^4 x \left( \frac{1}{2} \partial_{\mu} \phi \partial^{\mu} \phi - \frac{1}{2} m^2 \phi^2 \right), $$ $$ \frac{\delta S}{\delta \phi(x)} = - \left( \partial_{\mu} \partial^{\mu} + m^2 \right) \phi(x).$$

Plug this into the quantum action principle and you'll get

$$ \left( \frac{\partial}{\partial x^{\mu}} \frac{\partial}{\partial x_{\mu}} + m^2 \right) \left< \phi(x) \phi(y) \right> = i \delta^{(4)} (x - y). $$

That is exactly the distributional differential equation for the Klein-Gordon propagator.

Some notes:

  • The procedure above only works for free fields. The quantum action principle is still formally valid for interacting fields, but you won't get a differential equation on the propagator from the interacting action. You would rather get a series of complicated recursive relations on $n$-point functions of the theory that have not been solved in general.
  • The procedure outlined above is highly abstract, in fact it doesn't even use the definition of the path integral. Instead it uses its expected properties, such as the invariance under translations in the integration variable. These properties should be proven for each model under consideration, but the proof requires the definition of the path integral. Luckily, for free fields this definition exists (look for Gaussian path integrals), and it satisfies the properties that are used in this derivation.
  • $\left< \phi(x) \phi(y) \right>$ is the propagator, because it is equal to $\left< 0 \right| T \hat{\phi}(x) \hat{\phi}(y) \left| 0 \right>$ in the Heisenberg picture of the canonical formalism.
  • Beware of formal, non-precise definitions. For example, $\left< \phi(x) \phi(y) \right>$ is not a function on all $\mathcal{M}^2$, it has singularities at $x = y$ and when $x$ and $y$ are light-like separated. Similarly, $\hat{\phi}(x)$ is not really an operator – because its square, for example, does not exist strictly speaking. The real quantum operators are smeared with rapidly decaying test functions: $\hat{\phi}(f) = \int d^4 x f(x) \hat{\phi}(x)$.
  • $\begingroup$ Your answer is excellent, but a doubt remains for me. When you write $\frac{\delta S}{\delta \phi(x)}$ for the Klein-Gordon action, the integral over $d^4x$ present in the previous line is missing. Is this a conventional way of writing, an oversight, or am I missing something? $\endgroup$
    – Green05
    Commented Sep 9, 2023 at 18:42
  • $\begingroup$ @Green05 there is an $x$ in the denominator, in this notation the variational derivative is taken at the point $x$. There would be an integral over $d^4 x$ in the expression for $\delta S$. $\endgroup$ Commented May 10 at 5:00

Consider a free/quadratic action $$S_2[\phi] ~=~\frac{1}{2} \phi^k (S_2)_{k\ell} \phi^{\ell}~=~\frac{1}{2}\int\!d^4x\int\!d^4y~\phi(x) S_2(x,y)\phi(y). \tag{1}$$ Here we are using DeWitt condensed notation. E.g. in OP's example $$(S_2)_{k\ell}~=~S_{2}(x,y)~=~-(\Box_x+m^2)\delta^4(x\!-\!y).\tag{2}$$ The Greens function is formally the inverse $(S_2^{-1})^{k\ell}$ up to normalization and boundary conditions. E.g. in OP's example $$(S^{-1}_2)^{k\ell}~=~S^{-1}_{2}(x,y)~=~-(\Box_x+m^2)^{-1}\delta^4(x\!-\!y).\tag{3}$$ The free path integral/partition function is an infinite-dimensional Gaussian integral $$\begin{align} Z[J]~=~~~& \int {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar}\left(S_2[\phi] +J_k \phi^k \right)\right\}\cr \stackrel{\text{Gauss. int.}}{\sim}&~ {\rm Det}\left(\frac{1}{i} (S_2)_{mn}\right)^{-1/2} \exp\left\{- \frac{i}{2\hbar} J_k (S_2^{-1})^{k\ell} J_{\ell} \right\}.\end{align} \tag{4} $$ We recognize the Greens function $(S_2^{-1})^{k\ell}$ in the exponential (4). It is also the 2-point function $$\begin{align} \langle \phi^k \phi^{\ell}\rangle_{J=0} ~:=~&\frac{1}{Z[0]}\left. \int \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\phi^k\phi^{\ell}\exp\left\{ \frac{i}{\hbar} \left(S[\phi]+J_m \phi^m\right)\right\}\right|_{J=0}\cr ~\stackrel{(4)}{=}~&\frac{1}{Z[0]} \left.\left(\frac{\hbar}{i}\right)^2\frac{\delta^2 Z[J]}{\delta J_k\delta J_{\ell}}\right|_{J=0}~\cr ~\stackrel{(4)}{=}~&i\hbar(S_2^{-1})^{k\ell}. \end{align} $$ cf. my related Phys.SE answer here.


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