I am following a QFT course using Peskin & Schroeder (1995): An introduction to Quantum field theory. We've started the functional methods. According to my professor, the vertex rule is the coefficient of the cubic and quartic terms in the Lagrangian density. And the propagator is the inverse of the quadratic terms. I cannot see that this is true. For example, for $\phi^4$ theory: $$ \mathcal{L} = \frac{1}{2}(\partial \phi)^2 - \frac{1}{2}m^2 \phi^2 - \frac{\lambda}{4!} \phi^4 \implies \text{Fourier transformed Propagator} = \frac{i}{p^2 - m^2 + i\epsilon}. $$ I cannot see that this implication is true. If it is true, can you tell me the missing steps, if it is false, can you help rectify the misunderstanding?


3 Answers 3


Feynman rules by functional derivatives

It is not in general, it just coincides that way for polynomials of fields, without any derivatives or other complications. In truth, one takes functional derivatives until no field is left.

For example, schematically,

$$-i\frac{\delta^4}{\delta \phi^4} \frac{\lambda}{4!} \phi^4 \to -i\lambda$$

which is the entire reason for the factor of $4!$ - it is a convenient convention, but not a necessary coefficient and the physics remains the same without it. More complexly, we could have,

$$-i\frac{\delta^2}{\delta \phi^2} \frac{\delta}{\delta A_\mu}g\phi A^\nu \partial_\nu \phi \to -ig(p_1^\mu + p_2^\mu)$$

which describes a vector coupling to a scalar, where $p_1$ and $p_2$ would be momenta labelling two of the legs attached to the $A\phi\phi$ vertex.

Quadratic terms

The kinetic and mass term is,

$$\mathcal L = \frac12 (\partial \phi)^2 - \frac12 m^2 \phi^2.$$

In Fourier space, $\partial_\mu \phi \to ip_\mu \phi$ and so we have $(\partial \phi)^2$ must go as $p^2 \phi^2$. Interpreting the inverse in Fourier space as the multiplicative inverse, we then have that the propagator goes as,

$$\Delta \sim \frac{1}{p^2+m^2}.$$

Note that for the counterterm Lagrangian (for when you move on to renormalisation), the kinetic and mass counterterms are typically interpreted as interactions, and thus functional derivatives are taken, but no inversion is performed. However, this is a matter of choice, one could absorb coefficients into the propagator instead - they lead to the same physics.

Green's function

Note that a propagator - other than being the inverse of the quadratic terms - can also be interpreted as the Green's function of the equations of motion. This is a function which can be used to solve the equations, via,

$$\phi(x) = \int \mathrm dx' \, G(x,x') f(x')$$

where $(\square + m^2)\phi(x) = f(x)$. Conceptually, in the same way we can think of a function as being built of delta functions, we can think of a solution built up as Green's functions since,

$$(\square + m^2)G(x) \sim \delta^{(n)}(x).$$


Being somewhat schematic, you can think of a quantum field theory as being a "free" (non-interacting) Lagrangian which has solutions that do not interact with each other, plus an interaction Lagrangian which tells the modes how to interact. The interaction terms in the Lagrangian create the vertex factors in Feynman rules, but the propagator is from the free, noninteracting theory.

The propagator is the Fourier transform of the Green's function for the free equations of motion; in your case, the equation of motion is the Klein-Gordon equation $$ \partial^2 \phi + m^2 \phi = 0.$$

The Green's function $G$ is the solution to $$ \partial^2 G + m^2 G = \delta^4(x).$$

If you Fourier transform the entire equation, you get $$ -p^2 \tilde G + m^2 \tilde G = 1,$$

where $\tilde G$ is the Fourier transform of $G$. If you solve this for $\tilde G$, you get something like your propagator. If my memory serves me, the factor $i$ is simply a convention for convenience. The $i\epsilon$ term is a bit more complicated; consider that to get the position-space Green's function, you have to integrate this $p$-space function over $d^4k$. This is usually performed as a complex contour integral, and the $i\epsilon$ factor controls which contour you are choosing.

I may edit when I have access to my notes to provide more specific comments, but this should give a rough overview of how the propagator results from the "quadratic" terms in the Lagrangian (the non-interacting ones, including the quadratic derivatives of the field).


The main trick is to introduce sources $J_k$. If the free quadratic action$^1$ $$S_2[\phi] ~:=~\frac{1}{2} \phi^k (S_2)_{k\ell} \phi^{\ell} \tag{1}$$ is non-degenerate, then the free partition function is a Gaussian integral $$\begin{align} Z_2[J] ~:=~& \int {\cal D}\frac{\phi}{\sqrt{\hbar}}~\exp\left\{ \frac{i}{\hbar}\left(S_2[\phi] +J_k \phi^k \right)\right\} \cr \stackrel{\text{Gauss. int.}}{\sim}~& {\rm Det}\left(\frac{1}{i} (S_2)_{mn}\right)^{-1/2}\cr &\exp\left\{- \frac{i}{2\hbar} J_{\ell}J_k (S_2^{-1})^{k\ell} \right\}. \end{align}\tag{2} $$ We can now proceed in 2 ways:

  1. On one hand, if we define the free propagator as the Greens function $G$ for the differential operator ${\cal D}$ of the free quadratic action (1), then OP's title question follows directly from the defining property ${\cal D}G=\delta$ of a Greens function.

  2. On the other hand, if we define the free propagator as the 2-point function, then we calculate $$\begin{align} \langle \phi^k \phi^{\ell}\rangle^{(0)}_{J=0} ~:=~&\frac{1}{Z_2[J]} \int \! {\cal D}\frac{\phi}{\sqrt{\hbar}}~\phi^k\phi^{\ell}\cr & \left.\exp\left\{ \frac{i}{\hbar} \left(S_2[\phi]+J_m \phi^m\right)\right\}\right|_{J=0} \cr ~\stackrel{(2)}{=}~&\left.\frac{1}{Z_2[J]} \left(\frac{\hbar}{i}\right)^2\frac{\delta}{\delta J_k}\frac{\delta}{\delta J_{\ell}}Z_2[J]\right|_{J=0}\cr ~\stackrel{(2)}{=}~&i\hbar (S_2^{-1})^{k\ell},\end{align} \tag{3} $$ which answers OP's title question.

The remainder of OP's question is a matter of Fourier transformation to momentum space.

Example: Let the free quadratic action be on Hamiltonian 1st-order form $$ \begin{align}S_2~=~&\frac{1}{2}\iint\!dt~dt^{\prime}~z^K(t) ~(S_2)_{KL}(t,t^{\prime})~z^L(t^{\prime})\cr ~=~&\frac{1}{2}\int\!dt~z^K\omega_{KL}\dot{z}^L,\cr (S_2)_{KL}(t,t^{\prime}) ~=~&\omega_{KL}\frac{d}{dt}\delta(t\!-\!t^{\prime}), \end{align}\tag{4}$$ where $$ \omega_{KL}~=~(-1)^{(|K|+1)(|L|+1)}\omega_{LK}\tag{5}$$ is (components of) a constant symplectic 2-form, and $$ \omega^{KL}~=~-(-1)^{|K||L|}\omega^{LK}\tag{6}$$ its inverse. Then the 2-point function is $$ \begin{align} \frac{1}{i\hbar}\langle T[z^K(t)z^L(t^{\prime})]\rangle ~\stackrel{(3)}{=}~& (S_2^{-1})^{KL}(t,t^{\prime}) \cr ~\stackrel{(4)}{=}~&\frac{1}{2}\omega^{KL}{\rm sgn}(t\!-\!t^{\prime}).\end{align}\tag{7}$$ Differentiation of eq. (7) wrt. time leads to the CCR $$ \{z^K(t),z^L(t)\}_{PB} ~=~ \frac{1}{i\hbar}[z^K(t),z^L(t)] ~\stackrel{(7)}{=}~ \omega^{KL}.\tag{8}$$ End of example.


$^1$ We use DeWitt condensed notation to not clutter the notation. If $\phi^k$ has Grassmann-parity $|k|$, then consistency requires the following symmetry properties: $$\begin{align} (S_2)_{k\ell}~=~&(-1)^{|k||\ell|+|k|+|\ell|}(S_2)_{\ell k} \cr (S_2^{-1})^{k\ell}~=~&(-1)^{|k||\ell|}(S_2^{-1})^{\ell k}. \end{align}\tag{9}$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.