Consider $(\phi^*\phi)^2$ theory of complex scalar field. The goal is to come up with Feynman rules from functional derivatives, and the emphasis is on how does the symmetry factors or the multiplicity come about.

For the generating function of the full interacting theory, we can first separate the interaction and the free theory. Then we discover that the interaction part can be seen as a result of applying a functional derivative operator via series expansion. And notice that the each of $\phi(x)$ can be replaced by the operator $\frac{\delta}{\delta J^*(x)}$, and each of $\phi(x)$ can be replaced by the operator $\frac{\delta}{\delta J(x)}$, on the generating function of free field $Z_0[J,J^*]$. Hence, upon an expansion in powers of $\lambda$, the generating function $Z[J,J^*]$ is \begin{align*} \newcommand{\fmD}{\mathcal D} \newcommand{\d}{\mathrm d} \newcommand{\fDelta}[2]{\frac{\delta #1}{\delta #2}} \newcommand{cL}{\mathcal L} Z[J,J^*] &= \exp\left(\int d^4 x \cL_\text{int}\left[\fDelta{}{J^*(x)},\fDelta{}{J(x)}\right] \right) Z_{0}[J,J^*]. \end{align*}

Now we should come up with Feynman rules. For example when we draw the diagram of a $2N$ point function to the $n$th order, we will need to expand 1) the interaction to the $n$th order, and 2) expand the $Z_0[J,J^*]$ to the $2N+n$th order. Such is the only term that survives. Any lower order expansions of $Z_0[J,J^*]$ will be killed by the derivatives, and any higher order expansions of of $Z_0[J,J^*]$ will be zero since we are setting $J=J^*=0$ in the end. Therefore at such level, we can write

$$ \begin{aligned} &\frac{1}{Z[0,0]}\left(\prod_{i=1}^N\fDelta{}{J^*(x_i)}\fDelta{}{J(x'_i}\right) Z[J,J^*]\\ =& \frac{1}{Z[0,0]}\left(\prod_{i=1}^N\fDelta{}{J^*(x_i)}\fDelta{}{J(x'_i}\right) \frac{1}{n!}\left(\prod_{j=1}^n\int d^4 y_j \cL_\text{int}\left[\fDelta{}{J^*(y_j)},\fDelta{}{J(y_j)}\right] \right) Z_{0}[J,J^*]\\ =& \frac{1}{Z[0,0]}\left(\prod_{i=1}^N\fDelta{}{J^*(x_i)}\fDelta{}{J(x'_i)}\right) \frac{1}{n!}\left(\prod_{j=1}^n\int d^4 y_j \left(\fDelta{}{J^*(y_j)},\fDelta{}{J(y_j)}\right)^2 \right)\\ &\times \frac{1}{(N+2n)!}\left(\prod_{k=1}^{N+2n}\int \d^4 z_k\d^4 z'_k J(z_k) G(z_k-z_k') J^*(z_k')\right). \end{aligned} $$
And I know the Feynman rules are:

  • A general graph for the $2N$-point function has $2N$ external points and $n$ internal interaction vertices, where $n$ is the order in perturbation theory.
    • Of the $2N$ external points, $N$ of them will have arrows pointing outwards (away from the point), corresponding to a $\fDelta{}{J}\sim\phi^*$. $N$ of them will have arrows pointing inwards (towards the point), corresponding to a $\fDelta{}{J^*}\sim\phi$.
    • Each vertex is a point with a coordinate label and 4 lines coming out of it. Two of the lines will have arrows pointing outwards, and two of the lines inwards.
    • Each connected "leg" must have arrows matching.
  • Draw all topologically distinct graphs obtained by connecting the external points and the internal vertices in all possible ways, respecting the arrow. Discard all graphs that contain sub-diagrams not linked to at least one external point. Only consider diagrams will all vertices and all external legs are linked, including those are "disconnected".
  • The following weight is assigned to each graph:
    • For every vertex a factor of $-i \lambda $ in Minkowski space and $- \lambda $ in Euclidean space.
    • For every line connecting a pair of points $z_{1}$ and $z_{2}$, a propagator factor.
    • An overall factor of $\frac{1}{n !}$.
  • A multiplicity factor that counts the number of ways in which the lines can be joined without changing the topology of the graph.
  • Integrate over all internal coordinates $\left\{z_{i}\right\}$.

My question is where does the overall factor of $\frac{1}{2n +N!}$ go? And how do we "prove" that counting the symmetry factors of the diagram gives exactly all the ways of taking the derivatives? Is there a way to show that the "multiplicity" of the diagrams are exactly the "ways of taking the derivatives"?

Also how should we count the multiplicity exactly? Are the vertices labeled and treated as distinct objects or are they allowed to be permuted? I just don't see a clear way of linking "taking the derivatives" and "different ways to connect the legs". Especially when there are external lines and internal vertices.

Any help or pointing to directions would be great.


1 Answer 1


OP's underlying main question seems to be the fact that (if we are using the standard convention where each term$^1$ in the Lagrangian is divided by its symmetry factor) then the numerical coefficient in front of the Feynman diagram is the reciprocal of its symmetry factor $S$. For a proof, see e.g. Ref. 1.


  1. P. Etingof, Geometry & QFT, MIT 2002 online lecture notes; Chapter 3.


$^1$ In particular, this means that OP's interaction term in the Lagrangian should be normalized as $-\frac{\lambda}{4}\phi^{\ast 2}\phi^2$.


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