In my Quantum Field Theory course, I am having some trouble with the derivation of the Feynman rules from a Lagrangian. While I am confident I can get it under control by the time the exams come around, in the mean time I'm stuck on a homework problem that gives the following Lagrangian density for a scalar field $\phi$:

$$ \mathcal{L} = \frac{1}{2} \, \partial _\mu \phi \, \partial ^\mu \phi - \frac{m^2}{2} \phi ^2 - \frac{\lambda_3}{3!} \, \phi ^3 - \frac{\lambda_4}{4!} \, \phi ^4 $$

As I understand it, the first two terms are simply the Klein-Gordon Lagrangian, so we're left with an interaction Lagrangian of

$$ \mathcal{L} _I = - \frac{\lambda_3}{3!} \, \phi ^3 - \frac{\lambda_4}{4!} \, \phi ^4 $$

I am then asked to give the interaction vertices for this Lagrangian, which I think are just all possible vertices with three or four arrowless lines connected to them (?).

The second part of the question is: 'Determine the Feynman rules for the vertices (warning: the combinatorics will be non-trivial)' and this is where I have a problem. I really do not understand how to go about doing this.

  • $\begingroup$ I added a more complete answer. I hope you like it :D $\endgroup$
    – drandran12
    Dec 16, 2020 at 11:29
  • 1
    $\begingroup$ I will try and take the time to have a look soon, thank you very much! I hope I'll still be able to make sense of it, though - I pivoted away from theoretical physics to computational/numerical subjects a few months after posting this question, and am currently working on a PhD involving weather models - I haven't so much as looked at a Feynman diagram in over three years now :) $\endgroup$
    – Drubbels
    Dec 16, 2020 at 11:33
  • $\begingroup$ Could you consider accepting my answer? $\endgroup$
    – drandran12
    Dec 3, 2021 at 16:55

2 Answers 2


You have two interaction vertices in this theory, a local three point and four point interaction with coupling constants given by $\lambda_3$ and $\lambda_4$ respectively. A fully fledged calculation in all gory detail using wick's theorem will allow you to see that for every interaction vertex, given the standard inclusion of $3!$ and $4!$ of the terms within the interaction lagrangian, we have a factor of the form $-\mathrm{i}\lambda$. The feynman rule associated with an interaction vertex is precisely this factor, with $-\mathrm{i}\lambda_3$ for a $3$-point interaction and $-\mathrm{i}\lambda_4$ for a $4$-point one.

Let's do the aforementioned calculation in all gory detail restricting to the $\phi^4$ vertex:

Wick's theorem allows us to extract the correct coordinate space representation of the process with the result that the net numerical factor in front gives rise to the symmetry factor. We will see here that the $1/4!$ present in the interaction lagrangian is exactly cancelled by the cardinality of the symmetry group of the four point vertex (this makes it possible to say that at tree level, we have no non trivial symmetry factors). This vertex is produced (amongst other diagrams) in $$\langle p_3 p_4 | T( \phi(x)^4) |p_1 p_2 \rangle.$$ The term of interest is the one with no contractions, that is $\langle p_3 p_4 | : \phi(x)^4 :|p_1 p_2 \rangle$. Computing this should generate terms of the form $e^{-ix(p_1 + p_2 - p_3 - p_4)}$ so that it is equal to $e^{-ix(p_1 + p_2 - p_3 - p_4)} + \dots$ In the Dyson expansion we have therefore $$ \mathcal A = -\frac{i\lambda}{4!} \int \text{d}^4 x ( e^{-ix(p_1 + p_2 - p_3 - p_4)} + \dots) .$$

Using the result derived in my answer to the question here Symmetry factor via Wick's theorem, the result for $$\langle p_3 p_4 | : \phi(x)^2 \phi(y)^2 : | p_1 p_2 \rangle = 4(e^{-i(p_1+p_2)y} e^{i(p_3+p_4)x} + e^{i(p_4-p_1)y} e^{i(p_3-p_2)x} + e^{i(p_4-p_2)y} e^{i(p_3-p_1)x} + (x \leftrightarrow y)).$$ Specialising this to the case of $x=y$ as we have here, the above reduces to $4! \, e^{-ix(p_1 + p_2 - p_3 - p_4)}$ so that indeed the factor of $1/4!$ is cancelled.

The combinatoric argument without all this tedious calculation is that in the correlator we can contract four $\phi(x)$ fields with $p_1$, leaving three for $p_2$, two for $p_3$ and one for $p_4$. This gives $4!$ permutations of the prongs, again cancelling the $1/4!$.

  • $\begingroup$ +1. I'll just add that I find it easier to see how the vertex factors come about by trying to calculate an n-point function like $\langle 0|T(\phi(z_1)\phi(z_2)\phi(z_3)\phi(z_4))|0\rangle$ rather than an S-matrix element, since S-matrix elements have the additional complication of contracting $\phi$ with kets $|p\rangle$ $\endgroup$ Dec 10, 2016 at 20:17

We are given a scalar field $\phi$ with a Lagrange density \begin{equation} \mathcal{L} = \frac{1}{2} \, \partial _\mu \phi \, \partial ^\mu \phi - \frac{m^2}{2} \phi ^2 - \frac{\lambda_3}{3!} \, \phi ^3 - \frac{\lambda_4}{4!} \, \phi ^4. \end{equation} The first two terms in the Lagrange density are simply the Klein-Gordon Lagrangian, which we already solved in chapter 3 of Mandl and Shaw (1). So we are left with an interaction Lagrangian of \begin{equation} \mathcal{L} _I = - \frac{\lambda_3}{3!} \, \phi ^3 - \frac{\lambda_4}{4!} \, \phi ^4. \end{equation}

Interaction vertices

Since this interaction Lagrangian contains two terms, there will be two categories of interaction vertices as well. These are described by the second term in the S-matrix expansion of the transition element, described by \begin{equation} i \int d^4 x \ S^{(1)}_{fi} = i \int d^{4} x\left\langle f\left|\mathcal{T}\left\{-\frac{\lambda_{3}}{3 !} \phi^{3}-\frac{\lambda_{4}}{4 !} \phi^{4}\right\}\right| i\right\rangle. \end{equation} We start by looking at the cubic term. Using Wick's theorem, this reduces to \begin{equation} S_{f i}^{(1)}\left(\lambda_{3}\right)=\left\langle f\left|\mathcal{T}\left\{-\frac{i \lambda_{3}}{3 !} \phi^{3}\right\}\right| i\right\rangle=\left\langle f\left|-\frac{i \lambda_{3}}{3 !}: \phi \phi \phi:\right| i\right\rangle \end{equation} where all operators are evaluated at the same point. Since $\phi$ is a scalar field, it can be decomposed in a positive and negative frequency part $\phi = \phi_+ + \phi_-$ where the positive frequency part contains annihilation operators and the negative frequency part creation operators. We then get \begin{equation} S_{f i}^{(1)}\left(\lambda_{3}\right)=\left\langle f\left|-\frac{i \lambda_{3}}{3 !}:\left(\phi_{+}+\phi_{-}\right)\left(\phi_{+}+\phi_{-}\right)\left(\phi_{+}+\phi_{-}\right):\right| i\right\rangle \end{equation} where again all operators are evaluated at the same point. We will now expand the normal ordering, which gives \begin{equation} \begin{aligned} :\left(\phi_{+}+\phi_{-}\right)\left(\phi_{+}+\phi_{-}\right)\left(\phi_{+}+\phi_{-}\right):=&:\left(\phi_{+} \phi_{+}+\phi_{-} \phi_{+}+\phi_{+} \phi_{-}+\phi_{-} \phi_{-}\right)\left(\phi_{+}+\phi_{-}\right): \\ =&: \phi_{+} \phi_{+} \phi_{+}+\phi_{-} \phi_{+} \phi_{+}+\phi_{+} \phi_{-} \phi_{+}+\phi_{-} \phi_{-} \phi_{+} \\ &+\phi_{+} \phi_{+} \phi_{-}+\phi_{-} \phi_{+} \phi_{-}+\phi_{+} \phi_{-} \phi_{-}+\phi_{-} \phi_{-} \phi_{-} \\ =& \phi_{+} \phi_{+} \phi_{+}+3 \phi_{-} \phi_{+} \phi_{+}+3 \phi_{-} \phi_{-} \phi_{+}+\phi_{-} \phi_{-} \phi_{-} \end{aligned} \end{equation} Each of the terms appearing will give an interaction vertex. We will now discuss the vertices corresponding to each term

  • The term $\phi_+ \phi_+ \phi_+$ contains no creation operators. It will therefore correspond to a diagram with three out-going lines and no in-going lines.
  • The term $\phi_- \phi_+ \phi_+$ contains one creation operator and two annihilation operators. It therefore corresponds to a diagram with one in-going line and two out-going lines.
  • The term $\phi_- \phi_- \phi_+$ contains two creation operators and one annihilation operators. It therefore corresponds to a diagram with two in-going line and one out-going lines.
  • The term $\phi_- \phi_- \phi_-$ contains no creation operators. It will therefore correspond to a diagram with no out-going lines and three in-going lines.

This processes will be unphysical. Hence, in the first and the last process, energy is clearly either created or destroyed. Furthermore, analysing the third process in the center of mass frame, we have initial momentum 4-vectors $(E, \vec{p})$ and $(E, -\vec{p})$ and final one $(2E, 0)$. From the final momentum 4-vector, we find $p^2=2E\geq2m$. Consequently, the particle can never be on shell and is thus not physically allowed. The same analyses can be done for the second process.

Let us now do the same for the quartic term. Again using Wick's theorem and expanding the normal order yields \begin{equation} \left\langle f\left|\mathcal{T}\left\{\frac{-i \lambda_{4}}{4 !} \phi^{4}\right\}\right| i\right\rangle=\left\langle f\left|\frac{-i \lambda_{4}}{4 !}\left(\phi_{+}^{4}+4 \phi_{-} \phi_{+}^{3}+6 \phi_{-}^{2} \phi_{+}^{2}+4 \phi_{-}^{3} \phi_{+}+\phi_{-}^{4}\right)\right| i\right\rangle \end{equation} where again all operators are evaluated at the same point. The following Feynman diagrams correspond to each term in the same order. Feynman diagram of the quartic term. We can examine these analogously to the cubic interaction vertex. All terms, except for the third, will turn out to be unphysical for the same reasons as before. However, the third one is completely allowed when the size of the spatial momentum of the outgoing particles in the center of mass frame is equal to the size of the spatial momentum of the ingoing particles.

Feynman rules

To find the Feynman rule corresponding to adding a vertex we consider the S-matrix element given as \begin{equation} S_{f i}=\left\langle f\left|\sum_{n=0}^{\infty} S^{(n)}=\sum_{n=0}^{\infty} \frac{(-i)^{n}}{n !} \int \cdots \int d^{4} x_{1} \ldots d^{4} x_{n} T\left\{\mathcal{H}_{1}\left(x_{1}\right) \ldots \mathcal{H}_{n}\left(x_{n}\right)\right\}\right| i\right\rangle \end{equation} where we have $\mathcal{H} = -\left(\frac{\lambda_3}{3!}\phi^3 +\frac{\lambda_4}{4!}\phi^4\right)$. Now using Wick's theorem we can write $S^{(n)}$ as a sum of normal ordered products with all possible contractions. We will now investigate one such term and later generalise the discussion to all possible terms. We start by taking the term without any contractions. For $S^{(n)}$ this gives \begin{equation}\tag{1} \begin{array}{l} \left\langle f\left|N\left\{\mathcal{H}_{1}\left(x_{1}\right) \ldots \mathcal{H}_{n}\left(x_{n}\right)\right\}\right| i\right\rangle \\ =\left\langle f\left|N\left\{(-1)^{n}\left(\frac{\lambda_{3}}{3 !} \phi^{3}+\frac{\lambda_{4}}{4 !} \phi^{4}\right)_{x_{1}}\left(\frac{\lambda_{3}}{3 !} \phi^{3}+\frac{\lambda_{4}}{4 !} \phi^{4}\right)_{x_{2}} \ldots\left(\frac{\lambda_{3}}{3 !} \phi^{3}+\frac{\lambda_{4}}{4 !} \phi^{4}\right)_{x_{n}}\right\}\right| i\right\rangle \end{array} \end{equation} (Note that we wrote a product of the form $(\phi^3)_{x_1}(\phi^3)_{x_2}\dots (\phi^3)_{x_n}$ as $(\phi^3)^n$, this is only to facilitate notation. In this notation, the power on $(\phi^3)$ refers to how many terms of the form $(\phi^3)_{x_i}$ there are in the product, where the $x_i$'s are different for each factor.)

Expanding the product in the normal ordering gives rise to a number of terms. A general form of these terms is \begin{equation}\tag{2} \left\langle f\left|N\left\{(-1)^{n}\left(\frac{\lambda_{3}}{3 !} \phi^{3}\right)^{p}\left(\frac{\lambda_{4}}{4 !} \phi^{4}\right)^{n-p}\right\}\right| i\right\rangle \end{equation} The prefactor stating how many different terms in (1) give the term (2) after normal ordering is omitted, since this will be exactly what we calculate in the following discussion. Each $\phi$ appearing in (2) must either annihilate a particle in the initial state or create a particle in the final state. Since all particles are identical, the different diagrams that can be constructed from (2) are the diagrams where $k$ particles in the final state (and are thus created) and $l$ particles in the initial state (and are thus annihilated). We can further write this as \begin{equation} \left\langle f\left|N\left\{(-1)^{n}\left(\frac{\lambda_{3}}{3 !}\right)^{p}\left(\frac{\lambda_{4}}{4 !}\right)^{n-p}\left(\phi^{3}\right)^{p}\left(\phi^{4}\right)^{n-p}\right\}\right| i\right\rangle \end{equation}

For a certain diagram there will now be $n$ vertices. At each of these vertices. Each of the factors $\phi^3$ or $\phi^4$ correspond to one such vertex. At every vertex one is then free to choose which of the $\phi$'s appearing in the factor corresponding to the vertex will annihilate (or create) a certain particle. For the $\phi^3$ vertices, one has therefore $3$ options for the first particles, $2$ options for the second and only $1$ option for the third. This gives a factor $3!$ for each $3$-vertex. Similarly, for each $4$-vertex, there will be a factor $4!$ introduced. This gives a total factor of $(3!)^p (4!)^{n-p}$, exactly the inverse of the factorials appearing in (2). The constant factor in (2) will therefore be $\lambda_3^p \lambda_4^{n-p}$.

Apart from being able to choose which $\phi$ annihilates (or creates) a certain particle in a vertex. One can also choose which particles interact in which vertices. Again, since all particles are identical, this gives a factor of $p!$ for the $3$-vertices and $(n-p)!$ for the $4$-vertices. When all vertices are the same type, this exactly cancels out the factor $\frac{1}{n!}$ appearing in the Dyson expansion.

This discussion is not altered when internal lines are present. Since one has again the freedom to choose which line in a vertex forms an internal line (just like one could choose which line corresponded to a certain particle), the previous discussion can also be applied to diagrams with internal lines.

Now that is cleared out, we can construct the Feynman rules. To do this, we use exercise 3 and our understanding of the Feynman rules of QED, stated on p119 in Mandl and Shaw (1). This all yields the following rules to find the Feynman amplitude $\mathcal{M}$:

  1. Write a factor $\tfrac{(-i)^v}{v!}$ where $v$ is the number of vertices.
  2. For every vertex with 3 lines write a factor $\lambda_3$. For every vertex with 4 lines write a factor $\lambda_4$. The 3! and 4! terms gets cancelled as discussed above.
  3. For every internal line, labelled by momentum $k$, write a factor \begin{equation} i\frac{1}{k^2-m^2+i\epsilon}, \end{equation} determined by the meson propagator given on p. 52 in Mandl and Shaw.
  4. The external lines will not contribute a new factor in the Feynman amplitude. They contribute only to a pre-factor, see later.
  5. For each four-momentum $q$ which is not fixed by energy momentum conservation, carry out out an integration \begin{equation} \int d^4q. \end{equation}
  6. Add a factor $(2\pi)^{v-i-1}$ where $v$ is the number of vertices and and $i$ the number of internal lines.

Remember that we are working with boson particles. Therefore, in contrast to QED, we do not need to add a factor of -1 if the normal ordering takes an odd amount of permutations. Now, at the end of these procedure, one must add to the Feynman amplitude $\mathcal{M}$ a pre-factor equal to \begin{equation} (2\pi)^4 \delta^{(4)}(\sum_f k_f - \sum_i k_i)\prod_{\text{ext.}}\sqrt{\frac{1}{2V\omega_k}}.\\ \end{equation}


(1) Mandl Franz, and Graham Shaw. Quantum field theory. John Wiley & Sons, 2010.


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