I am trying to understand the diagrams that comes from a two-point correlation function, $$\langle \Omega|T\{\phi(x)\phi(y)\}|\Omega\rangle$$, in $\phi^4$-theory. The zeroth order contribution, i.e. $\lambda^0$, is simply $D_F(x-y)$, and in $\lambda^1$ we get an additional internal point such that $$-\underbrace{12}_{\text{possible connected contractions}}\frac{i\lambda}{4!}\int d^4z \phi(x)\phi(y)(\phi(z))^4\\ = -12\frac{i\lambda}{4!}\int d^4z D_F(x-z)D_F(z-z)D_F(y-z)$$ As for the contributions from the $\lambda^2$ term I can understand 2 of the 3 connected contractions of $\phi(x)\phi(y)(\phi(z))^4(\phi(w))^4$. The first one is $$-P\frac{\lambda^2}{4!4!}\int d^4z d^4w D_F(x-z)D_F(y-z)D_F(z-w)D_F(z-w)D_F(w-w),$$ and the second one is $$-P\frac{\lambda^2}{4!4!}\int d^4z d^4w D_F(x-z)D_F(y-w)D_F(z-w)D_F(z-z)D_F(w-w),$$ where $P$ is the possible configurations of the contraction. The third contraction yields the last diagram in the photo belowenter image description here, while the other diagrams correspond to the contractions above. My question is really how the last diagram would look in terms of contractions when using Wick's theorem. What does the expression look like for that diagram? There's clearly something i have misunderstood!


1 Answer 1


Although the third $\lambda^2$ diagram is a little bit unusual in that there are three lines going between the two vertices, the rule continues to apply that there is a factor of $D_F$ for each line, meaning that the third $\lambda^2$ term is $$-P\frac{\lambda^2}{4!4!}\int d^4z d^4w D_F(x-z)D_F(y-w)[D_F(z-w)]^3\ .$$ (I've continued using your notation of identifying the vertices as $z$ and $w$ in the expression, even though they are identified as $z_1$ and $z_2$ in the diagram.)

You've actually already correctly used multiple identical factors of $D_F$ when multiple lines connect two vertices; your expression for the first $\lambda^2$ term contains two factors of $D_F(z-w)$.

For calculating $P$, there are eight ways that the $\phi(x)$ can contract with one of the four $\phi(z)$'s or four $\phi(w)$'s, four ways that the $\phi(y)$ can contract with whichever of $\phi(z)$ or $\phi(w)$ that $\phi(x)$ didn't contract with, and $3!$ ways that the remaining three $\phi(z)$'s can contract with the remaining three $\phi(w)$'s, so $$P=8\cdot 4\cdot 3!=192\ .$$

  • $\begingroup$ Thank you very much! I actually realized this a few moments before i read your answer and I was ready to make a comment with my explanation, but you beat me to it! Ah, I didn't notice that the labels for the internal points were $z_{1,2}$ in the photo. $\endgroup$ Oct 3, 2017 at 8:27
  • $\begingroup$ You're missing a factor of two in your last expression ($4 \times 4 \times 3! = 96$). I think the reason is that for the first contraction of $\phi(x)$ with whether one of the four $\phi(z)$ or one of the four $\phi(w)$ you have eight possible contractions in total. $\endgroup$ Mar 12, 2021 at 3:31
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    $\begingroup$ @DavidLeonardoRamos I think you're right. I fixed my answer. Thank you for pointing out my error. $\endgroup$
    – Red Act
    Mar 13, 2021 at 4:27

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