Can a Feynman graph be an empty set?

Question

I was reading the section about Feynman graphs from the book Renormalization - An Introduction and this question arose. To set the notations, let $p \in \mathbb{N}_{+}$ and, for each $k \in \{1,...,p\}$, let $m_{k} \in \mathbb{N}_{+}$. Define the set: $$B_{0} = \{(k,i_{k}): \hspace{0.1cm} \mbox{$k \in \{1,...,p\}$ and $i_{k} \in \{1,...,m_{k}\}$}\}$$ The next step is to define a partial order in $B_{0}$ by setting $(k, i) < (l,j)$ iff $k < l$. For each $\emptyset \neq B \subset B_{0}$, we introduce: $$L_{0}(B) = \{(b,b') \in B\times B: \hspace{0.1cm} b < b'\}$$ Finally, also for $\emptyset \neq B \subset B$, we define $\mathcal{L}(B)$ to be the set of all sets $L$ of the form: $$L = \{(b,b') \in L_{0}(B): \hspace{0.1cm} \mbox{$\pi_{1}|_{L}(b,b')$ and $\pi_{2}|_{L}(b,b')$ are both bijections}\}$$ The notation is as follows. $\pi_{i}|_{L}(b,b')$ denotes the projection into the $i$-th entry, $i=1,2$. More precisely, $\pi_{1}|_{L}(b,b') = b$ and $\pi_{2}|_{L}(b,b') = b'$. In my opinion, this condition is a bit confusing and this is probably the heart of the issue. It is not clear what is the range of $\pi_{i}|_{L}$ where these maps are bijections, but I assume that the condition means that every $b \in B$ appears once either as a first or second entry of $(b,b') \in L$. In any case, the definition of a Feynman graph is:

Definition: Let $E \subset B_{0}$. A Feynman graph is an element of $\mathcal{L}(B_{0}\setminus E)$.

Because I was a bit confused with these definitions, I decided to work out some explicit examples. Suppose $p=2$ and $m_{1} = m_{2} = 2$, so that $B_{0} = \{(1,1),(1,2),(2,1),(2,2)\}$. Let $E = \{(1,1)\}$. Then, $B_{0}\setminus E = \{(1,2),(2,1),(2,2)\}$ and $L_{0}(B_{0}\setminus E) = \{((1,2),(2,1)),((1,2),(2,2))\}$. Now, the Feynman graphs are elements of $\mathcal{L}(B_{0}\setminus E)$ and these are subsets of $L_{0}(B_{0}\setminus E)$ where each element occurs only once in some entry. Because of the partial ordering condition, we see that for the above case $L_{0}(B_{0}\setminus E)$, it is impossible to form such a set because we would have to repeat the entry $(1,2)$ in order to include both $(2,1)$ and $(2,2)$ in the second entry. Hence, $\mathcal{L}(B_{0}\setminus E) = \emptyset$. In particular, this will happen everytime that $E$ removes more points of the form $(k,\cdot)$ than points $(l,\cdot)$, for $k \neq l$.

So, my question is: is my understanding of the definition wrong or are there simply many empty Feynman graphs, according to this definition? And, if so, what is the interpretation of an empty Feynman graph? Is it just a graph that does not contribute to the expansion of the underlying physical object?

Answer 1

Consolidating from comments:

• The definitions given in the question are a bit unusual, so it is not straightforward to interpret them.

• However, best as I understand the question, it is not about whether a Feynman graph can be empty, but about whether the set of Feynman graphs for a given theory (given by the $m_k$), order (given by $p$) and set of external legs (given by $E$) can be empty.

• The answer to that question is a resounding "yes", since e.g. there are no Feynman graphs with an odd number of external legs in $\phi^4$ theory, or more generally no Feynman graphs contributing to an odd-$n$ $n$-point function in a theory with only even-order vertices due to symmetry considerations (even vs odd under $\phi\mapsto-\phi$).

The example given appears to be (best as I can interpret it) about a theory with a quadratic interaction (that's already a bit unusual, since this would normally be treated as a mass instead), for which you are looking at the second-order contribution to the one-point function (one external leg). This is indeed empty, as explained above, since one is odd and two is even.