# Zee QFT – Part I.7

##### Chapter problems

We can start with problems that are not fully solved in the text.

###### Some terms of $\tilde{Z}$

We have

$\displaystyle \tilde{Z}(J) = e^{-\frac{\lambda}{4!}\left(\frac{d}{dJ}\right)^4}e^{\frac{1}{2m^2}J^2}$

as the full expression of $\tilde{Z}$. Expanding in powers of $\lambda$ and $J$ we get:

$\displaystyle \tilde{Z}(J) = \left[\sum_{j=0}^\infty \frac{(-1)^j}{j!} \left(\frac{\lambda}{4!}\right)^j\left(\frac{d}{dJ}\right)^{4j} \right]\left[\sum_{k=0}^\infty\frac{1}{k!}\left(\frac{1}{2m^2}\right)^k J^{2k}\right]$.

Now we can get the asked for terms. To get the $\lambda^1$ and $J^4$ term, we need to remember that the left side will apply four derivatives over the power of $J$. For this reason, we need to get the $j = 1$, $k = 4$ term:

$\displaystyle \frac{(-1)^1}{1!} \left(\frac{\lambda}{4!}\right)^1\left(\frac{d}{dJ}\right)^{4\cdot 1} \frac{1}{4!}\left(\frac{1}{2m^2}\right)^4 J^{2\cdot4} = \displaystyle -\frac{\lambda}{4!}\left(\frac{d}{dJ}\right)^4 \frac{1}{2^4 4! m^8} J^8$

$= \displaystyle -\frac{\lambda}{4!} \frac{1}{2^4 4! m^8} \frac{8!}{4!}J^4$

$= \displaystyle -\frac{8!}{2^4 (4!)^3 m^8} \lambda J^4$.

Let’s do now the $\lambda^2$ and $J^6$ term ($j = 2$ and $k = (6 + 4\cdot j) / 2 = 7$):

$\displaystyle \frac{(-1)^2}{2!} \left(\frac{\lambda}{4!}\right)^2\left(\frac{d}{dJ}\right)^{4\cdot 2} \frac{1}{7!}\left(\frac{1}{2m^2}\right)^7 J^{2\cdot7} = \frac{1}{2}\frac{\lambda^2}{(4!)^2}\left(\frac{d}{dJ}\right)^8 \frac{1}{7!}\frac{1}{(2m^2)^7}J^{14}$

$\displaystyle = \frac{\lambda^2}{2(4!)^2} \frac{1}{7!(2m^2)^7}\frac{14!}{6!}J^6$

$\displaystyle = \frac{14!}{2(4!)^2 7!6!(2m^2)^7}\lambda^2 J^6$

Now we can try to get the general rules for $\lambda^M$ and $J^N$ term. This requires setting $j = M$ and $k = (N + 4M)/2$:

$\displaystyle \frac{(-1)^M}{M!} \left(\frac{\lambda}{4!}\right)^M\left(\frac{d}{dJ}\right)^{4M} \frac{1}{(N/2 + 1M)!}\left(\frac{1}{2m^2}\right)^{N/2 + 1M} J^{2(N/2 + 2M)} =$

$\displaystyle =\frac{(-1)^M}{M!} \frac{\lambda^M}{(4!)^M}\left(\frac{d}{dJ}\right)^{4M} \frac{1}{(N/2 + 2M)!}\frac{1}{(2m^2)^{N/2+2M}} J^{N+4M}$

$\displaystyle = \left(-\frac{\lambda}{4!}\right)^M \frac{1}{M!(N/2 + 2M)!(2m^2)^{N/2+2M}} \frac{(N+4M)!}{N!}J^N$

$\displaystyle = \frac{(N+4M)!}{M!N!(N/2 + 2M)!(2m^2)^{N/2+2M}} \left(-\frac{\lambda}{4!}\right)^M J^N$

Checking for $M = 2$ and $N = 6$:

$\displaystyle \frac{(6+4\cdot 2)!}{2!6!(6/2 + 2\cdot 2)!(2m^2)^{6/2+2\cdot 2}} \left(-\frac{\lambda}{4!}\right)^2 J^6 = \frac{14!}{2!6!7!(2m^2)^7} \left(-\frac{\lambda}{4!}\right)^2 J^6$,

matching the previously obtained result.

Let’s think this value in terms of an associated graph, with $M$ internal vertices of degree 4 and $N$ external vertices of degree 1. The number of edges will be $E = (N + 4M) / 2 = N/2 + 2M$, giving this formula for the $\lambda^M J^N$ term:

$\displaystyle \frac{(2E)!}{M!N!E!(2m^2)^E} \left(-\frac{\lambda}{4!}\right)^M J^N$.

###### Wick contractions

Expanding $Z(J)$ in powers of $J$:

$\displaystyle Z(J) = \int_{-\infty}^{+\infty}dq\,e^{-\frac{1}{2}m^2 q^2-\frac{\lambda}{4!}q^4+Jq}$

$\displaystyle = \int_{-\infty}^{+\infty}dq\,e^{-\frac{1}{2}m^2 q^2-\frac{\lambda}{4!}q^4}e^{Jq}$

$\displaystyle = \int_{-\infty}^{+\infty}dq\,e^{-\frac{1}{2}m^2 q^2-\frac{\lambda}{4!}q^4}\left(\sum_{s=0}^\infty\frac{1}{s!}J^s q^s\right)$

$\displaystyle = \sum_{s=0}^\infty\frac{1}{s!}J^s \int_{-\infty}^{+\infty}dq\,e^{-\frac{1}{2}m^2 q^2-\frac{\lambda}{4!}q^4}q^s$

$\displaystyle = Z(0, 0) \sum_{s=0}^\infty\frac{1}{s!}J^s G^{(s)}$,

where $Z(0, 0)$ is a constant, independent from $\lambda$ and $s$.

The $G^{(s)}$ terms cannot be exactly evaluated, but they can be expanded as a series in $\lambda$. Following the book, let’s calculate the $\lambda^1$ term of $G^{(4)}$:

$\displaystyle G^{(4)} = \frac{1}{Z(0, 0)}\int_{-\infty}^{+\infty}dq\,e^{-\frac{1}{2}m^2 q^2-\frac{\lambda}{4!}q^4}q^4$

$\lambda^1$ term:

$\displaystyle \frac{1}{Z(0, 0)}\int_{-\infty}^{+\infty}dq\,e^{-\frac{1}{2}m^2 q^2}\left(- \frac{\lambda}{4!}q^4\right)q^4 =$

$\displaystyle = -\frac{\lambda}{4! Z(0, 0)}\int_{-\infty}^{+\infty}dq\,e^{-\frac{1}{2}m^2 q^2}q^8$

$\displaystyle = -\frac{\lambda}{4!}\frac{\int_{-\infty}^{+\infty}dq\,e^{-\frac{1}{2}m^2 q^2}q^8}{\int_{-\infty}^{+\infty}dq\,e^{-\frac{1}{2}m^2 q^2}}$.

Applying the equation I.2.18, we get:

$\displaystyle -\frac{\lambda}{4!}\frac{(8-1)!!}{m^8} = -\frac{7!!\lambda}{4!m^8}$