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}

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