# Sign of counterterm vertex factor (Srednicki)?

My question is about a mere minus sign which, although irrelevant in my specific problem (as will be shown), I fear may bite me later on.

In Srednicki chapter 14, the author is computing the 1-loop correction to the propagator in a renormalized $\phi^3$ theory:

$$\mathcal{L}=\mathcal{L}_0+\mathcal{L}_I+\mathcal{L}_{ct}$$

with:

\begin{align} \mathcal{L}_0&=-\frac{1}{2}(\partial\phi)^2-\frac{1}{2}m^2\phi^2\\ \mathcal{L}_I&=-\frac{1}{3!}Z_{g}g\,\phi^3\\ \mathcal{L}_{ct}&=-\frac{1}{2}\underset{A}{\underbrace{(Z_{\phi}-1)}}(\partial\phi)^2-\frac{1}{2}\underset{B}{\underbrace{(Z_m-1)}}m^2\phi^2 \end{align}

Srednicki is careful with every factor of $\pm i$ when writing the full propagator: a factor of $+i$ for every vertex, and a factor of $1/i$ for every propagator.

$$\frac{1}{i}\Delta(k^2)_{\text{full}}=\frac{1}{i}\Delta(k^2)+\frac{1}{i}\Delta(k^2)\left(i\Pi(k^2)\right)\frac{1}{i}\Delta(k^2)+\ldots\tag{14.2}$$

which is represented diagrammatically in Fig. 14.2 below. The lowest-order vertex-correction to the propagator, $i\Pi (k^2)$, is given by:

$$i\Pi(k^2)=\underset{\text{loop}}{\underbrace{\frac{1}{2}(ig)^2\int \frac{d^dl}{(2\pi)^d}\left(\frac{1}{i}\right)^2\Delta (l^2) \Delta ((l+k)^2)}}\,\underset{\text{counterterm}}{\underbrace{\, \color{red}{-i}(Ak^2+Bm^2)}}\tag{14.4}$$

I have colored red my issue. Why is the counterterm factor $-i=1/i$ instead of $+i$? It's simply a (quadratic) vertex, so it should come with a factor of $+i$, right? On the LHS, we have the generalized vertex, which comes with a factor of $i$. On the RHS, look at the loop term - we have the symmetry factor $1/2$, a factor of $i$ for each vertex, and a factor of $1/i$ for each propagator. Why is this not true for the counterterm part?

In this particular situation it actually doesn't matter because in the end we define $A$ and $B$ to satisfy certain field and mass normalization conditions (ideally cancelling infinities appearing in the loop integral).

From

$$\mathcal {L}_{ct} = -\frac{1}{2}A\,(\partial\phi)^2 - \frac{1}{2}B\, m^2 \phi^2$$

we get

\begin{align} i \int d^4x\; \mathcal{L}_{ct} &= i \int d^4x\; \left( -\frac{1}{2}A\,(\partial\phi)^2 - \frac{1}{2}B\, m^2 \phi^2 \right) \\ &= i \int d^4x\; \left( -\frac{1}{2}A\, \partial(\phi\partial\phi) + \frac{1}{2}A\,\phi\,\partial^2\phi - \frac{1}{2}B\, m^2 \phi^2 \right) \\ &= i \int d^4x\; \left( \frac{1}{2}A\,\phi\,\partial^2\phi - \frac{1}{2}B\, m^2 \phi^2 \right) \\ &= i \int d^4x\; \phi\left[ \left(\frac{1}{2}\right)\left(A\,\partial^2 - B\, m^2 \right) \right]\phi \\ &= i \int d^4x\; \phi\left[ \left(\frac{1}{2}\right)(-1)\left(-A\,\partial^2 + B\, m^2 \right) \right]\phi \\ \end{align}

The second line uses integration by parts, the third line drops the total divergence.

This corresponds to a vertex of

$$-i (A\,k^2 + B\, m^2)$$

in momentum space (remember that $$-\partial^2e^{ikx} = k^2\, e^{ikx}$$ ).

• If that was the origin of the minus sign, wouldn't it also apply to other vertex factors? Shouldn't we then also have $-ig$ for each 3-point vertex, and correspondingly a $+i$ for each propagator? – Arturo don Juan Sep 28 '18 at 18:34
• @ArturodonJuan the minus sign is not related to anything else, you may also write the vertex as $i (-A k^2 - B m^2)$. I edited my answer to clearify it. – Xiangyang Chen Sep 29 '18 at 2:14