# A question concerning the effective quantum action for a scalar field

Define the quantum action $\Gamma[\varphi]$ by:

$$\Gamma[\varphi] := -\frac{1}2\int \frac{d^Dk}{(2\pi)^D} \varphi(-k)\Big(k^2 + m^2 - \Pi(k^2)\Big)\phi(k) \\+ \sum_{n=3}^\infty \frac{1}{n!}\int \frac{d^Dk_1}{(2\pi)^D} \cdots \frac{d^Dk_n}{(2\pi)^D} \ (2\pi)^D \delta^D(k_1 + \dots + k_n) \ V_n(k_1, \dots, k_n) \varphi(k_1) \dots\varphi(k_n) \,.$$

Define the generating functional for the quantum action $\Gamma[\varphi]$ as:

$$Z_\Gamma [j] : = \int \mathcal D\varphi\ exp\Big(\frac{i}\hbar (\Gamma[\varphi] + j\cdot\varphi)\Big) = exp\Big(\frac{i}\hbar W_\Gamma [j] \Big)$$

where $j\cdot\varphi = \int d^Dx\ j(x)\varphi(x)\,.$

In Srednicki, Ch. 21, it reads:

$W_\Gamma [j]$ is given by the sum of connected diagrams (with sources) in which each line represents the exact propagator, and each $n$-point vertex represents the exact $1$PI vertex $V_n$.

$W_\Gamma [j]$ would be equal to $W [j]$ if we included only tree diagrams in $W_\Gamma [j]$.

I do not understand this last comment. Does that mean $W [j]$ contains no vacuum diagrams at all, and that $W_\Gamma [j] - W [j]$ generates all connected vacuum diagrams? Why?

Here, $iW [j]$ is the generator of connected diagrams with sources defined by:

$$Z [j] =: exp(iW [j])\,,$$

where $$Z [j] = \int \mathcal D\varphi\ exp\Big(\frac{i}\hbar (S[\varphi] + j\cdot\varphi)\Big), \quad S \text{ is the classical action.}$$

Kindly help. Thank you very much.

## Edit:

The following information may prove useful in answering this question. If we denote by $j_\varphi$ the solution to $$\frac{\delta}{\delta j} W[j] = \varphi\,,$$

then we have that

$$Z_\Gamma [j_\varphi] = exp\Big(\frac{i}\hbar W_\Gamma [j_\varphi] \Big) = \int \mathcal D\varphi\ exp\Big(\frac{i}\hbar W [j_\varphi] \Big) \,.$$

• Haven't you forgotten to define $W[j]$? – user154997 Jun 15 '17 at 18:39
• I am sorry. I have edited the post now. – Nanashi No Gombe Jun 15 '17 at 18:48

In Srednicki, Ch. 9, it reads:

This makes it easy to impose the normalization $Z_1(0) = 1$: we simply omit the vacuum diagrams (those with no sources), like those of figs. 9.1 and 9.2. We then have $$Z_1(J) = \exp [iW_1(J)]$$ where we have defined $$iW_1(J) \equiv \sum_{I \neq {0}} C_I$$ and the notation $I \neq {0}$ means that the vacuum diagrams are omitted from the sum, so that $W_1 (0) = 0$.

According to his definition, $W[j]$ do contain no vacuum diagrams at all, and $W(j)$ is the generator of all connected diagram. The vacuum diagrams is only a constant factor before $Z$ and can always be absorbed into the measure of the path integral $D[\phi]$. And in the calculation of correlation function, they can always be canceled.

There is another equivalent definition of quantum effective action in section 11.3 of Peskin's book. And in section 11.5, he proves that $W(j)$ is the generator of connected diagram and $\Gamma[\phi]$ is the generator of 1pI diagrams. It is another view on the same thing which may be helpful for your understanding.

• Thanks Eric. I'll check out Peskin. I'd like to ask anyway: is there a proof that the two viewpoints are equivalent or is it somehow obvious? – Nanashi No Gombe Jun 16 '17 at 3:22
• In Srednicki's book, he uses a constructive way to get the expression for effective action and you can see that effective action is the generator of 1PI diagram directly. However, in Peskin's book, he defines the effective action by a method similar to Legendre transformation in statistical physics in section 11.3. And he proves in section 11.5 that effective action is the generator of 1PI diagram. So we can conclude that their definition is equivalent. – Eric Yang Jun 16 '17 at 5:20