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I'm trying to understand M.Schwartz's description on his own QFT & SM book, which is about cancellation of disconnected diagrams so called bubbles when we compute two point correlation function with $$L_{int} = \frac{g}{3!}\phi^3\tag{7.65}.$$

Now the textbook says at P.91, eq 7.71,

$$\begin{align} <\Omega|T\{\phi(x_1)\phi(x_2)\}|\Omega> = \frac{1}{<0|T\{\ e^ {i\int L_{int}}\}|0>}\big\{D_{12}-g^2\int d^4x \int d^4y \big[\frac{1}{8}D_{12}D_{xx}D_{xy}D_{yy} + \frac{1}{12}D_{12}D_{xy}^3 + \frac{1}{2}D_{1x}D_{2x}D_{xy}D_{yy} + \frac{1}{4}D_{1x}D_{xx}D_{yy}D_{y2} + \frac{1}{2}D_{1x}D_{xy}^2D_{y2}\big] \big\} \end{align}\tag{7.71} $$ where the $D_{ij}$ is Feynman propagator between $x_i$ and $x_j$.

The book says, the first two terms, $$ {D_{12}-g^2\int d^4x \int d^4y \big[\frac{1}{8}D_{12}D_{xx}D_{xy}D_{yy}}\big] $$ will cancel with the denominator since the denominator expresses only disconnected diagram, which is,

$$ \begin{align} <0|T\{\ e^ {i\int L_{int}}\}|0> = 1-g^2\int d^4x \int d^4y \big[ \frac{1}{8}D_{xx}D_{xy}D_{yy} + \frac{1}{12}D_{xy}^3\big] \end{align}\tag{7.74} $$ up to $O(g^2)$

So that the equation become

$$\begin{align} <\Omega|T\{\phi(x_1)\phi(x_2)\}|\Omega>= \\ \frac{D_{12}-g^2\int d^4x \int d^4y \big[\frac{1}{8}D_{12}D_{xx}D_{xy}D_{yy} + \frac{1}{12}D_{12}D_{xy}^3 + \frac{1}{2}D_{1x}D_{2x}D_{xy}D_{yy} + \frac{1}{4}D_{1x}D_{xx}D_{yy}D_{y2} + \frac{1}{2}D_{1x}D_{xy}^2D_{y2}\big]}{1-g^2\int d^4x \int d^4y \big[ \frac{1}{8}D_{xx}D_{xy}D_{yy} + \frac{1}{12}D_{xy}^3\big]} \end{align}$$

The book argues that the denominator, which describes disconnected bubbles will cancel out disconnected diagrams on the numerator exactly, and i don't get how is such thing can be done. Even if i can factor out $D_{12}$ of first two terms, i still have few more terms after that so the denominator is still there. And i thought is it because we can exponentiate the bubble terms out of those connected diagrams (term such as $D_{12}$ or $D_{1x}D_xx$. Those are connected with external lines) and can express numerator as sum of connected diagrams multiplied with sum of bubble diagrams, but then come to think about it gives me some problems because others terms like $D_{1x}D_{xx}D_{yy}D_{y2} ...$ have nothing to do with bubble diagrams so that i can not factor out bubble terms to exponentiate this.

$$\begin{align} <\Omega|T\{\phi(x_1)\phi(x_2)\}|\Omega> = D_{12} + \frac{-g^2\int d^4x \int d^4y \big[\frac{1}{2}D_{1x}D_{2x}D_{xy}D_{yy} + \frac{1}{4}D_{1x}D_{xx}D_{yy}D_{y2} + \frac{1}{2}D_{1x}D_{xy}^2D_{y2}\big]}{1-g^2\int d^4x \int d^4y \big[ \frac{1}{8}D_{xx}D_{xy}D_{yy} + \frac{1}{12}D_{xy}^3\big]} \end{align}$$

I've read some descriptions about this on P&S book but still i don't get how exactly it can be done.

About this, the Schwartz's QFT book says

Since $\frac{1}{1+g^2x} = 1-g^2x + O(g^2)$, we can invert the denominator in perturbation theory to see that the bubbles exactly cancel. More generally, the bubbles will always cancel. Since the integrals in the expansion of the numerator corresponding to the bubbles never involve any external point, they just factor out. The sum over all graphs, in the numerator, is then the sum over all graphs with no bubbles multiplying the sum over the bubbles.

Can someone please explain me how exactly this bubble diagrams cancel out among the numerator/denominator as written by Schwartz's QFT book?

Qmechanic
  • 201,751
hwan
  • 159

2 Answers2

2

You just have to do an expansion in terms of $g^2$ as follows.

First, let me rewrite more suggestively what Schwartz's argument is with his "$\frac{1}{1+g^2x}=1-g^2x+\mathcal{O}(g^4)$". I would rather write it as $F(g^2)\equiv\frac{y+g^2z}{1+g^2x}=y+g^2(z-xy)+\mathcal{O}(g^4)$. Identifying $F(g^2)$ with your expression, we identify: \begin{align} y=&D_{12} \\ z=&\int d^4x \int d^4y \left[\frac{1}{8}D_{12}D_{xx}D_{xy}D_{yy}-\frac{1}{12}D_{12}D_{xy}^3 +\frac{1}{2}D_{1x}D_{2x}D_{xy}D_{yy}\right. \\ &\left.+ \frac{1}{4}D_{1x}D_{xx}D_{yy}D_{y2}+\frac{1}{2}D_{1x}D_{xy}^2D_{y2} \right] \\ x=&\int d^4x \int d^4y \left[\frac{1}{8}D_{xx}D_{xy}D_{yy}-\frac{1}{12}D_{xy}^3\right] \end{align}

Now, just note that $D_{12}$ is a constant with respect to the integration variables $x$ and $y$. So our contribution of order $\mathcal{O}(g^2)$ (which is $(z-xy)$) contains two terms, one that is the cancellation of the bubbles, and the other which is the remaining part: \begin{align} &\underbrace{\int d^4x \int d^4y \left[\frac{1}{8}D_{12}D_{xx}D_{xy}D_{yy}-\frac{1}{12}D_{12}D_{xy}^3\right]}_{\displaystyle=z_\text{bubble}} \\ &-\underbrace{D_{12}\int d^4x \int d^4y \left[\frac{1}{8}D_{xx}D_{xy}D_{yy}-\frac{1}{12}D_{xy}^3\right]}_{\displaystyle = xy} \\ &+\underbrace{\int d^4x \int d^4y\left[\frac{1}{2}D_{1x}D_{2x}D_{xy}D_{yy}+ \frac{1}{4}D_{1x}D_{xx}D_{yy}D_{y2}+\frac{1}{2}D_{1x}D_{xy}^2D_{y2} \right]}_{\displaystyle = z_\text{remaining}} \end{align} Of course, $z_\text{bubble}-xy=0$, so the bubbles indeed cancel at the order of two loops.

Finally, I think that the argument of Schwartz after the "more generally" in your quote is correct. Intuitively, just think about the correlator $\langle \mathcal{T}\{\phi_1 \phi_2\} \rangle$ as adding two different legs to each term in the partition function's expansion in terms of bubbles. You obviously will have non-trivial contributions, such as $z_\text{remaining}$. But you have a special choice to add (and not attach) two legs to a bubble: just attach them together and leave the bubbles alone. This will result in a term of the form $\text{bubble} \times D_{12}$. Doing this order-by-order in the expansion of the propagator will lead you to such cancellations, without worrying about the higher order terms (Because this is an order-wise method).

Jeanbaptiste Roux
  • 2,522
1

Sure, one may in principle check to any order in perturbation theory that the vacuum bubble diagrams in the numerator and denominator cancel, but a more systematic proof is as follows.

  1. Ref. 1 starts by using eq. (7.64) to calculate the 2-point function.

  2. It was shown in eq. (5) of my Phys.SE answer here that the 2-point function is$^1$ $$ \begin{align} \langle \phi^k \phi^{\ell}\rangle_J ~\stackrel{(7.64)}{=}&~ \frac{1}{Z[J]} \left(\frac{\hbar}{i}\right)^2\frac{\delta^2 Z[J]}{\delta J_k\delta J_{\ell}}\cr ~=~&\frac{\hbar}{i} \frac{\delta^2 W_c[J]}{\delta J_k\delta J_{\ell}} + \langle \phi^k \rangle_J \langle \phi^{\ell} \rangle_J .\end{align} \tag{5}$$

  3. It is implicitly assumed in Ref. 1 that there are no tadpoles $$ \langle \phi^k \rangle_{J=0}~=~0.$$

  4. This means that the 2-point function $$\langle \phi^k \phi^{\ell}\rangle_{J=0}~=~\langle \phi^k \phi^{\ell}\rangle^c_{J=0} $$ is equal to the connected 2-point function, i.e. only connected diagrams contribute. We also note that the denominator $Z[J\!=\!0]$ of vacuum bubbles has disappeared from the final expression. Altogether this answers OP's question.

References:

  1. M.D. Schwartz, QFT & the standard model, 2014; Subsection 7.2.5.

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$^1$ Here we use DeWitt's condensed notation.

Qmechanic
  • 201,751