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I am self-studying QFT using the book "A modern introduction to quantum field theory" by Maggiore. On page 124-125 he's doing the calculation in the interaction picture for a process with two initial particles $p_1$,$p_2$ and two final particles $k_1$,$k_2$ and $$\mathcal{H}_I={{\lambda}\over{4!}}\phi^4.\tag{5.68}$$ To get the LSZ formula he has to calculate the following integral at first order in $\lambda$:

$$\int d^4x_1 d^4x_2 d^4x_3 d^4x_4 e^{i(p_1x_1+p_2x_2+k_1x_3+k_2x_4)}$$ $${{\langle 0|T \{\phi(x_1)\phi(x_2)\phi(x_3)\phi(x_4) \exp\left[ -i\lambda/4!\int d^4x\phi^4\right]\}| 0 \rangle}\over{\langle 0|T\{\exp\left[ -i\lambda/4!\int d^4x\phi^4\right]\}|0\rangle}}\tag{5.87}$$

He calculates the numerator using Wick's theorem and get the following (the disconnected graphs can't cancel pole factors and they give zero on mass shell, the only connected Feynman graphs are obtained contracting each $\phi(x_i)$ with one of the four $\phi(x)$):

$$(-i\lambda)(2\pi)^4\delta^{(4)}(p_1+p_2-k_1-k_2)\left(\prod_{i=1}^{2}{{i}\over{p_i^2-m^2}} \right) \left(\prod_{j=1}^{2}{{i}\over{k_j^2-m^2}} \right).\tag{5.89}$$

So far so good, then on p.125 he has to consider the denominator $\langle 0|T\{\exp\left[ -i\lambda/4!\int d^4x\phi^4\right]\}|0\rangle$ and he states:

"The denominator gives only vacuum-to-vacuum graphs, i.e. Feynman diagrams with no external lines. However each contribution from numerator can be "dressed" with all possible vacuum-to-vacuum graphs, considering all possible disconnected graph made with the original graph plus all possible vacuum-to-vacuum graphs"

It's not clear to me mathematically where such "dressing" comes from, since he has already calculated the numerator.

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3 Answers 3

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Your second equation is not the full contribution resulting from the Wick decomposition of the numerator up to order $\lambda$ but only the (physically relevant) tree contribution of order $\lambda$. If you fully work out $$\langle 0 |{\rm T} \left\{\phi(x_1) \phi(x_2) \phi(x_3) \phi(x_4) \left(\mathbf{1} -(i\lambda/4!) \int\! d^4x \, \phi(x)^4 \right) \right\}|0\rangle,$$ you will find four different classes of contributions:

(i) A linear combination of products of two disconnected two-point functions like $$\langle |{\rm T }\phi(x_1) \phi(x_2) |0 \rangle \langle0 | {\rm T} \phi(x_3) \phi(x_4) |0\rangle $$ (in all possible combinations).

(ii) The connected tree diagram $$\sim -i \lambda \int\! d^4x \, \langle 0|{\rm T}\phi(x_1) \phi(x)|0\rangle \langle 0 |{\rm T}\phi(x_2) \phi(x)|0\rangle \langle 0 |{\rm T} \phi(x_3) \phi(x) |0\rangle \langle0|{\rm T} \phi(x_4) \phi(x) |0\rangle$$corresponding to the expression shown in your question (prior to Fourier integration).

(iii) A linear combination of the product of a tree propagator with a two-point function with a one-loop insertion like $$\sim \langle 0|{\rm T} \phi(x_1) \phi(x_2) |0\rangle\, \lambda \!\int \! d^4x \, \langle 0|{\rm T}\phi(x_3) \phi(x)|0\rangle \langle0 |{\rm T}\phi(x_4) \phi(x) |0\rangle \langle0|{\rm T}\phi(x) \phi(x) |0\rangle, $$ again in all possible combinations. (Again, a disconnected diagram.)

(iv) Graphs consisting of two diconnected tree propagators (like in (i)) and a two-loop vacuum bubble like $$\sim \langle 0|{\rm T}\phi(x_1) \phi(x_2) |0\rangle \langle 0|{\rm T} \phi(x_3) \phi(x_4) |0\rangle \, \lambda \!\int \! d^4x \langle 0| {\rm T}\phi(x) \phi(x) |0\rangle^2.$$

On the other hand, computation of the denominator up to the same order in $\lambda$ yields $$\langle 0 | {\rm T} \left\{ \mathbb{1} -(i \lambda/4!) \int \! d^4x \, \phi(x)^4 \right\}|0\rangle =1- \frac{i \lambda}{8}\int d^4x \langle 0 | {\rm T} \phi(x)\phi(x) |0\rangle^2.$$ Note that the contribution from the disconnected vacuum bubble exhibits an infrared divergence. Because of translation invariance, $\langle 0 |{\rm T} \phi(x) \phi(x) |0\rangle=\langle 0 |{\rm T}\phi(0) \phi(0)|0\rangle$ is a constant and the integration $$\int \! d^4x \, \langle 0 |{\rm T} \phi(0) \phi(0) | 0\rangle^2$$ requires an infrared regularization of space-time, $\int \! d^4x \to V T$. You can easily convince yourself that the contributions from the disconnected vacuum bubbles cancel in the ratio of your first equation to order $\lambda$ (remember to use the first two terms of the geometric series for the expansion of the denominator). In fact, contributions with disconnected vacuum bubbles cancel to all orders in $\lambda$ rendering the ratio infrared finite. (This does, of course, not remove the ultraviolet divergences present in $\langle 0 |{\rm T} \phi(0) \phi(0)|0\rangle$ or in the loop integrals.) This explains also, why diagrams with disconnected vacuum bubbles are never considered in actual calculation. On top of that, disconneted diagrams like (i) or (iii) do not contribute to $S$-matrix elements and (ii) remains the only tree contribution to the scattering cross section.

Of course, you can find this in all good text-books on quantum field theory. Nevertheless, you are strongly encouraged to work out the calculation sketched above in all details (keeping track of signs and factors) to familiarize yourself with the whole "machinery".

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  • $\begingroup$ It's much clearer now, thanks. I knew about the other terms of the Wick decomposition but I thought they should be ignored (in the book he says disconnected graph are unable to cancel all pole factors and they give zero on mass shell) $\endgroup$
    – Andrea
    Apr 18 at 6:31
  • $\begingroup$ @Andrea Yes, contributions from disconnected graphs do not survive the LSZ-procedure and only connected graphs contribute to $S$-matrix elements. $\endgroup$
    – Hyperon
    Apr 18 at 6:41
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This is a subtle point, illustrated far better than I can in some books linked below. First, I'll try to sketch the cancellation of the vacuum diagrams - Consider the corrections up to second order:

enter image description here

($s_1$ and $s_2$ are some symmetry factors that are not important for this discussion).

As you go to higher and higher orders, you end up with copies of the original diagram multiplied by ever more elaborate vacuum bubbles (disclaimer: the symmetry factors may be wrong):

enter image description here

Some subtle combinatorics can then be used to show that this series of vacuum bubbles is exactly

$$ \left\langle \exp \int -S[\phi] \right\rangle $$

which precisely cancels the denominator. It follows that the expansion of $$\frac{\left\langle \frac{\delta^n }{\delta \phi(j_1) ... \delta \phi(j_n)}\mathcal{Z}[j_n, \lambda] \Big|_{j=0} \right\rangle}{\langle \mathcal{Z}[0,\lambda] \rangle}$$ in $\lambda$ corresponds to only the connected components of the diagrams.

For a more detailed answer, see discussions of the "linked cluster theorem" in

  • Section 5.1.3 of Altland and Simons, "Condensed Matter Field Theory"
  • Section 4.4 of Peskin and Schroeder, (page 95 in my version)
  • Section 11.3 of Fradkin's "Quantum Field Theory" (brief)
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  1. The factorization between the numerator and denominator is perhaps best appreciated by remembering that an $n$-point function in the Heisenberg picture is $$ \langle \phi^{k_1}\ldots \phi^{k_n}\rangle_J~=~\frac{1}{Z[J]}\frac{\hbar}{i}\frac{\delta}{\delta J_{k_1}}\ldots \frac{\hbar}{i}\frac{\delta}{\delta J_{k_n}}Z[J],$$ and that $$Z[J]=\exp[\frac{i}{\hbar}W_c[J]],$$ where $W_c[J]$ is the generating functional of connected diagrams, cf. the linked-cluster theorem.

  2. A very similar factorization takes place in the interaction picture, cf. e.g. eq. (5.67) and this related Phys.SE post.

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  • $\begingroup$ Unfortunately I can't appreciate it yet, cause I have not studied path integral formulation yet, but I will come back to it. Btw, which is the related post? $\endgroup$
    – Andrea
    Apr 18 at 7:18
  • $\begingroup$ I updated the answer. $\endgroup$
    – Qmechanic
    Apr 18 at 7:26

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