Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,355 answers , 22,793 comments
1,470 users with positive rep
820 active unimported users
More ...

  What is the sense of introducing generating functional to the summands of expansion of S-matrix?

+ 4 like - 0 dislike
1050 views

Let's have generating functional $Z(J)$: $$ Z(J) = \langle 0|\hat {T}e^{i \int d^{4}x (L_{Int}(\varphi (x)) + J(x) \varphi (x))}|0 \rangle , \qquad (1) $$ where $J(x)$ is the functional argument (source), $\hat {T}$ is the chronological operator, $\varphi (x)$ - some field.

I want to understand the reasons for its introduction for the summands of expansion of S-matrix. As I read in the books, it helps to consider only the vacuum expectation values​​, forgetting about in- and out-states. But in $(1)$ appear summands like $\int \frac{J(p)dp}{p^2 - m^2 + i0}$ instead of the contributions from external lines. It may refer to the internal lines. So what to do with them and are there some other reasons to introducing $(1)$ except written by me?

This post imported from StackExchange Physics at 2014-03-24 04:01 (UCT), posted by SE-user Andrew McAddams
asked Nov 27, 2013 in Theoretical Physics by Andrew McAddams (340 points) [ no revision ]
Comment ot the question (v2): For a connection between off-shell correlation functions and on-shell S-matrix elements, see LSZ reduction formula.

This post imported from StackExchange Physics at 2014-03-24 04:01 (UCT), posted by SE-user Qmechanic

2 Answers

+ 4 like - 0 dislike

The primary utility in introducing the generating functional is in using it to compute correlation functions of the given quantum field theory.

Let's restrict the discussion to that of a theory of a single, real scalar field on Minkowski space, and let $x_1, \dots, x_n$ denote spacetime points. Of central importance are time-ordered vacuum expectation values of field operators evaluated at such points; \begin{align} \langle0|T[\phi(x_1)\cdots\phi(x_n)]|0\rangle. \end{align} It can be shown that these objects can be obtained from the generating functional by taking functional derivatives with respect to the $J(x_i)$ as follows: \begin{align} \langle0|T[\phi(x_1)\cdots\phi(x_n)]|0\rangle = \frac{1}{Z[0]}\left(-i\frac{\delta}{\delta J(x_1)}\right)\cdots \left(-i\frac{\delta}{\delta J(x_n)}\right)Z[J]\Bigg|_{J=0}. \end{align} This standard fact is proven in many books on QFT. It's often proven using the path integral approach which makes it pretty transparent why it's true. The crux of the argument is that every time you take a functional derivative with respect to the source $J(x_i)$, it pulls down a factor of the field $\phi(x_i)$. Dividing by $Z[0]$ is an important normalization relating to vacuum bubbles, and setting $J=0$ after computing the appropriate functional derivatives eliminates terms with more than $n$ factors of the field and renders the final result source-independent as it should be.

This post imported from StackExchange Physics at 2014-03-24 04:01 (UCT), posted by SE-user joshphysics
answered Nov 27, 2013 by joshphysics (835 points) [ no revision ]
+ 3 like - 0 dislike

If you can calculate vacuum-to-vacuum transition amplitudes, you can calculate S-matrix elements, because the two are related by the LSZ reduction formula. The LSZ will in any case chop off the propagators for external lines that the generating functional inserts, so you will end up only needing to compute amputated diagrams.

This post imported from StackExchange Physics at 2014-03-24 04:01 (UCT), posted by SE-user lionelbrits
answered Nov 27, 2013 by lionelbrits (110 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysics$\varnothing$verflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...