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
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Weinberg Volume I Equation (10.5.10) on page 451

+ 3 like - 0 dislike
1535 views

I originally asked in physics stackexchange, getting no answers within 2 days. http://physics.stackexchange.com/questions/194355/weinberg-volume-i-equation-10-5-10-on-page-451

I copy it here: $$\Delta'_{\mu\nu}(q)=\Delta_{\mu\nu}(q)+\Delta_{\mu\rho}(q) M^{\rho\sigma}(q) \Delta_{\sigma\nu}(q)$$.

$\Delta_{\mu\nu}(q)$ is the bare photon propagator, and $\Delta'_{\mu\nu}(q)$ is the complete photon propagator. Weinberg says the $M^{\rho\sigma}$ is the matrix element of two currents between vacuum states ($\alpha$ and $\beta$ are taken to be vacuum states in the following), proportional to equation (10.5.1): $$M^{\mu\mu'...}_{\beta\alpha}(q,q',...)=\int d^4x\int d^4x' e^{-iqx}e^{-iq'x'}\times(\Psi^-_b,T\{J^{\mu}(x)J^{\mu'}(x')...\}\Psi^+_a)$$, for arbitrary transition $\alpha\rightarrow\beta$.

It seems related to the form of photon propagator in the Heisenberg picture form. Could you explain this to me? Thank you in advance!

asked Jul 17, 2015 in Theoretical Physics by ruifeng14 (65 points) [ no revision ]

Isn't it a matter of definition?

Do you know how to draw the Feynman diagrams (full propagators) of $\langle A^\mu (x_1) A^\nu (x_2)\rangle$ and $\langle J^\rho (y_1) J^\sigma (y_2)\rangle$? The relation is rather transparent in the diagrams.

@Jia Yiyang I can understand the perturbative form in terms of one-photon-irreducible diagrams, but not aware of the full form in the question. Could you explain it to me? Thanks!

@ruifeng14, I've written you an answer. By the way to properly @ a user you need to eliminate all the spaces in his/her name. For example to @ me you need to use "@JiaYiyang" instead of "@Jia Yiyang".  

@JiaYiyang Thanks for your advice.

1 Answer

+ 3 like - 0 dislike

For a time-ordered product like $\langle T\{O(x_1)O(x_2)\cdots O(x_n)\}\rangle$, the Feynman diagram of it is like this: external lines are the lines that represent $O(x_1)$,$O(x_2)$...$O(x_n)$, and you try to connect these external lines with the internal lines and vertices available, and whenever you have a valid graph, this graph contributes to $\langle T\{O(x_1)O(x_2)\cdots O(x_n)\}\rangle$. Hence $\langle T\{O(x_1)O(x_2)\cdots O(x_n)\}\rangle$ is a sum of all such graphs, and we may graphically represent the sum of all such graphs in a single graph with the summed internal process as a blob. If you apply the above described picture to  $\langle T\{A^\mu (x_1) A^\nu (x_2)\}\rangle$ and $\langle T\{ J^\rho (y_1) J^\sigma (y_2)\}\rangle$, you immediately see the relation between the two:

Namely, the graphs that represent $\langle T\{A^\mu (x_1) A^\nu (x_2)\}\rangle$ can be divided into two classes:

(1)A direct Wick contraction between $A^\mu (x_1)$ and $A^\nu (x_2)$;

(2) $A^\mu (x_1)$ and $A^\nu (x_2)$ are connected to two internal vertices $\int \text{d}y_1 A_\rho(y_1) J^\rho(y_1)$ and $\int \text{d}y_2 A_\sigma(y_2) J^\sigma(y_2)$, and hence all the graphs that come with these two prescribed vertices(as external lines of the graph representing $M^{\rho \sigma}$). Note that $J^\mu(x)=\bar{\psi}(x)\gamma^\mu \psi(x)$, so one external line of $J^\mu$ is really two lines of $\psi(x)$ pinched to the same point $x$. 

answered Jul 18, 2015 by Jia Yiyang (2,640 points) [ revision history ]
edited Jul 18, 2015 by Jia Yiyang

@JiaYiyang Very clear answer! Thank you very much!

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$\hbar$y$\varnothing$icsOverflow
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
...