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Weinberg Volume I Equation (10.5.10) on page 451

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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

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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 (2,465 points) [ revision history ]
edited Jul 18, 2015 by Jia Yiyang

@JiaYiyang Very clear answer! Thank you very much!

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