1
$\begingroup$

I am studying Peskin and Schroeder's textbook of quantum field theory.

I have proceeded to Ward-Takahashi identity and have one question.

Eq.(7.66) and Eq.(7.67) are the two cases involved. Then the textbook proceeds to discuss the Ward-Takahashi identity by summing all possibilities for different Feynman diagrams (with external photon line striped) as well as different ways of inserting photons, where $M^{\mu}$ is the correlation function for $n$ inserting electrons and $n$ out-going electrons. $$k_{\mu}M^{\mu}(k;p_1...p_n;q_1...q_n)=-e\sum_i[M_0(k;p_1...p_n;q_1...(q_i-k)...q_n)-M_0(k;p_1...(p_i+k)...p_n;q_1...q_n)].$$

It is understood that if ($M^{\mu}$) has all its external electrons on-shell, then the amplitudes on the right-hand side of this identity each have only one external particle off-shell, and therefore they do not contribute to S-matrix elements. As a result, the corresponding S matrix is zero if all the external electrons in the left hand side are on shell.

I understood that the discussions related to Eq.(7.66) only involve the singularity factor due to the on-shell external electron lines, and therefore applies to any individual Feynman diagram which possesses such external electron line structure (while enumerating all different ways of inserting the additional photon line). On the other hand, Eq.(7.67) gives zero by itself.

However, the textbook mentioned on P.238:

The identity is generally not true for individual Feynman diagrams; we must sum over the diagrams for $M(k)$ at any given order.

My question is (1) Why one still need the summation of all possible Feynman diagrams. It seems individual Feynman diagram with all possible ways of insertion of photon lines do the trick. (2) P.242 in the first plot, the incoming and outgoing Fermi lines are paired ($p_i$ to $q_i$). Why one does not consider the case when $p_i$ is paired to $q_j$ ($i\ne j$), it seems to me that this also contributes to the same physical process. Many thanks for the answer or comment!

PS: This question is a follow-up of a question by Brioschi.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.