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I'm a learner of Peskin and Schroeder's textbook of quantum field theory.

I have proceeded to Ward-Takahashi identity and have one question when I look for Wikipedia for reference.

The following is the Ward-Takahashi identity, where $M^{\mu}$ is the correlation function for $n$ inserting electrons and $n$ out-going electrons. $$k_{\mu}M^{\mu}(k;p_1...p_n;q_1...q_n)=-e\sum_i[M_0(k;p_1...p_n;q_1...(q_i-k)...q_n)-M_0(k;p_1...(p_i+k)...p_n;q_1...q_n)].$$

The wiki says that

Note that if ($M^{\mu}$) has its external electrons on-shell, then the amplitudes on the right-hand side of this identity each have one external particle off-shell, and therefore they do not contribute to S-matrix elements.

Does on-shell means a divergent contribution according to LSZ reduction formula?

Besides, can you tell me why the S matrix is zero if all the external electrons in the left hand side are on shell?

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The S matrix is zero for all left handed external electrons is due to the property of gamma matrices: $(1-\gamma^5)\gamma^{\mu}(1-\gamma^5)=0$. – soliton Apr 19 '13 at 13:40
@soliton Can you give some explanation about your equation? Thank you! – Brioschi Apr 20 '13 at 3:28
For all left handed external electrons in QED, the amplitude has the form as $\bar{u}(p')(1-\gamma^5)\gamma^{\mu}(1-\gamma^5)u(p)$ since $\psi_L=\frac{1}{2}(1-\gamma^5)\psi$. – soliton Apr 20 '13 at 3:58
You can refer to Peskin & Schroeder's book Chapter 5.2, page 142. – soliton Apr 20 '13 at 4:07
Not sure if this was in jest or not, but OP asked about the left-hand side of the equation, not left-handed particles. – gn0m0n Dec 22 '13 at 3:41

With "on-shell" in QFT one usually refers to the mass-shell in momentum space defined by $$ p_\mu p^\mu = m^2$$ Generally while performing calculations the momenta of the fields do not need to fulfill this relation. Still, any external particles that are identified with actual physical particles have to fulfill it. This is why the $M$ must have the external electrons on-shell in order to contribute to physical processes.

To the second question: $k^\mu$ is the four-momentum of a photon, i.e. it fulfills $$k_\mu k^\mu = 0$$ If you subtract this from an electron's on-shell momentum it no longer fulfills the correct mass relation: $$(p-k)_\mu (p-k)^\mu = p^2 + k^2 - 2 p \cdot k = m^2 + 0 + 2 p \cdot k \neq m^2$$ If the electrons were on-shell on the left hand side of your equation, they are not on-shell for any of the $M_0$ on the right-hand side and therefore the sum is zero.

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Thank you for your answer. Besides, when defining mass shell, $p^{\mu}p_{\mu}=m^2$, is m the bass mass or physical mass? – Brioschi Apr 19 '13 at 11:13
Well, if you define the mass-shell for physical particles, I'd say it's the pole mass. Generally though, afaik this is not defined and the value of $m^2$ may change depending on the renomalization scale. – Neuneck Apr 22 '13 at 9:13

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