Contact terms in proof of Ward identity

I'm confused about how a contact term vanishes when proving the Ward identity, i.e. the spot immediately following equation 5.52 in Weigand's notes. Spelling out everything concretely, we consider a process with an external photon and some external fermions, giving an $$S$$-matrix element $$\langle f | i \rangle \sim \xi^\mu \int dx dx_1\ldots \, \partial_x^2 \not\partial_1 \ldots \langle 0 | T A_\mu(x) \psi(x_1)\ldots | 0 \rangle$$ and apply the Schwinger-Dyson equations to turn this into $$\langle f | i \rangle \sim \xi^\mu \int dx dx_1 \ldots \, \not\partial_1 \ldots \langle 0 | T j_\mu(x) \psi(x_1) \ldots | 0 \rangle$$ while throwing away a contact term, as explained here. Now we set the photon polarization $$\xi^\mu = k^\mu$$ and integrate by parts for $$\langle f | i \rangle \sim \int dx dx_1 \ldots\, \not\partial_1 \partial^\mu \ldots \langle 0 | T j_\mu(x) \psi(x_1) \ldots | 0 \rangle.$$ We now apply the Ward-Takahashi identity, generating more contact terms which are supposed to be zero; however, I don't understand why they are. Directly applying Ward-Takahashi gives a contact term for every external fermion, one of which has the form $$\langle f | i \rangle \sim \int dx dx_1 \ldots \, \not\partial_1 \langle 0 | T e\psi(x_1) \ldots | 0 \rangle \delta(x - x_1) \propto \int dx_1 \ldots \, \not\partial_1 \ldots \langle 0 | T \psi(x_1) \ldots| 0 \rangle.$$ Unlike the previous case, this seems to have precisely the right kind of pole structure; it's exactly what you would get if you applied LSZ.

What am I missing here? Why doesn't this contact term contribute?

• Isn't $\langle f\vert i\rangle$ still supposed to be a connected S-matrix element so the reasoning is still the same? – ACuriousMind Dec 9 '17 at 18:12
• @ACuriousMind Could you spell out the reasoning in a little more detail? – knzhou Dec 9 '17 at 18:17
• @ACuriousMind In particular I don't see why this is disconnected. If I had to translate this contact term in a diagram, it would look like 'incoming fermion absorbs the incoming photon, and then <rest of diagram>, and the outgoing fermion leaves' which is not necessarily disconnected. – knzhou Dec 9 '17 at 18:31
• The $\not\!\partial\not\partial\langle\bar\psi\psi\rangle$ term does not have the correct pole structure. As per Källén–Lehmann, $\langle\bar\psi\psi\rangle\sim \frac{1}{\not p-m_e}$. OTOH, the expected pole structure is $\frac{1}{\not p-m_e}\frac{1}{\not k-m_e}\frac{1}{k_\gamma^2}$. – AccidentalFourierTransform Dec 9 '17 at 20:19
• @AccidentalFourierTransform Ah, I'm regretting not learning LSZ better. I thought the LSZ reduction formula for two fermions really did look like $\not\partial_1 \not \partial_2 \langle \bar{\psi} \psi \rangle$. What's the key difference between what I wrote and the actual formula? – knzhou Dec 9 '17 at 22:27

• The number of fields is not incorrect. There are exactly as many fermion fields as we started with, and exactly as many $\not\partial$ factors.
The real issue is that the location of the pole has been shifted by the contact term. To see this, we restore the exponential factors for $$\langle f | i \rangle \sim \int dx dx_1 \ldots \, e^{ikx} e^{ik_1 x_1} \ldots \not\partial_1 \langle 0 | T e\psi(x_1) \ldots | 0 \rangle \delta(x - x_1).$$ Evaluating the delta function, we find $$\langle f | i \rangle \sim \int dx_1 \ldots \, e^{i(k+k_1) x_1} \ldots \not\partial_1 \langle 0 | T e\psi(x_1) \ldots | 0 \rangle.$$ This is exactly the LSZ reduction formula for a situation with an external fermion with momentum $k+k_1$. The issue is that $k + k_1$ is never on-shell since we required $k_1$ to be on-shell and $k^2 = 0$, so we don't get a $1/(k+k_1)$ pole from the correlation function. In other words, while the correlation function does have the right number of poles, it doesn't have them in the right places.