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(This post is a bit long; the key question is between the horizontal lines, and an example follows it.)

I am following the derivation of the Ward identity in Schwartz's QFT book, and there is a key step that I just cannot understand. We begin with the Ward-Takahashi identity for correlation functions involving the QED current:

$$\partial_\mu \langle j^\mu(x) \psi(x_1) \bar{\psi}(x_2) \cdots \rangle = (-\delta(x-x_1) + \delta(x-x_2) + \cdots) \langle \psi(x_1) \bar{\psi}(x_2) \rangle,$$

where we get more contact terms for each $\psi$ or $\bar{\psi}$ in the correlation function; I am using the shorthand $\langle \cdots \rangle = \langle \Omega | T \cdots | \Omega \rangle$.

Now we want to deduce the Ward identity, which says that if $\epsilon_\mu M^\mu$ is an amplitude involving an external photon with momentum $k$ and polarization $\epsilon$, then $k_\mu M^\mu = 0$. A crucial point is that while the external fermions involved in the amplitude must be on-shell, the photons can be off-shell, i.e., $k^2$ is not necessarily zero. This is important since (as I understand it) it lets us use the Ward identity for internal photon lines in a bigger diagram.

For this, we use the Schwinger-Dyson equation for the photon (in Feynman gauge for simplicity, though it doesn't matter):

$$\partial_{x_1}^2 \langle A^\mu(x_1) A^\nu(x_2) A^\rho(x_3) \psi(y_1) \bar{\psi}(y_2) \cdots \rangle = e \langle j^\mu(x_1) A^\nu(x_2) A^\rho(x_3) \psi(y_1) \bar{\psi}(y_2) \cdots \rangle + i \delta(x_1-x_2) g^{\mu\nu} \langle A^\rho(x_3) \psi(y_1) \bar{\psi}(y_2) \cdots \rangle + \cdots$$

where again we get a contact term for each $A$ field in the correlation function. It is claimed that these contact terms don't contribute:

The second term represents a contraction of two external photons with each other. (...) Since the contraction of two external photons gives a disconnected Feynman diagram, it does not contribute to the S-Matrix.

(Schwartz, Quantum Field Theory and the Standard Model, p. 281)

This seems slightly sketchy. I don't see why we can throw away a term because of its structure; we should be able to prove that it is zero in the context we're working in. Still, there's also this:

However, the contact terms cannot generate singularities in the $k^2$s of the other particles [i.e., not the one we turned into a current], and so they do not contribute to the left-hand side. (Remember that, for each of the other particles, there is still an appropriate wave operator (...) acting on the correlation function. These wave operators kill any term that does not have an appropriate singularity.

(Srednicki, Quantum Field Theory, p. 410, square brackets by me)

So it seems like the argument is that when we insert this correlation function into the LSZ formula and put the other particles (i.e., not the photons) on-shell, the contact terms won't have the right poles to be cancelled and hence will be zero. After this, we use the Ward-Takahashi identity (the first formula in the post) to turn $k_\mu M^\mu$ into more contact terms, which by the LSZ formula vanish when the fermions go on-shell.


My problem is that it is not clear that the first contact terms, the ones that appear in $\partial^2 \langle A \cdots \rangle = e \langle j \cdots \rangle + \text{contact terms}$, indeed vanish with on-shell electrons but off-shell photons.


For example, suppose we start with a function with three photons and two fermions:

$$\langle A^\mu(x_1) A^\nu(x_2) A^\rho(x_3) \psi(y_1) \bar{\psi}(y_2) \rangle.$$

The LSZ formula will act on this with operators of the form $\partial_{x_i}^2$ and $-i \not \partial_{y_i}+m$. Call the photon momenta $p_i$ and the fermion momenta $q_i$. Then we will get factors of $p_i^2$ and $\not q_i - m$. Now consider the first contact term from above:

$$i \delta(x_1-x_2) g^{\mu\nu} \langle A^\rho(x_3) \psi(y_1) \bar{\psi}(y_2) \rangle$$

The books claim that the three point function doesn't have the right pole structure. But it does! After Fourier transforming, it is

$$\frac{-i}{p_3^2} \frac{i}{\not q_1 - m} \gamma^\rho \frac{i}{\not q_2 - m}$$

which has exactly the right structure to be cancelled by the operators in the LSZ formula. Now, if the photons are put on-shell then the part with the delta function will indeed give zero, so for on-shell S-matrix elements everything is fine. But the Ward identity is supposed to be valid for off-shell photons (Peskin and Schroeder do prove it, in a different way). So what happens to the contact terms? Do they vanish like Srednicki says? Have I made a mistake? Or do we throw them away because they are disconnected, like Schwartz says? How do we justify that?

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  • $\begingroup$ I think I asked and self-answered a very similar question here, with almost the exact same title even. Not sure if it's the same issue since I've only skimmed your question, but hopefully it helps! $\endgroup$
    – knzhou
    Commented Aug 9, 2018 at 14:07
  • $\begingroup$ @knzhou Well, my question is about the part where you say "while throwing away a contact term" and link to another question. The answer seems to be that throwing the terms away because they are disconnected is the correct reasoning. TBH, I really hate it when books don't say things like these explicitly. $\endgroup$
    – Javier
    Commented Aug 9, 2018 at 14:12
  • $\begingroup$ I know, right? It wouldn’t even take that many extra pages to do that, for a whole book. $\endgroup$
    – knzhou
    Commented Aug 9, 2018 at 14:23
  • $\begingroup$ Just a minor comment: contact terms do not affect physical amplitudes. The amplitude for which you want to prove the Ward Identity $p_\mu \mathcal{M}^\mu$ is the amplitude of three photons and two fermions. Apply the LSZ on the LHS of $\partial^2 \langle A \cdots \rangle = e \langle j \cdots \rangle + \text{contact terms}$. Then, the contact terms has also $p_1^2$ and $p_2^2$ which, on-shell, vanish. In other words, it seems to me the contact terms does not have the right pole structure. $\endgroup$
    – apt45
    Commented Aug 10, 2018 at 7:31
  • $\begingroup$ @apt45 But the photons don't need to be on-shell to prove the Ward identity. That's the point. $\endgroup$
    – Javier
    Commented Aug 10, 2018 at 11:06

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