Suppose we have an amplitude $M^\mu\equiv\langle p|J^\mu(q)|0\rangle$, where $J$ is a conserved current so$$q_\mu M^\mu=0,$$and $\langle p|$ is a particle of momentum $p$ corresponding to the renormalized field $\varphi$ $$\langle0|\varphi(0)|p\rangle=1.$$

We can consider the off-shell amplitude

$$A^\mu(p,q)\equiv \int dx^4 dy^4 e^{ipx}e^{-iqy}\langle 0|T\varphi(x)J^\mu(y)|0\rangle,$$ which by LSZ reduction should be dominated by a pole when $p$ goes on mass shell, with a residue equal to $M$ $$A^\mu(p^0\rightarrow E_p,q)=\frac{i}{p^2-m^2}M^\mu(q)+\dots$$

Now the problem is if you dot $A$ with $q$, the RHS should vanish as above, but the LHS has a contact term due to the time ordering $$q_\mu A^\mu = i\int dx^4 dy^4 e^{ipx}e^{-iqy}\delta(x^0-y^0)\langle 0|[\varphi(x),J^0(y)]|0\rangle.$$

So the LHS does not equal zero in general (I did ignore boundary terms when I did integration by parts), but the RHS does (I did ignore the non-pole terms). What am I missing?


1 Answer 1


Contact terms vanish in the on-shell limit $k^2\to m^2$. See M. Srednicki's book, chapters 67, 68 for a more or less detailed discussion.

In a nutshell, a Dirac delta $\delta(x_1-x_2)$, in Fourier space, becomes a function of $k_1+k_2$, but it is independent of $k_1-k_2$. Therefore, a contact term cannot have a pole $$ \frac{1}{k_1^2-m^2}\frac{1}{k_2^2-m^2} $$ and therefore does not contribute to the pole of the correlation function; this in turn implies that such a term does not contribute to the $S$-matrix element.

  • $\begingroup$ Thank you for the reference and your answer. I think I follow the reasoning, but I'm a little puzzled because I have seen examples involving pions where the contact terms lead to commutators involving the axial current that help you to calculate the on-shell amplitude. $\endgroup$
    – octonion
    Feb 3, 2017 at 3:07

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