Consider an operator product $\bar{\psi}(x+\epsilon)\psi(x)$ in which $\psi(x)$ is the fermion field operator in QED. We want to compute the singular term in $\bar{\psi}(x+\epsilon)\psi(x)$ in the limit $\epsilon\rightarrow0$. In Peskin's book, an introduction to quantum field theory (page 660), the leading singular term ``is given by contracting the two operators using a free-field propergator'' as follows,

\begin{eqnarray*} \text{contraction of }\bigg(\psi(x+\epsilon)\bar{\psi}(x)\bigg) & = & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}}e^{-ik\cdot\epsilon}\\ & = & -\frac{i}{4\pi^{2}}\gamma\cdot\partial\bigg(\frac{1}{\epsilon^{2}}\bigg)\\ & = & -\frac{i}{2\pi^{2}}\frac{\gamma\cdot\epsilon}{\epsilon^{4}} \end{eqnarray*}

Question: why the leading singular term is given by contracting the two operators ?


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.