If we have two Observables (bilinears of Diracfield $\psi(x)$) $O_1(x)=\bar{\psi}(x)\Gamma_1\psi(x)$ and $O_2(y)=\bar{\psi}(y)\Gamma_2\psi(y)$ and if we calculate their time ordered product $T(O_1(x) O_2(y))$ via Wick's theorem , why do we then only consider contractions where the variables $x$ and $y$ are different? why don't we also consider contraction between the same variables like $\bar{\psi}(x)\psi(x)$?

The contraction of two Diracfields $\psi(x)\bar{\psi}(y)$ is given by the Feynman-propagator $D_F(x-y) = \int\frac{d^4p}{(2\pi)^4}\frac{i}{\not{p} -m}e^{-ip(x-y)}$.

  • $\begingroup$ i dont know how to make a slash over p, which is equal to $\gamma^\mu p_\mu$ $\endgroup$ – laguna Jan 20 '18 at 10:56
  • 1
    $\begingroup$ \not{p} is the only way I know of, though it looks a little "off" to me $\endgroup$ – Chris Jan 20 '18 at 11:00
  • $\begingroup$ \not\!p looks lightly better, $\not\!p$. $\endgroup$ – AccidentalFourierTransform Jan 20 '18 at 15:11
  • $\begingroup$ Normal ordering seems to be implicit. $\endgroup$ – AccidentalFourierTransform Jan 20 '18 at 15:12

If you have a series of a group of operators defined at the same space time point (as you might find in e.g interaction terms in a lagrangian consisting of local operator terms for maintenance of locality) then one can derive that $$T((AB \dots)_{x_1} \dots (PQ \dots)_{x_n})_{\text{No E.T.C}} = T(N(AB \dots)_{x_1} \dots N(PQ \dots)_{x_n})$$

There is a nice argument for this in Mandl and Shaw where, within the $r^{\text{th}}$ group on the $\text{r.h.s}$ the operators have their time location shifted by $\pm \epsilon$ depending on whether they are a creator or an annihilator. As the groups are in normal ordering already, with this shift they become time ordered so a naive Wick expansion can be applied without considering contractions within a group (because there $T(AB \dots)_{x_{r\pm \epsilon}} \equiv N(AB \dots)_{x_{r\pm \epsilon}}$). The limit as $\epsilon \rightarrow 0$ of such an expansion gives the result on the $\text{l.h.s}$.

Why would one do this? Defining the $S$-matrix as the time ordered product of an exponentiated normal ordered interaction hamiltonian removes tadpole contributions from the game completely which are formally infinite, to be removed with e.g renormalisation counterterms in the path integral formalism but are otherwise set conveniently to zero in this Wick-ordered operator formalism from the outset.

Ref: 'Quantum Field Theory', Mandl & Shaw, 2nd edition, chp.6, P.97


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