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Let's have s-matrix: $$ S_{\alpha \beta} = \langle \alpha | \hat {S} | \beta \rangle , $$ $$\hat{S} = \hat{T}e^{-i\int \hat{L}(x)d^{4}x}, \quad \hat{T} \left( \hat{\Psi}(t) \hat{\Psi}(t') \right) = \theta (t - t')\hat{\Psi}(t)\hat{\Psi}(t') \pm \theta (t' - t)\hat{\Psi}(t')\hat{\Psi}(t). $$ How to show that matrix elements diverges because time ordering is not well-defined operation when $t=t'$?

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    $\begingroup$ I have not time enough to properly answer. However the $T$ product is ill-defined when you consider Wick products, as they involve products of Feynman propagators (i.e. loops). These products are not well defined and the convolutions of them arising in computing the terms in the $S$-matrix expansion are consequently ill-defined. The problem shows up when arguments coincide ($x=x'$). Renormalization is a procedure to adjust these distributions, defining the $T$ product, before computing the convolutions. This procedure is ambiguous, giving rise to the well known renormalization counterterms. $\endgroup$ – Valter Moretti Jun 25 '14 at 20:17
  • $\begingroup$ Thank you, but I want to make another one detail clear. I But why do we think that ill definition of $T$ product leads to infinities? It may be the problem of theory which arises independently of ill definition of $T$ product, and by redefinition $T$-product we were lucky (!) to get rid of this problem. $\endgroup$ – Andrew McAddams Jun 26 '14 at 16:51
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    $\begingroup$ Infinities are not the problem just a symptom if one decide to follow a way. The properties of the T product do not completely fix it in view of the fact that the interactions are supposed to be described by Wick polynomials. So the T products must be fixed and ambiguities remain. These ambiguities are the finite renormalization counterterms. This approach is the so-called Epstein-Glaser's one. No infinities arise this way and it encompasses all possible definitions of T product. The procedure extends to QFT in curved spacetime. $\endgroup$ – Valter Moretti Jun 26 '14 at 17:11
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Mathematically one way to see it is that the (combined) step functions become Dirac pulses i.e $\theta(t) \to \delta(t)$, which diverge. When $t \to t'$, the fields have nearly identical values, thus the time-ordered product involving step functions degenerates into dirac pulse (a dirac pulse is the derivative of the step function)

Physically another way to see this is: S-matrix describes interactions between states and particles, as such a matrix element for $t \to t'$, requires an interaction to take place instanteneously, thus it would require infinite energy at that point (also related to time-energy unceratinty).

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