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How to replace T-product with commutator in LSZ formula

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502 views

I am reading Itzykson and Zuber's Quantum Field Theory book, and am unable to understand a step that is made on page 246:

Here, they consider the elastic scattering of particle $A$ off particle $B$:

$$A(q_1) + B(p_1) \rightarrow A(q_2) + B(p_2)$$

and proceed to write down the S-matrix element using the LSZ formula, with the $A$ particles reduced:

$$S_{fi}=-\int d^4x\, d^4y e^{i(q_2.y-q_1.x)}(\square_y+m_a^2)(\square_x+m_a^2)\langle p_2|T \phi^\dagger(y) \phi(x)|p_1 \rangle\qquad \text{(5-169)}$$

Then they say that because $q_1$ and $q_2$ are in the forward light cone, the time-ordered product can be replaced by a retarded commutator:

$$T \phi^\dagger(y) \phi(x) \rightarrow \theta(y^0-x^0)[\phi^\dagger(y),\,\phi(x)]\,.$$

This justification for this replacement completely eludes me. What is the mathematical reason for this?


This post imported from StackExchange Physics at 2014-07-08 08:19 (UCT), posted by SE-user QuantumDot

asked Jul 6, 2014 in Theoretical Physics by QuantumDot (195 points) [ revision history ]
edited Jul 8, 2014 by suresh

1 Answer

+ 4 like - 0 dislike

Try writing the Feynman propagator in terms of the retarded and advanced Green functions. terms i.e.,  $T(\phi^\dagger(y)\phi(x))=\theta(y^0-x^0) \phi^\dagger(y)\phi(x) + \theta(x^0-y^0) \phi(x)\phi^\dagger(y)$. Now show that (i) the second term vanishes and (ii) it is possible to replace the first term by the commutator.

answered Jul 8, 2014 by suresh (1,535 points) [ no revision ]
edited Jul 12, 2014 by Arnold Neumaier

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