LSZ reduction formula relates the matrix element of the scattering operator to the n-point Green's function $$\langle 0|\phi(x_1)\phi(x_2)...\phi(x_n)|0\rangle$$ My question is:

  1. Is the vacuum on the left same as that of the right of this expression? Or these are different in the sense that one is the vacuum of "in" Fock space and the other is the vacuum of the "out" Fock space? These are the vacuum of the free fields. Right?

  2. How is the above expression related to

$$\langle \Omega|\phi(x_1)\phi(x_2)...\phi(x_n)|\Omega\rangle$$

where $|\Omega\rangle$ is the vacuum of the interacting theory. Are these to quantities same? Which of the above expression really called n-point Green's function. In Peskin and Schroeder, they used $$\langle \Omega|\phi(x_1)\phi(x_2)|\Omega\rangle$$ as the 2-point Green's function and not $$\langle 0|\phi(x_1)\phi(x_2)|0\rangle.$$ I'm little confused.

  • 2
    $\begingroup$ 1. The same. In the notations of P&S, $| 0 \rangle$ is the vacuum of free field theory. "In" and "Out" states are in the same Fock space. Like Weinberg's QFT vol I p113 "Perhaps it should be stressed that 'in' and 'out' states do not inhabit two different Hilbert spaces. They differ only in how they are labelled: by their appearance either at $t \rightarrow -\infty$ or $t \rightarrow + \infty$. 2. The vacua and correlation functions in free and interacting theories are in-general different. Ref. (4.31) in P&S. $\endgroup$
    – user26143
    Mar 9, 2014 at 13:41
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    $\begingroup$ Be careful Roopam, the LSZ formulas concerns the $T$-ordered n-point functions and not the the "simple" n-point (Wightman) functions you wrote... $\endgroup$ Mar 9, 2014 at 18:03

1 Answer 1


$|\Omega\rangle$ is the vacuum of the full interacting theory and $|0\rangle$ is the vacuum of the free theory. They are related in the following way $$ |\Omega\rangle = \lim_{T\rightarrow(1-i\epsilon)\infty} (e^{-iE_0(T+t_0)}\langle \Omega|0\rangle)^{-1}e^{-iHT}|0\rangle $$ and the correlators $$ \langle \Omega|\phi(x_1)\phi(x_2)...\phi(x_n)|\Omega\rangle = \lim_{T\rightarrow(1-i\epsilon)\infty} \frac{\langle 0|\phi(x_1)_I\phi(x_2)_I...\phi(x_n)_Ie^{i\int_{-T}^T\mathcal{L}d^4x}|0\rangle}{\langle 0|e^{i\int_{-T}^T\mathcal{L}d^4x}|0\rangle} $$ where the fields at the RHS are in the interaction picture. This relation is exact, you can follow the derivation in Peskin & Schroeder section 4.2. Moreover expanding the exponential we may express this as a perturbative series which can be computed with the aid of Feynman diagrams. The disconnected (vacuum) diagrams factor and cancel with the denominator. This gives the usual way to calculate the correlator as the sum of all connected diagrams (when you renormalise you "get rid" of the non-amputated ones too).

  • $\begingroup$ But isn't this expression of Pesskin and Schroeder contradicts expression 5.67 of Michele Maggiore? Still one confusion is left. Which of above expressions appears in LSZ reduction formula? The correlator with $|\Omega\rangle$ or the other with $|0\rangle$? $\endgroup$
    – SRS
    Mar 9, 2014 at 13:58
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    $\begingroup$ About the Maggiore reference, I think he just uses the same notation for both vacuums. Withe the Peskin notation the correct expression is the one with the $|\Omega\rangle$ since you are interested in S-matrix elements of the interacting theory. In the free theory the particles just propagate and "cross" their ways. $\endgroup$
    – jpm
    Mar 9, 2014 at 14:16
  • $\begingroup$ It is irrelevant to use two different vacua rather than the same for free and interacting theory (and you find the two versions in the literature), in view of the so called "Haag's theorem" that proves that the formalism is non completely consistent. On the other hand, it does not matter because of the overall renormalization problem that makes all these issues very academic... $\endgroup$ Mar 9, 2014 at 18:06

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