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Considering $2 \rightarrow 2$ scattering in $\phi^4$, this loop diagram gives a contribution of

$$\int{dx_{1}dx_{2}dy_{1}dy_{2}dk_{1}dk_2 dp_1 dp_2 dq_1 dq_2 e^{-ik_1 x_1}e^{-ik_2 x_2}e^{ip_1 y_1}e^{ip_2 y_2}\frac{i}{q_{1}^2 - m^2}\frac{i}{q_{2}^2 - m^2} \\\ \delta(k_1 + k_2 - q_1 - q_2)\delta(p_1 + p_2 - q_1 - q_2)e^{-ik_1 x_1}e^{-ik_2 x_2}e^{ip_1 y_1}e^{ip_2 y_2}}.$$

$k_1$ and $k_2$ are the momenta of incoming particles, $q_1$ and $q_2$ are the loop momenta, and $p_1$ $p_2$ are the moments of the outgoing particles.

The delta functions imply that momentum is conserved at each vertex. I see that integrating over $q_1$ will give an overall delta function of $\delta(k_1 + k_2 - p_1 - p_2)$ implying that momentum is conserved overall. However, I’m not sure what we are supposed to do with the plane waves like $e^{-ik_1 x_1}$ in the integral. Naively it looks like the plane waves that come from LSZ will cancel with the plane waves that come from the external propagators since they have different signs, but this doesn’t seem right to me.

So how are we supposed to simplify the expression further?

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  • $\begingroup$ Also, it looks like there should be double plane waves due to each plane wave in the LSZ formula $\endgroup$
    – user310742
    Jul 8, 2023 at 16:42
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    $\begingroup$ Why is there an integral over k1, k2, p1, p2 at all? Are these not the external momenta? $\endgroup$ Jul 8, 2023 at 17:07
  • $\begingroup$ Because each external propagator contains an integral over momenta. It does seem that that should drop out, but I don’t see how it does $\endgroup$
    – user310742
    Jul 8, 2023 at 17:35
  • $\begingroup$ LSZ will tell you how to go from a vacuum expectation value (VEV) of fields to a scattering amplitude of on-shell particles corresponding to these fields. It is unclear to me what your expression is supposed to be. Since it has no external propagators I think it can't be a VEV. As was already pointed out, it seems strange that there are integrations over external momenta as well as external positions (e.g. x1 and k1). Also why are the exp-factors double? $\endgroup$
    – drfk
    Jul 10, 2023 at 23:27

2 Answers 2

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You're not quite using the LSZ reduction formula properly. As part of LSZ, you're supposed to take the residue of the poles as all external momenta go on-shell. If you start by including propagators for the external momenta, like $\int d^4k_1\frac{i}{k_1^2-m^2}e^{-ik_1x_1}$, then you can ignore any part of the $k_1$ integral that is not singular (in other words, any part where $k_1$ is off-shell). You removed the $\frac{i}{k_1^2-m^2}$ part of the propagator correctly (this is part of taking the residue), but you should have also removed the integral in the process. It also looks like you included redundant factors of $e^{-ik_1x_1}$ for each of the external momenta. These factors in the LSZ formula come from the propagators for the external legs.

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  • $\begingroup$ The KG operator acting on the plane wave will give a factor that cancels with the pole, but why would this also make the integration go away? I know that via residue theorem the contour integral will be sum of residue. But once the pole is canceled with the $-k^2 + m^2) from the action of the KG operator on the plane wave there should be no pole structure at all, no? $\endgroup$
    – user310742
    Jul 11, 2023 at 1:53
  • $\begingroup$ LSZ formula is $\int{dx_1 dx_2 dy_1 dy_2 e^{ikx_1}e^{ikx_2}e^{-iky_1}e^{-iky_2}(\partial^2 + m^2)^{4}<0|T\phi(x_1)\phi(x_2)\phi(y_1)\phi(y_2)|0>}$ so that’s where the plane waves came from, and then the vev of correlation function includes extra plane wave from each propagator that comes from each contraction. $\endgroup$
    – user310742
    Jul 11, 2023 at 1:59
  • $\begingroup$ I think I figured it out. Pg 94 of Schwartz qft has the external plane waves from LSZ as having different momentum labels than the ones that show up in the propagator. This gives a delta function that basically says k_1, k_2 are the initial moments and p_1 and p_2 are the final momenta. This makes the extra integration measures go away. Correct? $\endgroup$
    – user310742
    Jul 11, 2023 at 2:20
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The exponentials before the vev in LSZ did not necessarily have to have the same momenta as the exponentials that come from the propagators. Integrating gives a delta function that matches the external momenta to the propagators.

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