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I am trying to learn how to calculate scattering amplitudes in a Klein-Gordon theory. I am getting stuck with the simplest of the examples: $\phi\to\phi$ in a free scalar-field theory.

If I calculate the amplitude for the process $\phi\to\phi$, I get two different results depending on whether I use Feynman rules or the LSZ formula.

Let's say that the incoming particle has momentum $p_\mathrm{in}$ and that the outgoing one has momentum $p_\mathrm{out}$. Then a simple Feynman diagram gives amplitude = propagator, that is,

$$ p_\mathrm{in} ---\stackrel{p}{\blacktriangleright} ---p_\mathrm{out} \qquad=\qquad \frac{1}{p^2-m^2} \tag{1} $$ where $p$ is the momentum of a virtual particle connecting the in and out states. From momentum conservation, $p_\mathrm{in}=p=p_\mathrm{out}$.

From $(1)$ we get that the amplitude for $\phi\to\phi$ is $$ \mathcal A=\frac{1}{p^2_\mathrm{in}-m^2}=\frac{1}{p^2_\mathrm{out}-m^2} $$

If, on the other hand, we calculate the amplitude for the process with the LSZ reduction formula, we get

$$ \mathcal A(2\pi)^4\delta(p_\mathrm{in}-p_\mathrm{out})=i\int\mathrm dx\,\mathrm dy\ e^{ip_\mathrm{in}x} e^{-ip_\mathrm{out}y}\, \square_x \square_y \langle\phi(x)\phi(y)\rangle $$ where $\square_x\equiv \partial_x^2+m^2$, and $\langle \phi(x)\phi(y)\rangle$ is the propagator. As the KG derivative, when acting on the propagator, results in a delta, i.e., $i\square \langle\phi(x)\phi(y)\rangle=\delta(x-y)$, this integral evaluates to $$ \begin{align} \mathcal A (2\pi)^4\delta(p_\mathrm{in}-p_\mathrm{out})&=\int\mathrm dx\,\mathrm dy\ e^{ip_\mathrm{in}x} e^{-ip_\mathrm{out}y}\, \square_x \delta(x-y)\\ &=(2\pi)^4\delta(p_\mathrm{in}-p_\mathrm{out})(p_\mathrm{in}^2-m^2) \end{align} \tag{2} $$

From $(2)$ we get $$ \mathcal A=p_\mathrm{in}^2-m^2=p_\mathrm{out}^2-m^2 $$

Comparing this to the amplitude from the diagram, we find that the results disagree. Can anybody tell me where is my mistake? I believe that the right result is the first one, so maybe my formula for LSZ is wrong, or cannot be used in a one-particle process?

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    $\begingroup$ @Jen: this question has been asked again by OP and it has an answer. Maybe you could state if that answer is enough to you. AccidentalFourierTransform: If you don't get an answer you should edit your question or place a bounty, not ask the question again. $\endgroup$ – Javier Jan 17 '16 at 15:43
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    $\begingroup$ @Javier this happened six months ago, when I was just a couple of weeks old: as I got no answer the first time I asked, I thought I could ask a second time. Now I know this is not the correct procedure, but back then it felt natural. When I asked the question a second time (duplicate, I know) I didn't know how SE worked. Now I know it was wrong. Anyway, Jen placed a bounty on my question for some reason (BTW, I have nothing to do with that). [As a side note: IMHO the answer given in the other post is wrong, and the question remains open for me] $\endgroup$ – AccidentalFourierTransform Jan 17 '16 at 15:53
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    $\begingroup$ That's fine; I just wanted to make sure you knew. $\endgroup$ – Javier Jan 17 '16 at 15:56
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    $\begingroup$ It's cool. I've merged the other question into this one. Feel free to edit this one to reflect the wording of the other one, if you want. $\endgroup$ – David Z Jan 18 '16 at 9:25
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    $\begingroup$ @DavidZ thank you! I edited this a bit, using some of the notation from the other one (which I believe is better and more clear); added some bold text to emphasize the main points; and some details to the math. $\endgroup$ – AccidentalFourierTransform Jan 18 '16 at 21:28
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My own attempt at this: the first result is wrong, and the second one is right but incomplete.

  • Feynman: in the diagram, all the lines are external, so there is no propagator in the diagram. Therefore, Feynman rules give $\mathcal A=1$.

Check up of this:

In canonical quantisation, the amplitude is given by $\langle p|q\rangle=\langle 0|a_p a^\dagger_q|0\rangle=\langle 0|[a_p,a^\dagger_q]|0\rangle\propto\delta(\boldsymbol p-\boldsymbol q)\langle 0|0\rangle$. Therefore, $\mathcal A$ is momenta-independent, and can be made equal to $1$ by an appropriate choice of the one-particle states normalisation. So far so good.

  • LSZ: In the OP, I proved that LSZ gives $\mathcal A_\text{LSZ}= p_\mathrm{in}^2-m^2$; but as the external momenta are on-shell, this means that $\mathcal A_\text{LSZ}=0$. But $\mathcal A_\mathrm{LSZ}$ is not the total amplitude: it is just the connected contribution! In the process $\phi\to\phi$, the tree diagram is disconnected, which means that LSZ has to evaluate to zero, and this is expected behavior.

The total amplitude for any process is$^1$ $\mathcal A=\mathcal A_\mathrm{dis}+\mathcal A_\mathrm{LSZ}$. For this particular process, $\mathcal A=1$ is a disconnected contribution, and $\mathcal A_\mathrm{LSZ}=0$.


$^1$ e.g., see see Timo Weigand's notes on QFT, page 50.

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You are actually not calculating the same thing.

The momentum space Feynman of the propagator is used for internal momentum, which is off-shell. So your first calculation is correct. While in your second case, you are using the LSZ formula to force the momentum to be on-shell. One way to get the same result is as follow:

Consider the following diagram:

enter image description here

(1)Using momentum space Feynman rule, since there is no internal propagator, you should get 1. So the S-matrix element is $\sim \delta^4(p_{in}-p_{out})$

(2) Use LSZ formula: \begin{equation} (-i\int d^4xe^{-ip_{in}x}\Box_x)(-i\int d^4 ye^{ip_{out}y}\Box_y)\int d^4x_1<\phi(x)\phi(x_1)><\phi(x_1)\phi(y)>\sim \delta^4(p_{in}-p_{out}) \end{equation}

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    $\begingroup$ Actually, I have to say, the first calculation is not correct: the amplitude is not $\frac{1}{p^2-m^2}$, because the momenta are on-shell, which would make the amplitude divergent. The result from LSZ is right, but incomplete. For more details, see my own answer above (anyway, thank you for your answer :) ) $\endgroup$ – AccidentalFourierTransform Jan 18 '16 at 21:32

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