# Contradictory result for scalar-field propagator from Feynman rules and LSZ formula

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?

• @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. Jan 17, 2016 at 15:43
• @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] Jan 17, 2016 at 15:53
• That's fine; I just wanted to make sure you knew. Jan 17, 2016 at 15:56
• 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. Jan 18, 2016 at 9:25
• @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. Jan 18, 2016 at 21:28

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.

• How does one calculate these 'disconnected pieces in general (for situations greater than 1 in 1 out particle) - e.g. for a 8 in 8 out particle situation, with disconnected pieces where an ingoing external leg is directly connected to an outgoing external leg. Can you use Feynman rules, or do you always have to revert to commutations of creation/annihilation operators like you did here? Jan 26 at 22:57

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:

(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: $$(-i\int d^4xe^{-ip_{in}x}\Box_x)(-i\int d^4 ye^{ip_{out}y}\Box_y)\int d^4x_1<\phi(x)\phi(x_1)><\phi(x_1)\phi(y)>\sim \delta^4(p_{in}-p_{out})$$

• 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 :) ) Jan 18, 2016 at 21:32