I have read chapter 4 in Peskin & Schroeder on interacting fields and Feynman diagrams. I believe I have some fundamental misunderstanding, so I have outlined what I know so far below and included my two questions at the end.

Consider a $\phi^4$ interaction where $\phi$ is a Klein-Gordon field. To compute the probability that the field $\phi$ will create a point $x$ out of the vacuum |$\Omega \rangle$ at destroy it once it has propagated to another spacetime point $y$, we compute two-point correlation function $$\langle \Omega | T(\phi(x)\phi(y)| \Omega\rangle.$$ An important step to computing the above is to compute the two-point correlation function $$\Big\langle 0 \Big| T(\phi_I(x)\phi_I(y) \exp{\Big[-i\int_{-T}^T \int dtd^3x(t) \frac{\lambda}{4!}\phi_I^4\Big]}\Big| 0\Big\rangle \tag{1}$$ where $|0\rangle$ is the vacuum of the free theory and $\phi_I$ is the evolution of the field in the interaction picture.

To compute (1) we expand in powers of $\lambda$ and use Feynman diagrams and Wick's theorem. For example consider the following second order diagram:

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If we label the internal point closest to $x$ as $z$ and the other as $w$, I've computed the corresponding integral as $$-\frac{\lambda^2}{6}\iint dwdz \Delta_F(x-z)\Delta_F^3(z-w)\Delta(w-y) \tag{2}$$ where $\Delta_F$ is the Feynman propagator $$\Delta_F(x-y) = \begin{cases} D(x-y) \quad x^0 > y^0\\ D(y-x) \quad x^0 < y^0 \end{cases}$$ and the propagator $D(x-y)$ is calculated by taking the commutator of the field $$[\phi(x), \phi(y)] = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_{\textbf{p}}}} \int \frac{d^3q}{(2\pi)^3}\frac{1}{\sqrt{2E_{\textbf{q}}}} \Big[ (a_\textbf{p}e^{-ip \cdot x} + a^\dagger_\textbf{p}e^{ip\cdot x}), (a_\textbf{q}e^{-iq \cdot x} + a^\dagger_\textbf{p}e^{ip\cdot x}) \Big]\\ = \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_{\textbf{p}}} (e^{-ip\cdot(x-y)} - e^{ip \cdot (x-y)}) \\= D(x-y) - D(y-x). \tag{3}$$

Substituting the above in for (2) we finally obtain an amplitude for this particular diagram.

I have two questions about the above process:

  1. We never specified a momentum, so how are we carrying out the computations in (3)? Doesn't this imply we have specific 4-momenta $p$ and $q$ prescribed at the beginning? It seems to me that we have only specified positions $x$ and $y$ in spacetime. As a result, we do not know $E_\textbf{p}$ so how does one carry out the integration in (3) or actually compute a Feynman propagator? In other words, we said that the field $\phi(x)$ creates a particle at position $x$, but we never said what momentum it has. So how would we extract this information?

  2. Similarly, when considering Feynman diagrams in momentum-space one relabels each line connected to an external vertex by a 4-momentum. But how do we obtain this momentum? Peskin & Schroeder say that this can be done by taking the Fourier transform of the Feynman propagator, but how can we do this if we only know positions and not momentum?

  • $\begingroup$ For 1, you're integrating over p and q so they won't end up in your final answer. For 2, the momentum space rules are the Fourier transform of the real space rules and contain all the same information. It turns out that often we know the incoming and outgoing momenta rather than their positions and as such the Fourier transformed answers are better. $\endgroup$
    – Luke
    Commented Dec 28, 2022 at 0:42
  • $\begingroup$ @Luke Does that mean when we are calculating the amplitude for a particle to be created at $x$ and then annihilated at $y$ ($\langle \Omega | T(\phi(x)\phi(y)| \Omega\rangle$) we are letting the particle have any momentum? $\endgroup$
    – CBBAM
    Commented Dec 28, 2022 at 1:04
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    $\begingroup$ This interpretation of the two-point function $\langle \Omega | T\{\phi(x)\phi(y)\}|\Omega\rangle$ as a probability for a particle to be created at a point and propagate to another is, in my opinion, very misleading. In order to talk about probabilities of particles being at some point requires the existence of a position operator, whose eigenvalues would be the possible positions where the particles could be at and whose eigenstates would be states in which the particle is certainly at that point. Such position operators do not exist in relativistic QFT. $\endgroup$
    – Gold
    Commented Dec 28, 2022 at 2:04

1 Answer 1


About your first point, the computation you show in (3) is that of $[\phi(x),\phi(y)]$, so there is no momentum at all here. What happens is that to compute this you expand $\phi(x)$ and $\phi(y)$ in Fourier space in terms of creation and annihilation operators. This introduces momenta which are integrated over.

About your second point. Feynman diagrams compute contributions to $n$-point functions $$G_n(x_1,\dots, x_n)=\langle \phi(x_1)\cdots \phi(x_n)\rangle.$$ These are, initially, position space quantities: they depend on $n$ spacetime points. But as with any function, you can take a Fourier transform and trade the points $x_i$ by "momenta" $p_i$:

$$\widetilde{G}_n(p_1,\dots, p_n)=\int \prod_{i=1}^n d^4 x_i e^{ip_i\cdot x_i}G_n(x_1,\dots, x_n).$$

This is just one alternative representation of the correlator. It should be noticed, however, that the various $p_i$ here are not physical momenta at this stage, since they can be off-shell, $p_i^2\neq -m_i^2$. So these are just parameters of a function.

Later on you will relate $G_n(x_1,\dots,x_n)$ or equivalently $\widetilde{G}_n(p_1,\dots, p_n)$ to scattering amplitudes using the LSZ reduction formula. What one finds is that $\widetilde{G}_n(p_1,\dots, p_n)$ has poles as the momenta go on-shell, $p_i^2+m_i^2\to 0$, and the residues turn out to be the scattering amplitudes. When that happens you have specific values of the $p_i$ which correspond to the physical momenta of the external particles. In any case, since you seem to be getting started with QFT, a small piece of advice about LSZ. I think it is easier to understand first the derivation presented by Schwartz and Maggiore and later go back to see the one by PS. In any case, Schwartz and Maggiore do it early on, differently than PS.

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    $\begingroup$ Thank you, this was very helpful. I am indeed new to QFT and I am coming from it as a student of mathematics with no formal training in physics so I find some of it quite different (and difficult) than what I am used to. I have been using P&S and Folland's book for the most part. I tried skimming through Schwartz's book but it seems to be too suited for physicists than my math background allows, and as a result I find it tough to get through. $\endgroup$
    – CBBAM
    Commented Dec 28, 2022 at 2:55
  • 1
    $\begingroup$ You're welcome! I was a mathematician when I started studying QFT too. The first time I studied the subject was in Schwartz' book, but I felt like there were many things left unexplained. What worked for me was to study the first five chapters of Weinberg's The Quantum Theory of Fields. After that Schwartz became much clearer, and so did P&S. $\endgroup$
    – Gold
    Commented Dec 28, 2022 at 3:03
  • $\begingroup$ I'll take a look at Weinberg's book, thanks again for all your help! $\endgroup$
    – CBBAM
    Commented Dec 28, 2022 at 3:05
  • $\begingroup$ Weinberg is a bit tough, but it is really worth it. You may find useful questions here already on some of the points in these initial chapters, and you can always ask more anyway. One piece of advice for people with math background going into Weinberg. It is very important to observe that he has one criterium of simplicity in all that he does. $\endgroup$
    – Gold
    Commented Dec 28, 2022 at 3:09
  • $\begingroup$ This means that one must appreciate that his reasoning is basically to ask: "what is the simplest way to construct something?" This is important to realize because as a mathematician I was always trying to see everything he said as theorems stating that what he was writing down had to be implied by something, while in reality was just the simplest way to achieve a particular requirement. Viewing it like that it becomes much easier to follow his reasoning. $\endgroup$
    – Gold
    Commented Dec 28, 2022 at 3:11

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