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Given the Lagrangian $$\mathscr{L}=\bar{\psi}\left(i\partial\!\!\!/-m\right)\psi +\frac{1}{2}\left(\partial\phi\right)^2- \frac{1}{2}M^2\phi^2 - g\bar{\psi}\psi\phi^2,$$ calculate the propagator correction for the $\psi$ field at the first non-vanishing order in $g$.

I can't understand if this

really is a correction to the propagator: it's first order in $g$, but the vertex has one $\psi$ and one $\bar\psi$. If this is not a correction, do the first non-vanishing terms are

(second order in $g$)?

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  • $\begingroup$ If you're worrying whether the vertices are legal: as long as you can get the fermion arrows to match up, it's fine. Your first diagram is allowed. $\endgroup$
    – knzhou
    Commented Oct 6, 2018 at 10:21
  • $\begingroup$ @knzhou yes, this is exactly the point. Could you please briefly explain me why this is the case? Really thank you. $\endgroup$
    – Stig
    Commented Oct 6, 2018 at 11:06
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    $\begingroup$ Related: [Diagrams involved in 1-loop electron self energy in QED][1] Your interaction is $-g\bar{\psi} \psi\phi^2$. Practically speaking it means that every vertex must be touched by 2 fermionic lines and 2 bosonic lines. This is indeed the case for your first diagram. [1]: physics.stackexchange.com/questions/431609/… $\endgroup$
    – DavideL
    Commented Oct 6, 2018 at 12:14
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    $\begingroup$ To understand why, think in terms of Wick's theorem. Your notation may differ, but what you're looking for is $$\langle \psi(x) \bar{\psi}(y)e^{iS_{\text{interaction}}} \rangle = \langle \psi(x) \bar{\psi}(y)e^{-ig \int d^4z \bar{\psi}(z) \psi(z) \phi^2(z)} \rangle = \langle \, \psi(x) \bar{\psi}(y) \rangle + (-ig) \int d^4z \, \langle \, \psi(x) \bar{\psi}(y) \, \bar{\psi}(z) \psi(z) \phi^2(z) \, \rangle + O(g^2)$$ $\endgroup$
    – DavideL
    Commented Oct 6, 2018 at 12:15
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    $\begingroup$ To first order in the coupling you are dealing with the full contraction of 6 fields: 2 external fermions, 2 internal fermions and 2 inter. bosons. The two bosons must be contracted between each other. Considering that each $\psi$ can in principle be contracted with each $\bar{\psi}$ this leaves two possibilities: $\psi(x)-\bar{\psi}(y) \quad \psi(z)-\bar{\psi}(z)$, which is disconnected, and $\psi(x)-\bar{\psi}(z) \quad \psi(z)-\bar{\psi}(y)$, which is your first diagram. As for the non-vanishingness of the diagram, I got my own doubts (see related question, so I'm just commenting. $\endgroup$
    – DavideL
    Commented Oct 6, 2018 at 12:16

1 Answer 1

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Expanding from the comments: your first diagram has the correct vertex structure for the reasons explained in the comments, and it is actually non-zero. The vertex does not introduce any spinorial structure, and the propagator is the simple bosonic one. E.g. in dimensional regularization, after a Wick rotation:

\begin{equation} \begin{split} (-ig) \int \frac{d^dk}{2\pi^d} \, \frac{i}{k^2-m^2} &= -ig \int \frac{d^4k_E}{2\pi^4} \, \frac{1}{k_E^2+m^2} \\ &=\frac{-ig}{(4\pi)^{d/2}} \left(\frac{1}{m^2}\right)^{1-d/2}\Gamma(1-d/2) \end{split} \end{equation}

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