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For the zero-dimensional QFT with action $$S(\phi)=\frac{\alpha}{2}\phi^2+\frac{\lambda}{4!}\phi^4-J\phi,\tag{1}$$ we can perturbatively expand the partition function as $$Z_\lambda(J)=\int_{-\infty}^{\infty} d\phi~e^{-S}=\sum_{r=0}^{\infty}\frac{1}{r!}\left(\frac{-\lambda}{4!}\right)^r\left(\frac{\partial^4}{\partial J^4}\right)^r\exp\left(\frac{J^2}{2\alpha}\right),\tag{2}$$ which at first order in $\lambda$ gives $$-\frac{\lambda}{4!}\left(\frac{J^4}{\alpha^4}+\frac{6J^2}{\alpha^3}+\frac{3}{\alpha^2}\right).\tag{3}$$

This can be expressed via the diagrams

Three connected Feynman diagrams with one vertex

since we know we need one vertex and to contract in line with Wick's theorem. What I'm unsure about is why we can't also have the disconnected diagram

A disconnected Feynman diagram with one vertex

since it has the same number of sources, propagators and vertices as the second diagram above. This diagram does show up when calculating the second moment $\langle\phi^2\rangle$ using the partition function.

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  • $\begingroup$ The last diagram does not contribute to $\langle \phi^2\rangle$. It does contribute to $Z \langle \phi^2\rangle$, and dividing by $Z$ has the effect of removing such diagrams. This is basic diagrammatics, and you should read any introductory book on QFT for a start. $\endgroup$
    – Adam
    Feb 13, 2023 at 9:03
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    $\begingroup$ as answered by @Qmechanic, you need to series expand (or at least include) the exponential in (2) as well in order to get the full J dependence in (3). $\endgroup$ Feb 13, 2023 at 9:34
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    $\begingroup$ @Wakabaloola Thank you, I see it now! In the lecture course I'm taking at this point $\langle \phi^2\rangle$ was defined to include the partition function, and later on we redefine the normalisation by dividing by the partition function. I had completely missed that the exp term was in one of my expressions but not the other. $\endgroup$
    – acernine
    Feb 13, 2023 at 12:12

2 Answers 2

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Well, OP partition function $Z[J]$ does in fact contain disconnected diagrams, such as, e.g. OP's last diagram $8|$, cf. the linked cluster theorem $Z[J]=\exp(\frac{i}{\hbar}W_c[J])$.

Concretely, the propagator $|$ in OP's last diagram $8|$ comes from the bag $\exp\left(\frac{J^2}{2\alpha}\right)$ of propagators in OP's eq. (2).

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  • $\begingroup$ Thanks, I'll look into the linked cluster theorem. I'm confused however as to how there can be more diagrams contributing to the partition function at order $J^2$ as the second diagram in my post, when accounting for the symmetry factor of 4 agrees with the explicit calculation. Wouldn't any further diagrams give something that disagrees with this? $\endgroup$
    – acernine
    Feb 12, 2023 at 21:32
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Starting from $S(\phi,J) = S_0(\phi) + S_{\rm int}(\phi) - J \phi$ with $S_0(\phi) = \alpha \phi^2/2$ and $S_{\rm int}(\phi) = \lambda \phi^4\!/4!$, the generating function(al) of the full model (including the interaction term) is given by $$ Z(J) = \frac{\int \! d \phi \, e^{-S_0(\phi)} e ^{-S_{\rm int}(\phi)} e^{J \phi}}{\int \! d\phi \, e^{-S_0(\phi)}e^{-S_{\rm int}(\phi)}} =: \left\langle e^{J \phi} \right\rangle,$$ which can be rewritten as $$Z(J) = \frac{\int \! d\phi \, e^{-S_0(\phi)} e^{- S_{\rm int}(\phi)} e^{J \phi}}{\int \! d\phi \, e^{-S_0(\phi)} } \cdot \frac{\int \! d\phi \, e^{-S_0(\phi)} }{\int \! d\phi \, e^{-S_0(\phi)}e^{-S_{\rm int}(\phi)} } = \frac{\left\langle e^{-S_{\rm int}(\phi)} e^{J\phi} \right\rangle_0 }{ \left\langle e^{-S_{\rm int}(\phi)} \right\rangle_0}, $$ where $\left\langle f(\phi) \right\rangle_0$ denotes the Gaussian mean value defined by $$\left\langle f(\phi) \right\rangle_0 := \frac{\int \! d\phi \, e^{-S_0(\phi)} f(\phi)}{\int \! d\phi \, e^{-S_0(\phi)}}. $$ The perturbative expansion of the term $\left\langle e^{-S_{\rm int}(\phi)} e^{J \phi}\right\rangle_0$ is given by $$ \left\langle e^{-S_{\rm int}(\phi)} e^{J\phi} \right\rangle_0 = e^{-S_{\rm int}(\partial_J)} \left\langle e^{J\phi}\right\rangle_0 = e^{-S_{\rm int}(\partial_J)} Z_0(J), $$ with the generating function of the free theory $Z_0(J)= \left\langle e^{J \phi} \right\rangle_0 = e^{J^2/2 \alpha}$, leading to the compact expression $$ Z(J) = \frac{e^{-S_{\rm int} (\partial_J)}Z_0(J)}{e^{-S_{\rm int}(\partial_J)}Z_0(J) \Large|_{J=0}}. $$ Expanding up to terms linear in $\lambda$, one finds $$e^{-S_{\rm int}(\partial_J)}Z_0(J) = e^{J^2/2\alpha} \left[1-\frac{\lambda}{4!}\left(\frac{J^4}{\alpha^4} +\frac{6 J^2}{\alpha^3}+\frac{3}{\alpha^2}\right) +\mathcal{O}(\lambda^2) \right].$$ With the help of the comments, you had realized yourself that you had simply forgotten the exponential term $e^{J^2/2 \alpha}$ in this expression. Note that the term in the denominator, $$e^{-S_{\rm int}(\partial_J)} Z_0(J) {\large|}_{J=0} = 1 -\frac{\lambda}{4!} \frac{3}{\alpha^2}+ \mathcal{O}(\lambda^2),$$ cancels all graphs with disconnected vacuum bubbles (like your last diagram), arriving at $$Z(J) = e^{J^2/2 \alpha} \left[ 1 - \frac{\lambda}{4!} \left( \frac{J^4}{\alpha^4} + \frac{6 J^2}{\alpha^3} \right)+ \mathcal{O}(\lambda^2) \right]$$ and $\left\langle \phi^2 \right\rangle =1/\alpha - \lambda/2 \alpha^3 +\mathcal{O}(\lambda^2)$.

Alternatively, employing the Wick theorem, you find $$\left\langle \phi^2 \right\rangle =\frac{\left\langle e^{-S_{\rm int}(\phi)} \phi^2 \right\rangle_0}{\left\langle e^{-S_{\rm int}(\phi)} \right\rangle_0 } = \frac{\left\langle \phi^2 \right\rangle_0 - \frac{\lambda}{4!} \left\langle \phi^6 \right\rangle_0+ \mathcal{O}(\lambda^2)}{1 - \frac{\lambda}{4!} \left\langle \phi^4\right\rangle_0 + \mathcal{O}(\lambda^2)} = \frac{\frac{1}{\alpha}- \frac{\lambda}{4!} \frac{15}{\alpha^3}+ \mathcal{O}(\lambda^2)}{1- \frac{\lambda}{4!} \frac{3}{\alpha^2}+ \mathcal{O}(\lambda^2)} = \frac{1}{\alpha} - \frac{\lambda}{2 \alpha^3} + \mathcal{O}(\lambda^2),$$ in agreement with the previous result.

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