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In a six-dimensional cubic scalar field theory, three labelled box Feynman diagrams give the one-loop local quartic vertex at zero external momentum. With an effective-action term $g_4{\varphi }^4/4!$, their contribution is $\delta g_4=-3g^4{I}_4$. To check its sign and multiplicity, expand the one-loop scalar effective action as $\frac{1}{2}\mathrm{Tr}\log (D+g\varphi )$, where $D=-{\partial }^2+m^2$. Its fourth-order term is $-g^4\mathrm{Tr}(D^{-1}\varphi {)}^4/8$, giving the stated coefficient. The associated amputated connected diagram insertion has the opposite sign.

For a connected Feynman diagram with cubic vertices with four external legs and one loop, $3V=2I+4$ and $1=I-V+1$ give $V=I=4$. In a one-particle-irreducible Feynman diagram, every internal edge lies on the single cycle. Each of the four cubic vertices therefore has two internal and one external leg, giving a box Feynman diagram. With labelled external momenta there are three inequivalent cyclic orderings, modulo rotation and reversal, represented by $1234$, $1243$ and $1324$:

Take all momenta incoming with ${\sum }_i{p}_i=0$. The Feynman rules for the Euclidean action give a propagator $(q^2+m^2{)}^{-1}$ and a cubic insertion $-g$. For order $1234$, the amputated connected insertion iswhere $\chi (q)=1_{\{\Lambda <|q|<{\Lambda }_0\}}$ is the high-mode projector for a strict Wilsonian effective action. Each labelled box has Feynman-diagram symmetry factor one. A common fixed-loop-cutoff convention instead restricts only the chosen $q$ integration to the shell; both prescriptions give the same zero-external-momentum value. The effective-action vertex has the opposite sign to this connected insertion. Thus $\boxed{\text{three labelled boxes, with one cyclic topology}.}$

A leading short-distance Standard Model contribution to neutral D-meson mixing is a box Feynman diagram with two W bosons and two internal down-type quark lines:

The incoming $c\bar{u}$ pair becomes $u\bar{c}$; the internal labels run over the down quark, strange quark and bottom quark. Each corner is a weak charged current vertex. Four such vertices make the contribution of order $g^4$. Summing the internal flavours produces Cabibbo-Kobayashi-Maskawa matrix factors and the Glashow-Iliopoulos-Maiani mechanism: the flavour-independent term cancels by ${\sum }_i{V}_{ci}^{\ast }{V}_{ui}=0$. This box Feynman diagram is one contribution; long-distance intermediate hadron states can also contribute to neutral D-meson mixing.

With the valence quark assignments $K^0=d\bar{s}$ and ${\bar{K}}^0=s\bar{d}$, neutral kaon mixing changes strangeness by two units. A representative leading Standard Model contribution is a box Feynman diagram with two W bosons and two internal up-type quarks:The two lines exchange the two W bosons. There are four weak charged current interaction vertices, so the scattering amplitude is fourth order in the weak gauge coupling, or second order in the Fermi constant. The full result sums the internal quark flavors with their Cabibbo-Kobayashi-Maskawa matrix factors; one box alone illustrates the mechanism.

Use $B^0=d\bar{b}$ and ${\bar{B}}^0=\bar{d}b$. A box Feynman diagram with two W bosons and internal up-type quarks changes one neutral flavour into the other. The top quark contribution dominates the dispersive mixing amplitude and carries the squared CKM combination involving $V_{td},V_{tb}$. With a consistent meson phase convention,The illustration in b shows this box and the dominant decay topology. A suitable common final state is $f=J/\psi K_S$, reconstructed for example through $J/\psi \to {\mu }^{+}{\mu }^{-}$ and $K_S\to {\pi }^{+}{\pi }^{-}$. The leading $\bar{b}\to \bar{c}c\bar{s}$ tree transition gives $B^0\to J/\psi K^0$, while the conjugate gives ${\bar{B}}^0\to J/\psi {\bar{K}}^0$. Neutral kaon mixing projects both kaon flavours onto the observed $K_S$.

The final state is approximately a CP eigenstate with ${\eta }_f=-1$: $J/\psi$ and $K_S$ each have CP approximately $+1$, and their relative orbital angular momentum is $L=1$ to couple the vector charmonium to the spin-zero parent. Neglecting the small kaon CP impurity and subleading decay amplitudes,There are two indistinguishable paths, direct decay and mixing followed by decay, so their amplitudes interfere. For $\Delta {\Gamma }_d\simeq 0$ and negligible direct CP violation, the time-dependent neutral-B CP asymmetry isMeasure the production flavour and proper decay time, fit the oscillation coefficient, and extract $\sin 2\beta$. This determines $\beta$ up to the usual trigonometric ambiguities unless further information is used. The cancellation of the dominant decay phase is why this channel isolates the angle involving $V_{td}$ without needing $V_{ub}$.