Past exam of the quantum field theory course of the University of Oxford / 2018 / 4

The displayed oscillator expansion and the displayed action require two separate corrections to describe one conventional positive-energy theory. First compute what the oscillator algebra literally implies. Assume the two oscillator species commute with each other and annihilate the vacuum. Same-charge contractions vanish, soFor the two mixed contractions the printed phases give the same kernel, with $z=x-y$:Thus $⟨T\varphi (x){\varphi }^{\ast }(y){⟩}_0=⟨T{\varphi }^{\ast }(x)\varphi (y){⟩}_0=C_{\mathrm{lit}}(z)$. In $D$ dimensions replace $d^{3}p$ and $d^{4}p$ by $d^{D-1}p$ and $d^{D}p$; the source's explicit three-dimensional spatial measure is the $D=4$ specialization. The covariant integral is manifestly Lorentz invariant. Withwe have $C_{\mathrm{lit}}(z)=D_F(z{)}^{\ast }$. This reversed-frequency scalar contraction has the opposite pole prescription to the usual positive-energy Feynman propagator. For a single zero-momentum mode at positive time separation, the printed contraction is $e^{+imt}/(2m)$, while $D_F$ is $e^{-imt}/(2m)$: this is a concrete phase counterexample, rather than an immaterial change of integration variable.

There is also a normalization of a complex scalar propagator conflict independent of those phases. The displayed Lagrangian density gives the canonical momentum ${\pi }_{\varphi }={\dot{\varphi }}^{\ast }/2$. Substituting the printed modes yieldsinstead of the canonical $+i{\delta }^{(3)}$. Reversing the mode phases repairs the sign but leaves $+i{\delta }^{(3)}/2$, so it does not repair the normalization. The complex scalar field convention with unit mixed propagator normally has no $1/2$ in its quadratic action.^[ref]

To retain the actual action in the paper, use $\varphi ={\varphi }_1+i{\varphi }_2$ with canonically normalized real components, or equivalently replace its mode expansion byThe corresponding path integral with the usual Feynman i-epsilon prescription haswith the two same-charge propagators still zero. This action-consistent kernel is used in b and c. Keeping conventional unit modes instead would require removing the quadratic factors $1/2$ from the action as well as reversing the phases. Those are distinct theories unless the field and its coupling are rescaled together.

Let $C(u,v)=2D_F(u-v)$ be the action-consistent free propagator established in a. Write $S=S_0+S_I$ and putThe perturbative expansion isDivide numerator and denominator by the same free vacuum functional $Z_0$. For $\mathcal{O}=\Phi (y_1)\cdots \Phi (y_N)$, they becomeFor $\mathcal{O}=\varphi (y_1){\varphi }^{\ast }(y_2)$, abbreviate $C_{12}=C(y_1,y_2)$ and $C_{zz}=C(z,z)$. Since only opposite-charge fields contract, Wick's theorem givesandThe first multiplicity is $2!$ internal pairings. In the connected term, there are two choices for the vertex's ${\varphi }^{\ast }$ contracted to the external $\varphi$ and two for its $\varphi$ contracted to the external ${\varphi }^{\ast }$; the remaining pair forms a tadpole diagram. ThereforeExpanding the inverse denominator cancels the free-line times vacuum bubble term, leavingThe local quartic interaction vertex has factor $-i\lambda$, since $2!2!$ cancels its denominator 4. The oriented tadpole diagram has Feynman-diagram symmetry factor 1; the isolated two-loop vacuum vertex has factor $1/2$ in its coefficient. Coincident kernels and the integrals require a common regularization; none is being assigned a finite numerical value here.

More generally, vacuum bubbles exponentiate and factor from every numerator. Here global U(1) symmetry also makes all one-point functions zero. Thus a surviving graph with the two external insertions in separate connected components would have a component with one unbalanced charge and vanish. This proves that only connected graphs contribute to this particular two-point function to all perturbative orders. Vacuum normalization alone would not remove disconnected components containing external legs in a general four-point function. If one formally chooses a different free kernel, the displayed Wick multiplicities are unchanged; replacing $C$ by $C_{\mathrm{lit}}$ gives a formal oscillator series, not the Gaussian integral of the action as printed.

Work in the symmetric vacuum and continue with the action-consistent kernel $C(u,v)=2D_F(u-v)$. Under global U(1) symmetry, a correlator with $r$ fields $\varphi$ and $4-r$ fields ${\varphi }^{\ast }$ picks up a phase $e^{i\alpha (2r-4)}$. Invariance for every $\alpha$ requires $r=2$. Consequently ten of the sixteen ordered field-type choices vanish, and the six placements with two fields of each type can be nonzero. This is the charge selection rule for complex scalar correlators, not an assertion about a symmetry-breaking vacuum with a nonzero one-point function.

Choose external fields $\varphi (x_1)\varphi (x_2){\varphi }^{\ast }(y_1){\varphi }^{\ast }(y_2)$. Its zeroth-order full correlator is $C(x_1,y_1)C(x_2,y_2)+C(x_1,y_2)C(x_2,y_1)$ and is disconnected; its connected part starts at order $\lambda$. DefineThe tree term has Feynman-diagram symmetry factor 1 and is $-i\lambda \int d^{D}zP(z)$.

At second order, a connected graph has two quartic vertices, two internal lines and one loop. The three bubble integrals areThe oriented complex-scalar quartic graph factors distinguish the channels. In $B_s$ the two internal lines have the same orientation and can be interchanged, giving symmetry factor $S_s=2$. In each mixed channel the two orientations are opposite; interchanging them is not an oriented graph automorphism, so $S_t=S_u=1$. Equality of the numerical scalar kernels $C(z,w)=C(w,z)$ does not erase the field-slot distinction used in counting contractions.

There are also four connected external-leg tadpole diagrams. They must be included because a connected correlation function is neither amputated nor restricted to one-particle-irreducible Feynman diagrams. Representative integrals, with the other two obtained by the indicated external permutations, areEach oriented external-leg graph has symmetry factor 1: a directed self-loop has no independent reversal symmetry. Altogether,These expressions retain every external propagator. The physical normalization is explicit: each $C$ is $2D_F$, not $D_F$. Equivalently set $\chi =\varphi /\sqrt{2}$; its quadratic action has unit mixed propagator and its interaction is $-\lambda |\chi {|}^4$, giving vertex $-4i\lambda$. Multiplying a four-$\chi$ correlator by four converts it to the four-$\varphi$ correlator, confirming the powers of two.

Completeness can also be seen without a picture. With two vertices and four external legs, $I=(8-4)/2=2$. Connectivity allows either two intervertex lines with a two-and-two external split, or one bridge and one self-loop with a one-and-three split. The charge assignments give exactly the three bubble attachments and four leg choices above. Direct labelled Wick contraction counting gives respectively $16,32,32,32,32,32,32$ contractions. Dividing by the expansion denominator $2!4^2=32$ reproduces $1/2,1,1,1,1,1,1$. Vacuum factors cancel, and there are no additional connected topologies at this order.