Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 46 / 2

The cubic scalar field theory has a one-particle-irreducible Feynman diagram with two cubic vertices, one external leg at each vertex, and two internal lines joining them:

The Euclidean Feynman rule for each cubic interaction vertex is $-g{\mu }^{(6-d)/2}$: the minus sign comes from expanding $e^{-S_{\mathrm{int}}}$, and $1/3!$ in the action cancels the permutations of its three fields. The mass dimension of the cubic coupling is $(6-d)/2$, so ${\mu }^{(6-d)/2}$ permits $g$ to remain dimensionless. Two vertices supply $(-g{)}^2{\mu }^{6-d}$.

Each internal scalar propagator contributes $1/(p^2+m^2)$ with its own momentum. Conservation leaves one independent loop momentum; choose the two propagator momenta to be $p$ and $p+k$. The remaining Fourier integration measure is $d^{d}p/(2\pi {)}^d$. Finally, the factor $1/2$ is the Feynman-diagram symmetry factor for exchanging the two identical internal lines. Direct Wick contraction counting gives the same factor: two choices for which vertex receives a labeled external leg, $3^2$ choices of the incident fields and $2!$ pairings of the remaining fields, divided by $2!(3!{)}^2$, yield $1/2$.

The displayed loop integral is the amputated two-point insertion. For the correction to the full propagator, multiply it by the external scalar propagators, giving $D_0(k{)}^{2}I(k)$ with $D_0(k)=1/(k^2+m^2)$.

Apply a Feynman parameter and shift the loop momentum to $\ell =p+xk$. The common denominator becomes $[{\ell }^2+M_x^2{]}^2$, whereFor positive $M_x^2$, the Gamma-integral representation $a^{-2}={\int }_0^{\infty }dss{e}^{-sa}$ and a Gaussian integral giveThis formula initially converges for $\mathrm{Re}d<4$ and defines the Euclidean massive loop integral at other dimensions by analytic continuation. Consequently,With $d=6-ϵ$, the Gamma function factor is $\Gamma (-1+ϵ/2)=-2/ϵ+O(1)$. Only its pole matters: the other factors can be evaluated at $ϵ=0$ when extracting that pole. Since ${\int }_0^{1}x(1-x)dx=1/6$,This is the one-loop two-point divergence in six-dimensional cubic scalar theory. Its polynomial momentum dependence is precisely what permits subtraction by local counterterms.

Use a mass counterterm and a wavefunction renormalization counterterm, defining their additive coefficients byThe sign follows from how the Euclidean path integral expands. A quadratic counterterm inserts $-\delta K$ into the propagator, whereas the loop defined in the question inserts $+I$. To this order,Thus cancellation requires $\delta K=I_{\mathrm{div}}$, rather than its negative. In the minimal subtraction scheme, with no finite parts added,These are the minimal-subtraction two-point counterterms in cubic scalar theory. They absorb respectively the $k^2$ and constant terms in the two-point pole.

The additive mass coefficient is distinct from the shift of a bare mass when the bare field also includes wavefunction renormalization. If ${\varphi }_B=\sqrt{1+\delta Z}\varphi$ and $(1+\delta Z)m_B^2=m^2+\delta m^2$, then to one-loop orderThis last relation specifies the convention; the boxed coefficients are those multiplying the local counterterms displayed above.