Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 48 / 1

Use position eigenstates normalized by $⟨q|q^{\prime }⟩=\delta (q-q^{\prime })$, and the free Hamiltonian operator $\hat{H}={\hat{p}}^2/(2m)$. The time-sliced path integral is the matrix element of the time-evolution operator:Insert momentum eigenstates, with $⟨q|p⟩=(2\pi \hslash {)}^{-1/2}{e}^{ipq/\hslash }$. The free-particle propagator becomesComplete the square in the exponent:The Fresnel integral, defined by the damping prescription $T\to T-i0$, gives, for $T>0$,The square-root branch is $i^{-1/2}=e^{-i\pi /4}$; it is fixed by the regulated Gaussian integral and the short-time Dirac delta distribution limit, rather than an arbitrary phase.

The Euler-Lagrange equation for the free particle is $m\ddot{q}=0$. Its unique classical path satisfying the endpoint boundary conditions is $q_{\mathrm{cl}}(t)=q^{(i)}+\Delta qt/T$. Therefore its classical action isSubstituting into the operator result yields $\mathcal{I}=C(T)e^{i{S}_{\mathrm{cl}}/\hslash }$, with $C(T)=(m/(2\pi i\hslash T){)}^{1/2}$ independent of the endpoint positions.

The same independence is transparent directly in the path integral. Write $q=q_{\mathrm{cl}}+\eta$, with $\eta (0)=\eta (T)=0$. The cross term in the action is $m{\dot{q}}_{\mathrm{cl}}{\int }_0^T\dot{\eta }dt=0$, so $S[q]=S_{\mathrm{cl}}+(m/2){\int }_0^T{\dot{\eta }}^{2}dt$. The fluctuation integral depends only on $m,T,\hslash$. Because the action is quadratic, the factorization is exact; the semiclassical propagator already equals the full answer.

Put $I_0=\sqrt{2\pi /m^2}$ and assume $m^2>0$. The normalized free Gaussian integral has covariance $\Delta =1/m^2$. Expanding the interaction exponential and using Wick's theorem gives the Feynman rules: a quartic vertex contributes $-\lambda$, each internal propagator contributes $\Delta$, and each vacuum diagram is divided by its Feynman-diagram symmetry factor. Multiply the resulting sum by $I_0$. Because we are expanding $I$, rather than $\log I$, disconnected vacuum diagrams must also be included.

At order ${\lambda }^2$ there are two quartic vertices and four internal propagators. Let $r$ be the number of edges joining the two vertices. The remaining half-edges must pair at their own vertex, so $r$ is even. The possibilities $r=0,2,4$ exhaust the vacuum diagrams:

For $r=0$, each vertex has three possible pairings, giving $N_0=9$ Wick contractions. For $r=2$, choose the two half-edges at each vertex, then pair the chosen half-edges across the vertices: $N_2={\left(\genfrac{}{}{0ex}{}{4}{2}\right)}^{2}2!=72$. The remaining two half-edges at each vertex form a self-loop. For $r=4$, all half-edges pair across the vertices, giving $N_4=4!=24$. The common denominator from the expansion is $2!(4!{)}^2=1152$, so the Feynman-diagram symmetry factors areEach vacuum diagram has the same factor $I_0{\lambda }^2{\Delta }^4$ before division by its Feynman-diagram symmetry factor. Consequently the coefficient of ${\lambda }^2$ isAs an independent check, the eighth moment obtained from the Gaussian moment pairing theorem is $⟨x^8{⟩}_0=7!!/m^8=105/m^8$, and $105/[2!(4!{)}^2]=35/384$. The first terms of the zero-dimensional quartic perturbation series are thereforeStrictly, the printed phrase “Taylor expansion” means a formal expansion or a right-sided asymptotic expansion. The general coefficient has absolute successive ratio $(4v+3)(4v+1)/[24(v+1)m^4]\to \infty$, so the power series has zero radius of convergence. Nevertheless, for $\lambda \ge 0$ the remainder after any fixed truncation of $e^{-\lambda x^4/4!}$ is bounded by the next absolute term, and its Gaussian integral is finite. This proves the displayed asymptotic expansion as $\lambda \downarrow 0$. For negative real $\lambda$ the defining integral diverges.