Past exam of the theoretical physics major option course of the University of Oxford / 2016 / 3

Use $\hslash =1$. Taking the trace of the imaginary-time time-evolution operator identifies the initial and final position, giving the path integralThe measure includes integration over the common endpoint. The paths are real periodic bosonic paths on the imaginary-time circle, not paths with fixed zero endpoints and not only classical trajectories. The kinetic measure is defined by time slicing; typical contributing paths need not be differentiable. For the free expansion assume $m,\kappa >0$. The interacting partition function requires a potential bounded below; otherwise the following formulas have the meaning of a formal perturbation series about the free harmonic oscillator.

Here the imaginary-time Heisenberg picture position operator is $\bar{x}(\tau )=e^{\tau H_0}\hat{x}{e}^{-\tau H_0}$, where $H_0=H(0,0)$. Bosonic time ordering $T_{\tau }$ places the larger imaginary time to the left without an exchange sign. The brackets denote the normalized thermal expectation valueThe trace is cyclic and $H_0$ is time independent, so the two-point correlation function depends only on the time difference and extends periodically with period $\beta$. This periodicity is a property of the thermal correlation function, not the operator identity $\bar{x}(\tau +\beta )=\bar{x}(\tau )$.

The free Euclidean action has inverse propagator $D=-m{\partial }_{\tau }^2+\kappa$. Therefore the thermal Green's function obeyswhere ${\delta }_{\beta }$ is the periodic Dirac delta function. It is continuous, has continuous first derivative away from its source, and satisfies $G^{\prime }(0^{+})-G^{\prime }(0^{-})=-1/m$. These conditions follow by integrating the differential equation across the source and using the bosonic thermal trace.

For $\Omega =\sqrt{\kappa /m}$, solving the homogeneous equation on $0<\tau <\beta$ and imposing these conditions givesIn particular $g_0\equiv G^{(0)}(0)=\coth (\beta \Omega /2)/(2m\Omega )$. Equivalently $G^{(0)}={\beta }^{-1}{\sum }_n{e}^{-i{\omega }_n\tau }/(m{\omega }_n^2+\kappa )$ with bosonic Matsubara frequencies ${\omega }_n=2\pi n/\beta$.

Keep the same measure normalization as in the free generating functional and defineHere $W$ is the source-dependent partition function, despite the common alternative convention in which that letter denotes its logarithm. A functional derivative of $e^{\int Jx}$ brings down one factor $x(\tau )$ at the differentiated point. Expanding the interaction exponential and replacing each such factor by $\delta /\delta J(\tau )$ therefore givesCompleting the square in the free Gaussian integral gives $W_0[J]=W_0[0]\exp (\frac{1}{2}\int J{G}^{(0)}J)$ because $G^{(0)}=D^{-1}$ with periodic boundary conditions. Restoring $\hslash$ replaces the upper limit by $\hslash \beta$ and the interaction coefficients by $\lambda /(4!\hslash )$ and $\mu /(6!\hslash )$, exactly as in the source.

Set the external source to zero in solution and apply Wick's theorem. At a single quartic vertex, the four position insertions have three pairings; at a single sextic vertex, six insertions have $5!!=15$ pairings. ThusThermal time-translation invariance makes this $1-\lambda \beta g_0^2/8-\mu \beta g_0^3/48+\cdots$, with $g_0$ from solution. If desired the absolute corrections follow by multiplying by the free partition function $Z_0=[2\sinh (\beta \Omega /2){]}^{-1}$.

The first Vacuum Feynman diagram is one quartic vertex carrying two self-loops, with weight $-\lambda /8$ before time integration. The second is one sextic vertex carrying three self-loops, with weight $-\mu /48$. No disconnected subtraction is appropriate for $Z$ itself. The top row of the following figure shows these first-order terms and the conventions for solution.

Write $G=G^{(0)}(\tau -{\tau }^{\prime })$ and $g_0=G^{(0)}(0)$. With two identical degree-$n$ vertices, let $r$ lines join them. The remaining $n-r$ half-edges on each vertex form self-loops. Both $n$ and $r$ are even. The number of Wick contractions issince one chooses and matches the joining half-edges and then pairs the rest independently. Dividing by the expansion denominator $2!(n!{)}^2$ gives every symmetry factor weight without guessing graph multiplicities.

For two quartic vertices, $r=0,2,4$ givesThe graphs are respectively the product of two first-order two-loop graphs; two vertices joined by two lines with one self-loop on each; and two vertices joined by four lines with no self-loops.

For two sextic vertices, $r=0,2,4,6$ givesThe corresponding numbers of self-loops on each vertex are $3,2,1,0$. All seven second-order graphs are drawn in solution; in each series the $r=0$ graph is disconnected and the others are connected.

For completeness, a mixed quartic-sextic pair, not needed for the two pure second-order coefficients, contributesTaking $\log Z$ removes the products of first-order graphs, that is, the $r=0$ terms. The requested partition function retains them.