Past exam of the theoretical physics II course of the University of Cambridge / 2013 / Paper 1 / 2

Put $T=t-t^{\prime }$. momentum eigenstates evolve with energy $p^2/(2m)$, so the retarded momentum kernel is $\theta (T)\delta (p-p^{\prime })e^{-i{p}^{2}T/(2m\hslash )}$. Its Fourier transform reduces toCompleting the square and defining the oscillatory Gaussian integral by a convergent limiting prescription givesFor $T>0$ the square root has quantum phase $e^{-i\pi /4}$. As $T\downarrow 0$, this kernel tends to $\delta (x-x^{\prime })$ in the distributional sense. Differentiating the Heaviside step function consequently supplies the required $i\hslash \delta (T)\delta (x-x^{\prime })$ source; an extra $-i/\hslash$ factor would correspond to a different Green-function convention.

For the momentum-space propagator in a uniform force, the stated Fourier convention makes $x$ act on momentum wavefunctions as $i\hslash {\partial }_p$. HenceFor $T>0$, the characteristic has $p(s)=p^{\prime }-\alpha (s-t^{\prime })$. Along it the amplitude acquires quantum phase $-{\int }_{t^{\prime }}^{t}p(s{)}^{2}ds/(2m\hslash )$. momentum is translated without a Jacobian, givingThe cubic expression is evaluated on the delta-function support. There it equals $-i[p^{\prime 2}T-\alpha p^{\prime }{T}^2+{\alpha }^2{T}^3/3]/(2m\hslash )$, making both its sign and its continuous $\alpha \to 0$ limit clear. The jump at $T=0$ is the identity momentum kernel.

Use the Dirac delta function to set $p^{\prime }=p+\alpha T$, and write $X=x-x^{\prime }$. The remaining momentum-dependent quantum phase isShifting the integration variable by $\alpha T/2$ reduces the integral to the free Gaussian integral. ThereforeThe force is $-\alpha$, not $+\alpha$. Both the linear endpoint term and the cubic-time term disappear in the zero-force limit. The prefactor remains the free one.

Measure intermediate time $u$ from $t^{\prime }$. The classical path with the fixed endpoints isIt satisfies $m{\ddot{x}}_c=-\alpha$. Direct integration of $m{\dot{x}}_c^2/2-\alpha x_c$ givesFor an arbitrary path write $x=x_c+\eta$ with $\eta (0)=\eta (T)=0$. Integrating the cross term by parts gives $-\int (m{\ddot{x}}_c+\alpha )\eta du=0$. Thus $S[x]=S_c+\int m{\dot{\eta }}^{2}du/2$ exactly. The fluctuation path integral is the free endpoint-fixed one, so its determinant supplies $\sqrt{m/(2\pi i\hslash T)}$. Multiplication by $\theta (T)e^{i{S}_c/\hslash }$ reproduces the previous kernel. This exact classical-path factorization of a quadratic path integral is possible because a linear potential introduces no additional quadratic fluctuation term.