Past exam of the advanced quantum theory: path integrals and many-particle physics course of the University of Oxford / 2024 / 1

Put $\mathcal{T}=\beta \hslash >0$, the interval of imaginary time, and assume $m,\omega >0$. For a variation $x_0+ϵ\eta$ with $\eta (0)=\eta (\mathcal{T})=0$, the first variation of the Euclidean action isThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationIts sign differs from the real-time harmonic oscillator equation because the Euclidean action contains positive kinetic energy and positive potential energy. The general solution is $A\cosh (\omega \tau )+B\sinh (\omega \tau )$. Applying the two endpoint conditions gives $A=x$ and $B=-x\tanh (\omega \mathcal{T}/2)$. Equivalently, the equal-endpoint Euclidean oscillator path isThis form displays reflection about the midpoint and satisfies both endpoints immediately. On this path,Thus the Euclidean action reduces to a boundary term. Since ${\dot{x}}_0(\mathcal{T})=\omega x\tanh (\omega \mathcal{T}/2)$ and ${\dot{x}}_0(0)=-\omega x\tanh (\omega \mathcal{T}/2)$,The positive quadratic form $m{\int }_0^{\mathcal{T}}({\dot{\eta }}^2+{\omega }^2{\eta }^2)d\tau$ is the second variation, so this stationary path is the unique minimum, rather than merely an arbitrary stationary point.

After the shift $x(\tau )=x_0(\tau )+y(\tau )$, the endpoint values of $y$ are both zero. Expanding the Euclidean action givesBy integration by parts, the cross term isThe first term vanishes by the homogeneous Dirichlet boundary conditions, and the second by the Euler-Lagrange equation. The translation has unit Jacobian determinant in the time-sliced configuration-space path integral, so $\mathcal{D}x=\mathcal{D}y$. Consequently classical-path factorization of a quadratic path integral is exact here:All dependence on the common endpoint $x$ lies inThe endpoint domain and the integrand defining $Z$ are the same for every $x$. This $Z$ is a zero-endpoint Gaussian path integral; the canonical partition function instead integrates the entire diagonal kernel over $x$. The final sentence of the printed part omits $/\hslash$ from its exponential: the physical Boltzmann factor has the dimensionless exponent $-S/\hslash$.

An optional normalization check uses the Dirichlet oscillator fluctuation determinant. Its eigenvalues are $m[(n\pi /\mathcal{T}{)}^2+{\omega }^2]$, $n\ge 1$. Relative to the free-particle Gaussian path integral, the hyperbolic-sine infinite product givesThus $Z=\sqrt{m\omega /[2\pi \hslash \sinh (\beta \hslash \omega )]}$. Integrating the resulting diagonal thermal kernel of a quantum harmonic oscillator by a Gaussian integral yields $\mathrm{Tr}{e}^{-\beta H}=[2\sinh (\beta \hslash \omega /2){]}^{-1}$, including the zero-point energy.

Because $S[x_0]=0$ at $x=0$, the factor $Z$ cancels from the ratio. Write $z=\beta \hslash \omega$ and $q=\sqrt{m\omega /\hslash }x$. The exact answer isAt high temperature, $z\ll 1$. Expanding the hyperbolic tangent givesThe displayed leading expression is the ratio of classical Boltzmann factors for potential energy $V(x)=m{\omega }^2{x}^2/2$ and $V(0)=0$. Indeed the classical canonical ensemble givesThe momentum Gaussian integral is independent of $x$ and cancels in the ratio. This is the classical phase-space limit of the thermal kernel. Here $\beta =1/(k_{B}T)$, with $k_B$ the Boltzmann constant; the associated classical position variance is $1/(\beta m{\omega }^2)$. At fixed $x$, $R$ ultimately tends to one as temperature tends to infinity, while its width grows.

At low temperature, $z\gg 1$. Now $\tanh (z/2)=1-2e^{-z}+O(e^{-2z})$, so at fixed $q$Unlike the classical high-temperature expression, this limiting width remains nonzero at zero temperature, reflecting the ground state of the quantum harmonic oscillator.

Insert the complete orthonormal basis of energy eigenstates on either side of the Boltzmann factor. The spectral theorem givesAt low temperature, the gap $\hslash \omega$ suppresses excited energy eigenstates. The harmonic oscillator eigenfunction of the ground state isThereforeThe common zero-point energy factor cancels. Dividing the kernel by its trace would give the position probability density of the canonical ensemble; at zero temperature this tends to $|{\phi }_0(x){|}^2$, whose variance is $\hslash /(2m\omega )$. Thus zero temperature projects onto the ground state and retains its quantum spread, rather than concentrating at the minimum of the classical potential energy.

There is also a check on the first correction. With $q=\sqrt{m\omega /\hslash }x$, the first excited harmonic oscillator eigenfunction satisfies $|{\phi }_1(x){|}^2=2q^2|{\phi }_0(x){|}^2$ and ${\phi }_1(0)=0$. Hence the spectral theorem gives $R=e^{-q^2}[1+2q^2{e}^{-z}+O(e^{-2z})]$, in agreement with part (c).