Past exam of the theoretical physics II course of the University of Cambridge / 2014 / Paper 1 / 4

With $\beta =(k_{B}T{)}^{-1}$, the normalized canonical density operator isIts eigenvalues are the Boltzmann probabilities and its trace is one.

The density operator is diagonal in the oscillator number basis, so $⟨a^2⟩=⟨a^{†2}⟩=0$. With $\bar{n}=⟨a^{†}a⟩$ and $a{a}^{†}=a^{†}a+1$,HenceEquality holds even at zero temperature, when each contribution is $\hslash \omega /4$. Classical equipartition alone would miss this zero-point energy.

Set $q=e^{-\beta \hslash \omega }$. The oscillator populations are $(1-q)q^n$, and differentiating the geometric series gives $\bar{n}=q/(1-q)$. The canonical oscillator coordinate variance follows from $2\bar{n}+1=(1+q)/(1-q)=\coth (\beta \hslash \omega /2)$, andIt approaches $\hslash /(2m\omega )$ at zero temperature and $k_{B}T/(m{\omega }^2)$ at high temperature, agreeing respectively with ground-state fluctuations and classical equipartition.

Write $T_p=p^2/(2m)$ and $V=m{\omega }^2{x}^2/2$. Dropping commutators of order ${\beta }^2$ permits $e^{-\beta H}\simeq e^{-\beta V}{e}^{-\beta T_p}$. With $⟨x|p⟩=e^{ipx/\hslash }/\sqrt{2\pi \hslash }$, the momentum Gaussian givesIntegrating the diagonal yields $Z\simeq 1/(\beta \hslash \omega )$, so the leading high-temperature density kernel isThis unsymmetrized product has an apparent off-diagonal asymmetry at the omitted order; reversing the product puts $x^{\prime }$ in the potential factor. A manifestly Hermitian approximation uses $e^{-\beta V/2}{e}^{-\beta T_p}{e}^{-\beta V/2}$, replacing $\beta m{\omega }^2{x}^2/2$ by $\beta m{\omega }^2(x^2+x^{\prime 2})/4$ with the same leading normalization. The condition is $\beta \hslash \omega \ll 1$; the Gaussian off-diagonal width is of order $\hslash \sqrt{\beta /m}$.