Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 304 / 1

For $\omega >0$, the quantum harmonic oscillator has energies $E_r=\omega (r+1/2)$, $r=0,1,\dots$, in units $\hslash =1$. For positive inverse temperature $\beta$, taking the trace in the energy basis and summing a geometric series givesThis is the thermal partition function of a quantum harmonic oscillator, including its zero-point energy. The assumptions $\beta ,\omega >0$ ensure convergence; at $\omega =0$ the free particle on the noncompact line instead has an infinite spatial-volume factor.

The trace of the Euclidean evolution kernel identifies its initial and final positions. Thus the Euclidean worldline path integral is over periodic paths $x(\tau +\beta )=x(\tau )$, with $\tau \in [0,\beta ]$:The circle has circumference $\beta$, not an arbitrary unit length. The Euclidean action follows by Wick rotation of the unit-mass oscillator action. Equivalently, integration by parts with periodic boundary conditions gives $S_E=\frac{1}{2}⟨x,L_{\omega }x⟩$ for $L_{\omega }=-{\partial }_{\tau }^2+{\omega }^2$. The path integral measure is formal until a regularization and its normalization are specified; the next part supplies a finite-dimensional version.

Use a real orthonormal basis of Fourier modes on the time circle:The eigenvalue of $L_{\omega }$ on the constant mode is ${\omega }^2$, and each sine/cosine pair has eigenvalue ${\omega }^2+{\nu }_r^2$. Retaining these modes giveswhere $C_N$ can depend on $N,\beta$ but is independent of $\omega$. Each real Gaussian integral contributes $\sqrt{2\pi /{\lambda }_r}$. Consequently the Fourier-cutoff oscillator functional determinant isA spectral cutoff between successive eigenvalue pairs selects precisely this finite subspace. In comparing two frequencies, the displayed regularization keeps the same number of modes $N$; holding an absolute spectral threshold fixed need not give the same $N$ at finite cutoff.

The frequency-independent measure factor cancels in the ratio:Upon analytic continuation in frequency, this has poles at ${\omega }_1=0,\pm 2\pi ir/\beta$ and zeros at ${\omega }_2=0,\pm 2\pi ir/\beta$, for $1\le r\le N$, apart from cancellations when numerator and denominator vanish together. They match those of $\sinh (\beta {\omega }_2/2)/\sinh (\beta {\omega }_1/2)$. To justify equality, rather than merely matching this divisor, use the hyperbolic-sine infinite productIt follows from pairing the Weierstrass product for the reciprocal gamma function at opposite imaginary arguments and using the Gamma reflection formula. Normal convergence on compact sets follows from $\sum r^{-2}<\infty$. Thus, for positive real frequencies,The cutoff ratio determines the frequency dependence, agreeing with the thermal partition function of a quantum harmonic oscillator.

The printed assertion about an absolute ${\lim }_N{\mathcal{Z}}_N$ needs a normalization qualification: independence of $A_N$ from $\omega$ alone does not ensure existence of a nonzero limit. Set $B_N=A_N{\prod }_{r=1}^N{\nu }_r^{-2}$. Then ${\mathcal{Z}}_N=B_N{\omega }^{-1}{\prod }_r(1+{\omega }^2/{\nu }_r^2{)}^{-1}$, so if $B_N\to B\ne 0$, its limit is $B\beta /[2\sinh (\beta \omega /2)]$. The choice $B_N=1/\beta$ gives the required limit. Conversely, $B_N=(1+(-1{)}^N/2)/\beta$ is a positive frequency-independent normalization with no absolute limit, though every ratio above still converges. Matching zeros and poles alone also permits nonvanishing entire functions; the convergent normalized product is what rules them out here.