= Solution
Write $z=z_c+\eta$, with $\eta(0)=\eta(T)=0$. Independent variations of $z$ and $z^*$ give the <Euler-Lagrange equation> $\ddot z_c+\omega^2z_c=0$. Away from $\sin\omega T=0$ its unique endpoint solution is
$$
z_c(t)=\frac{z_i\sin\omega(T-t)+z_f\sin\omega t}{\sin\omega T}.
$$
<Integration by parts> cancels the linear fluctuation terms and gives $S[z]=S[z_c]+\int_0^T\eta^*\Delta_\omega\eta\,dt$. On the <classical solution>, the <action> is the boundary term $[z_c^*\dot z_c]_0^T$. Substituting the endpoint derivatives therefore gives
$$
S[z_c]=\frac{\omega}{\sin\omega T}\bigl[(|z_f|^2+|z_i|^2)\cos\omega T-z_f^*z_i-z_i^*z_f\bigr].
$$
The remaining <Gaussian path integral> contains two real fluctuation coordinates per mode, so it contributes an inverse <functional determinant>, rather than its inverse square root. Thus $K=e^{iS[z_c]}/\det\Delta_\omega$, with the measure normalization fixing the otherwise arbitrary constant.
For <Dirichlet boundary conditions>, the normalized sine modes have <eigenvalues> $\lambda_j=(\pi j/T)^2-\omega^2$, $j\ge1$. Their <determinant> ratio is the convergent <Dirichlet oscillator determinant ratio>
$$
\frac{\det\Delta_\omega}{\det\Delta_0}
=\prod_{j\ge1}\left(1-\frac{\omega^2T^2}{\pi^2j^2}\right)
=\frac{\sin\omega T}{\omega T}.
$$
The last equality is the <sine infinite product>. With the prescribed free <determinant> this gives
$$
\boxed{\det\Delta_\omega=\frac{\pi i\sin\omega T}{\omega},\qquad K(z_f,z_i;T)=\frac{\omega}{\pi i\sin\omega T}e^{iS[z_c]}.}
$$
The kernel uses the <Feynman i-epsilon prescription>; at a caustic it is a distributional limit, not an ordinary finite function. Its $\omega\to0$ limit is $(\pi iT)^{-1}\exp(i|z_f-z_i|^2/T)$.
There is a sign error in the printed complex-integral hint. For a positive damping parameter $\alpha$, polar integration gives the <regulated complex Fresnel integral>
$$
\int_{\mathbb C}e^{(i\lambda-\alpha)|z|^2}\,d^2z=\frac{\pi}{\alpha-i\lambda}\longrightarrow\frac{i\pi}{\lambda}=-\frac{\pi}{i\lambda}.
$$
This regulated value, together with the stated free-kernel normalization, fixes the phase consistently.
For $\omega>0$ and positive imaginary-time length $\beta$, <Wick rotation> gives
$$
K_E(z_f,z_i;\beta)=\frac{\omega}{\pi\sinh\omega\beta}\exp\left[-\frac{\omega}{\sinh\omega\beta}\bigl((|z_f|^2+|z_i|^2)\cosh\omega\beta-z_f^*z_i-z_i^*z_f\bigr)\right].
$$
Put $z_f=sz_i$ with $s=\pm1$ and use the convergent real <Gaussian integral>. Writing $r=e^{-\omega\beta}$ gives
$$
\int d^2z\,K_E(sz,z;\beta)=\frac1{2(\cosh\omega\beta-s)}=\frac r{(1-sr)^2}=\sum_{n\ge1}s^{n-1}nr^n.
$$
This is the <parity-twisted oscillator thermal trace>. The complex coordinate describes two independent real <quantum harmonic oscillators>, each with mass two in these units. Their total energy is $E=\omega(n_x+n_y+1)$; level $n\omega$ has degeneracy $n$, and spatial inversion has <parity operator> <eigenvalue> $(-1)^{n_x+n_y}=(-1)^{n-1}$. The plus sign is $\operatorname{Tr}e^{-\beta H}$; the minus sign is $\operatorname{Tr}(Pe^{-\beta H})$. The alternating <trace> inserts parity into a bosonic system; it does not change the oscillators into fermions.
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