Past exam of the quantum Condensed Matter Field Theory course of the University of Cambridge / 2016

Original source: Cambridge FOI 2023/399 released PDF, physical pages 797–799.

For the XXZ spin chain, write $\Delta =1+\alpha$ and replace each spin by a classical vector of length $S$. A bond is $J{S}^2{\mathbf{n}}_m^{\mathsf{T}}\mathrm{diag}(1,1,\Delta ){\mathbf{n}}_{m+1}$, whose minimum is $-J{S}^2\max (1,|\Delta |)$. For an even periodic chain all bonds can attain that bound simultaneously. The classical ground states of an XXZ spin chain therefore give $\alpha >0$ gives the two easy-axis Néel states ${\mathbf{S}}_m=(-1{)}^{m}S\hat{\mathbf{z}}$ and its reversal, with energy $-JN{S}^2(1+\alpha )$.

For $-2<\alpha <0$ the ground states are planar Néel configurations, ${\mathbf{S}}_m=(-1{)}^{m}S(\cos \phi ,\sin \phi ,0)$, with arbitrary $\phi$: their degeneracy is a circle and their energy is $-JN{S}^2$. This includes the easy-plane interval used later. Part (a), however, does not restrict negative $\alpha$ to that interval. For $\alpha <-2$, $\Delta <-1$ and the lowest states are instead the two uniform $z$-polarized ferromagnetic configurations, with energy $JN\Delta S^2$. At $\alpha =-2$, a rotation by $\pi$ about $z$ on alternating sites maps the classical energy to an isotropic ferromagnet: ${\mathbf{S}}_m=S((-1{)}^m{n}_x,(-1{)}^m{n}_y,n_z)$ for any unit vector $\mathbf{n}$ is a ground state. This special degeneracy is a sphere. At the other boundary, $\alpha =0$, the ordinary Néel direction is arbitrary on a sphere. These statements concern classical minimizing configurations, not the degeneracy of the exact finite-chain quantum ground state.

Let $[a,a^{†}]=1$, $n=a^{†}a$, and take $S^{+}=(S^{-}{)}^{†}=\sqrt{2S-n}a$. The physical Holstein–Primakoff occupation constraint limits the Hilbert space, which contains only $n=0,1,\dots ,2S$. Acting in this number basis givesThe upper and lower endpoints are annihilated as required. ConsequentlyThe change in $n$ also gives $[S^z,S^{\pm }]=\pm S^{\pm }$. These matrix elements prove the spin algebra on the full physical spin-$S$ space. Extending the oscillator space to arbitrary occupation would not give a physical finite spin: the occupation constraint and operator ordering in the square root matter. bosons on distinct sites have $[a_m,a_n^{†}]={\delta }_{mn}$ and all other site commutators zero.

Rotate the spin on one sublattice through $\pi$ about its $x$ axis, using the unitary product of the corresponding spin-rotation operators. On a rotated site the old components obey $S_{\mathrm{old}}^x=S^x$, $S_{\mathrm{old}}^y=-S^y$, $S_{\mathrm{old}}^z=-S^z$. Every bond joins opposite sublattices, soThe longitudinal product changes sign. Hence the rotated bond is $-J\Delta S_m^z{S}_{m+1}^z+\frac{J}{2}(S_m^{+}{S}_{m+1}^{+}+S_m^{-}{S}_{m+1}^{-})$, giving the requested form. The canonical spin algebra is unchanged because the transformation is unitary. A staggered easy-axis reference configuration is now the uniform all-up state. At $\alpha =0$ one first chooses a Néel direction, here $z$, before performing the same rotation.

For occupations small compared with $S$, use $S^z=S-n$, $S^{-}\simeq \sqrt{2S}{a}^{†}$ and $S^{+}\simeq \sqrt{2S}a$. The longitudinal bond contributes $-J\Delta S^2+JS\Delta (n_m+n_{m+1})+O(S^0)$; the transverse pair terms contribute $JS(a_m{a}_{m+1}+a_m^{†}{a}_{m+1}^{†})+O(S^0)$. Thus the direct real-space expansion isTake $a_m=N^{-1/2}{\sum }_k{e}^{ikm}{a}_k$, with $k=2\pi j/N$ over one full Brillouin zone and the lattice spacing set to one in $k$. Symmetrizing $k$ and $-k$ changes the phase in the pair sum to $\cos k$, givingnormal ordering the second Nambu spinor component then yieldsThe printed matrix has a factor-of-two error on its diagonal. With a full-zone sum its printed $2\Delta$ would produce $4\Delta a_k^{†}{a}_k$, and even its vacuum expectation would differ from the direct expansion by $J\Delta NS$. A half-zone convention cannot repair both coefficients: one must then group each $k,-k$ pair and double its off-diagonal pair coefficient as well. The formula above follows from the specified Hamiltonian and has the expected gapless isotropic limit; subsequent physical results use this consistent expansion. Terms quartic in bosons first enter at order $S^0$.

Write $u_k=\cosh {\theta }_k$, $v_k=\sinh {\theta }_k$, taking real even ${\theta }_k$. The transformation givesThe second equality uses $u_{-k}=u_k$ and $v_{-k}=v_k$, while the first uses $u_k^2-v_k^2=1$. The creation-creation commutator vanishes by conjugation. Therefore the $b$ modes are canonical bosons. In the paired basis,Its determinant is one and its inverse reverses the signs of the off-diagonal entries. Hence $\boxed{a_k=u_k{b}_k-v_k{b}_{-k}^{†}}$. At self-inverse momenta $k=0,\pi$ this is the corresponding one-mode squeezing transformation, still canonical for finite ${\theta }_k$.

For the easy-axis XXZ antiferromagnetic spin-wave spectrum, the consistent quadratic Hamiltonian in part (d) has $A=\Delta$ and $B_k=\cos k$. The inverse transformation acts on the Nambu matrix by a congruence. Its diagonal and off-diagonal entries become $A\cosh 2{\theta }_k-B_k\sinh 2{\theta }_k$ and $B_k\cosh 2{\theta }_k-A\sinh 2{\theta }_k$. The pair term vanishes whenFor $\alpha >0$ the transformation is real and finite for every $k$. Summing the two Nambu number operators over the full zone givesHere $E_0$ names the constant in the requested oscillator form; the actual quadratic ground-state energy is $E_{\mathrm{gs}}^{(2)}=E_0+\frac{1}{2}{\sum }_k{ϵ}_k=-J\Delta N{S}^2+JS{\sum }_k({\Omega }_k-\Delta )$. This distinguishes the normal-ordering constant from the total zero-point contribution. bosonic Bogoliubov diagonalization is a canonical, not an ordinary unitary, diagonalization of the Nambu matrix.

At $\alpha =0$ and $k\ne 0,\pi$, the same formula holds with ${\Omega }_k=|\sin k|$. The exact zero modes have $|\tanh 2{\theta }_k|=1$, so no finite squeezing angle diagonalizes them as positive-frequency oscillators. Treat collective rotations separately or retain a small anisotropy regulator before taking the limit. Finally, if one instead diagonalizes the literal erroneous printed matrix, it gives $\tanh 2{\theta }_k=\cos k/(2\Delta )$ and ${ϵ}_k^{\mathrm{print}}=2JS\sqrt{4{\Delta }^2-{\cos }^{2}k}$. That result remains gapped at $\alpha =0$ and is not the spin-wave spectrum of the Hamiltonian stated at the start.

For $\alpha >0$, the dispersion is periodic and has minima at $k=0,\pi$ in the rotated full-zone convention, withThe maximum is $2JS(1+\alpha )$ at $k=\pm \pi /2$. The easy-axis state has no continuously broken symmetry, so a gap is allowed. Near a minimum the energy behaves as $2JS\sqrt{\alpha (2+\alpha )+(\delta k{)}^2}$. At $\alpha =0$ it becomes ${ϵ}_k=2JS|\sin k|$, linear near its zeros, with small-$k$ slope $2JS$ in the chosen units. Restoring lattice spacing $a$ and $\hslash$ gives spin-wave speed $2JSa/\hslash$. Zone folding reorganizes the same modes into the magnetic Brillouin zone.

These are linear spin-wave predictions, not a statement that the exact one-dimensional quantum chain always has long-range Néel order. At the isotropic point the zero-point spin reduction contains $N^{-1}{\sum }_k(|\sin k{|}^{-1}-1)/2$, which diverges logarithmically as the infrared cutoff vanishes. This failure of the assumed ordered reference must not be mistaken for a calculation of the exact interacting spectrum.

In this part $0<\Delta =1+\alpha <1$, so the ordered reference lies in the $xy$ plane. Choosing one direction breaks the continuous rotation symmetry within that plane. A gapless, linearly dispersing phase mode is therefore expected in the spin-wave approximation, while tilting the Néel axis out of the plane costs anisotropy energy. It is incorrect to continue the easy-axis expansion about an unstable $z$-ordered reference and interpret its imaginary frequencies as physical modes.

For the easy-plane XXZ antiferromagnetic spin-wave spectrum, take the Néel direction along old $x$, then rotate alternating local spins so that all longitudinal components point up. One such local frame gives the full-zone quadratic coefficientsand thereforeNear $k=0$, ${ϵ}_k\simeq JS\sqrt{2(1+\Delta )}|k|$. The zone-edge mode has gap ${ϵ}_{\pi }=2JS\sqrt{2(1-\Delta )}$. A different sublattice phase convention shifts the zero by $\pi$; folding into the magnetic zone displays an acoustic and a gapped polarization. The figure uses this explicitly chosen frame. One-dimensional long-wavelength fluctuations again limit the ordered-reference interpretation.

Take a uniform nearest-neighbour harmonic chain with displacement $u_j$, conjugate momentum $p_j$, mass $m_0$, spring stiffness $\kappa$, lattice spacing $a$ and periodic length $\ell =Na$. These specify a standard model for the requested notes, not additional numerical exam data. Its Lagrangian and Hamiltonian areVariation gives $m_0{\ddot{u}}_j=\kappa (u_{j+1}-2u_j+u_{j-1})$. For a mode $e^{i(kja-\omega t)}$, periodicity requires $k=2\pi n/\ell$ andThe lattice makes $k$ periodic modulo $2\pi /a$, so choose one Brillouin zone. At long wavelength ${\omega }_k\simeq v|k|$, where $v=a\sqrt{\kappa /m_0}$ is the sound speed. The continuum limit of a harmonic chain expands $u_{j+1}-u_j\simeq a{\partial }_{x}u$ and replaces ${\sum }_j$ by $a^{-1}\int dx$. Defining mass density $\rho =m_0/a$ and one-dimensional elastic coefficient $K=\kappa a$ givesThe momentum density is $\Pi =\rho {\partial }_{t}u$, and the Hamiltonian density is ${\Pi }^2/(2\rho )+K({\partial }_{x}u{)}^2/2$. Translational invariance of $u$ makes total momentum conserved. The continuum approximation describes acoustic wavelengths large compared with $a$; it is not a justification for extending the lattice dispersion and zero-point sums to arbitrarily large wavevector.^[ref]

canonical quantization promotes the coordinates and momenta to operators with $[u_j,p_l]=i\hslash {\delta }_{jl}$ and the other equal-time commutators zero. In the continuum, $[u(x),\Pi (x^{\prime })]=i\hslash {\delta }_{\mathrm{per}}(x-x^{\prime })$. The nonzero normal modes can be represented by independent phonon operators obeying $[b_k,b_q^{†}]={\delta }_{kq}$. A useful lattice expansion iswhere $X=N^{-1}{\sum }_j{u}_j$ is the centre-of-mass displacement. The corresponding momentum isIncluding the centre-of-mass term restores the full site commutator: the nonzero-mode sum alone gives $i\hslash ({\delta }_{jl}-1/N)$. Substitution diagonalizes the Hamiltonian,A phonon is one quantum of a collective normal mode, carrying energy $\hslash {\omega }_k$ and lattice momentum $\hslash k$ modulo a reciprocal-lattice vector. Reality of $u_j$ relates its Fourier coordinates by $u_k^{†}=u_{-k}$; it does not remove the independent oppositely travelling phonon modes. Using one complete Brillouin zone avoids double-counting real standing-wave coordinates.^[ref]

The free centre of mass has no restoring force and must not be treated as an oscillator of zero frequency. Fixing it, weakly pinning it, or specifying a finite centre-of-mass volume defines the corresponding zero-mode factor. With a fixed centre of mass, the internal thermal partition function and occupations areThe internal energy is ${\sum }_{k\ne 0}\hslash {\omega }_k(⟨n_k⟩+1/2)$; its high-temperature heat capacity approaches $(N-1)k_B$. For a large one-dimensional chain at temperatures below the lattice cutoff but above its finite-size mode spacing,This one-dimensional acoustic phonon heat capacity shows that quantization turns classical normal waves into bosonic excitations with a linear low-temperature heat capacity in one dimension. The Euclidean field action $S_E=\frac{1}{2}{\int }_0^{\beta \hslash }d\tau dx[\rho ({\partial }_{\tau }u{)}^2+K({\partial }_{x}u{)}^2]$ is Gaussian, with periodic imaginary-time boundary conditions; its Fourier modes reproduce these oscillators. Absolute displacement correlations require the zero-mode convention, and the lattice regulates short-distance fluctuations.

For one bosonic mode, define the unnormalized bosonic coherent state $|z⟩=e^{z{a}^{†}}|0⟩$. The commutator $[a,a^{†}]=1$ gives $a|z⟩=z|z⟩$ and $⟨z^{\prime }|z⟩=e^{{\bar{z}}^{\prime }z}$. The normalized state is $e^{-|z{|}^2/2}|z⟩$; it is not orthogonal to a different coherent state. The coherent-state resolution of identity follows from Gaussian moments,For many modes take products of these states and measures. The complex amplitudes have the natural scale of square root occupation; coherent states turn normal-ordered matrix elements into classical symbols without replacing the quantum trace by a single classical configuration.

For fermions, $\{c_i,c_j^{†}\}={\delta }_{ij}$ and a nilpotent annihilation operator cannot have an arbitrary nonzero ordinary complex eigenvalue. Introduce independent Grassmann variables ${\eta }_i,{\bar{\eta }}_i$, anticommuting among themselves and with the fermionic operators, and define $|\eta ⟩=\exp (-{\sum }_i{\eta }_i{c}_i^{†})|0⟩$. Then $c_i|\eta ⟩={\eta }_i|\eta ⟩$ and $⟨{\bar{\eta }}^{\prime }|\eta ⟩=e^{{\sum }_i{\bar{\eta }}_i^{\prime }{\eta }_i}$. With an ordered Berezin integral measure satisfying $\int d\bar{\eta }d\eta \bar{\eta }\eta =-1$ for one pair,for an even operator $O$; this is the fermionic coherent-state trace. In one mode the trace integrand is $e^{-\bar{\eta }\eta }(O_{00}-\bar{\eta }\eta O_{11})$, whose integral is $O_{00}+O_{11}$. Without the sign reversal it would be the supertrace, not the thermal trace. These formal Grassmann-valued coherent states do not represent ordinary complex-valued fermion amplitudes.

To construct the many-body thermal path integral, write $\mathcal{Z}=\mathrm{Tr}{e}^{-\beta K}$ with $K=H-\mu \hat{N}$ and divide the imaginary-time interval $\mathcal{T}=\beta \hslash$ into slices of width $\delta \tau$. Insert the appropriate coherent-state resolution of identity between factors. For a normal-ordered Hamiltonian, the adjacent-slice kernel iscoherent-state time slicing makes the overlap and Gaussian measure combine into the first-order time derivative. The grand-canonical coherent-state action in the regulated continuum limit isFor bosons the integration variables are complex, with the conjugate labels on the original integration contour; for fermions the barred and unbarred Grassmann variables are independent. The continuum notation retains the time-sliced ordering prescription. For example, $n(n-1)=a^{†2}{a}^2$ has normal symbol ${\bar{z}}^2{z}^2$, whereas $n^2$ has that symbol plus $\bar{z}z$. Dropping that distinction changes a theory. The coherent-state thermal boundary conditions follow because the bosonic trace closes paths periodically and the fermionic sign in the trace closes them antiperiodically:The barred fields obey the same respective boundary conditions. Hence Matsubara frequencies are $2\pi n/(\beta \hslash )$ and $(2n+1)\pi /(\beta \hslash )$. The derivative term encodes the canonical structure; omitting it would remove the quantum dynamics.^[ref]

As a check, a free diagonal mode with $K=\xi a^{†}a$ gives $Z_B=(1-e^{-\beta \xi }{)}^{-1}$ for $\xi >0$, whereas $K=\xi c^{†}c$ gives $Z_F=1+e^{-\beta \xi }$. Gaussian integration produces an inverse determinant for bosons and a determinant for independent complex fermions, with their respective temporal boundary conditions. Differentiating $\log Z$ recovers Bose-Einstein and Fermi-Dirac occupations. Interactions become higher powers in the normal-ordered symbols, and sources generate time-ordered correlations. The coherent-state construction preserves statistics, ordering and boundary conditions together; a formal continuum expression without its measure and temporal prescription is incomplete.

Use the contact-interaction Bose-gas field integral for a homogeneous three-dimensional weakly interacting Bose gas with repulsive short-range interaction as the standard model. Let particle mass be $m_b$, scattering length $a_s>0$, physical coupling $g=4\pi {\hslash }^2{a}_s/m_b$ and gas parameter $n{a}_s^3\ll 1$. These are symbolic model definitions, not supplied numerical parameters. The grand-canonical coherent-state integral has periodic fields and actionThe contact term is the normal symbol of $g{\hat{\psi }}^{†2}{\hat{\psi }}^2/2$. In three dimensions a contact interaction needs a regulator and coupling matched to the measured scattering length; the displayed $g$ is the low-energy physical coupling. This field integral is useful because an ordered saddle and its quantum/thermal fluctuations follow from the same action.

Stationarity with respect to $\bar{\psi }$ gives the imaginary-time Gross–Pitaevskii equation. A uniform saddle exists at $\mu >0$ with ${\psi }_0=\sqrt{n_0}{e}^{i{\theta }_0}$ and condensate density $n_0=\mu /g$. Its arbitrary phase reflects global $U(1)$ symmetry; a selected condensate fixes a phase in the thermodynamic symmetry-breaking description. Write $\psi =\sqrt{n_0}+\eta$ after choosing that phase. Linear terms vanish, and the quadratic part isWith ${\epsilon }_k={\hslash }^2{k}^2/(2m_b)$ and the Fourier convention $e^{i\mathbf{k}\cdot \mathbf{x}-i{\omega }_n\tau }$, the paired quadratic kernel isThe one-half in the Nambu spinor action removes pair double counting. Analytic continuation of the inverse kernel gives the Bogoliubov dispersionAt high momentum it approaches the free-particle kinetic energy plus its mean-field shift. The crossover scale is the healing length $\xi =\hslash /\sqrt{2m_{b}g{n}_0}$. Gaussian integration adds a fluctuation determinant to the saddle free energy. Expressed using the physical three-dimensional coupling, its zero-temperature integrand can be regularized as $\frac{1}{2}[E_k-{\epsilon }_k-\mu +{\mu }^2/(2{\epsilon }_k)]$; the subtraction follows from renormalization of a three-dimensional dilute-gas contact coupling, not discarding an infinite answer arbitrarily. The thermal contribution is $k_{B}T\int d^{3}k/(2\pi {)}^3\log (1-e^{-\beta E_k})$. A calculation beyond mean field must keep that matching and its thermodynamic derivatives consistent.

The density-phase action of a Bose gas makes the physical content clear, $\psi =\sqrt{n_0+\delta n}{e}^{i\theta }$. Up to a periodic density total derivative, the long-wavelength quadratic action containsnumber-phase conjugacy pairs the density fluctuation and phase in the quadratic density-phase action. Dropping the density-gradient term only in the long-wave limit and completing the density Gaussian square gives the Bose-gas phase-only actionThis has the same sound pole. A term $i\hslash n_0{\partial }_{\tau }\theta$ is also present before restricting to the smooth zero-winding sector; although locally a total derivative, it cannot in general be ignored in topologically nontrivial sectors. The density and phase variables describe the same coupled Bogoliubov branch, not a second independent massive particle simply because the density has a static restoring cost.^[ref]

A phase gradient carries superflow ${\mathbf{v}}_s=(\hslash /m_b)\nabla \theta$. Its energy cost defines the superfluid number density. At leading mean-field order the stiffness uses $n_0$, but condensate density and superfluid number density are different observables: for a Galilean-invariant continuum gas at zero temperature the latter is the full density even when interactions deplete the condensate. Bogoliubov coefficients give $v_k^2=[({\epsilon }_k+g{n}_0)/E_k-1]/2$ and a small zero-temperature depletion of relative order $\sqrt{n{a}_s^3}$ in three dimensions. Large depletion or strong long-wavelength phase fluctuations would invalidate the small-fluctuation saddle approximation.

The Landau criterion for creating excitations is $v_L={\inf }_{k>0}{E}_k/(\hslash k)=c_s$ for this homogeneous dispersion. Thus repulsion produces a linear sound mode and a positive critical velocity from the Landau criterion, whereas an ideal quadratic gas has zero value of this criterion. Vortices, obstacles and nucleation can reduce an actual flow threshold below $c_s$; the criterion alone is not a complete theory of metastable flow. At finite temperature thermal quasiparticles reduce the superfluid number density. Dimensionality and finite size also matter: the uniform three-dimensional weak-depletion condensate expansion must not be transferred unchanged to a one- or two-dimensional thermodynamic system. The field integral provides systematic corrections only within its dilute, stable and sufficiently weak-fluctuation regime.

Keep the electromagnetic units of the supplied Lagrangian, including its factor $c$, and regard $e$ as the signed coupling used there. Set $a_B=e{\Phi }_B/(2\pi c)$. The canonical momentum is $p_{\varphi }=I\dot{\varphi }+a_B$, so the Legendre transform givesSingle-valued periodic wavefunctions have normalized eigenstates $e^{in\varphi }/\sqrt{2\pi }$ with integer $n$. Defining dimensionless flux $f=e{\Phi }_B/(hc)$ gives $E_n={\hslash }^2(n-f{)}^2/(2I)$, and the trace is thereforeThe flux-threaded quantum rotor spectrum is periodic under $f\mapsto f+1$ after relabelling $n$, corresponding to flux period $hc/|e|$ in these units. A local gauge change can remove the magnetic vector potential from the differential operator but then imposes a twisted boundary condition with phase $e^{-2\pi if}$; it cannot remove the flux dependence from both the Hamiltonian and boundary condition simultaneously. In SI conventions the flux coupling has no $c$, so one must not mix the two unit systems.

The trace integrates the diagonal thermal kernel over a single fundamental interval, $Z={\int }_0^{2\pi }d{\varphi }_0⟨{\varphi }_0|e^{-\beta \hat{H}}|{\varphi }_0⟩$. On the circle a closed path can wind an integer number $m$ of times. Lift it to the real covering coordinate so its endpoints are $\varphi (0)={\varphi }_0$ and $\varphi (\mathcal{T})={\varphi }_0+2\pi m$, with $\mathcal{T}=\hslash \beta$.

Under Wick rotation $t=-i\tau$, $\dot{\varphi }=i{\partial }_{\tau }\varphi$, so the real-time phase $iS/\hslash$ becomes minus the Euclidean action divided by $\hslash$, withThus sum the path integral $e^{-S_E/\hslash }$ over these lifted endpoint conditions and integrate ${\varphi }_0$ from zero to $2\pi$. This gives the requested winding-sector representation of the trace. The flux term is a total derivative, so it leaves the local classical equation unchanged, but its weight is $e^{i2\pi fm}$ and distinguishes different winding sectors. This is the Aharonov-Bohm phase in a thermal path integral, rather than a force from a magnetic field on the ring itself.

The Euclidean saddle in winding sector $m$ is a constant-speed lifted path, ${\varphi }_{\mathrm{cl}}(\tau )={\varphi }_0+2\pi m\tau /\mathcal{T}$. Write $\varphi ={\varphi }_{\mathrm{cl}}+\eta$ with Dirichlet fluctuation endpoints $\eta (0)=\eta (\mathcal{T})=0$. The cross term in the kinetic action is proportional to $\int {\partial }_{\tau }\eta d\tau =0$, and the flux term of $\eta$ vanishes for the same reason. ThereforeDefine the fluctuation prefactor byIt is the free-line diagonal Euclidean propagator for the angular coordinate with inertia $I$, or equivalently the Gaussian bridge amplitude about a given classical lifted path. It is independent of $m,{\varphi }_0$ and flux. It is not the complete zero-flux ring partition function, which still includes every winding and the initial-angle integral. Translational independence supplies the factor ${\int }_0^{2\pi }d{\varphi }_0=2\pi$, leavingThe momentum-winding duality of a quantum rotor uses a Gaussian expansion that is exact here because the kinetic action is quadratic and the flux coupling is a total derivative. It is not merely a leading semiclassical approximation.

Continue the free-particle propagator to $t=-i\mathcal{T}$ and replace its coordinate mass by the angular inertia $I$. The Euclidean free-particle bridge amplitude at coincident endpoints isThis normalization includes the time-sliced Gaussian measure, not just a formal determinant with an unspecified multiplicative constant. Set $A=2{\pi }^{2}I/({\hslash }^2\beta )>0$. The correct Poisson summation formula is ${\sum }_{m}F(m)={\sum }_n{\int }_{-\infty }^{\infty }dxF(x)e^{-2\pi inx}$; the source PDF omits the integral sign before its $d\theta$. For $F(x)=e^{-A{x}^2+i2\pi fx}$ the Gaussian Fourier integral givesThe overall prefactor cancels exactly: $2\pi Z_0\sqrt{\pi /A}=1$, and ${\pi }^2/A=\beta {\hslash }^2/(2I)$. Hencein agreement with the angular-momentum trace. Pairing $m$ and $-m$ makes the winding expression real, and the momentum expression shows positivity. Both are periodic in flux and invariant under $f\mapsto -f$. At high temperature the zero-winding term gives $Z\simeq \sqrt{2\pi I/(\beta {\hslash }^2)}$, with exponentially small flux dependence; at low temperature the nearest integer angular-momentum state dominates, with two equal lowest terms at half-integer flux. These independent limits also check the normalization and the flux phase.