Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 81 / 4

The exponential acts on the Fock vacuum: $|\psi ⟩=e^{\zeta {\sum }_n{\psi }_n{a}_n^{†}}|0⟩$. For bosons, ${\psi }_n$ are commuting complex numbers. Since $[a_n,{\sum }_m{\psi }_m{a}_m^{†}]={\psi }_n$, commuting $a_n$ through the exponential gives $a_n|\psi ⟩={\psi }_n|\psi ⟩$. These are unnormalized bosonic coherent states.

For fermions, the labels are independent odd Grassmann variables, which anticommute with one another and with the fermionic operators. For one mode, the fermionic coherent state is $|\psi ⟩=(1-\psi a^{†})|0⟩$. The canonical anticommutation relations give $a(-\psi a^{†})|0⟩=\psi |0⟩=\psi |\psi ⟩$, using ${\psi }^2=0$. The even factors $1-{\psi }_n{a}_n^{†}$ commute between modes, so the same argument applies to every $n$. A Grassmann eigenvalue is a formal extension of the state space, not an ordinary complex eigenvalue of the nilpotent annihilation operator.

With dual state $⟨\psi |=⟨0|e^{\zeta {\sum }_n{a}_n{\bar{\psi }}_n}$, both cases obey $⟨\psi |{\psi }^{\prime }⟩=e^{{\sum }_n{\bar{\psi }}_n{\psi }_n^{\prime }}$.

Write $O=\int d[\bar{\psi },\psi ]e^{-\sum \bar{\psi }\psi }|\psi ⟩⟨\psi |$. For bosons the measure is ${\prod }_{n}d\mathrm{Re}{\psi }_{n}d\mathrm{Im}{\psi }_n/\pi$ over $\mathbb{C}$ in each mode, with ${\bar{\psi }}_n={\psi }_n^{\ast }$. For fermions it is an ordered Berezin integral over independent ${\bar{\psi }}_n,{\psi }_n$; choose $\int d\bar{\psi }d\psi \bar{\psi }\psi =-1$, so $\int d\bar{\psi }d\psi e^{-\bar{\psi }\psi }=1$.

For one bosonic mode, $a|\psi ⟩=\psi |\psi ⟩$ and $⟨\psi |a={\partial }_{\bar{\psi }}⟨\psi |$. Integration by parts in the Gaussian measure gives $aO=Oa$; the conjugate argument gives $a^{†}O=O{a}^{†}$. Boundary terms vanish because of the Gaussian weight.

For one fermionic mode, put $n=a^{†}a$ and move Grassmann coefficients to the left. The weighted projector isIts ordinary commutators are $[a,W]=-a+\psi I$ and $[a^{†},W]=a^{†}-\bar{\psi }I$. Their Berezin integrals vanish, so again $[a,O]=[a^{†},O]=0$. Equivalently the sole surviving coefficient in $\int W$ is $-\bar{\psi }\psi I$, giving $I$ directly.

The modes factorize. In the irreducible Fock space representation, commuting with every creation and annihilation operator makes $O$ a scalar multiple of the identity. Its vacuum matrix element is the normalized Gaussian integral, equal to one. Thus the coherent-state resolution of identity isFor infinitely many modes, this argument first uses a finite-mode regulator.

The bosonic coherent-state resolution of identity gives the ordinary trace by insertion into a number-state basis. For fermions, the fermionic coherent-state trace requires a sign in the bra. In one mode, for an even operator $A$,Without that sign the result would be the supertrace $A_{00}-A_{11}$. Applying this identity mode by mode to the even operator $A=e^{-\beta (H-\mu N)}$ gives the grand canonical partition function

Let $K=H-\mu N$, take $M$ slices of width $ϵ=\beta /M$, and insert the coherent-state resolution of identity between factors of $e^{-ϵK}$. If $H$ is in normal ordering, each short-time matrix element isCombining the overlap with the Gaussian weight leaves ${\sum }_{j,n}{\bar{\psi }}_{n,j}({\psi }_{n,j}-{\psi }_{n,j-1})$ in the action. The coherent-state time slicing prescription fixes which adjacent labels appear in $H$, rather than allowing an arbitrary ordering change after taking the continuum limit.

The twist $\zeta$ in the thermal trace closes the path with the coherent-state thermal boundary conditions: periodic for bosons, antiperiodic for fermions. Taking the regulated continuum limit givesThe action is dimensionless because $\tau$ has inverse-energy units. For fermions, $\bar{\psi }$ and $\psi$ remain independent Grassmann fields. If the original Hamiltonian is not normally ordered, first express it in normal order, retaining all constants.

For the quantum harmonic oscillator, set $\mu =0$ and $\Omega =\hslash \omega$. The normal symbol is $H_N(\bar{\psi },\psi )=\Omega (\bar{\psi }\psi +1/2)$, so the thermal coherent-state time slicing givesThe connection with a phase-space path integral uses the canonical real coordinatesTheir derivative term satisfies $\bar{\psi }\dot{\psi }=-ip\dot{q}/\hslash$ up to total derivatives that vanish for periodic paths. The normal and Weyl symbols of a harmonic oscillator must be distinguished: in midpoint phase-space time slicing, the Weyl ordering symbol of $a^{†}a$ is $\bar{\psi }\psi -1/2$. Thus the appropriate midpoint Hamiltonian is $H_W=p^2/(2m)+m{\omega }^2{q}^2/2$, with no additional constant. The change from the adjacent-label normal prescription to the midpoint prescription includes this ordering correction.

In physical imaginary time $u=\hslash \tau$, the resulting phase-space path integral isGaussian momentum integration in a phase-space path integral produces the usual oscillator configuration-space path integral. An exact check follows directly from the coherent kernel $⟨{\psi }^{\prime }|e^{-\beta \Omega a^{†}a}|\psi ⟩=\exp (e^{-\beta \Omega }{\bar{\psi }}^{\prime }\psi )$:This is the thermal partition function of a quantum harmonic oscillator. Keeping the $\Omega /2$ normal-ordering constant a second time after switching to the Weyl symbol would double count the zero-point energy.