Quantum Field Theory: the Partition Function as a Path Integral

Face 3 of the Interlude recast ${\mathrm{e}}^{-\beta H}$ as evolution through an interval $\beta$ of imaginary time. To make a path integral of it we need a basis of states to propagate between, and the natural one is the basis of field-configuration eigenstates. Just as a single particle has position eigenstates $\widehat{x}|x⟩=x|x⟩$, a field has eigenstates of the (Schrödinger-picture) field operator,

$$\widehat{\varphi }(𝐱)|\varphi ⟩=\varphi (𝐱)|\varphi ⟩,\widehat{\varphi }(𝐱)=\sum _k\frac{1}{\sqrt{2{\omega }_k}}(a_k{u}_k(𝐱)+a_k^{†}{u}_k^{\ast }(𝐱)),$$

one for every classical profile $\varphi (𝐱)$, built from the mode functions $u_k$ and the ladder operators $a_k,a_k^{†}$ of Lecture 2; they resolve the identity, $\int 𝒟\varphi |\varphi ⟩⟨\varphi |=\mathrm{𝟏}$.

These $|\varphi ⟩$ are vectors in the very Fock space assembled in §2-the symmetric or exterior algebra on the modes. That space is graded: it splits into subspaces of definite energy, $\mathscr{H}={⨁}_n{\mathscr{H}}_n$ (and, independently, into the particle-number sectors carried by the fugacity $z^N$ of §4). A number eigenstate lives in a single grade; a field eigenstate $|\varphi ⟩$ is a superposition spread across all of them, as a position eigenstate spreads over the levels of an oscillator. Keep the word grading in view: this same energy grading returns as the $q^{L_0}$ of the torus (L4) and the graded dimension of a character (L5).

Now compute the trace by sandwiching, exactly as in §2 but in the field basis. Split ${\mathrm{e}}^{-\beta H}={({\mathrm{e}}^{-ϵH})}^N$ into $N$ steps of size $ϵ=\beta /N$ and insert the resolution of the identity between every pair:

$$Z=\mathrm{Tr}{\mathrm{e}}^{-\beta H}=\int 𝒟\varphi ⟨\varphi |{\mathrm{e}}^{-\beta H}|\varphi ⟩=\int \prod _{i=1}^N𝒟{\varphi }_i\prod _{i=1}^N⟨{\varphi }_{i+1}|{\mathrm{e}}^{-ϵH}|{\varphi }_i⟩,{\varphi }_{N+1}\equiv {\varphi }_1.$$

The trace has already done its work in setting ${\varphi }_{N+1}={\varphi }_1$: the chain of configurations closes into a loop. Each infinitesimal factor is the Boltzmann weight of a single imaginary-time step,

$$⟨{\varphi }_{i+1}|{\mathrm{e}}^{-ϵH}|{\varphi }_i⟩\simeq \exp (-ϵL_E[{\varphi }_i,{\dot{\varphi }}_i]),{\dot{\varphi }}_i=\frac{1}{ϵ}({\varphi }_{i+1}-{\varphi }_i),$$

with $L_E$ the Euclidean Lagrangian. Here is the rhyme with the Interlude: the exponent is additive over the steps, so the weights multiply-the very homomorphism of Face 2 (additivity $\to$ multiplicativity) that made $Z$ factorize over independent systems, now applied to successive slices of imaginary time. Their product assembles the exponential of the action, ${\prod }_i{\mathrm{e}}^{-ϵL_E}={\mathrm{e}}^{-{\sum }_iϵL_E}\to {\mathrm{e}}^{-{\int }_0^{\beta }d\tau L_E}={\mathrm{e}}^{-S_E[\varphi ]}$, and the sum over the intermediate configurations ${\varphi }_i$ becomes an integral over field histories $\varphi (\tau ,𝐱)$, their two ends glued by the trace into a circle:

$$\boxed{Z=\mathrm{Tr}{\mathrm{e}}^{-\beta H}={\int }_{\varphi (\tau +\beta )=\pm \varphi (\tau )}𝒟\varphi {\mathrm{e}}^{-S_E[\varphi ]}.}$$

(3.1)

The thermal partition function is a path integral over fields on a Euclidean time circle ${𝖲}_{\beta }^1$ of circumference $\beta$. Inverse temperature has become a length.

The same construction runs for a fermion, with one twist. Because its ladder operators anticommute ($\{c,c^{†}\}=1$, Lecture 2), the useful basis is the fermionic coherent states $|\psi ⟩$, eigenstates of the annihilator $c|\psi ⟩=\psi |\psi ⟩$ whose labels $\psi$ are anticommuting Grassmann numbers. Tracing over them carries an extra sign,

$$\mathrm{Tr}A=\int \mathrm{d}\overline{\psi }\mathrm{d}\psi {\mathrm{e}}^{-\overline{\psi }\psi }⟨-\psi |A|\psi ⟩,$$

and it is the $⟨-\psi |$-the fermionic trace closing on minus the starting configuration-that will force the antiperiodic boundary condition of the next section.

What is the $\pm$ in (3.1)? Tracing means gluing $\tau =0$ to $\tau =\beta$, and the gluing condition is fixed by the statistics of the field:

	bosons:	$\varphi (\tau +\beta )=+\varphi (\tau )$	$\text{(periodic)},$		(3.2)
	fermions:	$\psi (\tau +\beta )=-\psi (\tau )$	$\text{(antiperiodic)}.$		(3.3)

The minus sign for fermions is exactly the ${(-1)}^F$ of §4: the trace of ${\mathrm{e}}^{-\beta H}$ over a fermionic mode equals the antiperiodic path integral, while inserting ${(-1)}^F$ would flip it to periodic. This is the path-integral form of the Kubo–Martin–Schwinger (KMS) condition, whose proper home-correlation functions-is the business of the box below.

Before refining further, pause on what kind of choice this $\pm$ is. A fermion is a spinor, and to put spinors on a spacetime at all one must equip it with a spin structure: a consistent rule for the sign a spinor picks up when carried around each closed loop. In vacuum QFT on ${ℝ}^d$ this never comes up, and for good reason-${ℝ}^d$ is simply connected, its spin structure is unique, and there is no choice to make. But taking the trace changed the topology under our feet: thermal spacetime is ${𝖲}_{\beta }^1\times {ℝ}^{d-1}$, and a circle admits exactly two spin structures-the two gluings $\psi (\tau +\beta )=\pm \psi (\tau )$ just written. The partition function cannot be agnostic between them: $\mathrm{Tr}{\mathrm{e}}^{-\beta H}$ is the antiperiodic structure, and inserting ${(-1)}^F$ switches to the periodic one. “Statistics dictates boundary conditions” is, said geometrically, “taking a trace selects a spin structure.”

Nor is this exotic; it has been hiding in plain sight in every finite-temperature calculation. Fourier analysis on the time circle assigns a periodic (bosonic) field the Matsubara frequencies ${\omega }_n=2\pi n/\beta$, but an antiperiodic (fermionic) field the odd ones, ${\omega }_n=(2n+1)\pi /\beta$. Those odd integers, familiar from any first pass through thermal field theory, are a choice of spin structure in action.

That $\pm$ is only the coarsest twist. In §I.2 we refined the trace by a conjugate potential ${\mathrm{e}}^{\mathrm{i}\theta Q}$ for any conserved charge $Q$; in the path integral each such insertion becomes a twisted boundary condition-the field returns to itself only up to the symmetry group element $g={\mathrm{e}}^{\mathrm{i}\theta Q}$ that the insertion generates,

$$\mathrm{Tr}\left({\mathrm{e}}^{\mathrm{i}\theta Q}{\mathrm{e}}^{-\beta H}\right)={\int }_{\varphi (\tau +\beta )={\mathrm{e}}^{\mathrm{i}\theta }\varphi (\tau )}𝒟\varphi {\mathrm{e}}^{-S_E[\varphi ]}(\text{unit charge}).$$

(3.4)

The twist is precisely a flat background connection for the global symmetry generated by $Q$: a holonomy $g$ around ${𝖲}_{\beta }^1$, switched on by hand and not summed over. What it looks like depends on the group $Q$ generates.

If that group is finite-fermion parity ${(-1)}^F$ generating ${\mathbb{Z}}_2$ is the case in hand-the holonomy $g$ takes only finitely many values, each labelling a distinct twisted sector, with no continuum in between: antiperiodicity is the nontrivial element of ${\mathbb{Z}}_2$ and periodicity the identity. Summing over the sectors, instead of fixing one, gauges the symmetry-the orbifold construction.¹¹ 1 More precisely, a twist is a flat $G$-bundle on the spacetime: on ${𝖲}_{\beta }^1$ a holonomy $g\in G$ (a conjugacy class), on the torus of Lecture 4 a commuting pair $(g,h)$ of holonomies for the two cycles. When one sums over sectors to gauge $G$, the phases that may be attached to them form a torsor over $H^2(G,U(1))$-Vafa’s discrete torsion-while the obstruction to gauging at all is a ’t Hooft anomaly in $H^3(G,U(1))$.

If instead $Q$ generates a continuous group such as $U(1)$, the holonomy ${\mathrm{e}}^{\mathrm{i}\theta }$ runs continuously and the twist is a background $U(1)$ gauge field wrapped on the circle. This is the chemical potential exactly: $\mu$ enters as an imaginary time-component gauge field through ${\partial }_{\tau }\to {\partial }_{\tau }-\mu$, a Wilson line ${\mathrm{e}}^{\beta \mu }$ around ${𝖲}_{\beta }^1$, and one dials it rather than summing. Either way $Q$ must generate a global symmetry, acting on the states the trace runs over, and the connection is a background: a gauged (local) symmetry grades nothing-its charged states are projected out-and its connections are integrated over, not fixed. The holonomy $\theta$ of §I.2 is now literal, the phase a field acquires on going once around ${𝖲}_{\beta }^1$-every knob of the Interlude a boundary condition.

For a free (Gaussian) theory the path integral is computed exactly by a functional determinant,

(3.5)

The opposite signs of the exponent are the determinant-level shadow of Bose vs. Fermi, and the halves are the shadow of reality: a real (Majorana) fermion’s Gaussian integral is a Pfaffian, $\mathrm{Pf}𝒪={det}^{1/2}𝒪$, just as the real scalar gives ${det}^{-1/2}$; their complex cousins-Dirac fermion, complex scalar-carry the powers $\pm 1$. Writing the determinant as a product over the modes on ${𝖲}_{\beta }^1$ and doing the Matsubara sum reproduces (5) and (6), along with the zero-point factor the next section takes up: the path integral and the trace agree, as they must. Meanwhile $\log Z$ is the sum of connected vacuum diagrams-the field-theory free energy.

The determinant carries a zero-point piece $\frac{1}{2}{\sum }_n{\omega }_n$, the sum of the ground-state energies of all the modes (Example 1.2, summed). It diverges. Make it finite the cheap way: put the system in a periodic box of length $L$, an infrared regulator (a lattice length, nothing geometric yet), so a tower of modes discretizes to ${\omega }_n=2\pi n/L$ and the zero-point sum is $\frac{\pi }{L}{\sum }_{n\ge 1}n$. (Reading $L$ as a genuine spatial circle-one cycle of a torus-is the business of the 2d conformal field theory of Lecture 4; here it is only a cutoff.) Zeta-regularize:

$$\sum _{n\ge 1}n\stackrel{𝜁}{=}\zeta (-1)=-\frac{1}{12},⟹E_0=-\frac{\pi }{12L}.$$

(3.6)

The finite leftover is the Casimir energy.²² 2 The textbook Casimir energy of a 2d field on a circle, $-\pi c/6L$, counts the left- and right-moving towers together; our single tower ${\omega }_n=2\pi n/L$ is one chiral half of it.

What does the fermion contribute? Its sign is fixed by the mode algebra, before any geometry enters. Quantizing a bosonic mode symmetrizes the classical product: $H=\frac{\omega }{2}\left(a^{†}a+a{a}^{†}\right)=\omega \left(a^{†}a+\frac{1}{2}\right)$, the $+\frac{1}{2}\omega$ paid to the commutator $[a,a^{†}]=1$. Quantizing a fermionic mode antisymmetrizes: $H=\frac{\omega }{2}\left(c^{†}c-c{c}^{†}\right)=\omega \left(c^{†}c-\frac{1}{2}\right)$, and now the anticommutator $\{c,c^{†}\}=1$ pays $-\frac{1}{2}\omega$. The same one bit of statistics that set the $\pm$ of §3.2 flips the zero-point energy: every bosonic mode contributes $+\frac{1}{2}\omega$ to the vacuum, every fermionic mode $-\frac{1}{2}\omega$.

That opposite sign is an invitation. Suppose a theory supplies one fermionic mode for every bosonic mode of the same frequency-matched mode for mode, not component for component. (The distinction matters: a fermion’s action is first order, so the field is its own conjugate momentum and each component carries only half a degree of freedom; a naive component count makes fermions look twice as numerous as they are. Which multiplets match is then dimension-dependent bookkeeping-in four dimensions, for instance, the four real components of a Majorana fermion supply the same two modes per momentum as a complex scalar.) Then $+\frac{1}{2}\omega$ meets $-\frac{1}{2}\omega$ at every single frequency and the vacuum energy cancels mode by mode: divergence, finite part, everything, with no regulator invoked at all. This is historically one of the cheapest ways to motivate supersymmetry: a symmetry exchanging bosons and fermions forces exactly such a pairing of modes, and a vanishing vacuum energy is its signature-the SUSY algebra makes $H\ge 0$, saturated by an invariant ground state. What survives when the pairing is imperfect-the net count of unpaired ground states-is the Witten index of Lecture 5, a partition function that counts with signs.

When the spectra do not match, the regulator exposes the mismatch-and here the box itself asks the fermion the same question the time circle asked in §3.2: periodic or antiperiodic around the compact direction? Both are consistent; as with the two spin structures there, the regulated theory must simply pick one. (On the torus of Lecture 4 the two choices become the R and NS sectors.) Periodic, the fermion is integer moded like the boson, and its zero-point sum is the boson’s with the sign reversed:

$$E_0^{\mathrm{per}}=-\frac{1}{2}\sum _{n\ge 1}{\omega }_n=-\frac{\pi }{L}\sum _{n\ge 1}n\stackrel{𝜁}{=}+\frac{\pi }{12L},$$

(3.7)

equal and opposite to (3.6)-the finite shadow of the mode-by-mode cancellation above. Antiperiodic, it is half-integer moded, ${\omega }_r=2\pi r/L$ with $r=\frac{1}{2},\frac{3}{2},\mathrm{\dots }$, and

$$E_0^{\mathrm{anti}}=-\frac{1}{2}\sum _r{\omega }_r=-\frac{\pi }{L}\sum _{k\ge 0}\left(k+\frac{1}{2}\right)\stackrel{𝜁}{=}-\frac{\pi }{24L},$$

(3.8)

the shifted sum ${\sum }_{k\ge 0}(k+\frac{1}{2})$ $\zeta$-regularizing to $+\frac{1}{24}$. Two boundary conditions, two vacuum energies. Notice, finally, what did not spoil the cancellation: Euclidean time. The zero-point sum is a statement about the Hamiltonian on space, untouched by the Wick rotation, and the thermal antiperiodicity of §3.2 lives on the time circle, which never enters it. What halved the antiperiodic answer-$-\pi /24L$ against the boson’s $-\pi /12L$-is the shifted moding alone, $\sum (k+\frac{1}{2})$ against $\sum n$. On the torus these two numbers reappear as the leading powers of the theta-function blocks of Lecture 4.

• •

  $\beta \leftrightarrow$ length of the Euclidean time circle.

• •

  ${(-1)}^F$ insertion $\leftrightarrow$ flipping the fermion from antiperiodic to periodic $\leftrightarrow$ the choice of spin structure on ${𝖲}_{\beta }^1$ (§3.2), one bit of the four of L4.

• •

  $det\leftrightarrow$ the one-loop free energy; its zero-point part $\leftrightarrow$ the $-c/24$ shift.