Coherent-State Path Integrals for Many-Body Systems

A normal-ordered Hamiltonian becomes a coherent-state path integral by inserting resolutions of identity at finite time step. The discrete kernel fixes which field is evaluated on each side of a time link, the endpoint terms, and the bosonic or fermionic thermal boundary condition. Only after those data are fixed is the familiar continuum action $\psi^*(\partial_\tau-\mu)\psi+H(\psi^*,\psi)$ legitimate shorthand.

Required background. Second-Quantized Bosons and Fermions fixes the operator algebra; Regulated Bosonic Field Integrals supplies finite-dimensional measures; Grassmann Functional Integrals for Free Fermions supplies Grassmann Gaussian integration.

Helpful background. Thermal Density Operators and the KMS Condition explains the thermal trace, and Retarded, Advanced, and Keldysh Bases is the continuation route for real-time contours.

Coherent states and the finite-time kernel

For one bosonic mode, $|z\rangle=e^{-|z|^2/2}e^{za^\dagger}|0\rangle$ satisfies $a|z\rangle=z|z\rangle$ and

$$1=\int\frac{\mathrm d^2z}{\pi}\,|z\rangle\langle z|.$$

Fermionic coherent states use Grassmann numbers $\eta,\bar\eta$ and the Berezin resolution of identity. Their eigenvalue property is analogous, but moving Grassmann variables past one another carries signs.

For the grand-canonical operator $K=H-\mu N$ and $\epsilon=\beta/N_\tau$, insert an identity between the factors of $e^{-\epsilon K}$. If $K$ is normal ordered, then to first order

$$\langle z_{n+1}|e^{-\epsilon K}|z_n\rangle =\langle z_{n+1}|z_n\rangle \exp\!\left[-\epsilon K(z^*_{n+1},z_n)+O(\epsilon^2)\right].$$

The unequal labels $z^*_{n+1}$ and $z_n$ are the ordering prescription. Replacing both immediately by a single $z(\tau)$ erases information that can produce finite counterterms for nonlinear Hamiltonians Wilson and Galitski 2011, pp. 110401-1–110401-4.

After combining overlaps, a convenient discrete bosonic action is

$$S_E^{(N_\tau)} =\sum_{n=0}^{N_\tau-1} \left[ z^*_{n+1}(z_{n+1}-z_n) +\epsilon K(z^*_{n+1},z_n) \right],$$

with $z_{N_\tau}=z_0$ for the trace. Equivalent discretizations redistribute overlap terms and measures; they must agree before the continuum limit.

Bosonic and fermionic boundary conditions

For many modes, replace labels by fields and sums by regulated spatial integrals. Thermal bosons are periodic,

$$\phi(\beta,\mathbf x)=\phi(0,\mathbf x),$$

while thermal fermions are antiperiodic,

$$\psi(\beta,\mathbf x)=-\psi(0,\mathbf x).$$

The minus sign follows from closing the Grassmann coherent-state trace, not from a convention chosen after Fourier transformation. It produces Matsubara frequencies $2\pi nT$ for bosons and $(2n+1)\pi T$ for fermions.

With the discrete definition understood, the continuum Euclidean notation is

$$Z=\int_{\rm BC}\mathcal D(\psi^*,\psi)\, e^{-S_E}, \qquad S_E=\int_0^\beta\mathrm d\tau\, \left[ \int\mathrm d^d x\,\psi^*\partial_\tau\psi +K(\psi^*,\psi) \right].$$

For fermions, $\psi^*$ is an independent Grassmann variable conventionally written $\bar\psi$; it is not complex conjugation inside the integration algebra. For real-time evolution the Berry term acquires the corresponding factor of $i$, the contour and initial density matrix must be stated, and Euclidean antiperiodicity alone does not define an in–in functional.

Exact single-mode checks

Take $K=\xi a^\dagger a$ with $\xi=\varepsilon-\mu$. The discrete Gaussian determinant gives

$$Z_B=\frac{1}{1-e^{-\beta\xi}}, \qquad \xi>0,$$

for a boson, and

$$Z_F=1+e^{-\beta\xi}$$

for a fermion. These equal the operator traces exactly. The bosonic convergence condition is physical: when $\xi\le0$ for an isolated unbounded oscillator, the grand-canonical sum does not converge.

The occupation follows from $-\beta^{-1}\partial_\xi\ln Z$:

$$n_B(\xi)=\frac1{e^{\beta\xi}-1}, \qquad n_F(\xi)=\frac1{e^{\beta\xi}+1}.$$

This simple fixture simultaneously tests the sign of $\mu$, the temporal boundary condition, the determinant power, and normalization. A continuum formula that fails it is not repaired by taking more time slices.

Interactions and normal symbols

For a normal-ordered bosonic interaction

$$H_{\rm int}=\frac U2 a^\dagger a^\dagger aa,$$

the discrete normal symbol is $\tfrac U2(z^*_{n+1})^2z_n^2$. It is not $\tfrac U2|z_n|^4$ until the continuum identification has been justified. Moreover,

$$a^\dagger a^\dagger aa=N(N-1),$$

not $N^2$. A naive classical replacement can therefore generate an $O(U)$ ordering error even as a formal time derivative looks correct.

For spatial fields, contact products require a spatial regulator in addition to the temporal one. Time slicing controls operator ordering; it does not renormalize ultraviolet spatial loops. These are independent limits.

Sources and observables

Add sources on the same time links as the operators they represent. A density source coupled to $a^\dagger a$ uses $z^*_{n+1}z_n$. Differentiating $\ln Z$ before taking the continuum limit fixes equal-time contact terms. If sources are introduced only after writing a smooth continuum action, those terms can be missed and Ward identities may fail.

The safe workflow is therefore:

1. normal order the operator Hamiltonian relative to a declared reference;
2. construct the finite-$N_\tau$ coherent-state kernel;
3. impose endpoint or thermal boundary data;
4. add sources with their operator ordering;
5. verify a finite system or free determinant; and
6. only then take continuum notation and the temporal limit.

Common pitfalls

Dropping the one-step offset. $z^*_{n+1}z_n$ records normal ordering. Setting the labels equal before checking the discrete integral can shift nonlinear actions.

Using periodic Grassmann fields. A fermionic thermal trace is antiperiodic. Periodic Grassmann boundary conditions compute a different graded object.

Letting time slicing stand in for spatial renormalization. The temporal prescription fixes ordering; a contact interaction still needs spatial regularization and matching.

Exercises

Recover the fermionic determinant

Discretize one fermionic level and show that antiperiodicity gives $Z_F=1+e^{-\beta\xi}$.

Solution

The discrete bilinear matrix represents one-step transport by $e^{-\epsilon\xi}$ and an antiperiodic closure. Its determinant is $1+(e^{-\epsilon\xi})^{N_\tau}=1+e^{-\beta\xi}$. A periodic closure would give $1-e^{-\beta\xi}$ and fail the operator trace.

Locate an ordering error

Compare the spectra obtained from $H=\tfrac U2N(N-1)$ and the incorrectly ordered $H'=\tfrac U2N^2$.

Solution

On $|n\rangle$, the exact energies are $E_n=Un(n-1)/2$, whereas $E'_n=Un^2/2$. Their difference is $Un/2$, an $O(U)$ one-body shift. It does not vanish with the temporal step and must be corrected at the operator-symbol stage.

Continue

Hubbard–Stratonovich Fields and Collective Channels uses these regulated measures to decouple interactions. Densities, Currents, and Nonrelativistic Ward Identities differentiates source-coupled functionals. Quantum-Matter Correlators and Observable Conventions translates the resulting Euclidean correlators to real-time observables.

References

• Wilson, Justin H., and Victor Galitski. “Breakdown of the Coherent State Path Integral: Two Simple Examples.” Physical Review Letters 106 (2011): 110401. DOI. Open PDF.

Further reading

• Kochetov, Evgueny A. “SU(2) Coherent-State Path Integral.” Journal of Mathematical Physics 36 (1995): 4667–4679. DOI.
• Negele, John W., and Henri Orland. Quantum Many-Particle Systems. Boca Raton, FL: CRC Press, 2018; originally published 1988. DOI.