Unitarity, fluctuations, and counting fields on the closed time path

Key claim

The two Schwinger–Keldysh branches are an amplitude and its conjugate, not two physical copies of the system.

Equal sources make the branches cancel:

A controlled difference between the sources prevents that cancellation. Derivatives with respect to the difference source then generate expectation values, response functions, noise, and charge-transfer cumulants.

Theme

Closed-time-path unitarity, the fluctuation–dissipation theorem, and nonequilibrium quantum-dot transport.

Guiding question

Why does setting the two branch sources equal erase the generating functional, while an opposite phase twist on the branches reveals the full probability distribution of transferred charge?

Setup

Consider a spinless resonant level coupled to two noninteracting reservoirs:

Use a partitioned initial state,

and switch on the coupling before taking a long-time stationary limit. The reservoirs must be macroscopic enough to absorb energy and phase information without returning it on the observation timescale.

For branch-dependent sources, define

If $\mathrm{Tr}{\rho }_0=1$ and $J_{+}=J_{-}=J$, cyclicity of the trace and unitary evolution give

Source convention

We use $S_J=\int \mathrm{d}t(J_{+}{O}_{+}-J_{-}{O}_{-})$, $J_r=(J_{+}+J_{-})/2$, and $J_a=J_{+}-J_{-}$. Other sign conventions move minus signs or factors of two between sources and derivatives.

With $W=-\mathrm{i}\ln Z$, this convention gives

The first identity is exact. The second states how an observable is extracted before the physical limit $J_a\to 0$ is imposed.

Analysis

Derivation: causal rotation and the stationary dot

Introduce Grassmann fields $d_{+}$ and $d_{-}$ on the forward and backward branches. For fermions, a useful Larkin–Ovchinnikov rotation is

with the conjugate fields rotated as

In this convention the Green function has the causal form

Its components are

and

$G^{R/A}$ describes spectral support and causal response. $G^K$ carries occupation information. The triangular arrangement depends on basis convention, but the separation between response and state does not. A standard derivation is given by Kamenev and Levchenko.

The leads are Gaussian and can be integrated out exactly. Define

In the wide-band approximation,

where $F_{\alpha }=1-2f_{\alpha }$. Absorb the approximately constant real shift $\delta {ϵ}_{\alpha }$ into ${ϵ}_d$. Then

and

The effective retarded level ${ϵ}_d-\mathrm{i}\mathrm{\Gamma }/2$ fixes the resonance and decay rate. It does not fix the dot occupation.

Steady-state limit

The exact Keldysh Dyson equation also contains a term propagated from the initial dot Green function. Writing $G^K=G^R{\mathrm{\Sigma }}^K{G}^A$ assumes that this memory has decayed in the open-system long-time limit.

After the initial local memory is lost,

Using

one finds

Equivalently,

The retarded propagator can therefore remain unchanged while the bias reverses and the current changes sign. The missing information sits in $G^K$.

Scope and assumptions: fluctuation–dissipation is an equilibrium constraint

At equilibrium, the density matrix is grand canonical and the fermionic KMS relation gives

Since

their ratio is fixed:

This is the fermionic fluctuation–dissipation theorem. It says that, in thermal equilibrium, the fluctuation sector is not independent of the spectral sector.

For the resonant level, the same result follows directly when

because then $F_L=F_R=\tanh [(\omega -\mu )/(2T)]$ and therefore $F_d=F_L$.

KMS is additional input

Closed-contour unitarity gives $Z[J,J]=1$ for any normalized state. FDT also requires a thermal KMS state. In effective Schwinger–Keldysh theory this extra condition is encoded by a dynamical KMS symmetry, subject to the relevant microscopic reversal assumptions.

Under voltage or temperature bias,

is generally not a single Fermi function. No universal pair $(T_{\mathrm{e}\mathrm{f}\mathrm{f}},{\mu }_{\mathrm{e}\mathrm{f}\mathrm{f}})$ reconstructs it at all frequencies. A fitted low-frequency effective temperature may depend on the observable and should not be mistaken for a KMS temperature.

The same distinction appears in current fluctuations. When $f_L=f_R=f$, the long-time counting formula below gives

Linearizing the particle current in $V={\mu }_L-{\mu }_R$ gives

so that

This is the zero-frequency thermal-noise form of FDT in our counting convention. Electrical noise restores factors of charge, while one-sided versus two-sided spectral conventions can supply another factor of two.

Physical interpretation: the counting field resolves histories

To count particles transferred through the left contact during a time window $\tau$, attach opposite phases to both terms of the left hopping Hamiltonian:

For a two-point measurement of the left-lead particle number, with the initial density matrix compatible with the first measurement,

Measurement protocol

A phase-deformed contour is not by itself a classical probability distribution. The formula for $P_{\tau }(N)$ assumes a specified detector or, here, a two-projective-measurement protocol. Different protocols can differ through initial coherences and finite-time boundary terms.

For the stationary, noninteracting, two-terminal resonant level at long measuring time, the Levitov–Lesovik result is

with

For the resonant-level model, the long-time formula and its measurement setting can be established directly in the nonequilibrium steady state Bernard and Doyon.

The cumulants follow from

The first two rates are

and

The two terms are thermal noise and partition noise. At zero temperature, for a bias window wide enough to cover the resonance,

Symmetric barriers give $F_{\mathrm{F}\mathrm{a}\mathrm{n}\mathrm{o}}=1/2$.

Units and regime

Here $\hslash =k_B=e=1$, and $I$ is particle current. The determinant formula assumes a spinless Gaussian conductor, a stationary state, and long measurement time; the displayed Lorentzian transmission also uses wide-band leads.

If both reservoirs have the same inverse temperature $\beta$, then

The integrand is invariant under

which yields the steady-state fluctuation symmetry

This relation assumes microreversibility and the long-time steady-state limit. A systematic account of quantum measurement protocols and fluctuation theorems is given by Esposito, Harbola, and Mukamel.

Interactions change the calculation. For an Anderson impurity, the average current can be written in terms of interacting Green functions under the conditions of the Meir–Wingreen formula. Noise and higher cumulants generally require counting-field-dependent self-energies and vertex information. They do not follow from one dressed transmission probability.

False claim to diagnose

Once the retarded effective Hamiltonian is known, all nonequilibrium transport observables follow from its complex eigenvalues.

The complex pole gives the resonance position and lifetime. It contains no reservoir occupation factors and no correlations between transfer events.

Two steady states can therefore share the same $G^R$ while carrying opposite currents. Their $G^K$ functions and counting statistics are different.

What follows — and what does not

Statement	Status
$Z[J,J]=1$ for normalized unitary evolution.	Exact identity
The causal Keldysh matrix separates $G^{R/A}$ from $G^K$.	Exact structure, with basis-dependent placement
$G^R=(\omega -{ϵ}_d+\mathrm{i}\mathrm{\Gamma }/2{)}^{-1}$.	Noninteracting wide-band result
$G^K=G^R{\mathrm{\Sigma }}^K{G}^A$.	Long-time result after initial memory is lost
Equilibrium $G^K$ is fixed by $G^R-G^A$.	KMS/FDT constraint, not unitarity alone
A biased stationary state must have an effective temperature.	False in general
The displayed $\mathcal{W}(\chi )$ gives all charge cumulants.	Long-time Gaussian two-terminal result with a measurement protocol
The same transmission formula applies unchanged to an interacting dot.	False; vertices enter higher cumulants
The fluctuation symmetry holds without thermodynamic assumptions.	False; equal temperature, microreversibility, and steady-state conditions matter

The closed contour supplies a consistent language for normalization, response, fluctuations, and measurement. It does not make equilibrium identities or Gaussian determinant formulas universal.

Exercise

For the resonant-level model:

1. Prove $Z[J,J]=1$ from the trace definition.
2. Derive $G^K=(G^R-G^A)F_d$ in the wide-band stationary limit.
3. Use the KMS relation to derive the fermionic FDT.
4. Differentiate $\mathcal{W}(\chi )$ to obtain the current and second cumulant.
5. For equal reservoir temperatures, verify the fluctuation symmetry.

Hint 1

For the first part, cycle $U_J^{†}$ through the trace. For the third, express $G^K$ and $G^R-G^A$ through $G^{>}$ and $G^{<}$.

Hint 2

Set $a=f_L(1-f_R)$ and $b=f_R(1-f_L)$. At equal temperature, $a/b={\mathrm{e}}^{\beta V}$.

Oral check 1. At what step is initial-state memory discarded in the derivation of the stationary $G^K$?

Oral check 2. Why can a biased stationary state violate FDT without violating unitarity?

Solution

Equal sources give

$$Z[J,J]=\mathrm{Tr}(U_J{\rho }_0{U}_J^{†})=\mathrm{Tr}({\rho }_0{U}_J^{†}{U}_J)=1.$$Copy LaTeX

In the stationary open-system limit,

$$G^K=G^R{\mathrm{\Sigma }}^K{G}^A=-\mathrm{i}{G}^R{G}^A({\mathrm{\Gamma }}_L{F}_L+{\mathrm{\Gamma }}_R{F}_R).$$Copy LaTeX

Since $G^R-G^A=-\mathrm{i}\mathrm{\Gamma }{G}^R{G}^A$,

$$G^K=(G^R-G^A)\frac{{\mathrm{\Gamma }}_L{F}_L+{\mathrm{\Gamma }}_R{F}_R}{\mathrm{\Gamma }}.$$Copy LaTeX

At equilibrium, KMS gives $G^{>}=-{\mathrm{e}}^{\beta (\omega -\mu )}{G}^{<}$. Therefore

$$\frac{G^K}{G^R-G^A}=\frac{G^{>}+G^{<}}{G^{>}-G^{<}}=\tanh \left(\frac{\omega -\mu }{2T}\right).$$Copy LaTeX

For the counting problem, differentiation gives

$$\frac{⟨N⟩}{\tau }=\int \frac{\mathrm{d}\omega }{2\pi }\mathcal{T}(a-b)=\int \frac{\mathrm{d}\omega }{2\pi }\mathcal{T}(f_L-f_R),$$Copy LaTeX

and

$$\frac{⟨⟨N^2⟩⟩}{\tau }=\int \frac{\mathrm{d}\omega }{2\pi }[\mathcal{T}(a+b)-{\mathcal{T}}^2(a-b{)}^2].$$Copy LaTeX

This is algebraically equal to the thermal-plus-partition form in the main text.

At equal temperature, $a={\mathrm{e}}^{\beta V}b$. Replacing $\chi$ by $-\chi +\mathrm{i}\beta V$ exchanges the forward and backward terms inside the logarithm, proving

$$\mathcal{W}(\chi )=\mathcal{W}(-\chi +\mathrm{i}\beta V).$$Copy LaTeX

The steady-state $G^K$ step discarded the propagated initial-dot term. FDT required KMS in addition to stationarity and unitarity.

Check your understanding

Suppose two reservoirs have the same ${\mathrm{\Gamma }}_L$, ${\mathrm{\Gamma }}_R$, and density of states, but their chemical potentials are interchanged.

Which of $G^R$, $G^K$, $I$, and the equilibrium FDT relation remain unchanged? Explain each answer from the equations rather than from intuition.

You may also reply with “deeper,” “too easy,” “too hard,” or your derivation.

Further Reading

• Keldysh technique and non-linear sigma-model: basic principles and applications
• Full Counting Statistics in the Resonant-Level Model
• Nonequilibrium fluctuations, fluctuation theorems, and counting statistics in quantum systems
• Effective field theory of dissipative fluids
• Schwinger–Keldysh formalism I: BRST symmetries and superspace
• Landauer formula for the current through an interacting electron region

Connections and next step

• Unitarity: branch equality fixes normalization before any model approximation.
• Causality: the Keldysh rotation separates response from state information.
• Equilibrium: KMS relates fluctuations to dissipation; stationarity alone does not.
• Transport: a branch-antisymmetric phase resolves charge-transfer histories.
• Many-body limit: higher cumulants expose vertex information hidden from the average current.
• Next step: derive a counting-field-dependent Dyson equation and its charge-conservation Ward identity.
• Revisit: compare the thermal noise here with the fluctuation physics in the irreversibility entry.