Quantum Detailed Balance and Weighted Inner Products

A stationary density operator need not make the dynamics reversible. Quantum detailed balance adds a symmetry condition relative to that state, but several inequivalent noncommutative symmetry conventions exist. A mathematical statement must name its weighted inner product and its treatment of Hamiltonian motion. The canonical Detailed Balance article develops transition rates and thermal examples; this page fixes an operator-space convention and derives its consequences.

Required background. The GKSL–Lindblad Theorem gives the Heisenberg generator. Helpful background. Normal States explains faithful density operators in infinite dimension.

Detailed balance depends on the operator inner product

First take a finite-dimensional Hilbert space and a faithful density operator $\sigma>0$. Two common inner products on observables are

$$\langle A,B\rangle_{\sigma,\mathrm{GNS}} =\operatorname{Tr}(\sigma A^\dagger B),$$

and

$$\langle A,B\rangle_{\sigma,\mathrm{KMS}} =\operatorname{Tr}(\sigma^{1/2}A^\dagger\sigma^{1/2}B).$$

They agree for observables commuting with $\sigma$ but generally differ on coherences. In an eigenbasis $\sigma=\sum_ip_i|i\rangle\langle i|$, a matrix unit $E_{ij}$ has squared GNS norm $p_j$ and squared KMS norm $\sqrt{p_ip_j}$. Moving a factor of $\sigma^{1/2}$ through an arbitrary observable would erase precisely this distinction.

We use the following GNS detailed-balance convention with Hamiltonian part. A Heisenberg generator is written

$$\mathcal G(A)=i[H,A]+\mathcal D(A),\qquad [H,\sigma]=0,$$

where $\mathcal D$ generates a unital CP semigroup and is self-adjoint in the GNS inner product:

$$\langle A,\mathcal D(B)\rangle_{\sigma,\mathrm{GNS}} =\langle\mathcal D(A),B\rangle_{\sigma,\mathrm{GNS}}.$$

The Hamiltonian derivation is skew-adjoint in this inner product. The full generator therefore need not be self-adjoint; detailed balance concerns its dissipative symmetry after the specified reversible part is separated. Other definitions use KMS symmetry or an explicit time-reversal operation. They must not be treated as interchangeable without additional hypotheses.

Symmetry implies stationarity and nonnegative dissipation

Since $\mathcal D(I)=0$, symmetry gives

$$\operatorname{Tr}[\sigma\mathcal D(A)] =\langle I,\mathcal D(A)\rangle_{\sigma,\mathrm{GNS}} =\langle\mathcal D(I),A\rangle_{\sigma,\mathrm{GNS}}=0.$$

The Hamiltonian contribution has zero $\sigma$ expectation because $[H,\sigma]=0$. Thus $\sigma$ is stationary for the predual of $\mathcal G$.

Let $S_t=e^{t\mathcal D}$. The unital CP Schwarz inequality and stationarity imply

$$\|S_t(A)\|_{\sigma,\mathrm{GNS}}^2 \le\operatorname{Tr}[\sigma S_t(A^\dagger A)] =\|A\|_{\sigma,\mathrm{GNS}}^2.$$

Differentiating at zero shows $\operatorname{Re}\langle A,\mathcal D(A)\rangle_{\sigma,\mathrm{GNS}}\le0$. Self-adjointness makes this quantity real, so

$$\mathcal E(A,A)=-\langle A,\mathcal D(A)\rangle_{\sigma,\mathrm{GNS}}\ge0.$$

This is the associated Dirichlet form. In finite dimensions, the spectral theorem on this weighted Hilbert space gives real nonpositive eigenvalues for $\mathcal D$ and orthogonal relaxation modes. The statement does not imply that the full $\mathcal G$, including Hamiltonian motion, has only real eigenvalues.

The product-dissipation identity also gives $\mathcal E(A,A)=\frac12\operatorname{Tr}[\sigma\Gamma_{\mathcal D}(A)]$. The positive commutator-square expression for $\Gamma_{\mathcal D}$ makes the loss of conserved observable components explicit.

Worked example: reversible resetting

For any faithful $\sigma$, define

$$\mathcal D(A)=\gamma(\operatorname{Tr}(\sigma A)I-A),\qquad \gamma>0.$$

The map $P_\sigma(A)=\operatorname{Tr}(\sigma A)I$ is the orthogonal projection onto the span of $I$ for both the GNS and KMS inner products. Indeed, $\|I\|_\sigma^2=1$ and $\langle I,A\rangle_\sigma=\operatorname{Tr}(\sigma A)$ in either convention.

Therefore $\mathcal D=\gamma(P_\sigma-I)$ is self-adjoint in both inner products. It generates the dual of the reset channel semigroup, fixes $I$, and acts as $-\gamma$ on every observable of zero $\sigma$ mean. In the GNS convention,

$$\mathcal E(A,A)=\gamma\left( \operatorname{Tr}(\sigma A^\dagger A) -|\operatorname{Tr}(\sigma A)|^2 \right).$$

Thus the dissipation rate is $\gamma$ times the weighted variance of the observable. This example does not require $\sigma$ to be a Gibbs state; detailed balance is a symmetry relative to a specified faithful state.

Stationarity alone permits circulating probability

On three levels, take jumps $L_j=|j+1\rangle\langle j|$, with indices modulo three and unit rates. The uniform state $\sigma=I/3$ is stationary, since the incoming and outgoing total intensities both equal $I$.

On diagonal observables $f_j$, however,

$$(\mathcal D f)_j=f_{j+1}-f_j.$$

The nonconstant Fourier modes have eigenvalues $e^{\pm2\pi i/3}-1$, which are not real. Hence the generator is not self-adjoint in the uniform weighted inner product. Subtracting a Hamiltonian derivation cannot repair its asymmetric population form: the diagonal entries of $[H,A]$ vanish for every diagonal $A$. Probability circulates in one direction around the cycle; there are no balancing reverse transitions.

On a classical diagonal algebra, GNS symmetry reduces to $\pi_iq_{ij}=\pi_jq_{ji}$ for each pair of transition rates. Stationarity only requires total incoming and outgoing probability fluxes to cancel at each vertex. The latter is strictly weaker.

Infinite-dimensional weighted spaces require care

For a faithful normal density operator on an infinite-dimensional Hilbert space, the same inner-product formulas are defined on bounded observables. Their completions are weighted operator Hilbert spaces. But $\sigma$ has no bounded inverse in general, so these norms are not uniformly equivalent to an unweighted Hilbert–Schmidt norm.

For an unbounded generator, symmetry must be stated on a specified dense domain, and self-adjointness or closure of the associated form needs proof. A formal cyclic rearrangement of unbounded products is not such a proof. KMS symmetry, modular compatibility, and time-reversal variants have their own exact conditions; Fagnola and Umanità analyze these distinctions for norm-continuous semigroups on $B(\mathcal H)$.

Exercises

Two-state restriction. For a classical two-state generator with rates $a:0\to1$ and $b:1\to0$, both positive, find the stationary weights and verify detailed balance.

Solution

$\pi_0=b/(a+b)$ and $\pi_1=a/(a+b)$ give $\pi_0a=\pi_1b=ab/(a+b)$. There is only one edge, so stationary flux balance also balances that edge. The three-state directed cycle shows why this coincidence does not generalize to networks with cycles.

Faithfulness. What fails if $\sigma$ has a nontrivial kernel?

Solution

The weighted forms can assign zero norm to nonzero observables. They are then only semidefinite on the full algebra. One must restrict to the supported algebra or quotient by null vectors before invoking Hilbert-space self-adjointness. A statement using a faithful inner product cannot simply drop faithfulness.

References

• Alicki, Robert. “On the detailed balance condition for non-hamiltonian systems.” Reports on Mathematical Physics 10, 249–258 (1976). doi:10.1016/0034-4877(76)90046-X. Quantum generator symmetry and detailed balance.
• Fagnola, Franco, and Veronica Umanità. “Generators of KMS symmetric Markov semigroups on $B(h)$: symmetry and quantum detailed balance.” Communications in Mathematical Physics 298, 523–547 (2010). doi:10.1007/s00220-010-1011-1; preprint. KMS symmetry and distinct quantum detailed-balance conventions.