Open-System Applications

Symmetric noise can erase a conserved quantity of the isolated system. An isotropic depolarizing channel, for example, treats every spin direction equally while reducing every spin expectation value. Open-system symmetry therefore requires two separate questions: does the dynamical map commute with the symmetry action, and does its adjoint preserve the observable of interest?

Helpful background. Density operators describe mixed states. Kraus maps describe finite operations, and the Lindblad equation describes the semigroup approximation used below. The explicit examples are finite dimensional.

Covariance is a property of the channel

For a unitary representation $U_g$, write $\mathcal U_g(\rho)=U_g\rho U_g^\dagger$. A completely positive trace-preserving channel is covariant if

$$\mathcal E\circ\mathcal U_g=\mathcal U_g\circ\mathcal E$$

for every group element. Preparing a rotated input and then applying the channel gives the rotated version of the original output. This need not make the input and output equal.

For example, on qubit density matrices,

$$\mathcal E_p(\rho)=(1-p)\rho+pI/2,\qquad 0\leq p\leq1.$$

Because $UIU^\dagger=I$, the channel is covariant under every qubit unitary. Nevertheless its Bloch vector shrinks as $\mathbf r\mapsto(1-p)\mathbf r$. In particular, full rotational covariance does not conserve $\langle\mathbf S\rangle$.

A Kraus representation $\mathcal E(\rho)=\sum_aK_a\rho K_a^\dagger$ is not unique. Requiring each $K_a$ individually to commute with $U_g$ is sufficient for covariance but unnecessarily restrictive. A symmetry can mix Kraus operators without changing their complete map. The depolarizing and amplitude-damping examples illustrate different covariance groups.

The environment state can break the reduced symmetry

For a fixed initial environment state $\eta$ and an initially uncorrelated system, the joint unitary $W$ induces

$$\mathcal E(\rho)=\operatorname{Tr}_E[W(\rho\otimes\eta)W^\dagger].$$

Suppose $W$ commutes with $U_g\otimes V_g$ and $V_g\eta V_g^\dagger=\eta$. Moving this joint symmetry through $W$ and using invariance of the partial trace gives

$$\begin{aligned} \mathcal E(U_g\rho U_g^\dagger) &=\operatorname{Tr}_E[(U_g\otimes V_g) W(\rho\otimes\eta)W^\dagger (U_g^\dagger\otimes V_g^\dagger)]\\ &=U_g\mathcal E(\rho)U_g^\dagger. \end{aligned}$$

Thus a symmetric interaction alone is not the whole argument. A polarized environment can supply a preferred direction. For a concrete counterexample, a full swap of two qubits commutes with $U\otimes U$ for every $U$, but its reduced channel is $\mathcal E(\rho)=\eta$. This replacement channel has full rotational covariance only if $\eta=I/2$.

The fixed-product-state assumption also matters: arbitrary initial correlations do not generally define one completely positive map acting on every possible system input.

Conservation is tested with the adjoint generator

A time-homogeneous completely positive trace-preserving semigroup in finite dimension has the Gorini–Kossakowski–Sudarshan–Lindblad form

$$\mathcal L(\rho)=-\frac i\hbar[H,\rho] +\sum_\mu\left(L_\mu\rho L_\mu^\dagger -\frac12\{L_\mu^\dagger L_\mu,\rho\}\right).$$

Rates are absorbed into $L_\mu$, which have units of inverse square-root time. General reduced dynamics need not be a semigroup. Lindblad’s original result and the generator reference explain the assumptions; unbounded generators require additional domain analysis.

The adjoint is defined by $\operatorname{Tr}[Q\mathcal L(\rho)]=\operatorname{Tr}[\mathcal L^\dagger(Q)\rho]$. For a time-independent observable,

$$\mathcal L^\dagger(Q)=\frac i\hbar[H,Q] +\sum_\mu\left(L_\mu^\dagger QL_\mu -\frac12\{L_\mu^\dagger L_\mu,Q\}\right).$$

Its expectation is conserved for every state precisely when $\mathcal L^\dagger(Q)=0$. For explicit $Q(t)$, include $\partial_tQ$. Checking $[H,Q]=0$ alone ignores the dissipator. Checking only $\langle Q\rangle$ also does not in general establish preservation of the complete measurement distribution; one must test the spectral projectors or the corresponding functions of $Q$. The three-level example in Exercise 4 preserves the mean while increasing the variance.

The covariance test instead is $\mathcal L\circ\mathcal U_g=\mathcal U_g\circ\mathcal L$. These are different equations for different physical questions.

Dephasing and relaxation make the distinction concrete

Take Pauli operators $X,Y,Z$ and no Hamiltonian term. With $L=\sqrt{\gamma/2}\,Z$, $\gamma\geq0$,

$$\dot\rho=\frac\gamma2(Z\rho Z-\rho),\qquad \dot r_x=-\gamma r_x,\quad \dot r_y=-\gamma r_y,\quad \dot r_z=0.$$

The channel is covariant under rotations about $z$. It preserves both $Z$ populations and exponentially damps transverse coherence. Since $Z$ has only two outcomes, their probabilities are fixed by $\langle Z\rangle$ and normalization in this particular example.

Amplitude damping has $L=\sqrt\kappa\,|g\rangle\langle e|$. For the excited-state projector $P_e=|e\rangle\langle e|$,

$$\mathcal L^\dagger(P_e)=-\kappa P_e, \qquad p_e(t)=e^{-\kappa t}p_e(0).$$

It is also covariant under phase rotations $U_\phi=e^{-i\phi P_e}$: conjugating $L$ only multiplies it by a phase that cancels in the dissipator. Thus even covariance under the group generated by $P_e$ does not imply conservation of $P_e$.

Stable states and irreversibility need further assumptions

If a covariant generator has a unique stationary density operator $\rho_*$, then $U_g\rho_*U_g^\dagger$ is stationary too, so uniqueness forces it to equal $\rho_*$. If several stationary states exist, symmetry may permute them. Covariance alone does not prove uniqueness, relaxation to a stationary state, or a preferred pure state.

For a simple dephasing model, suppose the complete joint evolution has controlled form $W=\sum_a|a\rangle\langle a|\otimes W_a$. Then

$$\rho_{ab}\longmapsto\rho_{ab} \operatorname{Tr}(W_a\eta W_b^\dagger).$$

Populations remain unchanged. Off-diagonal terms shrink only when the conditional environment evolutions reduce this overlap. If all $W_a$ coincide, there is no decoherence. Finite environments can also give revivals. A diagonal interaction is therefore not enough to guarantee irreversible decay, and competing system dynamics can change which states are stable. See decoherence for the interpretation of environmental records.

A dissipative semigroup generally has no channel inverse for negative time. For instance, formally reversing dephasing amplifies Bloch-vector components and can send valid states outside the Bloch ball. This effective irreversibility does not prove violation of microscopic antiunitary time reversal. Initial conditions, discarded environment information and coarse graining enter separately; detailed balance is another condition, not a synonym for covariance.

Exercises

1. Verify the dephasing equations using $ZXZ=-X$ and $ZYZ=-Y$. Starting from $|+x\rangle$, find the purity at time $t$.

Solution

The adjoint maps $X$ to $-\gamma X$, $Y$ to $-\gamma Y$ and $Z$ to zero. Hence $\mathbf r(t)=(e^{-\gamma t},0,0)$. A qubit has $\operatorname{Tr}\rho^2=(1+|\mathbf r|^2)/2$, giving $(1+e^{-2\gamma t})/2$. The channel loses purity while preserving the $z$ probabilities.

2. An amplitude-damping semigroup is phase covariant. Calculate $\mathcal L^\dagger(P_e)$ and explain why covariance does not forbid population decay.

Solution

Here $L^\dagger P_eL=0$ and $L^\dagger L=\kappa P_e$, so the anticommutator contributes $-\kappa P_e$. A phase rotation changes coherences but leaves populations unchanged; performing that rotation before or after decay gives the same answer. This compatibility places no requirement that the populations themselves remain constant.

3. Show why the reduced full-swap channel with a pure environment $\eta=|+z\rangle\langle+z|$ fails covariance under a half-turn about $x$.

Solution

Every input, rotated or otherwise, is replaced by $|+z\rangle\langle+z|$. Rotating the output by the half-turn instead gives $|-z\rangle\langle-z|$. The joint swap is symmetric, but the prepared environment is not.

4. Consider three orthonormal states $|-1\rangle,|0\rangle,|+1\rangle$ and the dimensionless observable $Q=\operatorname{diag}(-1,0,1)$ in that order. Let $H=0$ and

$$L_\pm=\sqrt\gamma\,|\pm1\rangle\langle0|, \qquad \gamma>0.$$

Here $\gamma$ has units of inverse time. Calculate $\mathcal L^\dagger(Q)$ and $\mathcal L^\dagger(Q^2)$. Starting from $|0\rangle$, find the three outcome probabilities and the variance of $Q$.

Solution

Each jump contributes $L_\pm^\dagger L_\pm=\gamma|0\rangle\langle0|$. Since $Q|0\rangle=Q^2|0\rangle=0$, the anticommutator terms vanish for both observables. The two gain terms have opposite signs for $Q$ and equal signs for $Q^2$:

$$\mathcal L^\dagger(Q)=0, \qquad \mathcal L^\dagger(Q^2)=2\gamma|0\rangle\langle0|.$$

Thus the mean of $Q$ is conserved for every initial state, but its second moment need not be. For the stated initial state, the population equations are

$$\dot p_0=-2\gamma p_0, \qquad \dot p_\pm=\gamma p_0.$$

With $p_0(0)=1$ and $p_\pm(0)=0$, they give

$$p_0(t)=e^{-2\gamma t}, \qquad p_\pm(t)=\frac{1-e^{-2\gamma t}}2, \qquad (\Delta Q)^2(t)=1-e^{-2\gamma t}.$$

The mean stays zero because the two outgoing populations balance. The measurement distribution changes from a certain zero outcome toward equal probabilities for $\pm1$. With three outcomes, normalization and one fixed mean leave room for this redistribution.

References

• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press, 2002.
• V. Gorini, A. Kossakowski, and E. C. G. Sudarshan, “Completely positive dynamical semigroups of N-level systems,” Journal of Mathematical Physics 17, 821–825, 1976. DOI.
• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Springer, 1983.
• G. Lindblad, “On the generators of quantum dynamical semigroups,” Communications in Mathematical Physics 48, 119–130, 1976. DOI.
• H. M. Wiseman and G. J. Milburn, Quantum Measurement and Control, Cambridge University Press, 2010.