|q⟩ Bad Qubits

advanced · Physics · Quantum Channels & Open Systems

Channel Composition

Noise rarely happens once. A qubit passes through gate after gate, idle period after idle period — each a channel. To reason about the cumulative effect we need to compose channels, and the good news is that the class of CPTP maps is closed under everything you would want to do: composing channels in series, in parallel, and convexly all yield channels again.

Series composition

Applying E1\mathcal{E}_1 then E2\mathcal{E}_2 gives the composite E2E1\mathcal{E}_2\circ\mathcal{E}_1 with (E2E1)(ρ)=E2(E1(ρ))(\mathcal{E}_2\circ\mathcal{E}_1)(\rho) = \mathcal{E}_2\big(\mathcal{E}_1(\rho)\big). If E1(ρ)=jAjρAj\mathcal{E}_1(\rho)=\sum_j A_j\rho A_j^\dagger and E2(ρ)=kBkρBk\mathcal{E}_2(\rho)=\sum_k B_k\rho B_k^\dagger, then

(E2E1)(ρ)=j,k(BkAj)ρ(BkAj),(\mathcal{E}_2\circ\mathcal{E}_1)(\rho) = \sum_{j,k} (B_k A_j)\,\rho\,(B_k A_j)^\dagger,

so the Kraus operators of the composite are all products {BkAj}\{B_k A_j\}. Completeness propagates:

j,k(BkAj)(BkAj)=jAj(kBkBk)Aj=jAjAj=1,\sum_{j,k}(B_kA_j)^\dagger(B_kA_j) = \sum_j A_j^\dagger\Big(\textstyle\sum_k B_k^\dagger B_k\Big)A_j = \sum_j A_j^\dagger A_j = \mathbb{1},

using kBkBk=1\sum_k B_k^\dagger B_k=\mathbb{1} first, then jAjAj=1\sum_j A_j^\dagger A_j=\mathbb{1}. The composite is CPTP. The number of Kraus terms can grow as the product of the two ranks, but the Kraus rank of the composite is often smaller and never exceeds d2d^2.

Composition need not commute

Order matters: in general E2E1E1E2\mathcal{E}_2\circ\mathcal{E}_1 \ne \mathcal{E}_1\circ\mathcal{E}_2. Dephasing then amplitude-damping is not the same process as amplitude-damping then dephasing. Two important families do commute among themselves — Pauli channels with each other, and dephasing channels with each other — because their Kraus operators share an eigenbasis.

Channels that compose cleanly

Some channel families are closed under composition with a simple parameter rule:

In each case the family forms a one-parameter semigroup under composition — the continuous-time, Markovian limit of which is the Lindblad master equation.

Parallel composition

Acting on two systems independently gives the tensor product channel E1E2\mathcal{E}_1\otimes \mathcal{E}_2, whose Kraus operators are the products {AjBk}\{A_j\otimes B_k\}. This is how noise on a multi-qubit register is built: independent single-qubit channels tensor together. Convex mixtures pE1+(1p)E2p\,\mathcal{E}_1 + (1-p)\,\mathcal{E}_2 are channels too — "with probability pp apply E1\mathcal{E}_1."

Try it

Pass the excited state ρ=11\rho=|1\rangle\langle1| through two amplitude-damping channels in series, γ1=0.2\gamma_1 = 0.2 then γ2=0.5\gamma_2 = 0.5. Apply the first channel, then the second, using the Kraus operators K0=diag(1,1γ),K1=γ01K_0=\operatorname{diag}(1,\sqrt{1-\gamma}),\,K_1=\sqrt{\gamma}\,|0\rangle\langle1|, and return the surviving excited population 1(Eγ2Eγ1)(ρ)1\langle1|(\mathcal{E}_{\gamma_2}\circ\mathcal{E}_{\gamma_1}) (\rho)|1\rangle. It must equal (1γ1)(1γ2)=0.8×0.5=0.4(1-\gamma_1)(1-\gamma_2) = 0.8\times0.5 = 0.4.

Run your code to see the quantum state.

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