Chapter 13

Quantum channels: discarded environments and noise

A qubit prepared in $|+\rangle$ can decay to its ground state by emitting a photon into its surroundings. If no one looks at the surroundings, what is left of the qubit’s state? What changes if the photon is detected and the result recorded? And does noise always make a state more mixed?

Before you start: Recorded and ignored outcomes, and measurement operators (Chapter 2); density matrices, the Bloch ball and the partial trace (Chapter 3). The companion guide’s measurement essay discusses decoherence in physical terms. Notation guide.

One interaction, three descriptions

Model the surroundings by one environment qubit $E$, starting in $|0\rangle_E$. If the system is in $|0\rangle$, nothing happens. If it is in $|1\rangle$, it emits with probability $\gamma$, and the emission flips $E$:

$$|0\rangle|0\rangle_E\mapsto|0\rangle|0\rangle_E,\qquad |1\rangle|0\rangle_E\mapsto\sqrt{1-\gamma}\,|1\rangle|0\rangle_E+\sqrt\gamma\,|0\rangle|1\rangle_E.$$

Take $\gamma=\tfrac34$ and the input $|+\rangle$. By linearity the joint state is

$$\frac1{\sqrt2}\Big(|0\rangle|0\rangle_E+\tfrac12|1\rangle|0\rangle_E+\tfrac{\sqrt3}2|0\rangle|1\rangle_E\Big).$$

The same physical situation now has three descriptions, and they answer different questions.

The joint state. System and environment together are in the pure state above. Nothing has been lost; the system has become entangled with $E$.

The environment ignored. If $E$ is out of reach, predictions for the system alone come from its reduced state (Chapter 3). Tracing out $E$ gives $\rho=\left(\begin{smallmatrix}7/8&1/4\\1/4&1/8\end{smallmatrix}\right)$, a mixed state with Bloch vector $(\tfrac12,0,\tfrac34)$.

The environment measured and recorded. If a detector reads $E$ in the basis $|0\rangle_E,|1\rangle_E$ and the result is kept, each reading has its own conditional state (Chapter 2). A photon is detected with probability $\tfrac12\cdot\tfrac34=\tfrac38$, leaving $|0\rangle$. No photon, with probability $\tfrac58$, leaves $(2|0\rangle+|1\rangle)/\sqrt5$. Both conditional states are pure.

The descriptions are linked by Chapter 2’s averaging rule: the two recorded branches, weighted $\tfrac38$ and $\tfrac58$, average to the ignored-outcome state. But they are not interchangeable. The mixed $\rho$ is what someone without the record must use; the pure branch states are what someone with the record may use. A different measurement of $E$ would give different branches with the same average, as the last section explains.

The rest of the chapter turns the ignored-outcome description into a general rule, a channel, and then returns to what a record adds.

Derive the operator-sum representation

Assume the system starts in state $\rho$ and the environment in a fixed pure state $|0\rangle_E$, with no initial correlations. Let them interact through a joint unitary $U$. If the environment is inaccessible, the output state is

$$\mathcal E(\rho)=\operatorname{Tr}_E[U(\rho\otimes|0\rangle\langle0|)U^\dagger].$$

The same result, derived term by term for a pure input

First take a pure system input $|\psi\rangle$. Whatever the joint unitary does, we can expand its output using the environment basis:

$$U(|\psi\rangle|0\rangle_E)=\sum_j|v_j\rangle_S|j\rangle_E.$$

The system vectors $v_j$ are not generally normalised or orthogonal. Taking the environment component with $I\otimes\langle j|$ selects $v_j$. Since both attaching the initial environment and applying $U$ are linear, $v_j$ depends linearly on $\psi$; this dependence defines a system operator $K_j$ with $v_j=K_j\psi$.

Now form the joint density matrix and take the partial trace. The reason that only one sum survives is visible term by term:

$$\begin{aligned}\rho'_{SE}&=\sum_{j,\ell}|v_j\rangle\langle v_\ell|\otimes|j\rangle\langle\ell|,\\ \rho'_S&=\sum_{j,\ell}|v_j\rangle\langle v_\ell|\,\langle\ell|j\rangle =\sum_j|v_j\rangle\langle v_j|\\ &=\sum_jK_j|\psi\rangle\langle\psi|K_j^\dagger. \end{aligned}$$

Different environment labels are distinguishable alternatives for the discarded subsystem, so their cross terms disappear in local predictions. If the initial state is a mixture, apply this derivation to each pure member and average its density matrix. Linearity gives the same formula with $|\psi\rangle\langle\psi|$ replaced by $\rho$. No assumption that the system's output remains pure has been made.

In the example, $K_0=\operatorname{diag}(1,\sqrt{1-\gamma})$ gave the no-emission branch and $K_1=\sqrt\gamma\,|0\rangle\langle1|$ the emission branch. In general, insert an orthonormal environment basis $\{|j\rangle\}$ in the partial trace and define $K_j=(I\otimes\langle j|)U(I\otimes|0\rangle)$, called Kraus operators. Then

$$\mathcal E(\rho)=\sum_jK_j\rho K_j^\dagger,\qquad\sum_jK_j^\dagger K_j=I.$$

The last identity follows from $U^\dagger U=I$ and implies trace preservation: $\operatorname{Tr}\mathcal E(\rho)=\operatorname{Tr}[\rho\sum_jK_j^\dagger K_j]=1$. A fixed mixed environment can be included by purifying it into a larger space. Initial system-environment correlations require more care and are outside this construction's assumptions.

Why positivity must hold with an untouched partner

Each term $K_j\rho K_j^\dagger$ is positive when $\rho$ is positive. The same argument holds on a larger state with operators $I_R\otimes K_j$, so $I_R\otimes\mathcal E$ remains positive for every reference system $R$. This is complete positivity. A quantum channel is a completely positive, trace-preserving linear map.

Ordinary positivity, that $\mathcal E$ maps every density matrix to a density matrix, is weaker. The standard counterexample is transposition, which is positive but gives a negative eigenvalue when applied to half of a Bell pair.

The partial transpose, and the negative eigenvalue

A partial transpose is a mathematical operation on a bipartite matrix: transpose only the indices belonging to the selected subsystem, in a specified product basis. On a basis operator the rule is

$$(|i\rangle\langle k|\otimes|j\rangle\langle\ell|)^{T_B} =|i\rangle\langle k|\otimes|\ell\rangle\langle j|.$$

For example, $|00\rangle\langle11|$ becomes $|01\rangle\langle10|$, while $|00\rangle\langle00|$ stays fixed. Unlike a partial trace, this does not discard a subsystem or reduce the matrix dimension. Applying the rule to the four terms of the Bell projector gives the matrix below. Its negative eigenvalue will show why this otherwise well-defined mathematical operation is not a valid local channel.

Matrix transposition $\rho\mapsto\rho^T$ preserves the eigenvalues of a single state, but transposing only the second half of a Bell projector gives

$$\left(|\Phi^+\rangle\langle\Phi^+|\right)^{T_B}=\frac12\begin{pmatrix}1&0&0&0\\0&0&1&0\\0&1&0&0\\0&0&0&1\end{pmatrix}.$$

The middle block has eigenvalues $+1/2,-1/2$. This cannot be an output density matrix. Transposition is positive but not completely positive, so it cannot act as a universal physical local channel on unknown inputs.

Amplitude damping: populations and coherence

Model a transition from excited $|1\rangle$ to ground $|0\rangle$ with probability $\gamma$, where $0\le\gamma\le1$. Choose

$$K_0=\begin{pmatrix}1&0\\0&\sqrt{1-\gamma}\end{pmatrix},\qquad K_1=\begin{pmatrix}0&\sqrt\gamma\\0&0\end{pmatrix}.$$

Completeness follows from $K_0^\dagger K_0=\operatorname{diag}(1,1-\gamma)$ and $K_1^\dagger K_1=\operatorname{diag}(0,\gamma)$. Multiplying out both branches gives

$$\mathcal E\begin{pmatrix}\rho_{00}&\rho_{01}\\\rho_{10}&\rho_{11}\end{pmatrix}=\begin{pmatrix}\rho_{00}+\gamma\rho_{11}&\sqrt{1-\gamma}\rho_{01}\\\sqrt{1-\gamma}\rho_{10}&(1-\gamma)\rho_{11}\end{pmatrix}.$$

Population flows to the ground state; coherence is attenuated by a square root. With $\gamma(t)=1-e^{-t/T_1}$, the excited population decays as $e^{-t/T_1}$ and off-diagonal entries as $e^{-t/(2T_1)}$. The corresponding $T_2=2T_1$ holds for this relaxation-only model; additional dephasing can shorten coherence time.

For input $|+\rangle$ and $\gamma=3/4$, the output is $\left(\begin{smallmatrix}7/8&1/4\\1/4&1/8\end{smallmatrix}\right)$. Its Bloch vector is $(1/2,0,3/4)$ and its purity is $(1+1/4+9/16)/2=29/32$. At full damping, every input becomes $|0\rangle\langle0|$, a pure state. Noise can make the system purer; it need not make it more mixed.

Phase noise behaves differently

A random phase flip has channel $\mathcal D_p(\rho)=(1-p)\rho+pZ\rho Z$. The diagonal entries stay fixed and the off-diagonal entries are multiplied by $1-2p$. At $p=1/2$, all computational-basis coherence disappears. At $p=1$, the operation is the known unitary $Z$, so it does not destroy purity. Increasing a parameter called “error probability” does not always increase mixedness monotonically.

On the Bloch ball this channel maps $(r_x,r_y,r_z)$ to $((1-2p)r_x,(1-2p)r_y,r_z)$. Amplitude damping instead maps it to $(\sqrt{1-\gamma}r_x,\sqrt{1-\gamma}r_y,(1-\gamma)r_z+\gamma)$. The displacement toward the north pole explains why damping can purify.

A recorded outcome supplies more than a channel

If the environment basis is actually measured and outcome $j$ retained, its probability is $p_j=\operatorname{Tr}(K_j\rho K_j^\dagger)$ and its conditional state is $K_j\rho K_j^\dagger/p_j$. The effects are $K_j^\dagger K_j$ (not to be confused with the environment $E$ or the channel $\mathcal E$). Forgetting the outcome gives the channel sum. If a recorded outcome $m$ groups hidden alternatives $j$, the conditional numerator is $\sum_jK_{mj}\rho K_{mj}^\dagger$.

A Kraus representation is not unique: replacing the $K_j$ by unitary linear combinations leaves the channel unchanged. Consequently, a Kraus index does not automatically identify a unique physical event. It becomes an experimentally meaningful outcome only when a particular environment measurement is specified.

Where this leaves us

With the surroundings ignored, the decayed $|+\rangle$ is described by the mixed state $\rho=\left(\begin{smallmatrix}7/8&1/4\\1/4&1/8\end{smallmatrix}\right)$: population has moved towards $|0\rangle$ and coherence has shrunk. If the emission is recorded, the qubit is in one of two pure states, $|0\rangle$ or $(2|0\rangle+|1\rangle)/\sqrt5$, and the record says which. The joint state of qubit and surroundings stayed pure throughout. Noise is not the same as mixing: full damping ends in a pure state, and a phase flip with $p=1$ is a known unitary.

A channel acting on a lone qubit generally cannot be undone, because the information has gone to a system we do not hold. The next chapter spreads one qubit's amplitudes over several physical qubits, so that the effect of a small error can be identified and reversed without looking at the amplitudes themselves.

Problems

1. Apply full dephasing to one half of $|\Phi^+\rangle$. Is the output entangled?

Worked answer

Full dephasing on $B$ is the equal mixture of doing nothing and applying $Z$ to $B$. The second operation converts $|\Phi^+\rangle$ to $|\Phi^-\rangle=(|00\rangle-|11\rangle)/\sqrt2$. Therefore

$$\rho'=\tfrac12|\Phi^+\rangle\langle\Phi^+|+\tfrac12|\Phi^-\rangle\langle\Phi^-|.$$

Each projector has diagonal terms $|00\rangle\langle00|/2$ and $|11\rangle\langle11|/2$. Their cross terms have opposite signs and cancel in the mixture. The result is $(|00\rangle\langle00|+|11\rangle\langle11|)/2$, a mixture of two explicit product states, hence separable. Measurement in the computational basis still gives equal bits on the two sides.

2. In amplitude damping, find the probability of outcome $K_1$ and its conditional output.

Worked answer

Multiply the input matrix by the jump operator on both sides:

$$\begin{aligned}K_1\rho&=\begin{pmatrix}\sqrt\gamma\rho_{10}&\sqrt\gamma\rho_{11}\\0&0\end{pmatrix},\\K_1\rho K_1^\dagger&=\begin{pmatrix}\gamma\rho_{11}&0\\0&0\end{pmatrix}.\end{aligned}$$

The trace gives $p_1=\gamma\rho_{11}$. If it is positive, dividing by it leaves $\left(\begin{smallmatrix}1&0\\0&0\end{smallmatrix}\right)=|0\rangle\langle0|$. If it is zero, this outcome cannot occur, so no conditional state is assigned by division.

References and further reading

Nielsen & Chuang, §8.2.2–8.2.4 (environments, operator sums, complete positivity), §8.3.3 (phase flips), §8.3.5 (amplitude damping), and §8.5 (limitations of the formalism).