Dephasing Channel

The dephasing channel is the basic channel for loss of phase coherence in a preferred basis. It suppresses off-diagonal density-matrix elements while leaving the corresponding populations unchanged.

For a qubit in the $Z$ basis, the defining action is

$$\begin{pmatrix} \rho_{00} & \rho_{01}\\ \rho_{10} & \rho_{11} \end{pmatrix} \mapsto \begin{pmatrix} \rho_{00} & \lambda\rho_{01}\\ \lambda^*\rho_{10} & \rho_{11} \end{pmatrix}, \qquad |\lambda|\le1.$$

The magnitude of the coherence factor $\lambda$ gives the surviving fraction of off-diagonal coherence; its phase produces a coherent $Z$ rotation. A single dephasing map does not say whether the lost coherence can revive at later times.

Helpful background. Kraus operators describe the map, while Dephasing vs Dissipation explains when coherence loss also changes energy.

Dephasing and the chosen basis

Dephasing preserves the populations of its chosen basis:

$$\rho_{00}'=\rho_{00}, \qquad \rho_{11}'=\rho_{11}.$$

Only coherences are damped:

$$\rho_{01}'=\lambda\rho_{01}.$$

This is why pure dephasing is the canonical model for:

• fluctuating energy splittings,
• unread which-path information,
• elastic scattering that distinguishes alternatives,
• nonselective projective measurement in a basis,
• slow classical phase noise,
• environment records that become correlated with basis states.

The basis matters. A channel that is pure dephasing in one basis may change populations when viewed in another basis.

For the broader physical distinction, see Dephasing vs Dissipation.

Energy is unchanged when the dephasing projectors commute with the Hamiltonian. For a counterexample, let $H=\hbar\omega X/2$ and start in $|+\rangle$. Complete $Z$ dephasing sends this state to $I/2$, changing $\langle H\rangle$ from $\hbar\omega/2$ to zero. Preserving computational-basis populations does not by itself mean preserving energy.

Dephasing and Amplitude Damping connects measured coherence and relaxation times to finite-duration noise models.

Phase Flip Kraus Form

For real $-1\le\lambda\le1$, the qubit dephasing channel can be written as a phase-flip mixture:

$$\Phi_\lambda(\rho) = \frac{1+\lambda}{2}\rho + \frac{1-\lambda}{2}Z\rho Z.$$

A Kraus representation is

$$K_0 = \sqrt{\frac{1+\lambda}{2}}\,I, \qquad K_1 = \sqrt{\frac{1-\lambda}{2}}\,Z.$$

The trace-preserving condition is

$$K_0^\dagger K_0+K_1^\dagger K_1=I.$$

Applying the channel to the density matrix gives

$$\Phi_\lambda(\rho) = \begin{pmatrix} \rho_{00} & \lambda\rho_{01}\\ \lambda\rho_{10} & \rho_{11} \end{pmatrix}$$

when $\lambda$ is real. If $\lambda$ is complex, write $\lambda=|\lambda|e^{-i\theta}$ and treat the phase as a $Z$ rotation together with damping by $|\lambda|$.

Equivalently, use an error probability

$$p=\frac{1-\lambda}{2}.$$

Then

$$\Phi_p(\rho) = (1-p)\rho+pZ\rho Z, \qquad 0\le p\le1.$$

This is the $Z$-axis special case of a Pauli channel.

The Markovian pure-dephasing semigroup has $0\le p(t)\le1/2$ and $\lambda(t)=e^{-\Gamma_\phi t}$; allowing $p>1/2$ is still a valid phase-flip channel but not the usual monotone Markovian decay convention.

Bloch Sphere Action

Write a qubit state as

$$\rho = \frac{1}{2} \left( I+r_xX+r_yY+r_zZ \right).$$

For real $\lambda$, the dephasing channel acts as

$$(r_x,r_y,r_z) \mapsto (\lambda r_x,\lambda r_y,r_z).$$

The Bloch sphere is contracted toward the $Z$ axis. The north and south pole states are fixed, while transverse superpositions lose contrast.

Repeated applications multiply coherence factors. For $\lambda\ne1$, the fixed states are exactly diagonal in the dephasing basis; at $\lambda=1$, every state is fixed. Convergence to a diagonal state under repeated uses requires $|\lambda|<1$: a nontrivial unit-modulus phase merely keeps rotating. See Channel Composition and Fixed Points for the general channel language.

Complete dephasing is the case $\lambda=0$:

$$(r_x,r_y,r_z) \mapsto (0,0,r_z).$$

It maps every state to its diagonal part in the $Z$ basis.

Choi Matrix and Complete Positivity

Using the output-input convention of Choi Matrix, the Choi matrix of the qubit dephasing channel is

$$J_{\Phi_\lambda} = \begin{pmatrix} 1 & 0 & 0 & \lambda\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ \lambda^* & 0 & 0 & 1 \end{pmatrix}.$$

The nonzero block has eigenvalues

$$1+|\lambda|, \qquad 1-|\lambda|.$$

Therefore the channel is completely positive exactly when

$$|\lambda|\le1.$$

The Choi rank is $2$ for $|\lambda|<1$, including complete dephasing, and $1$ for a unitary phase rotation with $|\lambda|=1$.

Nonselective Measurement

An unread ideal Lüders measurement in the $Z$ basis gives complete dephasing:

$$\Delta_Z(\rho) = P_0\rho P_0+P_1\rho P_1,$$

where

$$P_0=\lvert0\rangle\langle0\rvert, \qquad P_1=\lvert1\rangle\langle1\rvert.$$

In matrix form,

$$\Delta_Z \left( \begin{pmatrix} \rho_{00} & \rho_{01}\\ \rho_{10} & \rho_{11} \end{pmatrix} \right) = \begin{pmatrix} \rho_{00} & 0\\ 0 & \rho_{11} \end{pmatrix}.$$

Thus an unread ideal measurement is not “no measurement.” It is a channel that removes coherence in the measured decomposition. See Measurement Backaction for the operational distinction.

Environment-Record Picture

Suppose the environment becomes correlated with the dephasing basis:

$$\lvert0\rangle\lvert e\rangle \mapsto \lvert0\rangle\lvert e_0\rangle, \qquad \lvert1\rangle\lvert e\rangle \mapsto \lvert1\rangle\lvert e_1\rangle.$$

For an input coherence $\rho_{01}$, tracing out the environment gives

$$\rho_{01} \mapsto \langle e_1|e_0\rangle\,\rho_{01}.$$

Thus

$$\lambda=\langle e_1|e_0\rangle.$$

Orthogonal environment records give complete dephasing. Environment vectors differing only by a phase give $|\lambda|=1$, so they are physically indistinguishable and cause only a system phase rotation. Partially distinguishable records give $|\lambda|<1$.

This is the simplest channel version of Environment-Induced Decoherence.

Random Phase Picture

Classical random phase noise gives another representation. Let

$$U_\phi = e^{-i\phi Z/2}$$

and average over phases with probability density $p(\phi)$:

$$\Phi(\rho) = \int d\phi\,p(\phi)\, U_\phi\rho U_\phi^\dagger.$$

The off-diagonal element transforms as

$$\rho_{01} \mapsto \left[ \int d\phi\,p(\phi)e^{-i\phi} \right] \rho_{01}.$$

Therefore

$$\lambda = \langle e^{-i\phi}\rangle.$$

For a Gaussian random phase with mean $0$ and variance $\sigma_\phi^2$,

$$\lambda=e^{-\sigma_\phi^2/2}.$$

Random-unitary phase averaging is a useful model for classical frequency noise. A quantum bath can produce the same channel form while carrying quantum correlations rather than a classical random label.

Markovian Master Equation Limit

In a frame removing the commuting Hamiltonian evolution, a time-homogeneous pure-dephasing master equation with constant rate $\Gamma_\phi\ge0$ is

$$\frac{d\rho}{dt} = \frac{\Gamma_\phi}{2} \left( Z\rho Z-\rho \right).$$

The populations are constant:

$$\frac{d\rho_{00}}{dt}=0, \qquad \frac{d\rho_{11}}{dt}=0.$$

The coherence obeys

$$\frac{d\rho_{01}}{dt} = -\Gamma_\phi\rho_{01}.$$

Hence

$$\lambda(t)=e^{-\Gamma_\phi t}.$$

The corresponding phase-flip probability is

$$p(t) = \frac{1-e^{-\Gamma_\phi t}}{2}.$$

For the exact coherence-factor model behind nonexponential dephasing, see Pure Dephasing Model. For the generator-level assumptions, rate conventions, and $T_2$ relation, see Pure Dephasing Master Equation. In $T_2$ language, this pure-dephasing contribution is often written as $T_\phi^{-1}=\Gamma_\phi$, with relaxation contributions added separately.

Higher Dimensional Dephasing

For projectors $\{P_a\}$ onto a preferred orthogonal decomposition, complete dephasing is

$$\Delta(\rho) = \sum_a P_a\rho P_a.$$

More general partial dephasing can multiply blocks by coherence factors:

$$\Phi(\rho) = \sum_{a,b} \Gamma_{ab} P_a\rho P_b.$$

For nonzero mutually orthogonal projectors summing to $I$, this map is completely positive exactly when $\Gamma\ge0$; trace preservation additionally requires

$$\Gamma_{aa}=1.$$

This $\Gamma$ can be interpreted as a Gram matrix of environment states:

$$\Gamma_{ab} = \langle e_b|e_a\rangle.$$

This remains true for higher-rank projectors: coherences inside each block are preserved, while the interblock terms are multiplied by $\Gamma_{ab}$. For necessity, choose one unit vector in each nonzero block and restrict the map to their span. For sufficiency, factor the positive matrix $\Gamma$ into environment overlaps and use the corresponding isometry.

Pairwise bounds are not enough in three dimensions. Consider rank-one projectors and

$$\Gamma=\begin{pmatrix}1&0.9&0.9\\0.9&1&-0.9\\0.9&-0.9&1\end{pmatrix}.$$

Every individual coherence multiplier has magnitude at most one, but the eigenvalues are $1.9,1.9,-0.8$. Applying the map to $(|0\rangle+|1\rangle+|2\rangle)/\sqrt3$ would produce $\Gamma/3$, which has a negative eigenvalue. Different coherences cannot be assigned independently without checking the whole matrix.

Exercises

Kraus action

Using

$$K_0=\sqrt{\frac{1+\lambda}{2}}\,I, \qquad K_1=\sqrt{\frac{1-\lambda}{2}}\,Z,$$

show that $\rho_{01}\mapsto\lambda\rho_{01}$ for real $\lambda$.

Solution

The channel is

$$\Phi_\lambda(\rho) = \frac{1+\lambda}{2}\rho + \frac{1-\lambda}{2}Z\rho Z.$$

Since $Z=\operatorname{diag}(1,-1)$,

$$(Z\rho Z)_{01}=-\rho_{01}.$$

Therefore

$$(\Phi_\lambda(\rho))_{01} = \frac{1+\lambda}{2}\rho_{01} - \frac{1-\lambda}{2}\rho_{01} = \lambda\rho_{01}.$$

Bloch contraction

Show that real-$\lambda$ dephasing maps $(r_x,r_y,r_z)$ to $(\lambda r_x,\lambda r_y,r_z)$.

Solution

For

$$\rho = \frac12 \begin{pmatrix} 1+r_z & r_x-ir_y\\ r_x+ir_y & 1-r_z \end{pmatrix},$$

dephasing leaves the diagonal entries fixed and multiplies off-diagonal entries by $\lambda$. Hence

$$r_z' = r_z, \qquad r_x'-ir_y'=\lambda(r_x-ir_y).$$

For real $\lambda$ this gives

$$r_x'=\lambda r_x, \qquad r_y'=\lambda r_y.$$

Gaussian phase noise

Let $\phi$ be a zero-mean Gaussian random variable with variance $\sigma_\phi^2$. Show that $\langle e^{-i\phi}\rangle=e^{-\sigma_\phi^2/2}$.

Solution

The characteristic function of a zero-mean Gaussian with variance $\sigma_\phi^2$ is

$$\langle e^{ik\phi}\rangle = e^{-k^2\sigma_\phi^2/2}.$$

Set $k=-1$ to obtain

$$\langle e^{-i\phi}\rangle = e^{-\sigma_\phi^2/2}.$$

References

• H.-P. Breuer and F. Petruccione, The Theory of Open Quantum Systems, Oxford University Press (2002). Publisher.
• K. Kraus, States, Effects, and Operations: Fundamental Notions of Quantum Theory, Lecture Notes in Physics 190, Springer (1983). DOI.
• J. Watrous, The Theory of Quantum Information, Cambridge University Press (2018). Author’s text.