6.4.1 Decoherence#

Prompts

Explain decoherence: how does coupling to an environment turn a coherent superposition into a classical mixture? Walk through the partial trace calculation step by step.
What are pointer states, and why does the environment “select” them (einselection)? Give a specific example with a coupling Hamiltonian.
Why do macroscopic objects decohere almost instantaneously while small quantum systems maintain coherence? What determines \(T_2\)?
How is decoherence related to the dephasing quantum channel from §6.3.3? What does “measurement + forgetting” mean physically?

Lecture Notes#

When a qubit couples to its environment—a thermal bath, stray photons, or surrounding atoms—the environment becomes entangled with the system and effectively “measures” it. The off-diagonal elements of the density matrix decay, turning a coherent superposition into a classical mixture. This is decoherence: the mechanism behind the quantum-to-classical transition.

Overview#

Decoherence can be understood in two steps: the environment entangles with the system (like a measurement), then we trace out the environment (like forgetting the outcome). No new postulate is needed — decoherence is simply the dephasing quantum channel (§6.3.3) applied by the environment. Pointer states are the robust states that survive this process; all other superpositions decohere on a timescale \(T_2 = 1/\gamma\).

Discussion

For a qubit coupled to a thermal bath, the decoherence time satisfies \(T_2 \leq 2T_1\). This bound is saturated by pure dephasing. Express \(T_2\) in terms of the relaxation rate \(\Gamma_1 = 1/T_1\) and pure dephasing rate \(\Gamma_\varphi = 1/T_\varphi\): the relation is \(\frac{1}{T_2} = \frac{1}{2T_1} + \frac{1}{T_\varphi}\). Why does energy relaxation contribute to dephasing even without random phase kicks?

Decoherence Process#

Consider a qubit \(S\) initially in a superposition, coupled to environment \(E\) (bath of many modes). After interaction, the total state evolves to:

(154)#\[ \alpha\vert 0\rangle_S \vert E_0\rangle_E + \beta\vert 1\rangle_S \vert E_1\rangle_E \;\longrightarrow\; \alpha\vert 0\rangle_S \vert E_0(t)\rangle_E + \beta\vert 1\rangle_S \vert E_1(t)\rangle_E \]

where \(\vert E_0(t)\rangle\) and \(\vert E_1(t)\rangle\) are (nearly) orthogonal environmental states. The environment has become a which-path detector: it encodes information about which branch the system is in.

Pointer States and Einselection

Pointer states are the eigenstates of the system-environment interaction Hamiltonian \(\hat{H}_\text{int}\). They are the states robust under decoherence: the environment correlates with them without creating superpositions between them.

Einselection (Zurek): The environment selects a preferred pointer basis. Superpositions of pointer states decohere rapidly into mixtures; individual pointer states remain stable.

Example: For \(\hat{H}_\text{int} = \hat{\sigma}^z \otimes B_E\), the pointer basis is \(\{\vert 0\rangle, \vert 1\rangle\}\) (eigenstates of \(\hat{\sigma}^z\)). The superposition \(\vert+\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\) decoheres rapidly; \(\vert 0\rangle\) and \(\vert 1\rangle\) do not.

Reduced Density Matrix#

Tracing out the environment gives the reduced density matrix:

(155)#\[ \hat{\rho}_S(t) = \text{Tr}_E\!\left(\vert\Psi(t)\rangle\langle\Psi(t)\vert\right) \]

Computing explicitly:

(156)#\[ \hat{\rho}_S(t) = \vert\alpha\vert^2 \vert 0\rangle\langle 0\vert + \vert\beta\vert^2 \vert 1\rangle\langle 1\vert + \alpha^*\beta\,\langle E_1(t)\vert E_0(t)\rangle\,\vert 0\rangle\langle 1\vert + \beta^*\alpha\,\langle E_0(t)\vert E_1(t)\rangle\,\vert 1\rangle\langle 0\vert \]

As the environmental states become orthogonal (\(\langle E_0\vert E_1\rangle \to 0\)), the off-diagonal coherences vanish. The system transitions from a pure state to a mixed state — not from any measurement postulate, but purely from entanglement with the environment.

Decoherence Timescale#

In the weak coupling limit, the environmental overlap decays exponentially:

(157)#\[ \langle E_0(t) \vert E_1(t)\rangle \approx \mathrm{e}^{-t/T_2} \]

where \(T_2 = 1/\gamma\) is the dephasing time. The microscopic derivation:

Derivation: Dephasing Rate from Spin-Bath Coupling

For \(H_\text{int} = \hat{\sigma}^z \otimes \sum_k g_k (a_k \mathrm{e}^{-\mathrm{i}\omega_k t} + a_k^\dagger \mathrm{e}^{\mathrm{i}\omega_k t})\) (spin coupled to bosonic bath), the overlap integral gives:

\[ \langle E_0(t)\vert E_1(t)\rangle = \exp\!\left(-\int_0^\infty \frac{J(\omega)}{\omega^2}(1 - \cos\omega t)\coth\frac{\hbar\omega}{2k_BT}\,\mathrm{d}\omega\right) \]

where \(J(\omega) = \sum_k g_k^2\,\delta(\omega - \omega_k)\) is the spectral density of the bath. For an ohmic bath at high temperature \(k_BT \gg \hbar\omega_c\), this gives exponential decay \(\mathrm{e}^{-\gamma t}\) with rate \(\gamma \propto k_BT\,\sum_k g_k^2/\omega_k^2\).

Macroscopic objects: The decoherence rate scales extensively with system size \(N\): \(\gamma_\text{total} \sim N\gamma_\text{atom}\). For a macroscopic object (\(N \sim 10^{26}\)), \(T_2 \sim 10^{-26}\,\text{s}\) — decoherence is essentially instantaneous. This is why we never observe macroscopic superpositions.

Decoherence as a Quantum Channel#

Decoherence is not a new phenomenon — it is the dephasing channel (§6.3.3) derived from a microscopic model:

Entanglement (environment measures): \(|\psi\rangle \otimes |E_0\rangle \to \alpha|0\rangle|E_0(t)\rangle + \beta|1\rangle|E_1(t)\rangle\)
Partial trace (forgetting): \(\text{Tr}_E(\cdot)\) discards the measurement outcome.

The result is exactly the dephasing channel:

(158)#\[\begin{split} \mathcal{E}(\hat{\rho}) = \begin{pmatrix} \hat{\rho}_{00} & \hat{\rho}_{01}\,\mathrm{e}^{-t/T_2} \\\ \hat{\rho}_{10}\,\mathrm{e}^{-t/T_2} & \hat{\rho}_{11} \end{pmatrix} \end{split}\]

This identification — decoherence = quantum channel — means the full machinery of §6.3.3 applies to open-system evolution without any new framework.

Decoherence is not measurement collapse

Decoherence explains why off-diagonal elements of \(\hat{\rho}_S\) decay, making interference unobservable. But the full system+environment state remains pure — decoherence does not select a single outcome. The measurement problem (why we observe definite results) is a separate philosophical question.

Quantum-to-Classical Transition#

Under decoherence, a superposition of pointer states rapidly becomes an effective classical mixture:

\[\begin{split} \hat{\rho}_S(t) \approx \begin{pmatrix} 1/2 & \mathrm{e}^{-t/T_2}/2 \\\ \mathrm{e}^{-t/T_2}/2 & 1/2 \end{pmatrix} \xrightarrow{t \gg T_2} \frac{I}{2} \end{split}\]

For macroscopic objects, this happens in \(\sim 10^{-26}\) s, far below any observable timescale. This is why classical physics works: quantum coherence is perpetually destroyed by environmental coupling before it can manifest at macroscopic scales.

Decoherence-Free Subspaces#

Some states are immune to specific decoherence channels due to symmetry. For two qubits coupling identically to the environment (\(H_\text{int} = (\hat{\sigma}^z_1 + \hat{\sigma}^z_2)\otimes B_E\)), the singlet state is decoherence-free:

\[ (\hat{\sigma}^z_1 + \hat{\sigma}^z_2)\,\vert\Psi^-\rangle = 0, \qquad \vert\Psi^-\rangle = \frac{1}{\sqrt{2}}(\vert 01\rangle - \vert 10\rangle) \]

The singlet’s total \(z\)-spin is zero, so the common-mode dephasing leaves it unchanged. Decoherence-free subspaces are a foundation of quantum error correction (§6.4.3).

Summary#

Decoherence arises when the system entangles with its environment; the partial trace turns a pure state into a mixed state by destroying off-diagonal coherences.
The coherence factor \(\langle E_0(t)\vert E_1(t)\rangle = \mathrm{e}^{-t/T_2}\) decays on the dephasing timescale \(T_2 = 1/\gamma\).
Decoherence = measurement by the environment + forgetting the outcome; it is equivalent to the dephasing quantum channel (§6.3.3).
Pointer states are einselected by the environment — they are the eigenstates of \(H_\text{int}\) and remain stable under decoherence.
Macroscopic objects decohere in \(\lesssim 10^{-26}\) s, making quantum coherence unobservable at human scales.
Decoherence-free subspaces exploit symmetry to protect quantum information from specific noise channels.

Homework#

1. A qubit \(S\) is initially in \(\vert\psi_S\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\). After coupling to environment \(E\) (initially \(\vert 0_E\rangle\)), the joint state is \(\vert\Psi_{SE}(t)\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle_S \vert E_0(t)\rangle_E + \vert 1\rangle_S \vert E_1(t)\rangle_E)\).

(a) Compute the reduced density matrix \(\hat{\rho}_S(t) = \text{Tr}_E(\vert\Psi_{SE}(t)\rangle\langle\Psi_{SE}(t)\vert)\).

(b) Show that when \(\langle E_0(t)\vert E_1(t)\rangle = 0\), the off-diagonal coherence terms vanish completely. Explain physically why orthogonality of environmental states implies loss of quantum coherence.

(c) For \(\langle E_0(t)\vert E_1(t)\rangle = \lambda(t)\) with \(\vert\lambda(t)\vert < 1\), show that \(\hat{\rho}_{01}(t) = \frac{\lambda(t)}{2}\vert 0\rangle\langle 1\vert\).

2. In the weak coupling limit, \(\langle E_0(t)\vert E_1(t)\rangle = \mathrm{e}^{-t/T_2}\).

(a) Show that the coherence decays as \(\hat{\rho}_{01}(t) = \hat{\rho}_{01}(0)\,\mathrm{e}^{-t/T_2}\).

(b) Estimate the decoherence rate \(\gamma = 1/T_2\) for a spin-1/2 magnetic impurity in a crystal lattice, where the effective coupling to phonons is \(\lambda \sim 10^{-1}\) meV and the Debye temperature is \(\Theta_D = 300\) K. Use dimensional analysis.

(c) For a macroscopic object (\(m = 1\) mg, \(N \sim 10^{20}\) atoms) with each atom decohering at rate \(\gamma_\text{atom}\), estimate the total decoherence rate \(\Gamma = N\gamma_\text{atom}\) and the timescale \(T_2 = 1/\Gamma\). Why is macroscopic decoherence so much faster?

3. The interaction Hamiltonian is \(H_\text{int} = \hat{\sigma}^z \otimes B_E\).

(a) Show that if the system is in pointer state \(\vert 0\rangle\), the system-environment evolution has the form \(\vert 0\rangle \otimes \vert E_0(t)\rangle\) — the system state remains definite, only the environment evolves.

(b) Show that if the system starts in \(\vert+\rangle = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\), the evolution creates entanglement between system and environment, leading to decoherence.

(c) Explain in words: why are \(\vert 0\rangle\) and \(\vert 1\rangle\) pointer states for this Hamiltonian but \(\vert+\rangle\) and \(\vert-\rangle\) are not?

4. A spin-1/2 system couples to a fluctuating field \(H_\text{int} = \hat{\sigma}^z B_z(t)\), where \(\langle B_z(t)B_z(t')\rangle = \frac{\Delta^2}{2\tau_c}\mathrm{e}^{-\vert t-t'\vert/\tau_c}\) is colored noise.

(a) Show that \(\vert 0\rangle\) (a \(z\)-basis pointer state) commutes with \(H_\text{int}\) and undergoes no dephasing.

(b) A spin in \(\vert+\rangle\) decoheres as \(\hat{\rho}_{01}(t) = \frac{1}{2}\mathrm{e}^{-t/T_2}\) with \(T_2 = 2\tau_c/\Delta^2\). Show that at \(t \gg T_2\), the state is indistinguishable from the random mixture \(\frac{1}{2}\vert 0\rangle\langle 0\vert + \frac{1}{2}\vert 1\rangle\langle 1\vert\).

(c) Explain why \(\vert+\rangle\) is not a pointer state for \(B_z\) noise, but \(\vert+\rangle\) would be a pointer state for \(B_x\) noise (\(H_\text{int} = \hat{\sigma}^x B_x\)).

5. Consider the Schrödinger cat state \(\vert\Psi\rangle = \frac{1}{\sqrt{2}}(\vert\text{alive}\rangle + \vert\text{dead}\rangle)\) for a macroscopic object with \(N \sim 10^{26}\) atoms. Each atom decoheres at rate \(\gamma_\text{atom} \sim 10^{12}\) s\(^{-1}\).

(a) Estimate \(\Gamma = N\gamma_\text{atom}\) and \(T_2 = 1/\Gamma\). Compare to the Planck time \(t_P \sim 10^{-44}\) s.

(b) Explain why decoherence — not a fundamental postulate — is the reason we never observe macroscopic superpositions. What would have to change for us to observe a “cat state”?

(c) Discuss: the full system+environment remains in a pure state even after decoherence. Why then does the system appear to be in a classical mixture?

6. (Decoherence as a Quantum Channel.) A qubit \(S\) couples to an ancilla \(A\) via \(U = \mathrm{e}^{\mathrm{i}\theta\hat{\sigma}^z_S \otimes \hat{\sigma}^z_A}\). The ancilla starts in \(\vert+\rangle_A = \frac{1}{\sqrt{2}}(\vert 0\rangle + \vert 1\rangle)\).

(a) Apply \(U\) to \(\hat{\rho}_S \otimes \vert+\rangle\langle+\vert_A\) and trace out the ancilla. Show that the result is the dephasing channel \(\mathcal{E}(\hat{\rho}) = K_0\hat{\rho} K_0^\dagger + K_1\hat{\rho} K_1^\dagger\) with \(K_0 = \sqrt{1-p}\,I\), \(K_1 = \sqrt{p}\,\hat{\sigma}^z\), where \(p = \frac{1}{2}(1 - \cos 2\theta)\).

(b) Identify this as decoherence in the \(\hat{\sigma}^z\) pointer basis. What is the role of the ancilla in this model?

(c) Explain in words why this calculation demonstrates “decoherence = measurement + forgetting”: what is measured, and what is forgotten?

7. Two qubits couple to a common environment via \(H_\text{int} = g(\hat{\sigma}^z_1 + \hat{\sigma}^z_2)\otimes B_E\) (identical, correlated noise).

(a) Show that the singlet \(\vert\Psi^-\rangle = \frac{1}{\sqrt{2}}(\vert 01\rangle - \vert 10\rangle)\) is decoherence-free: \((\hat{\sigma}^z_1 + \hat{\sigma}^z_2)\vert\Psi^-\rangle = 0\).

(b) Show that the triplet \(\vert\Psi^+\rangle = \frac{1}{\sqrt{2}}(\vert 01\rangle + \vert 10\rangle)\) is not decoherence-free and will decohere under this coupling.

(c) Explain the physical mechanism: why does the singlet’s entanglement protect it from identical noise? What symmetry is responsible?

(d) Explain how you could encode one logical qubit using the singlet state and the triplet state to protect against this common-mode noise.