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 the timescale \(T_2\)?
How is decoherence related to the dephasing quantum channel? What does “measurement + forgetting” mean physically when you trace out the environment?
Lecture Notes#
Overview#
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.
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: bound between \(T_1\) and \(T_2\)
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:
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 \(\hat{H}_\text{int} = \hat{Z} \otimes B_E\), the pointer basis is \(\{\vert 0\rangle, \vert 1\rangle\}\) (eigenstates of \(\hat{Z}\)). 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:
Computing explicitly:
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:
where \(T_2 = 1/\gamma\) is the dephasing time. The microscopic derivation:
Derivation: Dephasing Rate from Spin-Bath Coupling
For \(\hat{H}_\text{int} = \hat{Z} \otimes \sum_k g_k (a_k \mathrm{e}^{-\mathrm{i}\omega_k t} + a_k^\dagger \mathrm{e}^{\mathrm{i}\omega_k t})\) (spin coupled to bosonic bath), the overlap integral gives:
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.
Poll: Decoherence and superposition suppression
In the spin-environment model, a two-level spin couples to many environmental modes. Starting from a superposition \(|\psi\rangle = (|0\rangle + |1\rangle)/\sqrt{2}\), the density matrix of the spin acquires off-diagonal terms (coherences) that decay: \(\rho_{01}(t) \propto \exp(-\gamma t)\). What is the physical origin of this suppression?
(A) The spin loses energy to the environment; dissipation destroys coherence.
(B) Entanglement with the environment causes the spin’s state to become mixed; off-diagonal terms depend on environmental correlations that are unobservable.
(C) The environment measures the spin, projecting it into an eigenstate.
(D) Quantum mechanics is fundamentally non-unitary on open systems.
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): \(\vert \psi\rangle \otimes \vert E_0\rangle \to \alpha\vert 0\rangle\vert E_0(t)\rangle + \beta\vert 1\rangle\vert E_1(t)\rangle\)
Partial trace (forgetting): \(\operatorname{Tr}_E(\cdot)\) discards the measurement outcome.
The result is exactly the dephasing channel:
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:
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 (\(\hat{H}_\text{int} = (\hat{Z}_1 + \hat{Z}_2)\otimes B_E\)), the singlet state is decoherence-free:
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 mechanism: Entanglement between system and environment creates entangled state; tracing out environment (since inaccessible) leaves reduced density matrix; off-diagonal (coherence) terms decay.
Pointer basis: Environment couples to specific system observable (e.g., position); only that basis is robust—superpositions in the pointer basis decohere exponentially (\(\tau_d \sim 1/\gamma\)).
Decoherence time: \(\tau_{d}\propto 1/(N\gamma)\) where \(N\) is the environment size and \(\gamma\) is the per-particle coupling rate (so the total rate \(\gamma_{\text{total}}\sim N\gamma\)). Macroscopic objects decohere in femtoseconds; microscopic quantum systems survive much longer.
Pure to mixed: Initial pure state \(\vert\psi\rangle\) evolves to mixed state \(\hat{\rho}(t)\) with vanishing off-diagonal terms in pointer basis; superposition becomes incoherent mixture.
Arrow of time: Decoherence is irreversible in practice (environment has huge dimension); explains classical behavior of macroscopic objects and the emergence of a classical world from quantum mechanics.
See Also
6.3.3 Quantum Channels: Completely positive trace-preserving maps—mathematical frame for noise and decoherence.
6.4.2 Lindblad Master Equation: Markovian open-system generators, Lindblad operators, and positivity.
6.1.1 Mixed States: Mixed reduced states from tracing out an environment—purity loss as decoherence signature.
Homework#
1. Environment entanglement. 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 \(\hat{\rho}_S(t) = \operatorname{Tr}_E(\vert\Psi_{SE}(t)\rangle\langle\Psi_{SE}(t)\vert)\).
(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. Weak coupling decoherence. 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. Dephasing Hamiltonian. The interaction Hamiltonian is \(\hat{H}_\text{int} = \hat{Z} \otimes B_E\).
(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. Fluctuating field decoherence. A spin-1/2 system couples to a fluctuating field \(\hat{H}_\text{int} = \hat{Z} B_z(t)\), 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 \(\hat{H}_\text{int}\) 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 (\(\hat{H}_\text{int} = \hat{X} B_x\)).
5. Schrödinger cat decoherence. 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 Quantum Channel. A qubit \(S\) couples to an ancilla \(A\) via \(U = \mathrm{e}^{\mathrm{i}\theta\hat{Z}_S \otimes \hat{Z}_A}\). 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 \(K_0 = \sqrt{1-p}\,I\), \(K_1 = \sqrt{p}\,\hat{Z}\), where \(p = \frac{1}{2}(1 - \cos 2\theta)\).
(b) Identify this as decoherence in the \(\hat{Z}\) 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. Collective decoherence and DFS. Two qubits couple to a common environment via \(\hat{H}_{\text{int}} = g(\hat{\sigma}^{z}_{1} + \hat{\sigma}^{z}_{2})\otimes\hat{B}_{E}\) (identical, correlated noise).
(a) Show that both the singlet \(\vert\Psi^{-}\rangle = (\vert 01\rangle - \vert 10\rangle)/\sqrt{2}\) and the \(S^{z}{=}0\) triplet \(\vert\Psi^{+}\rangle = (\vert 01\rangle + \vert 10\rangle)/\sqrt{2}\) are decoherence-free: both satisfy \((\hat{\sigma}^{z}_{1} + \hat{\sigma}^{z}_{2})\vert\psi\rangle = 0\).
(b) Show by contrast that the triplets \(\vert 00\rangle\) and \(\vert 11\rangle\) (eigenvalues \(\pm 2\) of \(\hat{\sigma}^{z}_{1}+\hat{\sigma}^{z}_{2}\)) acquire a relative phase under the noise and do decohere. Conclude that the decoherence-free subspace (DFS) under collective \(\hat{\sigma}^{z}\) noise is the entire \(S^{z}{=}0\) subspace, spanned by \(\{\vert\Psi^{+}\rangle, \vert\Psi^{-}\rangle\}\).
(c) Identify the symmetry responsible: the noise commutes with which collective operator, and hence preserves which conserved quantum number?
(d) Explain how to encode one logical qubit by identifying \(\vert 0\rangle_{L} = \vert\Psi^{+}\rangle\) and \(\vert 1\rangle_{L} = \vert\Psi^{-}\rangle\) — fully protected against this common-mode noise while requiring two physical qubits.