== Decoherence with Entanglement == Decoherence is a phenomenon of the quantum theory where the loss of coherence occurs and the interference effects are lost. As an example, consider a general pure state of a qubit (e.g. spin half particle)
∣ψ⟩ = α∣0⟩ + β∣1⟩
, where ∣α∣2 + ∣β∣2 = 1. If we are to measure the x component of the spin σx, then the expectation value is ⟨σx⟩ = 2Re(αβ * ) ≠ 0. However, if we consider the measurement apparatus as a quantum object and instead of measuring the x component, we let the qubit and the apparatus to interact along the z axis in a non-destructive way, then the final wave function may be written as
∣ψ⟩ = α∣0⟩ ⊗ ∣A0⟩ + β∣1⟩ ⊗ ∣A1⟩
.
Here we assume ⟨A0∣A1⟩ = 0 are completely orthogonal, meaning that they are completely distinguishable. We say the qubit gets "entangled" with the "environment" (or any macroscopic object). With this entangled wavefunction, if we are to measure the x component of the spin (using another measurement apparatus), then we must concluded that ⟨σx⟩ = 0. This is what we mean by the loss of coherence.
Generally, we would like to focus on the system instead of the environment. This could be done by considering the (reduced) density matrix of the system. The entangling process with the environment may be regarded as the vanishing of the off-diagonal matrix elements of the density matrix:
$$\rho= \left( \begin{array}{ccc}
\left|\alpha \right|^2& \alpha \beta^* \\
\alpha^* \beta & \left|\beta \right|^2 \\
\end{array} \right) \rightarrow
\left( \begin{array}{ccc}
\left|\alpha \right|^2 & 0 \\
0 & \left|\beta \right|^2 \\
\end{array} \right)$$
.
Note that the diagonal elements are not changed, and this type of decoherence is called "pure dephasing". The open system dynamics of this reduced density matrix can be formulated in terms of the master equation.
Decoherence without Entanglement
Here we should be careful that sometimes people would also describe an ensemble spins as a single 2 × 2 density matrix. The loss of coherence is also described by the vanishing of the off-diagonal matrix elements, but there is no entanglement with the environment. This approach is fundamentally different from that described above.
To elaborate further, let us consider an ensemble of N spin-1/2 particles initially prepared in the same state,
∣ψ⟩ = (α∣0⟩ + β∣1⟩) ⊗ (α∣0⟩ + β∣1⟩) ⊗ ... ⊗ (α∣0⟩ + β∣1⟩)
.
If we are to measure the total x component of the spins, then we have ⟨X⟩ ≡ ⟨σx1 + σx2 + ... + σxN⟩ = 2NRe(αβ * ). In reality, there may be some static but spatially varying magnetic field causing dephasing, i.e. each spin experience some unknown phase shift α∣0⟩ + βeiϕk∣1⟩. For sufficiently large N, one has ⟨X⟩ = 2Re[αβ * (e − iϕ1 + e − iϕ2 + ... + e − iϕN)] ∼ 0 due to the incoherent sum of the random phases. The main difference between the one described above and here is that the former used ⟨σx⟩ (for a single spin) as a measure of coherence, while the latter used ⟨X⟩/N (for an ensemble of spins). Consequently, the former involves entanglement but the latter does not.
One may argue that it is in principle possible to perform individual measurement for the spins, and since we do not know the precise quantum state (due to the random magnetic field), the system of spins may be described by a 2 × 2 matrix:
$$\rho= \left( \begin{array}{ccc}
\left|\alpha \right|^2& \langle e^{i \phi} \rangle \\
\langle e^{-i \phi} \rangle & \left|\beta \right|^2 \\
\end{array} \right) \sim \left( \begin{array}{ccc}
\left|\alpha \right|^2& 0 \\
0 & \left|\beta \right|^2 \\
\end{array} \right)$$
This seems to suggest that the system is in a mixed state and may be entangled with something. However, if we write down the full wavefunction describing all spins as did above, then we can conclude that no entanglement of any kind is involved in the "dephasing" process. In this case, it is also clear that the entanglement entropy is not suitable for quantifying entanglement. In physics, mixed state representation is conveniently used to described a single system or an ensemble of systems and ordinarily both give the same measurement results. However, we have to bear in mind the fundamental difference between them.