Let's say for a specific simple decoherence model you end up with the following density matrix for the "system" (I will use the common separation in "system" and "environment" usually used in decoherence literature):

$$\frac{1}{2} \begin{pmatrix}1&r(t)\\ r^*(t)& 1\end{pmatrix}$$

with $r(t)=e^{-2i\omega t}$ ($\omega$ being some parameter of the interaction hamiltonian standing for the strength of the interaction between system and environment).

Usually, when $r(t)\rightarrow 0$, one says that whatever system state has been considered, it is "fragile" to the decoherence process in consideration. Here although, $|r(t)|^2=1$. So what is the important parameter?

Since an arbitrary density matrix can be written as $$\rho=\begin{pmatrix}1-\rho_{00}&\rho_{01}\\ \rho_{01}^*&1-\rho_{00}\end{pmatrix}$$ and a measurement probability is calculated along $p(\lambda)=\text{Tr}(\rho P)$ with $P$ e.g. being $P=|+\rangle\langle+|$ (in this case), one finds that $p(\lambda)=\frac{1}{2}\left(1+2\text{Re}(\rho_{01})\right)$. Therefore, only the real part of $r(t)$ is important for whether an observer experiences the system to act quantum-mechanically (but only relative to the basis $|+\rangle,|-\rangle$.

Therefore one can say that for the considered model, the considered state of the system (before the interaction) is also fragile to this concrete decoherence process.

Is it correct that only the real part of $r(t)$ (see above) matters for whether "the system loses non classical properties" (in regard to a local observer)?

More generally: When using an arbitrary state $|\psi\rangle=a|0\rangle+b|1\rangle$ instead of $|+\rangle$ one arrives for the probability at $|a|^2\rho_{00}+ab^*\rho_{01}^*+a^*b\rho_{01}+|b|^2(1-\rho_{00})$ but from this one can again only say that the probability depends on the real part of $ab^*\rho_{01}$. So that doesn't support the suspicion that only the real part is important, since $Re(x)*Re(y)\neq Re(x*y)$.


1 Answer 1


The density matrix $$\rho(t)=\frac12\begin{pmatrix}1&e^{-2i\omega t}\\e^{2i\omega t}&1\end{pmatrix}$$ is the density matrix of a system evolving through the pure states $$|\psi(t)\rangle=\frac1{\sqrt2}\begin{pmatrix}1\\e^{2i\omega t}\end{pmatrix}=\frac1{\sqrt 2}(|0\rangle+e^{2i\omega t}|1\rangle),$$ because you can verify $\rho(t)=|\psi(t)\rangle\langle\psi(t)|.$ This last equation can be interpreted as saying that the system at time $t$ has a 100% classical probability of being found in the quantum state $|\psi(t)\rangle.$ In this (clearly idealized) system, there is absolutely no decoherence. All uncertainties in measurements have "purely quantum" origin.

A general mixed state $\rho$ can be written as a combination of projectors onto orthogonal states: $$\rho=\sum_i p_i|\psi_i\rangle\langle\psi_i|,$$ where the $p_i$ are classical probabilities (real, nonnegative, summing to $1$). $p_i$ is the probability of finding the system in quantum state $\psi_i.$ A perfectly coherent state is one where mostly $p_i=0$ except for one $p_i=1,$ which indicates a system that's certainly in a given quantum state. Decoherence happens when you become uncertain of the quantum state of the system: quantum uncertainty leaks into classical uncertainty. Decoherence is the resistance to writing $\rho=|\psi\rangle\langle\psi|.$ In your system, this hasn't happened.

Measurement outcomes being probabilistic is a bad indicator of decoherence, since, even if you know the quantum state of a system perfectly, basic QM tells you certain (most) measurements will be nondeterministic. Your suspicion at the end is on the right track: if you measure the observable $P(t)=|\psi(t)\rangle\langle\psi(t)|$ (measure the answer to "is the system in state $|\psi(t)\rangle$?") when the system is in the state $\rho(t),$ you will certainly get a "yes" and never a "no." This again suggests decoherence hasn't occurred. Your calculations may be correct, but you've interpreted them incorrectly by assigning too much importance to the chosen basis, while the physics might not care about the basis.


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