Is quantum decoherence basis-dependent?

Question

Is quantum decoherence basis-dependent? I've seen claims for both 'yes' and 'no'. Could someone provide a nice simply explained example (geared to an amateur quantum mechanic) that demonstrates the answer.

Answer 1

TL; DR: No, but yes.

While I concur with the basic quantum mechanical arguments given in the other answers, there are two reasons why decoherence will often appear as basis-dependent. As a reference model we can consider Bloch equations, describing relaxation and decoherence of a two-level system: $$ \frac{dM_x(t)}{dt}=\gamma\left(\mathbf{M}(t)\times\mathbf{B}(t)\right)_x - \frac{M_x(t)}{T_2},\\ \frac{dM_y(t)}{dt}=\gamma\left(\mathbf{M}(t)\times\mathbf{B}(t)\right)_y - \frac{M_y(t)}{T_2},\\ \frac{dM_z(t)}{dt}=\gamma\left(\mathbf{M}(t)\times\mathbf{B}(t)\right)_z - \frac{M_z(t)-M_0}{T_1}. $$ It is evident that there is a preferred basis here: the relaxation along the z-direction is governed by time $T_1$, while along the other directions it is governed by $T_2$. Decoherence is however independent on the choice of x and y axes. There are two things happening here:

• Problem already has a choses basis Decoherence results from the interaction with the environment and this interaction is often anisotropic. E.g., in case of the Bloch equations there is the implicit assumption that the magnetic field has a strong constant component along the z-axis, splitting the levels and producing non-zero residual magnetization $M_0$: $$\mathbf{B}(t)=\left(B_x(t), B_y(t),B_0 + \Delta B_z(t)\right).$$
• Approximations are not basis invariant The other reason is that the approximations used are often not aimed at being basis invariant - e.g., one may calculate the relaxation time $T_1$ using the Fermi Golden rule, whereas times $T_2$ can be evaluated assuming no energy transfer. In many derivations this asymmetry may be difficult to spot, but it is often present.

Answer 2

No, quantum decoherence is not basis-dependent.

First consider that a perfectly isolated system will maintain coherence indefinitely; it is through the interactions with other systems (or parts of the system that were previously not present or inaccessible) that leads to decoherence.

Second imagine that you have two basis sets that equally well represent the system and that you can transform between them. The measurement of physical properties (e.g. expectation of momentum, position, energy, spin operators) will not depend on the basis set.

Finally, the interaction effects / terms that lead to decoherence will be different for the different basis sets / their components, however the expectation of the physical interaction that causes the decoherence is the same regardless of the basis. Hence decoherence does not depend on the basis.

Answer 3

If you are familiar with QM formulated in terms of density matrices, it is very easy to see why decoherence basis independent.

If you have a system in a pure state $|\psi\rangle$, then the density matrix is $$\rho = |\psi\rangle\langle\psi|,$$ for which you can immediately check that not only $\rho$ had unit trace (properly normalised) but also its square has unit trace: $$\mathrm{trace~}\rho^2 = 1.$$

The effect of decoherence is to produce a mixed state, defined as a stochastic mixture of pure states, which is represented as sum of density matrices, weighted by probabilities:

$$\rho_m = \sum_i p_i\rho_i,$$

You can check that, even when $\rho_m$ is properly normalised, you will have

$$\mathrm{trace~}\rho_m^2 < 1.$$

Since the trace is a basis independent operation, this gives you a basis independent check on the purity of a state.

Hence decoherence is basis independent.

Maybe the text that said that decoherence is basis dependent meant to say that decoherence happens only if you have a superposition with respect to a certain basis.