Friday, February 18, 2011

Cute demonstration of how random phase shifts lead to decoherence

This blog needs more physics.

I've been looking around for a barebone example of quantum decoherence that doesn't needlessly complicate matters with spin states and the like. Finding no satisfactory example, I whipped one up myself.

In the quantum world, probabilities are described by the probability amplitude (let's call it ψ(x)), while in the classical world, it's enough to use probability densities, the absolute square of the probablity amplitude (i.e. $|\psi(x)|^2$).

Decoherence is basically the process in which the quantum description of probability becomes equivalent to the classical description of probability through random phase changes.

Let's consider a superposition of two wavefunctions:

\[\psi(x) = e^{i\xi}\psi_1(x) + e^{i\phi}\psi_2(x)\]

The probability density to measure the wavefunction at position x is now

\[|\psi(x)|^2 = |\psi_1(x)|^2 + |\psi_2(x)|^2 + e^{i(\xi-\phi)}\psi_1(x)\psi_2(x)^*\]\[+ e^{-i(\xi-\phi)}\psi_1^*(x)\psi_2(x)\]

The first two terms are phase independent and corresponds to classical, additive probabilities, and the last two are quantum terms that do not appear in classical physics. Also note that they depend on phase.

So let's suppose we introduce a mechanism that at each time step has a miniscule probability of introducing a small fixed phase shift in the wavefunctions.

Fiddling around a bit with expectation values it's straightfoward to show that this leads to a brownian walk in the phase difference δ=ξ-φ.

It's well known that Brownian motion is diffusive, and can suitably be described with the heat equation. So the probability density for the phase difference is governed by

\[\frac{dP(\delta)}{dt} = \alpha \frac{d^2P(\delta)}{d\delta^2}\]

This gives solutions on the form

\[P(\delta) = \exp(-\frac{\delta^2}{\alpha t})\]

The phase factor next to the attenuation terms must now be re-written to take into account the diffusion of the phase.

\[e^i(\delta) \rightarrow \int d\delta P(\delta) e^{i\delta} \]

This is a standard integral that is easy to evaluate, giving

\[ \int d\delta P(\delta) e^{i\delta} = e^{-\alpha t} \]

So the revised probability density is

\[|\psi(x)|^2 = |\psi_1(x)|^2 + |\psi_2(x)|^2 + e^{-\alpha t}(\psi_1(x)\psi_2(x)^* + \psi_1^*(x)\psi_2(x))\]

In summary, brownian-like phase shifts leads to linear attenuation of the interference terms, and a decoherence time of α-1. This is of course a very simplistic model, and great care must be taken to ensure that such a process actually is applicable in order to avoid pathological cases.

Take for example $\psi_1(x)=\psi_2(x)$, this is clearly an unphysical case as there is no way to independently phase-shift just one of the completely indistinguishable wavefunctions. If you do never the less feed this case to the model, it tells you very adamantly that either $\psi(x)=0$, or $2=1$.

Follow-up Feb 25: For a less hand-wavy and more intuitive theoretical demonstration, it's easy to show that mounting a device over one of the gaps in a double slit experiment that introduces random phase shifts to particles passing through it leads to the vanishing of the interference pattern. One can then move that action to a medium in the chamber, and get the same results (where attenuation depends on time-of-flight between slit and screen. For explicit time dependence, a moving screen could be introduced).

1 comment:

  1. This post explains how random phase shifts lead to decoherence. This is something new which I come to know with the help of your post. The concept is not very easy but the post help a lot in understanding the topic. Thanks for the post.