Polish School of Negativism · [Nov 27, 01:53 PM]

It all has to do with reading this paper again:

Can one build a quantum hard drive? A no-go theorem for storing quantum information in equilibrium systems
quant-ph/0603260

The paper is about existence of a quantum memory in thermal equilibrium. More precisely it states that they don’t. Maybe it is besides the point that a quantum memory need not operate in thermal equilibrium, I was thinking about the probabilistic argument the authors give towards the end of the paper about the Kitaev model.

The deceptively pedagogical argument is something like: Suppose individual fluctuations need to form a band of length $L$ around the system to kill the quantum memory, where $L$ is the size of the lattice. The probability of doing so (like a random walk) is proportional to $1/L^2$ and the number of starting points for such a band is $L^2$. Then we take the product of the probabilities and come up with a probability for forming a band that is given by a constant independent of the size of the system, hence no improvement in the error probability as a function of the size of the system. My problems: (1) This argument applies to almost everything and (2) How are all these random walks independent of each other? Should these bands cross, I guess they will cancel each other.

I am not sure about (1)... maybe that is exactly what the authors have in mind. And they do talk about (2):

In our case, we do not have usual random walk, because coming back to the origin means annihilating the pair, and the walk is ended. If the path was short, then the process does not affect quantum memory. Thus we want to estimate the probability that a long path occur. Since the process is Markovian (i.e. memoryless) and the average number of anyons is constant, we can imagine, that the walk of the just annihilated pair is continued by a new pair that has been just created (see Fig. 2) In this way we have reduced the problem to the usual random walk.

The number of anyons will be constant on the average but not over the microscopic scale of the walk. Maybe this is exactly the point: You will need to be far enough from equilibrium (or from Markovian-ness) for this argument to break.

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