This question is about self-averaging quantities in physics. Here is a link to the wikipedia page: https://en.wikipedia.org/wiki/Self-averaging.

I'm currently studying transmission through disordered systems (Anderson localization) and have some doubts with what is called "self averaging". Often the following claim comes up: the transmission probability $\langle T \rangle$ is not a self-averaging quantity, but $\langle \ln T \rangle$ is.

As far as I understand it has something to do with that the transmission amplitude for a system of length $L$ scales multiplicatively with the system. That is $\langle T \rangle_L = (\langle T \rangle_1)^L$ where $\langle T_1 \rangle$ is the transmission amplitude for a system with one disordered impurity and $L$ is the length of the system. On the other hand the logarithm scales additively $\langle \ln T \rangle_L = \langle \ln T_1 \rangle + \langle \ln T_1 \rangle + \dots + \langle \ln T_1 \rangle = N \langle \ln T_1 \rangle$.

My first question is

  1. Why does an additive and multiplicative scaling law lead to that the quantity is self-averaging and non self-averaging respectively?

My second question is related to the wikipedia article. To me it seems that one can check if the transmission probability is self averaging by computing the relative variance $\mathrm{Rvar(T)} = \frac{\mathrm{Var(T)}}{\langle T \rangle^2}$ and check whether or not it vanishes when we send the system size $L \rightarrow \infty$. However, this seems fishy to me because in a strongly disordered one-dimensional system we expect the transmission amplitude to decay exponentially and approach zero as we increase the system size. That is if we were to do this on a computer we would obtain both $\langle T \rangle = \mathrm{Var}(T) = 0$ if we let $L\rightarrow \infty$. As we see we obtain an indeterminate expression. However, we can get around this by looking at the reflection probability $\langle R \rangle$ instead:

$$\mathrm{Rvar(R)} = \frac{\mathrm{Var(R)}}{\langle R \rangle^2}$$

For increasing system size we expect $\langle R \rangle \rightarrow 1$ and $\mathrm{Var(R)} \rightarrow 0$ meaning that $\mathrm{Rvar}(R) \rightarrow 0.$ However, it seems very weird to me that $R$ can be self-averaging but $T$ is non self-averaging because they are connected through the identity $R + T = 1$.

To be concrete my second question/confusion is:

  1. Is it correct to say that if the relative variance (of some quantity) approaches zero with increasing system size the quantity is self averaging? If not what is a good statistical measure of self-averageness?

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.