Participation ratio ($P=({\sum }_n|{\psi }_n{|}^4{)}^{-1}$)

For an eigenvector normalized by ${\sum }_{n=1}^N|{\psi }_n{|}^2=1$, the participation ratio in its site basis isIt estimates how many sites carry appreciable probability. Equal weight on $K$ sites and zero elsewhere gives $P=K$, irrespective of their wave phases. A single occupied site gives $P=1$; uniform probability on all sites gives $P=N$. These are exact constructed examples, not claims about particular Hamiltonian operators.

To prove the bounds, write $p_n=|{\psi }_n{|}^2$. Since $0\le p_n\le 1$, ${\sum }_n{p}_n^2\le {\sum }_n{p}_n=1$, giving $P\ge 1$. The Cauchy-Schwarz inequality gives $1=({\sum }_n{p}_n{)}^2\le N{\sum }_n{p}_n^2$, giving $P\le N$. Thus the measure has useful limits without assuming an exponential shape. Comparing $P$ as $N$ increases helps distinguish a state filling the sample from one occupying a fixed region, provided energy selection and disorder statistics are controlled.