The Metropolis Algorithm and Detailed Balance

To understand this better, assume that we have two energies, \( E_i \) and \( E_j \), with \( E_i < E_j \). This means that the largest acceptance value must be \( A(j\rightarrow i) \) since we move to a state with lower energy. It follows from also from the fact that the probability \( w_i \) is larger than \( w_j \). The trick then is to fix this value to \( A(j\rightarrow i)=1 \). It means that the other acceptance probability has to be $$ \begin{equation*} A(i\rightarrow j)= \exp{(-\beta(E_j-E_i))}. \end{equation*} $$