Detailed balance

Now we have to show that this algorithm does satisfy the detailed balance condition:  

$$W_{1\to 2}\exp(-\beta U_1) = W_{2\to 1}\exp(-\beta U_2)$$ (2)

where $W_{i\to j}$ is a probability to go from the micro-state i to the micro-state j. In our case the micro-states 1 and 2 differ by the position of one particle, that is moved at a single MC step. The probability $W_{1 \to 2}$ is the product of

1.

probability to pick particle 1, and the displacement vector, $1\to2$, equal to $1/N\mathbf{x}_{max}^2$, and

2.

probability to accept the move by the Metropolis criterium, equal to $\text{max}(1,\exp(-\beta(U_2-U_1))$

The probability of the opposite step $W_{2 \to 1}$ is the product of

1.

probability to pick particle 2, and the displacement vector $2\to1$, equal to $1/N\mathbf{x}_{max}^2$, and

2.

probability to accept the move by the Metropolis criterium $\text{max}(1,\exp(-\beta(U_1-U_2))$

If the state 1 has higher energy than 2, then $\exp(-\beta(U_1-U_2))<1$, and $\exp(-\beta(U_2-U_1))\gt 1$, the first contributions are the same in both direct and reverse probabilities, and equation 2 is satisfied.