Metropolis Algorithm Transition-Proposal Probability

Question

I'm working my way through a short section on the Metropolis algorithm in the lecture notes on Computational Quantum Physics by Prof. Troyer.

However, I am not sure what probability distribution was used in the last term.

The partition function is derived to be of the form

$$ Z ( N , V , T ) = \int \cdots \int \prod _ { j = 1 } ^ { M } \mathrm { d } \boldsymbol { R } _ { j } \prod _ { j = 1 } ^ { M } \left\{ \rho ^ { \mathrm { free } } \left( \boldsymbol { R } _ { j } , \boldsymbol { R } _ { j + 1 } , \tau \right) \exp \left[ - \tau V \left( \boldsymbol { R } _ { j } \right) \right] \right\} $$

Where we are working with the primitive approximation, using only the linear terms in the BKH expansion of the Hamiltonian

$$ \exp ( - \tau \hat { H } ) \cong \exp ( - \tau \hat { T } ) \exp ( - \tau \hat { V } ) $$

In this notation, we have

\begin{align} M &\dots \text{number of imaginary time steps} \\ \tau &= \frac{\beta=(k_BT)^{-1}}{M} \\ \boldsymbol{R_j} &= (r_1,r_2, \dots, r_N) \\ &\dots \text{coordinate vector of system at time step $j$} \end{align}

We now want to sample this imaginary time path integral.

Now, the 1D case of the free-particle density matrix we can write as

$$ \rho \left( x , x ^ { \prime } , \beta \right) = \sqrt { \frac { m } { 2 \pi \beta \hbar ^ { 2 } } } \exp \left( - \frac { m } { 2 \beta \hbar ^ { 2 } } \left( x - x ^ { \prime } \right) ^ { 2 } \right) $$

Now, to sample this path integral, we use the Metropolis algorithm. In this notation, we have

\begin{align} X & \dots \text{system configuration} \\ \pi(X) & \dots \text{stationary distribution} \\ P(X,X') & \dots \text{transition probability for } X \rightarrow X' \\ T(X,X') &\dots \text{proposal probability for } X \rightarrow X' \\ A(X,X') &\dots \text{acceptance probability for } X \rightarrow X' \end{align}

The overall acceptance probability $A(X,X')=P(X,X')P(X,X')$ that fulfils the condition of detailed balance

$$ \pi ( X ) P \left( X , X ^ { \prime } \right) = \pi \left( X ^ { \prime } \right) P \left( X ^ { \prime } , X \right) $$

is defined for the Metropolis algorithm as

$$ A \left( X , X ^ { \prime } \right) = \min \bigg( 1, \frac { \pi \left( X ^ { \prime } \right) T \left( X ^ { \prime } , X \right) } { \pi ( X ) T \left( X , X ^ { \prime } \right)} \bigg)$$

which, for the case of the partition function $Z$ introduced above, is defined in the lecture notes as

$$ \chi \left( X , X ^ { \prime } \right) = \frac { \exp \left[ - \frac { {\left( r _ { j - 1 } ^ { i } - r _ { j } ^ { i \prime } \right) ^ { 2 }} + \left( r _ { j } ^ { i {\prime}} - r _ { j + 1 } ^ { i } \right) ^ { 2 } } { 2 \hbar ^ { 2 } \tau / m } \right] } { \exp \left[ - \frac { \left( \boldsymbol { r } _ { j - 1 } ^ { i } - \boldsymbol { r } _ { j } ^ { i } \right) ^ { 2 } + \left( \boldsymbol { r } _ { j } ^ { i } - \boldsymbol { r } _ { j + 1 } ^ { i } \right) ^ { 2 } } { 2 \hbar ^ { 2 } \tau / m } \right] } \exp \left[ - \tau \left( V \left( \boldsymbol { R } _ { j } ^ { \prime } \right) - V \left( \boldsymbol { R } _ { j } \right) \right) \right] $$

Nothing was said about the choice of the a-priori distribution governing the proposal of the next configuration. So what is T(X,X')?

Answer 1

Your quantity $\chi(X,X')$ depends on a time step $j$ and a particle number $i$. The simplest implementation of this Monte Carlo chain consists in choosing first one time step $j$ and one particle $i$ before proposing a change $\vec r_j^i\leftrightarrow {\vec r_j^i}'$. In this case, the proposal probability $T(X,X')$ is equal to $1/MN$ when the configurations $X$ and $X'$ differ only by the displacement of one particle at one time step. It should vanish otherwise. More complex algorithms (loop algorithms) may allow more transitions $X\rightarrow X'$ in a single Monte Carlo step.