Molecular dynamics and detailed balance

Question

In developing methods to perform Monte Carlo simulations one sufficient condition to preserve the stationarity of the target probability distribution is to impose detailed balance i.e.

[Gardiner page 148]

A Markov process satisfies detailed balance if, roughly speaking, in the stationary situation each possible transition balances with the reversed transition.

Where the transition is given by $$(r, v, t)\rightarrow (r', v', t+\tau)$$ and its reversal correspond to the time reversed transition which requires the velocities to be reversed because the motion from r' to r is in the opposite direction from that from r to r'. $$(r', - v', t)\rightarrow (r, - v, t+ \tau).$$

Suppose our transiotions correspond to the movement from a state of the Harmonic oscillator in the phase space $(q,p)$ to another state of the same H.O., $(q',p')$.

Here my question is: how are conservation of Energy and detailed balance linked?

The answer should be:

energy conservation $\Rightarrow$ detailed balance This should be clear because if energy is conserved that means that I am moving on the ellipse in phase space which is composed by all the microstates available to my system.

While the converse should not be always true, i.e. there could be jumps which satisfy detailed balance but violates conservation of energy, but I cannot find an example of them.

Answer 1

If you are sampling a canonical ensemble energy is not generally conserved. The detailed balance condition in such a setting reads $$ \tag{1} e^{-\beta \mathcal H_i} P(i \rightarrow j) = e^{-\beta \mathcal H_j} P(j \rightarrow i), $$ where $i,j$ label the states in the canonical ensemble at temperature $T=1/k\beta$ and $\mathcal{H}_i$ is the energy of the $i$-th state.

The stationary probability resulting from (1) can be shown to be $$ \tag{2} \Pi_i = \frac{e^{-\beta \mathcal H_i}}{\sum_j e^{-\beta \mathcal H_j}}, $$ and this is used for example when the Markov chain is used to calculate averages of the form $$ \tag{3} \langle \mathcal{O} \rangle = \frac{ \int \mathcal D \phi \,\, \mathcal O [\phi] e^{-\beta \mathcal{H}[\phi]}} {\int \mathcal D \phi \,e^{-\beta \mathcal{H}[\phi]}}. $$

Answer 2

I am not sure I fully understand your question, because in the title you talk about Molecular Dynamics and then in the body you talk about Monte Carlo. Anyway, here is my answer:

Molecular Dynamics does not satisfy detailed balance.

In a NVE molecular dynamics simulation, the transition

$$(\mathbf r(t), \mathbf v(t)) \to (\mathbf r(t+\tau), \mathbf v(t+\tau))$$

is deterministic, meaning that given a state $i$ there is one and only one a state $j$ such that

$$P(i\to j) =1 $$

the probability to go from $i$ to any other state is $0$:

$$P(i\to j') =0 \ \ \forall j'\neq j$$

The same is true also in a NVT simulation, but in a more subtle way, because the presence of the thermostat introduces "noise" in the dynamics of the system. Anyway, since Molecular Dynamics is based on the deterministic equations of motions of Newton, it will never satisfy detailed balance.

Conversely, Monte Carlo codes usually satisfy detailed balance, but they don't necessarily conserve energy: a well-known example is the Metropolis algorithm.