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The chance of occupying state with energy $E_i$ is given by the Boltzmann distribution:

$$ P(E_i)= \frac{1}{Z} \exp \left( \frac{-E_i}{kT} \right) $$

The exponential distribution is a common probability distribution and is known to be memoryless, see this derivation. Since the Boltzmann distribution is just a special version of the exponential distribution, this suggests that the Boltzmann distribution should also be 'memoryless' in some sense. What is the physical interpretation of this property for the Boltzmann distribution?

My guess: The microstate a system is in is independent of the microstate it was in before. However, this seems not completely right, since you cannot go immediately from all molecules in one side of a room to being spread out. There needs to be a path in between I would say.

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    $\begingroup$ You are mixing the concept of dynamics of the system (i.e. evolution from one state A at some time t to another state B at a later time t') with that of equilibrium, where essentially the concept of time is absent. In equilibrium, you can only assume that all microstates accessible correspond to a given probability distribution. In your case, because you are fixing the temperature of the system, this gives the Boltzmann distribution. However, if instead you would have fixed the energy, your distribution would be the constant distribution, meaning each state is equally likely. $\endgroup$
    – Zarathustra
    Commented Sep 10, 2023 at 14:27
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    $\begingroup$ "It is known by being 'memoryless'." -> please provide a reference for this concept ad clarify its meaning. What do you mean by "memoryless"? (My guess: it is a measure for an equilibrium system under certain constraints and equilibrium systems are the ones that mostly "forget" about their initial condition.) Maybe this helps: physics.stackexchange.com/a/389714/226902 $\endgroup$
    – Quillo
    Commented Sep 16, 2023 at 17:59
  • $\begingroup$ "Could somehow help me?" is not a clear question. Consider rewriting your question more clearly. $\endgroup$
    – hft
    Commented Sep 16, 2023 at 18:24
  • $\begingroup$ I adjusted it, thank you for your comment $\endgroup$
    – bananenheld
    Commented Sep 17, 2023 at 19:08
  • $\begingroup$ What's your experiment? For some experiments, you'll see correlations between successive measurements, for others you won't. The "memoryless" concept isn't insightful here. $\endgroup$
    – John Doty
    Commented Sep 17, 2023 at 19:13

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As with every probability density distribution in the Statistical Mechanics theory of the ensembles, the Boltzmann distribution assigns probabilities to the microstates without any reference to the underlying dynamics. This is the step forward introduced by the probabilistic methods of Statistical Mechanics: we do not have to know the details of the microscopic dynamics. To evaluate equilibrium averages, we need only the equilibrium probability distribution. It is the essence of the ergodic hypothesis that it is possible for any system amenable to a statistical mechanics description to substitute the time average of an observable $A$, $$ \lim_{\tau \rightarrow \infty}\frac{1}{\tau}\int_0^{\infty}A(q(t),p(t))dt, $$ (integral over time using the time evolution of the coordinates and momenta) with the ensemble average $$ \int \dots\int A(q,p) \rho(q,p) dq \dots dp $$ over the whole phase space.

Second part of the answer, after the clarifying edit of the question

The first part of the answer indirectly explains in part why a guess based on the temporal sequence of states cannot work. I'll try to address the question more directly.

A memoryless probability distribution is usually the distribution function of a random variable with the meaning of a time. The canonical distribution is a probability distribution in phase space. However, if we look at points in phase space as random variables, the canonical probability is not exponential in their values. It rather depends on them through the Hamiltonian, which could be a pretty complicated function. The only variable apparently appearing as an exponential is the energy of a microstate. However, in general, there is some degeneracy (many microstates have the same energy), and the canonical probability is not exponential as a function of the energy. This fact preempts any further attempt to find analogies with memoryless distributions except for the exceptional case of non-degenerate states where the behavior of the exponential distribution as memoryless distribution described in the Wikipedia article you cited can be adapted to a situation where the energy plays the same role as the time.

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  • $\begingroup$ Thank you for your answer. What do you exactly mean with substituting time averages with ensemble averages? $\endgroup$
    – bananenheld
    Commented Sep 17, 2023 at 19:07
  • $\begingroup$ He means that the ensemble average just looks at the average over every possible consistent configuration of the system without any regard for dynamics. The time average “follows” the system starting from some particular state as it evolves over time and then averages over all the states it visits over time. If you wait long enough, and if the system is capable of dynamically sampling all its relevant states, then the two averages agree. This is a key idea in statistical mechanics. $\endgroup$
    – Matt Hanson
    Commented Sep 18, 2023 at 0:16
  • $\begingroup$ @bananenheld Matt Henson clarified this point. I'll add the explicit formulae to my answer. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Sep 18, 2023 at 5:37
  • $\begingroup$ This is a good answer but has little to do with the original question (that is unclear). The OP is asking about the memoryless property of the exponential distribution in the context of time series (i.e. Poisson point process). Since The Gibbs weight contains an exponential, the OP is assuming that it should have this property as well (how and for what it's unclear, because in this case we are not in the context of time series). $\endgroup$
    – Quillo
    Commented Sep 18, 2023 at 6:42
  • $\begingroup$ @Quillo . I do not think you are interpreting OP's question correctly. Memoryless of distribution probabilities usually applies to the distribution of waiting times. That is not the case of the Boltzmann distribution. $\endgroup$
    – GiorgioP-DoomsdayClockIsAt-90
    Commented Sep 18, 2023 at 7:12

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