Boltzmann distribution has the form $$ p_i={1\over{Z}} e^{-\beta E_i} $$ for the probability of a system to be in the state $E_i$. $E_i$ is the total energy of the system. For example if in $i$th state all the particles have the same energy $\epsilon$, than $E_i=N\epsilon$, where $N$ is the total number of particles in the system as I've understood from this notes.

On the other hand the same formula is used for the probaility of finding the single particle of the system in the state $\epsilon_i$

$$ p'_i={{N_i}\over{N}}={1\over{Z}} e^{-\beta \epsilon_i} . $$

Are both of these statements correct? Dont they contradict one another? I also find it confusing that in the Wikipedia article both system and particle states are labeled by the same symbol.


4 Answers 4


As Seth pointed out, $$P_i=\frac{1}{Z}e^{-\beta E_i}$$ is the probability of a single particle to be in the state $i$. The state $i$ has energy $E_i$. The prefactor $\frac{1}{Z}$ is only a normalization constant for the probability -- it insures that $$\sum_{\textrm{all states $i$}} P_i = 1$$ in the discrete probability case. Just insert the upper formula here and we get \begin{align} 1 &= \sum_{\textrm{all states $i$}} P_i = \sum_{\textrm{all states $i$}} \frac{1}{Z}e^{-\beta E_i} = \frac{1}{Z} \sum_{\textrm{all states $i$}} e^{-\beta E_i} \\ \Rightarrow Z &= \sum_{\textrm{all states $i$}} e^{-\beta E_i} \end{align} Hence, by using the notation $N_i = e^{-\beta E_i}$ and $N = N_{tot} = \sum_{\textrm{all states $i$}} N_i = Z$ we get $$P_i = \frac{N_i}{N}$$ which has a intuitive interpretation as a probability. Note that here $N$ is the number of states and not the number of particles.

Now, if we take $N$ independent particles, then $$P_{tot} = P_{i_1} \cdot P_{i_2} \cdot \ldots \cdot P_{i_N} = \prod_{n=1}^N P_{i_n}$$ is the probability that the first particle is in state $i_1$, the second particle is in state $i_2$, ... If all the particles are in the state $j$, then this formula simplifies to $P_{tot} = (P_j)^N = \frac{1}{Z^N}e^{-\beta N E_j}$.

  • $\begingroup$ How can $E_i$ be probability of single particle to be in the state $i$, when it is explicitly said that it is the probability of the system to be in state $i$ ? $\endgroup$ Jun 20, 2017 at 15:35
  • $\begingroup$ Don't get your comment. Therefore, once again: $E_i$ is the energy of the $i$-th state and $P_i$ is the probability a particle occupies this state. $P_i$ is a function of $E_i$. $\endgroup$
    – Semoi
    Jun 23, 2017 at 14:14
  • $\begingroup$ State of the particle or of the system ? $\endgroup$ Jun 23, 2017 at 19:03

Seth and Semoi said the right thing but I would like to make some clarifications.

The Boltzmann (or Gibbs, or canonical) distribution describes ANY subsystem if it is:

  • quasi-isolated,
  • in the equilibrium with its surroundings.

This subsystem could be a single particle as well as the huge volume containing large amount of particles. For a single particle we usually talk about the ideal gas as in the ideal gas particles are far away from each other and don't interact.

Any quasi-isolated subsystem in the equilibrium has stationary states and so it has an energy spectrum labeled by $E_i$. The probability of system to be found with an energy $E_i$ is distributed according to the canonical distribution

$$p_i = \frac{1}{Z}e^{-\beta E_i}\tag{1}$$

Suppose you have two quasi-isolated subsystems in the equilibrium $A$ and $B$. Each of them has their own energy spectrum $E_i^{(A)}$ and $E_i^{(B)}$ with distributions $p_i^{(A)}$ and $p_i^{(B)}$ both of the form (1). As $A$ and $B$ are quasi-isolated their joint energy spectrum is just the sum $E_i^{(A)}+E_j^{(B)}$ for all $i$ and $j$ and the joint distribution is the product of $p_i^{(A)}$ and $p_i^{(B)}$ which again has the form of (1). From here we can see that it doesn't matter how small the subsystem is while we are able to describe it as quasi-isolated.

Note, that the canonical distribution could also describe the isolated system if it is huge enough that the big part of it is quasi-isolated.


In the formula $p_i=\frac{1}{Z}e^{-\beta E_i}$,$E_i$ is not the total energy of the system. It is the energy of a state $i$ which has a probability of occurrence is $p_i$. So in your case, $p_i=\frac{1}{Z}e^{-\beta \epsilon}$.

  • $\begingroup$ What is "state" ? $\endgroup$ Jun 19, 2017 at 18:46
  • $\begingroup$ state is here same as energy state/level where the particles have same energy. $\endgroup$ Jun 19, 2017 at 18:54

Short answer: Both of the statements can be correct. $E_i$ in Boltzmann distribution can both mean the total system energy and the energy of a single particle in the system.

Long answer: Two common situations in which the Boltzmann distribution occur are:

  • Microcanonical ensemble: The total system energy $E$ is fixed. One commonly tries to find a distribution over the possible particle states, each state having energy $\epsilon_i$, making sure that $\sum_i^N N_i\epsilon_i = E$ where $N_i$ is the number of particles having energy $\epsilon_i$. Note: If we instead were interested in the distribution over microstates, then by the Fundamental postulate of statistical mechanics, the probability of each microstate is equal.

  • Canonical ensemble: The total system energy is not fixed. One commonly tries to find a distribution over the possible microstates, each having a total system energy of $E_i$. The expected value for the total system energy is then $\sum_i^kp_iE_i = \langle E\rangle$ where the sum is over all microstates.


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