Editing Boltzmann distribution (section)

==The distribution==

The Boltzmann distribution is a [[probability distribution]] that gives the probability of a certain state as a function of that state's energy and  temperature of the [[system]] to which the distribution is applied.<ref name="McQuarrie, A. 2000">{{cite book |last=McQuarrie |first=A. |year=2000 |title=Statistical Mechanics |publisher=University Science Books |location=Sausalito, CA |isbn=1-891389-15-7 }}</ref> It is given as
<math display="block">
p_i=\frac{1}{Q} \exp\left(- \frac{\varepsilon_i}{kT} \right)  = \frac{ \exp\left(- \tfrac{\varepsilon_i}{kT} \right) }{ \displaystyle \sum_{j=1}^{M} \exp\left(- \tfrac{\varepsilon_j}{kT} \right) }
</math>

where:
*{{math|exp()}} is the [[exponential function]],
*{{mvar|p<sub>i</sub>}} is the probability of state {{mvar|i}},
*{{mvar|ε<sub>i</sub>}} is the energy of state {{mvar|i}}, 
*{{mvar|k}} is the [[Boltzmann constant]], 
*{{mvar|T}} is the [[absolute temperature]] of the system,
*{{mvar|M}} is the number of all states accessible to the system of interest,<ref name="McQuarrie, A. 2000"/><ref name="Atkins, P. W. 2010"/> 
*{{mvar|Q}} (denoted by some authors by {{mvar|Z}}) is the normalization denominator, which is the [[canonical partition function]]<math display=block>
Q = \sum_{j=1}^{M} \exp\left(- \tfrac{\varepsilon_j}{kT} \right)
</math> It results from the constraint that the probabilities of all accessible states must add up to 1.

Using [[Lagrange multipliers]], one can prove that the Boltzmann distribution is the distribution that maximizes the [[entropy]]
<math display=block>S(p_1,p_2,\cdots,p_M) = -\sum_{i=1}^{M} p_i\log_2 p_i</math>

subject to the normalization constraint that <math display="inline">\sum p_i=1</math> and the constraint that <math display="inline">\sum {p_i {\varepsilon}_i}</math> equals a particular mean energy value, except for two special cases.  (These special cases occur when the mean value is either the minimum or maximum of the energies {{mvar|ε<sub>i</sub>}}.  In these cases, the entropy maximizing distribution is a limit of Boltzmann distributions where {{mvar|T}} approaches zero from above or below, respectively.)

The partition function can be calculated if we know the energies of the states accessible to the system of interest. For atoms the partition function values can be found in the [[National Institute of Standards and Technology|NIST]] Atomic Spectra Database.<ref>[http://physics.nist.gov/PhysRefData/ASD/levels_form.html NIST Atomic Spectra Database Levels Form] at nist.gov</ref>

The distribution shows that states with lower energy will always have a higher probability of being occupied than the states with higher energy. It can also give us the quantitative relationship between the probabilities of the two states being occupied. The ratio of probabilities for states {{mvar|i}} and {{mvar|j}} is given as
<math display=block>\frac{p_i}{p_j} = \exp\left( \frac{\varepsilon_j - \varepsilon_i}{kT} \right)</math>

where:
*{{mvar|p<sub>i</sub>}} is the probability of state {{mvar|i}}, 
*{{mvar|p<sub>j</sub>}} the probability of state {{mvar|j}}, 
*{{mvar|ε<sub>i</sub>}} is the energy of state {{mvar|i}},
*{{mvar|ε<sub>j</sub>}} is the energy of state {{mvar|j}}.

The corresponding ratio of populations of energy levels must also take their [[Degeneracy (quantum mechanics)|degeneracies]] into account.

The Boltzmann distribution is often used to describe the distribution of particles, such as atoms or molecules, over bound states accessible to them. If we have a system consisting of many particles, the probability of a particle being in state {{mvar|i}} is practically the probability that, if we pick a random particle from that system and check what state it is in, we will find it is in state {{mvar|i}}. This probability is equal to the number of particles in state {{mvar|i}} divided by the total number of particles in the system, that is the fraction of particles that occupy state {{mvar|i}}.

:<math>p_i = \frac{N_i}{N}</math>

where {{mvar|N<sub>i</sub>}} is the number of particles in state {{mvar|i}} and {{mvar|N}} is the total number of particles in the system. We may use the Boltzmann distribution to find this probability that is, as we have seen, equal to the fraction of particles that are in state i. So the equation that gives the fraction of particles in state {{mvar|i}} as a function of the energy of that state is  <ref name="Atkins, P. W. 2010"/>
<math display=block>
\frac{N_i}{N} = \frac{ \exp\left(- \frac{\varepsilon_i}{kT} \right) }{ \displaystyle \sum_{j=1}^{M} \exp\left(- \tfrac{\varepsilon_j}{kT} \right) }
</math>

This equation is of great importance to [[spectroscopy]]. In spectroscopy we observe a [[spectral line]] of atoms or molecules undergoing transitions from one state to another.<ref name="Atkins, P. W. 2010"/><ref>{{cite book |last1=Atkins |first1=P. W. |last2=de Paula |first2=J. |year=2009 |title=Physical Chemistry |edition=9th |publisher=Oxford University Press |location=Oxford |isbn=978-0-19-954337-3 }}</ref> In order for this to be possible, there must be some particles in the first state to undergo the transition. We may find that this condition is fulfilled by finding the fraction of particles in the first state. If it is negligible, the transition is very likely not observed at the temperature for which the calculation was done. In general, a larger fraction of molecules in the first state means a higher number of transitions to the second state.<ref>{{cite book |last1=Skoog |first1=D. A. |last2=Holler |first2=F. J. |last3=Crouch |first3=S. R. |year=2006 |title=Principles of Instrumental Analysis |publisher=Brooks/Cole |location=Boston, MA |isbn=978-0-495-12570-9 }}</ref> This gives a stronger spectral line. However, there are other factors that influence the intensity of a spectral line, such as whether it is caused by an allowed or a [[forbidden transition]].

The [[softmax function]] commonly used in machine learning is related to the Boltzmann distribution:

:<math>
(p_1, \ldots, p_M) = \operatorname{softmax} \left[- \frac{\varepsilon_1}{kT}, \ldots, - \frac{\varepsilon_M}{kT} \right]
</math>