I wanted to estimate the probability of finding a molecule in the ground vibrational level using the Boltzmann distribution:

$$ p_i = \frac{e^{-\epsilon_i/kT}}{\sum_{i=0}^{N}e^{-\epsilon_i/kT}} $$

Using the quantum harmonic oscillator as a model for the energy

$$ \epsilon_i = h\nu (i+1/2) =/i=0/=\frac{h\nu}{2} $$

In the Boltzmann distribution, we have the state of interest divided by the sum of all possible states. But how should I treat the denominator?

Searching a bit I found that the analytical expression for this geometric series is ($i$ not imaginary number)

$$ \sum_{i=0}^{N}e^{-i h\nu/kT} = \frac{1}{1 - e^{h\nu/kT}} $$

However, is this using a shifted energy scale for the harmonic potential? In that the vibrational energies are $0$, $h\nu$, $2h\nu$, ..., and not $\frac{1}{2}h\nu$, $\frac{3}{2}h\nu$, $\frac{5}{2}h\nu$, ...? Should I make sure I use the same energy scale for the nominator and denominator in the Boltzmann distribution?

Doing what porphyrin suggested, I get

$$ \sum_{i=0}^{\infty} e^{-h\nu(i+\frac{1}{2})/kT} = e^{-h\nu/kT} \sum_{i=0}^{\infty} e^{- ih\nu/kT} $$

Expanding the four first terms

$$ e^{-h\nu/kT} \sum_{i=0}^{\infty} e^{- ih\nu/kT} = (e^{-h\nu/kT} \cdot 1) + (e^{-h\nu/kT} \cdot e^{-h\nu/kT}) + (e^{-h\nu/kT} \cdot e^{-2h\nu/kT}) + (e^{-h\nu/kT} \cdot e^{-3h\nu/kT}) \\ = e^{-h\nu/kT} + e^{-2h\nu/kT} + e^{-3h\nu/kT} + e^{-4h\nu/kT} = \sum_{1}^{\infty}e^{-n\cdot h\nu/kT} $$

which has an analytical expression for the converged value, right?

  • $\begingroup$ The denominator should be calculated exactly. Ever heard of a geometric progression and its sum? $\endgroup$ Commented Oct 17, 2016 at 12:30
  • $\begingroup$ Ah, yes. Looking at it, it should converge. $\endgroup$
    – Yoda
    Commented Oct 17, 2016 at 12:32
  • 1
    $\begingroup$ The summation is called the partition function and it always extends over all possible levels $n=0 ..,\infty$. You are assuming a harmonic oscillator which has an infinite number of levels. Its easier if the 1/2 is separated out first and treated separately, this just adds a constant to the energy as you realise. Expand the summation as $1+e^{-a}+e^{-2a}+....$ ($a=h\nu /(kT)$ which converges to the result you quote. $\endgroup$
    – porphyrin
    Commented Oct 17, 2016 at 12:54

1 Answer 1


You're on the right track.

Also, using $i$ as an index can be confusing some times because it can be confused with the imaginary number; however, here it should not present a problem. As a matter of habbit however, I like to use $j$ or $n$ or something else..there are only so many letters in the alphabet.

The sum in the denominator is called the partition function, and has the form

$$Z = \sum_{j}e^{-\frac{\epsilon_j}{kT}}$$

For the harmonic, oscillator $\epsilon_j = (\frac{1}{2}+j)\hbar \omega$ for $j \in \{ 0,1,2.. \}$

Note that $\epsilon_0 \neq 0$ there exists a zero point energy.

Let's write out a few terms $$Z = e^{-\frac{\hbar \omega/2}{kT}} + e^{-\frac{\hbar \omega3/2}{kT}} + e^{-\frac{\hbar \omega5/2}{kT}} +..... $$

factoring out $e^{-\frac{\hbar \omega/2}{kT}}$

$$Z = e^{-\frac{\hbar \omega/2}{kT}} \left( 1+ e^{-\frac{\hbar \omega}{kT}} + e^{-\frac{2\hbar \omega}{kT}} +.....\right) $$

The sum in the bracket takes the form of a geometric series whose sum converges as shown below $$1+x+x^2+... = \frac{1}{(1-x)} $$ herein, $ x \equiv e^{-\frac{\hbar\omega}{kT}} $

Putting all of this together

$$Z = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}$$

Now, $$p_0 = \frac{e^{-\epsilon_0/kT}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{Z} = \frac{e^{-\frac{\hbar \omega/2}{kT}}}{\frac{e^{-\frac{\hbar \omega/2}{kT}}}{(1-e^{-\frac{\hbar\omega}{kT}})}} = (1-e^{-\frac{\hbar\omega}{kT}}) $$

  • 1
    $\begingroup$ Great answer. But I strongly disagree with not using i as an index variable. The possible confusion with the imaginary unit only arises from typographic sloppiness. Variables should be italic and mathematical constants should be upright set. This includes e.g. the euler number, but no one would think that you exponented the variable e. From context it is clear that you exponented the euler number. The same holds true for sum symbols, it would make no sense to write the imaginary unit underneath them and even with typographic sloppiness it is clear from context that you mean the variable i. $\endgroup$
    – mcocdawc
    Commented Oct 18, 2016 at 13:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.