Solutions ¹ of the Schrödinger equation yield energy levels. You define the system, which ultimately defines the potential energies and the boundary conditions, and the Schrödinger scheme will generate (usually approximately) the relevant energy levels.

Now, given an aasembly of these systems, how many systems occupy each of these energy levels depends on the temperature and external conditions, and is governed by the Maxwell Boltzmann distribution. A derivation of that distribution without using the Lagrange method of undetermined multipliers is re-presented.

Knowing the Energy Levels of a system, i.e., $\{\epsilon_i\}$ and given a set of occupation numbers for these energy levels, $\{n_i\}$, where the two subscripts are linked together, i.e., the occupation number tells the number in the energy level with the same value of `i', we seek the maximum in the $\Omega$, or the minimum in the associated Helmholtz free energy, A.

First and foremost for this simplified derivation, we assume that the energy levels are non-degenerate².

Next, we know that

$$\Omega = \frac{N!}{\prod_i n_i!}$$ (1)

where, the standard constraint equations hold:

$$N = \Sigma_i n_i!$$ (2)

and

$$E = \Sigma_i n_i \epsilon_i$$ (3)

and, of course,

$$S = k \ln \Omega$$ (4)

and

 

A = E - TS (5)

From Equation 5 we have

$$A = \Sigma_i n_i \epsilon_i - k T \ln \frac{N!}{\prod_i n_i!}$$ (6)

(where we have used Equation 3 and Equation 4) which, for convenience, we rearrange to form

$$A = - kT \ln \prod_i e^{-n_i\epsilon_i/kT} - kT \ln \frac{N!}{\prod_i n_i!}$$ (7)

which is, combining logarithms,

$$A = - kT \ln \left ( N! \prod_i \frac{\left (e^{(-\epsilon_i/kT}\right )^{n_i}}{n_i!} \right )$$

For that state whose occupation numbers minimize A, i.e. $\{n^\ast_i\}$, we would have

$$A = kT \ln \left ( N! \frac{ \left ( e^{-\epsilon_0/kT}\rig... ...-\epsilon_i/kT}\right )^{n^\ast_i}}{n^\ast_i!} \cdots \right )$$ (8)

Choose any two of these `optimally' occupied levels, say 3 and 6, and, holding all other occupation numbers fixed, raise one and lower the other by exactly one particle (system). Then, one would have

$$A' = kT \ln \left ( N! \frac{ \left ( e^{-\epsilon_0/kT}\ri... ...-\epsilon_i/kT}\right )^{n^\ast_i}}{n^\ast_i!} \cdots \right )$$ (9)

where we have incremented the third (and decremented the sixth) occupation number ( $n^\ast_6 \rightarrow n^\ast_6-1$ and $n^\ast_3 \rightarrow n^\ast_3+1$ ). If A is a minimum, then A' (Equation 9 ) should be smaller than or equal to A (Equation 8) for any choice of incrementing any level's occupation number and decrementing another. (We will use this fact again in a few lines, vide infra.)

Therefore $A'\geq A$, i.e.,

$$\ln \left \{ \frac{ \left ( e^{-\epsilon_3/kT}\right )^{n^\as... ... e^{-\epsilon_6/kT}\right )^{n^\ast_6}}{(n^\ast_6)!} \right \}$$ (10)

or

$$\left \{ \frac{ \left ( e^{-\epsilon_3/kT}\right )^{n^\ast_3+... ...n_6/kT}\right )^{n^\ast_6}}{(n^\ast_6)(n^\ast_6-1)!} \right \}$$ (11)

or (expanding $(n^\ast_i+1)! \rightarrow (n^\ast_i+1) (n^\ast_i) !$), we have

$$\left \{ \frac{ \left ( e^{-\epsilon_3/kT}\right )^{+1}}{(n^\... ...t )^{-1} \right \} \geq \left \{ \frac{1}{n^\ast_6} \right \}$$ (12)

i.e.,

$$\left \{ \frac{ \left ( e^{-\epsilon_3/kT}\right )^{+1}}{(n^... ...)} { \left ( e^{-\epsilon_6/kT}\right )^{+1}} \right \} \geq 1$$ (13)

We repeat the above argument, this time incrementing level six's occupation number, and decrementing level three's ( $n^\ast_6 \rightarrow n^\ast_6+1$ and $n^\ast_3 \rightarrow n^\ast_3-1$ ). If A is a minimum, then A'

$$A'' = kT \ln \left ( N! \frac{ \left ( e^{-\epsilon_0/kT}\r... ...-\epsilon_i/kT}\right )^{n^\ast_i}}{n^\ast_i!} \cdots \right )$$ (14)

i.e.,

$$\ln \left \{ \frac{ \left ( e^{-\epsilon_3/kT}\right )^{n^\as... ... e^{-\epsilon_6/kT}\right )^{n^\ast_6}}{(n^\ast_6)!} \right \}$$ (15)

or, expanding and cancelling

$$\ln \left \{ \frac{ \left ( e^{-\epsilon_3/kT}\right )^{-1}}{... ...{ 1}{(n^\ast_3)(n^\ast_3-1)!} \frac{1 }{(n^\ast_6)!} \right \}$$ (16)

$$\left \{ \frac{ \left ( e^{-\epsilon_6/kT}\right )}{(n^\ast_6... ...ast_3)} { \left ( e^{-\epsilon_3/kT}\right )} \right \} \geq 1$$ (17)

$$\lim_{n^\ast_6 \rightarrow large} \left \{ \left ( \frac{1}{... ...rac {n^\ast_3} { e^{-\epsilon_3/kT}} \right ) \right \} \geq 1$$ (18)

$$\lim_{n^\ast_3 \rightarrow large} \left \{ \left ( \frac{1}{1... ...rac {n^\ast_6} { e^{-\epsilon_6/kT}} \right ) \right \} \geq 1$$ (19)

or, inverting,

$$\left \{ \left ( \frac {n^\ast_3} {e^ {-\epsilon_3/kT}} \ri... ...rac { e^{-\epsilon_6/kT}} {n^\ast_6} \right ) \right \} \leq 1$$ (20)

which seems to contradict Equation 18.

We have, combining the two (Equations 18 and 20),

$$1 \leq \left \{ \left ( \frac{ e^{-\epsilon_6/kT}}{n^\ast_6... ...rac {n^\ast_3} { e^{-\epsilon_3/kT}} \right ) \right \} \geq 1$$ (21)

which can only be true if the interior argument is, itself, one (1). This can be accomplished by defining one of these ratios as `a' and the other as `1/a'. Then

$$1 \leq \left \{ \left ( a\right ) \left ( \frac {1} {a} \right ) \right \} \geq 1$$ (22)

which will be true since a/a is, a priori, unity.

This would mean that

$$n^\ast_i = a e^{-\epsilon_i/kT}\,\forall i$$ (23)

and, since, if we add up the occupation numbers of all levels,

$$\sum_{all\,levels}n^\ast_i = \sum_i a e^{-\epsilon_i/kT} = a \sum_i e^{-\epsilon_i/kT} \equiv a Z \equiv N$$ (24)

where N is the total number of particles. Therefore, a=N/Z. Having defined Z this way, we finally obtain

$$n^\ast_i = \frac{N}{Z} e^{-\epsilon_i/kT}$$ (25)

which completes our derivation.