- #1

jostpuur

- 2,116

- 19

[itex]

E_1=\mathcal{E}_1

[/itex]

[itex]

E_2=\mathcal{E}_2

[/itex]

[itex]

E_3=\mathcal{E}_3

[/itex]

[itex]\vdots[/itex]

The probabilities defined by the Boltzmann distribution under a temperature [itex]T[/itex] will be

[itex]

p(1) = \frac{1}{Z(T)} e^{-\frac{\mathcal{E}_1}{T}}

[/itex]

[itex]

p(2) = \frac{1}{Z(T)} e^{-\frac{\mathcal{E}_2}{T}}

[/itex]

[itex]

p(3) = \frac{1}{Z(T)} e^{-\frac{\mathcal{E}_3}{T}}

[/itex]

[itex]

\vdots

[/itex]

where the partition function is

[itex]

Z(T) = e^{-\frac{\mathcal{E}_1}{T}} + e^{-\frac{\mathcal{E}_2}{T}} + e^{-\frac{\mathcal{E}_3}{T}}+ \cdots

[/itex]

Suppose we find out that the model was only an approximation of a more accurate model, and according to the new more accurate model the energy values are going to be [itex]\mathcal{E}_n[/itex] and [itex]\mathcal{E}_{2n}+\epsilon[/itex] with some small positive epsilon. Now the energy levels are

[itex]

E_1 = \mathcal{E}_1

[/itex]

[itex]

E_2 = \mathcal{E}_2

[/itex]

[itex]

E_3 = \mathcal{E}_2+ \epsilon

[/itex]

[itex]

E_4 = \mathcal{E}_3

[/itex]

[itex]

E_5 = \mathcal{E}_4

[/itex]

[itex]

E_6 = \mathcal{E}_4 + \epsilon

[/itex]

[itex]

E_7 = \mathcal{E}_5

[/itex]

[itex]

\vdots

[/itex]

Now the probabilities defined by

[itex]

p(n) = \frac{1}{Z(T)}e^{-\frac{E_n}{T}}

[/itex]

turn out to be

[itex]

p(1) = \frac{1}{Z(T)}e^{-\frac{\mathcal{E}_1}{T}}

[/itex]

[itex]

p(2) = \frac{1}{Z(T)}e^{-\frac{\mathcal{E}_2}{T}}

[/itex]

[itex]

p(3) = \frac{1}{Z(T)}e^{-\frac{\mathcal{E}_2+\epsilon}{T}}

[/itex]

[itex]

p(4) = \frac{1}{Z(T)}e^{-\frac{\mathcal{E}_3}{T}}

[/itex]

[itex]

p(5) = \frac{1}{Z(T)}e^{-\frac{\mathcal{E}_4}{T}}

[/itex]

[itex]

p(6) = \frac{1}{Z(T)}e^{-\frac{\mathcal{E}_4+\epsilon}{T}}

[/itex]

[itex]

p(7) = \frac{1}{Z(T)}e^{-\frac{\mathcal{E}_5}{T}}

[/itex]

[itex]

\vdots

[/itex]

where the partition function is

[itex]

Z(T) = e^{-\frac{\mathcal{E}_1}{T}} + e^{-\frac{\mathcal{E}_2}{T}} + e^{-\frac{\mathcal{E}_2 + \epsilon}{T}} + e^{-\frac{\mathcal{E}_3}{T}} + e^{-\frac{\mathcal{E}_4}{T}} + e^{-\frac{\mathcal{E}_4 + \epsilon}{T}} + e^{-\frac{\mathcal{E}_5}{T}} + \cdots

[/itex]

Suppose we decide that the epsilon is so small that it has not much significance, and we might as well simplify the formulas by taking the limit [itex]\epsilon\to 0[/itex]. This limit is going to give us a new probability distribution

[itex]

p(1) = \frac{1}{Z(T)}e^{-\frac{\mathcal{E}_1}{T}}

[/itex]

[itex]

p(2) = \frac{2}{Z(T)}e^{-\frac{\mathcal{E}_2}{T}}

[/itex]

[itex]

p(3) = \frac{1}{Z(T)}e^{-\frac{\mathcal{E}_3}{T}}

[/itex]

[itex]

p(4) = \frac{2}{Z(T)}e^{-\frac{\mathcal{E}_4}{T}}

[/itex]

[itex]

p(5) = \frac{1}{Z(T)}e^{-\frac{\mathcal{E}_5}{T}}

[/itex]

[itex]

\vdots

[/itex]

where the partition function is

[itex]

Z(T) = e^{-\frac{\mathcal{E}_1}{T}} + 2e^{-\frac{\mathcal{E}_2}{T}} + e^{-\frac{\mathcal{E}_3}{T}} + 2e^{-\frac{\mathcal{E}_4}{T}} + e^{-\frac{\mathcal{E}_5}{T}} + \cdots

[/itex]

Now we have two different probability distributions for the case [itex]\epsilon = 0[/itex]. Is one of them right, and the other one wrong? Which way around would be the right answer?