- #1
jostpuur
- 2,116
- 19
Suppose we have some model for some system, and that model has given us a sequence [itex]\mathcal{E}_1,\mathcal{E}_2,\mathcal{E}_3,\ldots[/itex], whose values are interpreted as the energy levels of the system. Denoting the energy levels slightly redundantly for future modification soon below, we state that the energy levels are
[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?
[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?