3
$\begingroup$

I am having problems understanding the formulas used for describing the partition functions and the probability distributions for canonical ensembles.

In the first case I have two formulas for the partition function: I can label each system microstate with $j$, associate it with an energy $E_j$, and state that:

$$p_j = e^{-\beta E_j/Z}\quad\text{with}\quad Z=\sum_je^{-\beta E_j/Z}$$

On the other hand, in my lecture notes, the canonical partition function and the probability distribution are given by:

$$\begin{align} Z &= \sum_j\Omega(N,Q)e^{-\beta E_j/Z} \\ P(Q) &=(1/Z)\Omega(N,Q)e^{-\beta Q} \end{align}$$

for a given system of $N$ weakly coupled oscillators and $Q$ quantas.

My question is: Why is the multiplicity $\Omega(N,Q)$ not taken into account in the first case?

$\endgroup$
  • $\begingroup$ Related: physics.stackexchange.com/q/247104 $\endgroup$ – Janus Boffin Apr 3 '16 at 11:11
  • 1
    $\begingroup$ you linked my question @JanusBoffin $\endgroup$ – SkyTalentz Apr 3 '16 at 11:44
  • $\begingroup$ my guess is that in the first case you are summing over states, whilst in the second you are summing over energies. $\endgroup$ – Quantum spaghettification Apr 3 '16 at 12:40
  • $\begingroup$ @Quantumspaghettification According to the formula both sums are for the energies $E_j$ $\endgroup$ – SkyTalentz Apr 3 '16 at 13:04
  • $\begingroup$ @SkyTalentz I have seen the notation you have given where $j$ denotes the state rather then purely an index for energy so $E_j$ would be the energy of the $j$th state and there is nothing stopping $E_j=E_i$ for two states $i$ and $j$. $\endgroup$ – Quantum spaghettification Apr 3 '16 at 13:08
2
$\begingroup$

(I am not sure this is an answer but it is to long to be a comment)

Let us create a simple example of a system of $3$ states, state $1$, state $2$ and state $3$. Let state $1$ and state $2$ both have an energy of $E$ and state $3$ have an energy of $E'\ne E$.

Your first summation is summing over individual states. I.e. it is saying 'let us call the energy of state $1$; $E_1$, the energy of state $2$; $E_2$ and the energy of state $3$; $E_3$. The sum then looks something like this: $$Z=\sum_{j=1}^3e^{-\beta E_j}$$ $$=e^{-\beta E_1}+e^{-\beta E_2}+e^{-\beta E_3}$$ $$=2e^{-\beta E}+e^{-\beta E'}$$

Whilst your second summation is summing over individual energies. I.e. it is saying 'let us call the energy $E$; $E_1$ and the energy $E'$; $E_2$. With $\Omega_1=2$ and $\Omega_2=1$ (i.e. the number of states with each energy). our sum now looks something like this: $$Z=\sum_{j=1}^2 \Omega_j e^{-\beta E_j}$$ $$=2e^{-\beta E}+e^{-\beta E'}$$ I hope this clears it up a bit, let me know if you have any further problems.

$\endgroup$
  • $\begingroup$ Oh ok so these two sum should always give the same partition function for a given system, it's just a question of "perspective"? $\endgroup$ – SkyTalentz Apr 3 '16 at 16:34
  • $\begingroup$ @SkyTalentz yep that's right, it simply depends on how you look at the problem. $\endgroup$ – Quantum spaghettification Apr 3 '16 at 16:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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