I have n distinguishable particles and m distinguishable boxes. If all particles are in the same box the system has an energy of -$\epsilon$ in all the other cases the energy is 0.

Now I want to compute the canonical partition function. The Helmhltoz energy and the entropy of the ground state.

To start with I would count the number of possibilities to put all particles in single boxes.

There are $n!$ ways to arrange n distinguishable particles. And I have m boxes. And last I have $m!$ ways to arrange the different boxes. Therefore I have $n!*m*m!$ possibilities. The next step is to compute all possibilities to distribute n distinguishable particles in m distinguishable boxes. To distribute n particles in m boxes I have $m^{n}$ possibilities for a single configuration. Then I can arrange each configuration in $n!$ ways and last I can arrange the boxes again in $m!$ different ways. Hence the number of remaining possibilities is

$$ m^{n}*n!*m!-n!*m*m!. $$

This gives the partition function:

$$ Z(m,n,T)=m^{n}*n!*m!-n!*m*m!+n!*m*m!*e^{\beta*\epsilon}=\\ m!*n!*m*(m^{n-1}-1+e^{\beta\epsilon}) $$

Since constant factors in the partition function do not matter when computing averages since they cancel I can neglect the prefactor m!*n!*m The partition function is therefore $$ Z(m,n,T)=m^{n-1}-1+e^{\beta\epsilon} $$

The Hemholtz free energy is then: $$ F=-k_bT*log(m^{n-1}-1+e^{\beta\epsilon}) $$

Now I am not sure if the counting I have done above is correct and the next thing is I am not completely sure how to compute the ground state entropy.

The entropy is defined as $S=-k_{b}\sum_{i}p_{i}log(p_i)$ and for as single state it should be $S=-k_{b}*p*log(p)$. The probability for the ground state would then be:

$$ p_{G}=\frac{e^{\beta\epsilon}}{m^{n-1}-1+e^{\beta\epsilon}} $$ and hence the entropy $$ S=-\frac{k_b*e^{\beta\epsilon}}{m^{n-1}-1+e^{\beta\epsilon}}*log\left(\frac{e^{\beta\epsilon}}{m^{n-1}-1+e^{\beta\epsilon}}\right) $$

But still I am not quite sure if this is correct especially with the counting I feel very insecure. Thanks for your help

  • $\begingroup$ I am not sure, but here is what I think could be the right way. I am not sure why you are bothered with arranging the boxes and particles after you have considered m^n possibilities. I think that this itself includes all possible arrangements of the particles and boxes. Rearranging the order in which you consider the boxes or particles should not create more microstates. Hence I think the partition function should be (m^(n) - m) + m * e^(beta * E) $\endgroup$ – Hari Sep 28 '18 at 15:30
  • $\begingroup$ @Hari. Okay but what is the criterion for creating a new microstate I think this is the point I don't quite get. So what is the criterion to be a microstate? Because I thought since the particles are distinguishable and so are the boxes any arrangement of particles and boxes would create a new microstate. Is this wrong $\endgroup$ – zodiac Sep 29 '18 at 10:09
  • $\begingroup$ Suppose there are 2 boxes (B1,B2) and 2 particles (P1, P2). The microstates will be : 1) B1 (P1,P2) B2 () ; 2) B1 () B2(P1,P2) ; 3) B1 (P1) B2(P2) ; 4) B1(P2) B2(P1). I think these are the only microstates, rearranging the order in which you write B1 and B2, in this case 5) B2(P1) B1(P2) or 6) B2(P2) B1(P1) are not microstates as they are physically the same as (4) and (3) respectively. $\endgroup$ – Hari Sep 29 '18 at 18:33
  • $\begingroup$ Thank but I fear I still don't get it right I guess. I mean what you are saying sounds completely reasonable. But if 5) and 6) are the same as 4) and 3) does't this mean those states have a higher entropy since they have more realizations. $\endgroup$ – zodiac Oct 1 '18 at 6:44
  • $\begingroup$ And also I can place the labeled particles in different orders to the boxes since they are distinguishable. Then for example your state 1) B1 (P1,P2) B2 () could also be written as 1')B1(P2,P1) , B2(). I mean for sure those states are physically indistinguishable but dont' they have a higher a priori probability since there are more possibilities to create those states. I am sorry for bothering you again. But this seems to be a very crucial point that I don't get. I think this is also shown on this wiki page. link: en.wikipedia.org/wiki/Microstate_(statistical_mechanics). $\endgroup$ – zodiac Oct 1 '18 at 6:44

Based on Harrys comments I would like to answer this question to close this thread. It does not matter in which order I put the labeled particles in to the boxes since there are no drawers in the boxes and hence in the end I am not able to differentiate between the order of the particles anymore. This means $B(1,2,3)$ is the same as B(2,1,3) because the balls are lose in this box and are able to roll around freely. The next thing is the boxes have labels this states that box one will always be box one no matter where it is put to on the shelf. Hence there is no need to take into account the permutations of the boxes. Therefore the canonical partition function is just

$$ Z(n,m,T)=(m^{n}-m)+me^{\beta\epsilon} $$

The Helmholtz free energy $$ F(n,m,T)=-k_{B}T=log\left( m^{n} - m + e^{\beta\epsilon} \right) $$

And the ground state probability and the entropy stay the same since one m factors out.


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