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In the Canonical Ensemble, given a quantum system with $N$ distingishable and non-interacting particles distributed amongst $r$ energy levels of energy $\epsilon _1,\epsilon _2,\epsilon _3,...,\epsilon _r$ and degeneracy $g_1,g_2,g_3,...,g_r$, the partition function of the single-particle is defined as

$$Z_{SP}=\sum_{i=1}^r g_ie^{-\epsilon_i\beta(T)}\tag{1}$$

with $\beta(T)=\frac{1}{K_BT}$, and the partition function of the whole system of $N$ particles is defined as

$$Z_{N}=\prod_{i=1} ^N (Z_{SP})_i =(Z_{SP})^N \tag{2}$$

(with the last equality holding if all particles are identical and indistinguishable and, if not, $Z_N=Z_{SP}^N/N!$).

  • Where does this definition, (2), come from? Why a product and not, let's say, a sum?

  • On the other hand, would then be correct to define $Z_N$ also in this way (3)?

$$Z_N=\sum_{j=1}^Sg_je^{-Ej\beta(T)}\tag{3}$$

  With $S$ the number of microstates of the whole system and $E_j$ the energy of the whole system at the microstate $j$.

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  • $\begingroup$ Not sure I understand the question - are you familiar with the idea of the partition function and only find this application to the specific system unclear, or do you need help with the basic concept of what is the partition function and why do we use it? $\endgroup$ – user245141 Dec 30 '19 at 13:47
  • $\begingroup$ I'm familiar with the partition function, but I'm a little confused with the definition of the partition function of the system, $Z_N$ as the product of the partition function of every single particle, $Z_{SP}$. Why a product and not, let's say, a sum? $\endgroup$ – Quaerendo Jan 3 at 9:42
  • $\begingroup$ started answering in a comment but it got long so posted an answer below $\endgroup$ – user245141 Jan 3 at 9:59
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When considering a partition function of a system composed of several distinguishable subsystems you never add the separate partition functions up, and always multiply them.

The reason is that the partition function covers the possible states of a system, and when for a system composed of subsystem we can set subsystem $A$ to a certain state and then we have to cover all of the states of subsystem $B$. Then change the state of subsystem $A$ and again sum over all states of subsystem $B$. This is multiplication $$ Z_{AB} = \sum_{A,B} e^{-\beta(E_A+E_B)} = \sum_{A} e^{-\beta E_A} \sum_B e^{-\beta E_B} = Z_A Z_B$$ and the generalization to more than two subsystems is immediate.

Note that for this to be valid the subsystems must be separate and distinguishable. If they are interacting, for example, then you might have $E_{AB} \neq E_A + E_B$. If the particles are identical and indistinguishable, then they cannot be separated into subsystems $A$ and $B$ to begin with.

By the way - this rule of multiplication of partition functions is valid for classical systems as well as quantum ones.

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  • $\begingroup$ Ok, I see. With regard to my second question (I was editing while you were answering, didn't notice it sorry), would definitions (1) and (3) be correct? Would They be linked by (2) as I wrote down them? I have gone across a lot of founts and I have found definitions slightly different, I'm already a little confused... $\endgroup$ – Quaerendo Jan 3 at 10:37
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    $\begingroup$ Yes. Your eq. (3) is the definition of the partition function, regardless of other details. The partition function is the sum over all states, with each state being weighted by $e^{-\beta E}$ where $E$ its energy. The question is how does the partition function of the entire system relate to the partition function of the subsystems that it is composed of. If each of the subsystems is described by $Z_{SP}$ of eq. (1), and the subsystems are independent and distinguishable, then the total $Z$ is the multiplication of them, given in your eq. (2). $\endgroup$ – user245141 Jan 3 at 10:43
  • $\begingroup$ And would it be possible to derive $Z_N$ from $Z_{SP}$ and the relation $<E_{SP}>=-\frac{\partial \ln Z_{SP}}{\partial\beta}$ this way: $<E_{N}>=-\frac{\partial \ln Z_{N}}{\partial\beta}=N<E_{SP}>=-N\frac{\partial \ln Z_{SP}}{\partial\beta}$, so $\frac{\partial \ln Z_{N}}{\partial\beta}=\frac{\partial \ln Z_{SP}^N}{\partial\beta}$, and therefore $Z_N=Z_{SP}^Ne^{\eta-N\xi}$, where $\eta$ and $\xi$ are some integrating constants depending on the rest of variables? How can they be determined? $\endgroup$ – Quaerendo Jan 8 at 12:29
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Let us take the two-particle case explicitly: for one particle to have energy $\epsilon_1$ and the other to have energy $\epsilon_2$, if they don't interact then the energy is $\epsilon_{(1,2)}=\epsilon_1+\epsilon_2$, further if they are distinguishable particles then the degeneracy is $g_{(1,2)}=g_1 g_2$. Both of these terms enter multiplicatively into the partition function and when you work it all out, the partition function (using your single-particle expression to sum over every pair $(i,j)$ of states) for Hamiltonian $H_A\otimes I+I\otimes H_B$ is $Z_A Z_B$ by the distributive rule of multiplication.

If the particles are indistinguishable, then this definition of the partition function, while it is still correct, turns out to be quite limited. It describes an ensemble with a fixed number of particles, but once they can be created or destroyed at will, it is useful to perform the same trick that we did with energy to get to the partition function in the first place: we connected it to a reservoir of energy and allowed energy to flow between the two systems. So when we have indistinguishable particles, the math is much easier if we connect the system to a larger system that is a reservoir of those particles, to allow particles to flow between the two systems. This is called the grand canonical ensemble, and it has a grand canonical partition function: now $i$ indexes a single particle state and it is helpful to renumber the states so that we give different numbers to each state with its different energy. Those states are occupied by $n_i$ particles and then the grand canonical partition function is $$\mathcal Z = \sum_{i, n_i} \exp\big({-\beta (\epsilon_i(n_i) - \mu n_i)}\big).$$ Then if the particles are not interacting $\epsilon_i(n_i)=\epsilon_i(1)~n_i =\epsilon_i^\text{sp}~n_i$ and we can do the sum over $n_i$ either directly for fermions $1+\exp\left({-\beta(\epsilon_i^\text{sp}-\mu)}\right)$ or with a geometric series for bosons, $$1\over 1-\exp\left({-\beta(\epsilon_i^\text{sp}-\mu)}\right)$$

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    $\begingroup$ I would like a more specific answer to the OP question, something like that: for the canonical ensemble the product formula $Z_{N}=(Z_{SP})^N$ (over particles) does not exist, instead we have product formula for the gand canonical ensemble (over energy states), right? $\endgroup$ – Aleksey Druggist Jan 2 at 13:08
  • $\begingroup$ I do not personally feel comfortable saying that the math is impossible for a canonical ensemble of indistinguishable particles. For example if there were only two energy levels possible for $N$ non-interacting bosons then you have a sum of $$Z= \sum_{k=0}^N e^{-\beta \epsilon k} = {1 - e^{-\beta \epsilon (N+1)}\over1 - e^{-\beta \epsilon}}$$or so...so the math does not seem categorically impossible; I think it just rapidly grows in complexity as you try to enforce $k_i\ge 0;~~\sum_i k_i = N.$ $\endgroup$ – CR Drost Jan 2 at 14:04
  • $\begingroup$ And the formula that you cited does exist for distinguishable non-interacting particles, just not for indistinguishable or interacting ones. $\endgroup$ – CR Drost Jan 2 at 14:07
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    $\begingroup$ Oh I see, OP asked about distinguishable non-interacting particles...but then it's better in answer to substantiate this product formula for distinguishable particles $\endgroup$ – Aleksey Druggist Jan 2 at 14:26

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