Say there are $N$ atoms of type $A$ in a box of volume $V$ and they are undergoing a reversible association-dissociation reaction $A + A = A_2$. Let an $A$ atom have mass $m$, and hence the molecule $A_2$ has mass $2m$. Assume that the temperature $T$ is high-enough that Boltzmann statistics applies.

Say $N_1$ be the number of $A$ atoms and say $N_2$ be the number of $A_2$ molecules at equilibrium. Then $N_1 + 2N_2 = N$.

If there had to be $N$ $A$-type atoms in the the box then I guess the partition function $Z$ would be given by,

$$Z = \frac{1}{N!} [ \int e^{-\frac{\beta p^2}{2m}} \frac{d^3p d^3q}{h^3} ] ^N = \frac{1}{N!} [ \frac{V}{h^3} (2\pi m kT)^{\frac{3}{2}} ] ^N .$$

Say that the dissociation energy of the reaction is $E_0$. Say $N_1$ be the number of $A$ atoms and say $N_2$ be the number of $A_2$ molecules at equilibrium. Then $N_1 + 2N_2 = N$.

I want to write the partition function for the above mixed system in equilibrium.

Using the expression for $Z$ as stated above, I am guessing that this partition function (say $Z_N$) would look like,

$$Z_N = \sum _ {N_1 + 2N_2 = N} e^{-\beta N_2 E_0} \frac{1}{N_1 !} [ \frac{V}{h^3} (2\pi m kT)^{\frac{3}{2}} ] ^{N_1} \frac{1}{N_2 !} [ \frac{V}{h^3} (4\pi m kT)^{\frac{3}{2}} ] ^{N_2}. $$

(In the above I am using the idea that for $N_2$ molecules to have formed $N_2$ $E_0$ (binding) energy would have been released and hence that much of extra energy has to be accounted for which is not captured in the partition functions over the $N_1$ atoms and $N_2$ molecules...)

  • Is the above expression for the partition function correct ?

One further introduces the grand canonical partition function (say $Z_{\mu}$ ) for the above system with a chemical potential $\mu$ to "take care of the constraint $N_1 + 2N_2 = N$",

$$Z_{\mu} = \sum _ {N=0} ^\infty e^{\beta \mu N} Z_N.$$

  • Firstly I don't understand the meaning of the $\mu$ in the above. (I can't relate it to what I would usually think as chemical potential) I mean..isn't the above expression for $Z_{\mu}$ independent of $N_1$ and $N_2$?

  • How does the $Z_{\mu}$ help find the expectation value for $N_1$ and $N_2$?

  • If my expressions for $Z_N$ is correct then does $Z_{\mu}$ have a closed form expression?

All this is to understand the claim that as expectation values the following relationship apparently holds,

$$\frac{N_1 ^2} {N_2} \sim e ^{-\frac{|E_0|}{kT}} .$$

I would be glad to know of other insights along the above.

  • $\begingroup$ Dear @Anirbit. I took the liberty to rename $Z_{N,\mu}$ in your question (v1) as just $Z_{\mu}$, since it does not depend on $N$. $\endgroup$
    – Qmechanic
    Commented Jun 1, 2011 at 18:50
  • $\begingroup$ The final equation (N1^2 / N2 equals the Boltzmann factor), is incorrect. This is a big deal, since you claim the whole point is to understand that equation. To see that it's wrong, note that if you double the size of your box and double the number of particles (while keeping the temperature and density constant), your formula would predict the ratio of atoms to molecules would change. That ain't physical. The correct answer involves the densities of the two species, the Boltzmann factor, and some hbars and m's and T's. $\endgroup$ Commented Jun 2, 2011 at 0:10
  • $\begingroup$ @Anonymous Coward: The "$\sim$" sign in the question (v2) is not an "$=$" sign. Here it just means "is proportional to", where the proportionality factor could be big or small; it could depend on $V$, $\hbar$, $T$, etc., but not $E_0$. $\endgroup$
    – Qmechanic
    Commented Jun 2, 2011 at 6:03
  • 1
    $\begingroup$ @Qmechanic: Normally "is proportional to" is written as "$\propto$", no? And if you're trying to express proportionality, you need some way of communicating that the ONLY proportionality included in the equation is the $E_0$ dependence; the full dependence on $T$ is omitted, etc. $\endgroup$ Commented Jun 2, 2011 at 16:39
  • $\begingroup$ @Anonymous Coward: I have seen both notations "$\sim$" and "$\propto$" to stand for "is proportional to". For instance, on a blackboard, the former is often used, as the latter looks like the Greek letter alpha "$\alpha$" when written by hand. $\endgroup$
    – Qmechanic
    Commented Jun 2, 2011 at 18:07

1 Answer 1


Yes, the Maxwell-Boltzmann partition functions of the two subsystems are

$$Z_1(N_1) ~=~ \frac{1}{N_1!}\left[\frac{V}{h^3} \left(\frac{2\pi m}{\beta}\right)^{\frac{3}{2}}\right] ^{N_1}, $$ $$Z_2(N_2) ~=~ \frac{1}{N_2!}\left[\frac{V}{h^3} \left(\frac{4\pi m}{\beta}\right)^{\frac{3}{2}} e^{-\beta E_0} \right] ^{N_2}, $$

where we ignore the rotational and vibrational modes of the molecule $A_2$. Here $E_0<0$ is the binding energy of the chemical reaction $2A \leftrightarrows A_2$. The full partition function, for a fixed number $N=N_1+2N_2$, is

$$ Z(N) ~=~ \sum_{N_2=0}^{[\frac{N}{2}]}Z_1(N-2N_2)Z_2(N_2). $$

The corresponding grand partition functions for the two subsystems are

$$Z_1(\mu)~:=~\sum_{N_1=0}^{\infty}Z_1(N_1)e^{\beta\mu N_1} ~=~\exp\left[\frac{V}{h^3} \left(\frac{2\pi m}{\beta}\right)^{\frac{3}{2}}e^{\beta\mu } \right] , $$

$$Z_2(\mu)~:=~\sum_{N_2=0}^{\infty}Z_2(N_2)e^{2\beta\mu N_2} ~=~\exp\left[\frac{V}{h^3} \left(\frac{4\pi m}{\beta}\right)^{\frac{3}{2}}e^{\beta(2\mu-E_0) } \right] , $$

where we have inserted a factor of $2$ to prepare for the implementation of the constraint $N_1+2N_2=N$ of the chemical reaction $2A \leftrightarrows A_2$. The full grand partition function, that is relevant for the chemical reaction $2A \leftrightarrows A_2$, is

$$Z(\mu)~:=~\sum_{N=0}^{\infty}Z(N)e^{\beta\mu N}~=~Z_1(\mu)Z_2(\mu).$$

The average occupation numbers are

$$\left< N_1\right>_{\mu} ~=~ \frac{1}{\beta}\frac{\partial\ln Z_1(\mu)}{\partial\mu} = \ln Z_1(\mu) , $$

$$\left< N_2\right>_{\mu} ~=~ \frac{1}{2\beta}\frac{\partial\ln Z_2(\mu)}{\partial\mu} = \ln Z_2(\mu) . $$

The following fraction is independent of the chemical potential $\mu$:

$$ \frac{\left< N_1\right>^2_{\mu=0}}{\left< N_2\right>_{\mu=0}}~=~\frac{\left< N_1\right>^2_{\mu}}{\left< N_2\right>_{\mu}} ~=~\frac{V}{h^3} \left(\frac{\pi m}{2\beta}\right)^{\frac{3}{2}}e^{\beta E_0} ~\sim~ e^{\beta E_0}~=~e^{-\beta |E_0|} ,$$

where the "$\sim$" sign means "is proportional to".

  • $\begingroup$ I see that as equations your $Z(N)$ = my $Z_N$. But I didn't understand how you defined $Z_1(\mu)$ and $Z_2(\mu)$. Is what you are calling as $\mu$ the same as what has been called as $\mu$ in the definition of $Z_{N,\mu}$ that I quoted ? (I am feeling that the $\mu$ in my reference is not the same as what you are calling as $\mu$) I didn't understand why you have $e^{2\beta \mu N_2}$ in your definition of $Z_2(\mu)$. Shouldn't the grand partition function for the $A_2$ molecules be insensitive to the constrain of $N_1 + 2N_2 = N$ ? $\endgroup$
    – Student
    Commented Jun 1, 2011 at 18:43
  • $\begingroup$ Also can you kindly explain why the $Z(\mu)$ should factor as $Z_1(\mu) Z_2(\mu)$ ? The atoms $A$ and the molecules $A_2$ are not really decoupled systems. Kindly correct me about what I am missing. $\endgroup$
    – Student
    Commented Jun 1, 2011 at 18:44
  • $\begingroup$ I like what you've done there, but I think you need some modification of your final equation. The "approximately equals sign" isn't appropriate here because the omitted prefactor could be a number that's many orders of magnitude different than 1. The prefactor also depends on the temperature and the volume, and the functional dependance can be important. But with just a smidge more work, I think you've got the right answer. $\endgroup$ Commented Jun 2, 2011 at 0:19
  • $\begingroup$ Thanks for modifying your answer to include the full answer. I'm always a bit uncomfortable with a "proportional to" that includes some of the functional dependence on temperature but not all of it. Your new version is much more clear. $\endgroup$ Commented Jun 2, 2011 at 21:21

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