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In my notes it states that the convention for summing over the classical states is

$$\sum_{\Gamma} \longrightarrow \frac{1}{N!}\int \prod_{i=1}^N \frac{d^3q_id^3p_i}{h_0^3} \tag1$$

Now I know that as classical microstates form a continuum (not sure whether this is the exactly correct definition) we must divide state space in small cubes of which points inside are considered to be the same. This is done by $$\delta q_i \delta p_i=h_0 \text{ for }i=1,\dots,3N$$ but I cannot see how to get the RHS of $(1)$. I tried $$\sum_{i=1}^{3N} \frac{\delta q_i \delta p_i}{h_0} = \sum_{i=1}^{N} \frac{(\delta q_i \delta p_i)^3}{h_0^3}$$

UPDATE (from By Symmetry's answer) we get $$\frac{1}{N!}\sum_{i=1}^{N} \frac{(\delta q_i \delta p_i)^3}{h_0^3}$$

I do not expect limits to be taken as rigourously as in mathematical analysis.

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I will only focus on the "measure" aspect of the problem you seem having trouble with. When it comes to the $N!$, let's say the argument given by By Symmetry is essentially correct.

Now, one way to look at it is to start directly from the actual quantum partition function of a system of $N$ distinguishable particles:

\begin{equation} Q_q(\beta, N, V) \equiv Tr\left(e^{-\beta \hat{H}} \right) = \sum_{i} \langle u_i | e^{-\beta \hat{H}} |u_i \rangle \end{equation}

where the $\{|u_i \rangle \}_{i=1,..}$ form a basis of the the quantum states with $N$ particles. We then replace $H$ by its more detailed low energy expression $\hat{H} = K(\hat{\mathbf{p}}) + V(\hat{\mathbf{x}})$ where $\hat{\mathbf{p}}$ and $\hat{\mathbf{x}}$ are collections of $3N-$momenta and $3N-$coordinates respectively.

The idea consists then in expanding the exponential in series:

\begin{equation} e^{-\beta \hat{H}} = \hat{1} -\beta[K(\hat{\mathbf{p}}) + V(\hat{\mathbf{x}})]+ \frac{1}{2}\beta^2 [K(\hat{\mathbf{p}}) + V(\hat{\mathbf{x}})]^2 + ... \end{equation}

and compare it to the operator

\begin{eqnarray} e^{-\beta K(\hat{\mathbf{p}})} e^{-\beta V(\hat{\mathbf{x}})} &=& [\hat{1} -\beta K(\hat{\mathbf{p}}) + \frac{1}{2}\beta^2 K(\hat{\mathbf{p}})^2][\hat{1} -\beta V(\hat{\mathbf{x}}) + \frac{1}{2}\beta^2 V(\hat{\mathbf{x}})^2] +...\nonumber \\ &=& \hat{1} -\beta[K(\hat{\mathbf{p}}) + V(\hat{\mathbf{x}})] + \frac{1}{2}\beta^2 [K(\hat{\mathbf{p}}) + V(\hat{\mathbf{x}})]^2 \nonumber \\ && -\frac{1}{2}\beta^2[K(\hat{\mathbf{p}})V(\hat{\mathbf{x}})+ V(\hat{\mathbf{x}})K(\hat{\mathbf{p}}) - 2K(\hat{\mathbf{p}})V(\hat{\mathbf{x}})] +... \end{eqnarray}

so that the difference

\begin{equation} e^{-\beta \hat{H}}-e^{\hat{K}}e^{\hat{V}} = \frac{1}{2}\beta^2[K(\hat{\mathbf{p}})V(\hat{\mathbf{x}})+ V(\hat{\mathbf{x}})K(\hat{\mathbf{p}}) - 2K(\hat{\mathbf{p}})V(\hat{\mathbf{x}})] +... \end{equation}

This seems probably very formal but basically if $[\hat{p},\hat{x}] \neq 0$, then $e^{-\beta \hat{H}}-e^{\hat{K}}e^{\hat{V}} \neq 0$ and there is no way to separate the contributions from the kinetic and the potential energies to the Boltzmann weight. This is one aspect of the quantum regime of statistical mechanics which has to do with the uncertainty relations (or the wave-like behaviour).

Now, in the last equation above, we have an additional factor that is very important which is that the non zero r.h.s is multiplied by $\beta^2$.

So, if $\beta$ is small enough (i.e. if the temperature is big enough), it is possible to have the r.h.s. non zero still but very very small.

In that case, at high temperature, it is then reasonable to approximate

\begin{equation} e^{-\beta \hat{H}} \approx e^{\hat{K}}e^{\hat{V}} \end{equation} This is a trivial application of what is called the Trotter formula/expansion.

The above formula is in fact enough to retrieve the (semi)classical partition function.

The idea is to say that

\begin{equation} Q_q(\beta,N,V) \approx Q_{sc}(\beta,N,V) \equiv \sum_{i} \langle u_i | e^{\hat{K}}e^{\hat{V}} |u_i \rangle \end{equation}

We can then insert two closure relations:

\begin{equation} \int d^{3N}p \: | \mathbf{p} \rangle\langle \mathbf{p} | = \hat{1} \end{equation} and \begin{equation} \int d^{3N}x \: | \mathbf{x} \rangle\langle \mathbf{x} | = \hat{1} \end{equation}

which give

\begin{eqnarray} Q_q(\beta,N,V) &\approx& \sum_{i} \langle u_i |\int d^{3N}p \: | \mathbf{p} \rangle\langle \mathbf{p} |\: e^{\hat{K}} \int d^{3N}x \: | \mathbf{x} \rangle\langle \mathbf{x} |\:e^{\hat{V}} |u_i \rangle \nonumber \\ &\approx& \int d^{3N}p d^{3N}x\: \langle \mathbf{x}| \left[\sum_i |u_i\rangle\langle u_i|\right] |\mathbf{p}\rangle \langle \mathbf{p}| \mathbf{x} \rangle e^{-\beta K} e^{-\beta V} \end{eqnarray}

By using the closure relation

\begin{equation} \sum_i |u_i\rangle\langle u_i| = \hat{1} \end{equation}

and the pseudo scalar product

\begin{equation} \langle \mathbf{x} | \mathbf{p} \rangle = \frac{1}{h^{3N/2}} \exp \left(-i\frac{\mathbf{p}\cdot \mathbf{x}}{\hbar} \right) \end{equation}

we finally find that, at high enough temperature:

\begin{equation} Q_q \approx Q_{sc} = \int \frac{d^{3N}p \:d^{3N}x}{h^{3N}} e^{-\beta [K(\mathbf{p})+ V(\mathbf{x})]} \end{equation}

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I think you're almost there. The only major point you are missing is that your $N$ particles are identical. This means if you exchange the position and momentum of two particles you get back the same state and your final equation is, therefore, double counting certain states.

Now since we are dealing with classical particles we can assume that there are vastly more available state than particles, so any correction due to what might happen if two particles tried to occupy the same state can be ignored. This means all we have to do to take care of the overcounting is divide through by the number permutations of $N$ distinct particles, which is $N!$.

After that it is simply a question of deciding how pedantic you are about taking limits.

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  • $\begingroup$ Yeah I cant see how you would roughly take this limit, Id ideally like an explanation of how this could be done $\endgroup$ – Permian Apr 20 '15 at 20:49
  • $\begingroup$ The argument of normalizing by N! is due to Gibbs, and is called correct Boltzmann counting in stat mech, covered in every good text and WP article. It leads to the correct Sackur-Tetrode equation. It is ad hoc, basically, until the introduction of quantum mechanics and Bose statistics, when it becomes a trivial necessity. $\endgroup$ – Cosmas Zachos Apr 2 '16 at 16:40

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