0
$\begingroup$

Other than that we need the $h^3$ to 1) make the units correct and to 2) account for the normalization of the probability distribution, what interpretation can we give to this $h^3$ term in

\begin{align} p(E_i) = \frac{1}{Z_1 h^3}\exp(-\beta E_i) \ , \end{align} with 1-molecule partition function $Z_1 = \frac{1}{h^{3}} \int_\mathbb{R^3} \int_\mathbb{R^3} \exp \Big(-\beta\frac{p^2}{2m}\Big) \ \mathrm{d\mathbf{r}} \ \mathrm{d\mathbf{p}} $ and energy $E_i$ of microstate $i$.

Namely, if the partition function is defined to be the normalization constant for obtaining the probability distribution over the microstate energies, how did the situation arise that we need to correct for the units?

My confusion lies in the following. $Z_1$ represents the total number of microstates of a single molecule, were we have used $h^3$ to denote the phase-space volume of a single state. The naive guess would then be to assume that the probability distribution of a microstate $i$ would then be given by \begin{align} p(E_i) = \frac{1}{Z_1}\exp(-\beta E_i). \end{align}

That is, without the $h^3$ factor explicitly appearing in the probability distribution.

Is there no better explanation for the appearence of the $h^3$ factor in $p$ other than that we need to cancel out the $h^3$ factor in $Z_1$?

$\endgroup$
0
$\begingroup$

The correct expression is

$$p(\mathbf p ,\mathbf q)= \frac{e^{-\beta H}}{\int \int d \mathbf p d \mathbf q \ e^{-\beta H}}$$

since the integral of this is $1$.

But you would also like to write $Z$ as a dimensionless quantity, since it should represent the number of microstates. Since the integral

$$\int \int d \mathbf p d \mathbf q \ e^{-\beta H}$$

has the dimension of action cube, you need a quantity with the dimension of action cube to make it dimensionless. You therefore define $Z$ as

$$Z \equiv \frac 1 {h^3} \int \int d \mathbf p d \mathbf q \ e^{-\beta H}$$

Therefore, you will write

$$p(\mathbf p ,\mathbf q)= \frac{e^{-\beta H}}{\int \int d \mathbf p d \mathbf q \ e^{-\beta H}} = \frac{e^{-\beta H}} {Z h^3}$$

In principle, nothing prevents you from defining a dimensional partition function $Z'$:

$$Z' \equiv \int \int d \mathbf p d \mathbf q \ e^{-\beta H}$$

and writing

$$p(\mathbf p ,\mathbf q)= \frac{e^{-\beta H}}{\int \int d \mathbf p d \mathbf q \ e^{-\beta H}} = \frac{e^{-\beta H}} {Z'}$$

So, in the end, it it is just a matter of being coherent, but you can in principle use the notation that you prefer.

$\endgroup$
  • $\begingroup$ Therefore, you will write $$p(\mathbf p ,\mathbf q)= \frac{e^{-\beta H}}{\int \int d \mathbf p d \mathbf q \ e^{-\beta H}} = \frac{e^{-\beta H}} {Z h^3}\ .$$ Yes, this is what a started off my question with; I already have the result. My question: other than to make the integral 1 and getting the correct units, what reason is there to divide by $h^3$ in the final expression? we divide by the phase-space integral by $h^3$ in order to get the number of states. Why do we need to correct for this $h^3$, other than to make the integral 1 or getting the units correct? $\endgroup$ – Mussé Redi Dec 21 '17 at 19:11
  • $\begingroup$ The only reason is that you want $Z$ to be dimensionless. There is no deeper reason, really. $\endgroup$ – valerio Dec 23 '17 at 9:58

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.