Talk:Partition function (statistical mechanics)

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Shouldn't the first title read 'Classical', instead of 'Classic'? Unless, of course, 'Classic' means something specific in this context. Gokul

Contents

[edit] Notation

Z=\frac{1}{N!h^{3N}} \int_{(R^3)^n} \, \left(\int_{(R^3)^n} \, \exp[-\beta H(p_1 \cdots p_N, x_1 \cdots x_N)] \; \mathrm{d}^3p_1 \cdots \mathrm{d}^3p_N \right)\, \mathrm{d}^3x_1 \cdots \mathrm{d}^3x_N [clarify]

I don't understand this d3 notation. How is it supposed to be interpreted?

Note that pi and xi are elements of 3 dimensional space R3. This notation serves to remind us that, in other words, it is a shorhand notation for something like d3x = dx1dx2dx3. (Igny 17:34, 20 September 2007 (UTC))
Thanks. Another question if you are inclined to answer it: are each of the pi and xi scalar fields on \mathbb {R}^3 or are they each discrete vectors?ChrisChiasson 16:03, 23 September 2007 (UTC)
Technically they are not fields. both pi(momentum) and xi(position) are 3 dimensional vectors for all i. The Hamiltonian basically determines the likelihood of the particles to have given momentums and positions, the likelihoods are proportional to
\exp[-\beta H(p_1 \cdots p_N, x_1\cdots x_N)]
When divided by the normalizing factor (the partition function) it gives the distribution of pi and xi over all possible values of the momentums/ positions.(Igny 16:31, 23 September 2007 (UTC))
So are they vector fields then?ChrisChiasson 17:04, 23 September 2007 (UTC)
Or, wait, you are saying they are plain vectors (which would only be functions of, say, time)? —Preceding unsigned comment added by ChrisChiasson (talk • contribs) 17:06, 23 September 2007 (UTC)

They are not functions of anything they are just dummy 3-D variables, indicating a possible state of the system, over which you integrate. The Hamiltonian may be a function of time and the state (t,xi,pi), thus resulting in the partitition function as well as the position/momentum distribution changing with time. (Igny 17:45, 23 September 2007 (UTC))

This may be a stupid question, but why would someone integrate on a domain like d3p1 that is, essentially, a point?ChrisChiasson 04:24, 24 September 2007 (UTC)
We integrate the likelihood of the state over all possible states, see my edit of the integral above.(Igny 05:25, 24 September 2007 (UTC))

[edit] For the canonical ensemble partition function: Q versus Z

For the canonical ensemble, should Q be used in place of Z? Z is actually a special case (the momentum component is factored out) often called the configuration integral. Ref "Understanding Molecular Simulation" 2nd Ed. Frenkel and Smit

No. There is no standard notation, so this change is not necessary. -- CYD

i think d/dbeta of minus average energy gives the average squared energy.

so the second derivative wrt beta of ln Z gives the second moment of energy, not the second central moment as indicated

[edit] A good book

I suggest a good book to read. Fedrick Beif's Statistical and Thermal Physics. Of course-by the way can someone discuss your comment about it between other authors' books.--GyBlop 10:50, 28 February 2006 (UTC)

I think that's Frederick Reif (or at least that's how it's spelled in my edition). Flutefreek 05:44, 2 November 2006 (UTC)

[edit] Sum and Integral

Hello..something is fuzzy in the article for example if we have a certain system with a Hamiltonian H and energies En my question is..what partition function should we use?..

-The expression given by the Sum Z_1 (\beta)=\sum_{n} g(n)e^{-\beta E_n}

-Or the expression given by the Integral Z_2 (\beta)=\int_{-\infty}^{\infty}dp\int_{-\infty}^{\infty}dq e^{-\beta H(q,p)}

But clearly Z1 and Z2 are different, the problem is how can you calculate the Partition function if you don,t know the energies of the system (Hamiltonian too much difficult to solve) and if approximately Z1˜Z2

En is energy of a certain state (indexed by n under assumption that number of states is countable). In the case of the integral, the state of the system is determined by p and q, and there is a continuum of possible states. In both cases the meaning of the partition function stays the same. There is however a disagreement in using the degeneracy factor, it is used in the definition but nowhere else, I think I ll try to rephrase the article now.(Igny 17:23, 18 July 2006 (UTC))


Those two are used in different contexts, shouldn't be confused. Z1 is the quantum mechanical partition function. The summation is over number of distinct energy eigenvalues. It is, assuming some technical conditions hold, the trace of exp(-β H), where H is the quantum mechanical Hamiltonian. If you sum over energy eigenvalues including multiplicity, then the degeneracy g(n) would be removed from the expression. On the other hand, Z2 describes the classical canonical ensemble, therefore is a normalization factor for a probability measure on the phase space. Mct mht 21:32, 18 July 2006 (UTC)

[edit] Error in article

BTW, there is an error in the article. The so called "coarse graining" refers to semiclassical treatment of quantum mechanical ensembles, not classical systems. For classical systems, the factor 1/h3N shouldn't be there. Mct mht 02:33, 19 July 2006 (UTC)

[edit] Calculating the thermodynamic total energy

In the Calculating the thermodynamic total energy section shouldn't it say "and then set λ to one," rather than "and then set λ to zero"? If I follow the argument correctly, it is essentially an application of perturbation theory, in which you expand the change in the Hamiltonian (A in this case) in small λ and then set λ to one at the end, not to zero, because this would cancel out the perturbative terms. --zipz0p 20:05, 11 October 2006 (UTC) yes you're right I believe. Also to the comment above about not needing to renormalize by h you still do in order to get the right units (well and also to get the right answer for the partition function!). —Preceding unsigned comment added by 99.246.56.142 (talk) 03:55, 25 October 2007 (UTC)

[edit] Derivation of <E>:

Maybe it would be noteworthy for the beginners that the third transformation ("\frac{1}{Z}\sum_j E_j e^{\beta E_j} = -\frac{1}{Z}\frac\{\partial}{\partial\beta}Z(\beta, E_1, E_2,\ldots)" ) comes from differentiating Z being a function of \beta over \beta, and the next transformation comes from differentiating the \ln function

Just wondering whether it would useful for beginners... —Preceding unsigned comment added by W2023 (talk • contribs) 17:31, 10 March 2008 (UTC)