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Most of the literature says that for a quenched average over disorder, an average over the log of the partition function must be taken:

\begin{equation} \langle \log Z \rangle, \end{equation}

while for the annealed average, it's

\begin{equation} \langle Z \rangle. \end{equation}

But a while ago, I came across a book that said that the annealed average is not $\langle Z \rangle$, though I don't remember what it said should be calculated instead.

Does anyone know which book this is, or what they might want to calculate instead of $\langle Z \rangle$ for the annealed average?

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Without the book reference, or a specific problem where you know what the answer should be, the issue is all about vocabulary--- it is about the precise definition of "annealed" vs. "quenched" disorder. It is likely that the book had a confusion of terms. –  Ron Maimon Nov 4 '11 at 15:51
    
Perhaps it is about the applicability of the "annealed average" to actual annealed systems? I don't think I've actually seen anyone apply the "annealed average" to an annealed system (but this is probably my own ignorance). –  BebopButUnsteady Nov 4 '11 at 19:04
    
@Ron I remember the book being very explicit that the usual way of calculating the annealed average as $\langle Z \rangle$ was wrong. I thought someone here might be aware of it. –  Calvin Nov 6 '11 at 13:59

3 Answers 3

If your disorder is not annealed but quenched, the distribution of impurities does not obey the thermal (Gibbs) distribution. Non-annealed, "quenched" disorder is an externally given background, and you need to average extensive quantities, such a free energy ($\propto \log Z$). It is hard for me to guess the exact context of your annealing, but I assume it to be thermalization of some kind. If it brings the disorder to thermal equilibrium with the rest of the system, then disorder averaging merges with thermodynamic averaging and you can use $\langle Z \rangle$.

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Thanks, I know that this is the usual argument, but I'm looking for something else. The book I referred to in the question explicitly stated that the usual method of finding the annealed average, $\langle Z \rangle$, is wrong, and then they provided an alternative. –  Calvin Nov 6 '11 at 14:04

I've found the book I was looking for: Ma - Modern Theory of Critical Phenomena (~ p366).

It says that rather the difference between a quenched and annealed average is the probability distribution it should be averaged over. Both types of average (if I understand it correctly) should be taken over the free energy:

\begin{equation} -T \langle \ln Z \rangle \end{equation}

but with different meanings of $\langle \cdots \rangle$ for the quenched and annealed averages.

It says that $-T \ln \langle Z \rangle$ should not be used.

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Hopefully, I can clear this stuff up. The difference between annealed impurities and quenched impurities isn't that complicated, yet the literature out there sure makes it seem that way!

Annealed Impurity: When the system has come into thermal equilibrium, we are interested in determining the probability that the system is in a particular configuration. This probability is a JOINT probability that the impurity is located somewhere and the matrix is located somewhere. In the example of Ma, his system is a magnet with random annealed impurities. The joint probability in this case is the probability of observing a set of spins for the annealed impurities and a set of spins for the matrix. One does not multiply the two probabilities because the probability that the impurities have some set of spins could (and DOES) depend on the probability that the matrix adopts a certain set of spins. If x and y are the annealed and host variables respectively, the JOINT probability of observing a configuration is

$P[x,y]=(1/Z)exp(-H[x,y]/kT)$

$Z=\int dxdyexp(-H[x,y]/kT)$

Free Energy is $-kTlnZ$

Note we do not take an average of this free energy!

Quenched Impurity In this case we don't allow the system to TOTALLY relax in that suppose the impurities are frozen in. A charged polymer (such as a protein!) is a good example of this. The charged monomers remain charged and their positions on the chain are fixed, the molecule can't relax into a state where it rearranges its monomers...its stuck with the order because the location of the monomers are fixed or QUENCHED. Any probability, like the probability of finding the polymer in a particular configuration, is CONDITIONAL on the location of the charged monomers. If x and y are now the quenched and host variables (x could be the location of the charged monomer say), the CONDITIONAL probability is:

$P[y | x]=1/Zexp(-H[y|x]/kT)$

$Z[x]=\int dxexp(-H[y|x]/kT)$

The Free Energy for this particular configuration of quenched impurity is $F[x]=-kTln(Z[x])$. As you prepare different samples of the protein, the charged monomers may end up on different parts of the molecule. This probability distribution $P[x]$ of the charged monomers is related to the way that they were made...it does NOT encode anything about thermal equilibrium. Now let's say you measured the end-to-end distance of the polymer. Theoretically the answer is an average of the end-to-end distances of each configuration of the charged monomer. In an equation it is:

$R^{2}=\sum P[y]R[y]^2$

This last equation is a manifestation of doing an "average of the $ln(Z)$" I did not mention this because it is confusing.

Final Thoughts If we allowed the charged monomers to somehow redistribute themselves (now we are pretending the charged monomer is an annealed variable) so that the free energy of the polymer was a minimum we would find, in general, a DIFFERENT probability distribution of the charge monomer than the quenched case. Hope this helps and sorry for the bad notation!

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