3
$\begingroup$

I am reading the book "Critical Dynamics - A Field Theory Approach to Equilibrium and Non-Equilibrium Scaling Behavior" (sections 1.1.2 and 1.1.3) and have been somewhat confused about the meaning of a mean-field theory (related questions: 1, 2).

To summarise, in deriving the Ising mean-field theory, the spin variable at site $i$ is written as $$\sigma_i = m + (\sigma_i - \langle \sigma_i \rangle),$$ where $m$ represents the mean magnetisation, and then the Ising Hamiltonian, $$ H = -\frac{1}{2} \sum_{i,j}J \sigma_i \sigma_j - h \sum_i \sigma_i$$ ($h$ is a magnetic field) is expanded by neglecting the term $(\sigma_i - \langle \sigma_i\rangle)(\sigma_j - \langle \sigma_j \rangle)$. Then the partition function is computed from which a consistency equation for $m$ is derived, namely $$ m = \tanh \big(\beta (h+J m) \big).$$ So far so good.

Next, the author proceeds to compute the correlation function via the response function by considering a site-dependent magnetic field by writing $$\chi_{ij} = \frac{\partial m_i}{\partial h_j} = \beta \langle (\sigma_i - \langle \sigma_i\rangle)(\sigma_j - \langle \sigma_j \rangle )\rangle.$$ The response function $\chi$ on the lhs can be computed by taking appropriate derivatives from the consistency relation for $m$, from which the so-called Ornstein-Zernicke form for the correlation funciton on the rhs is obtained. This form is also later obtained from the coarse-grained Landau free energy.

It appears to me that this is an inconsistent approach, since we first ignore the correlations between spatial fluctuations (by omitting the $(\sigma_i - \langle \sigma_i \rangle)(\sigma_j - \langle \sigma_j \rangle)$ ) in order to obtain the partition function, from which free energy and $m$ are evaluated; and then we go back to compute the correlations that we had assumed to be zero and get a non-zero value for it!

What am I missing here?

PS: I also wonder whether "mean-field" theories, in general, should fully ignore any correlations between the variables, or do they still account for the correlations in some sense?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.