I'm studying Lee-Yang theorem following Volume I, Section 3.2 of Itzykson and Drouffe famous book.

In doing so, I've stumbled upon a dilemma that - while almost insignificant at first glance - poses some real issues when carrying out critical limits afterwards.

They consider an Ising model on an arbitrary $N$-sites graph, with $L$ the total numbers of links. Having defined the following $$\rho_i = e^{-2 h_i} \qquad\qquad \tau=e^{-2\beta}$$ they immediately write the partition function as $$Z_N = \frac{1}{2^N}\,\exp{\{\beta L + \sum_i h_i \}}\,P(\tau,\{\rho_i\})$$ with $$P(\tau,\{\rho_i\})=\sum_{\sigma_i = \pm 1} \exp{\left\lbrace \beta\sum_{<ij>}(\sigma_i\sigma_j - 1) + \sum_i h_i (\sigma_i - 1) \right\rbrace}$$ and the summation $\sum\limits_{<ij>}$ runs over all links.

So, while they don't write this themselves, I guess it's pretty obvious that they're considering a system described by an Hamiltonian almost of the following type (read EDIT 1): $$H = -\frac{1}{2}\sum_{<ij>}\sigma_i \sigma_j - \sum_i h_i \sigma_i$$ Now, what I don't understand is the lack of $\beta$ in the external field's terms of the polynomial and polynomial's prefactor inside $Z_N$.

Infact, carrying out the calculation myself with the Hamiltonian above, I end up with something like this $$Z'_N = \frac{1}{2^N}\,\exp{\{\beta L + \beta\sum_i h_i \}}\,P'(\tau,\{\rho_i\})$$ with $$P'(\tau,\{\rho_i\})=\sum_{\sigma_i = \pm 1} \exp{\left\lbrace \beta\sum_{<ij>}(\sigma_i\sigma_j - 1) + \beta\sum_i h_i (\sigma_i - 1) \right\rbrace}$$

While I still don't understand the lack of $\beta$ in the book, one can find out that little to nothing changes (for many calculations they carry out) re-defining $$\rho_i = e^{-2\beta h_i}$$ and infact, from what I can read online, this is the usual definition of fugacity, or activity (the L-Y theorem states that all the zeros of the partition function are one the unit circle of the activity complex plane).

But, for example, a problem arises when calculating critical limits of the polynomial. When $T\rightarrow\infty$ ($\beta\rightarrow 0$, $\tau\rightarrow 1$), the book states that $P = P(1,\rho) = (1+\rho)^N$ (one zero, with $N$ multiplicity) with $\rho=e^{-2h}$ (having considered, for simplicity, a uniform external field). Infact, one can verify $$\lim_{\beta\to 0} P(\tau,\{ \rho \}) = \sum_{\sigma_i = \pm 1} \exp{\left\lbrace h\sum_i (\sigma_i^N - 1)\right\rbrace} = \prod_i^N \sum_{\sigma_i = \pm 1} \exp{\left\lbrace h (\sigma_i - 1)\right\rbrace} = (1+\rho)^N $$

But the result changes if one considers the polynomial $P'$ instead of $P$.

What am I doing wrong? Thanks in advance for any help!


If this Holy Grail of a manual features no errors, the actual Hamiltonian that the authors are considering has the following form: $$H = -\frac{1}{2}\sum_{<ij>}\sigma_i \sigma_j - \frac{1}{\beta}\sum_i h_i \sigma_i$$ But why and, more importantly, how one should define such an Hamiltonian? Isn't it uncorrect on dimensional grounds?


Indeed, a few pages above, they state

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But again, how is that correct? To arbitrarily omit a relevant variable ($\beta$) only from a term of the whole formula, a variable that is later crucial in the above mentioned critical limits $T\rightarrow\infty$ and $T\rightarrow 0$!

I'm sure I must be missing something but I can't really figure out what!


1 Answer 1


The constant of a spin interaction with an external magnetic field has the dimension of energy and is independent of temperature. Often (in other textbooks) it is denoted as $h$. And the constant of interaction of the nearest spins is like $J$ (often the dimensionless constant $K = \beta J$ is introduced). In this textbook (Itzykson and Drouffe), $\beta (Itzykson) = \beta J$, $h (Itzykson) = \beta h = \beta (Itzykson) h / J$, where $h,J$ have the dimension of energy and $\beta=1/kT$, $k$ is the Boltzmann constant. So "true external field is propotional to"... $h (Itzykson) / \beta (Itzykson) $

  • $\begingroup$ So $\beta (Itzykson)$ and $h(Itzykson)$ are adimensional? $\endgroup$ Jun 8, 2017 at 11:45
  • $\begingroup$ Yes, dimensionless $\endgroup$ Jun 8, 2017 at 11:54
  • $\begingroup$ That makes sense analytically but it raises the same theorical perplexities: $h(Itzykson)$ is not truly $\beta$-independent and this shouldn't be forgotten when performing the limits $\beta\rightarrow\infty$ and $\beta\rightarrow 0$, should it? $\endgroup$ Jun 8, 2017 at 11:55
  • $\begingroup$ I mean, one can re-define terms inside a formula as he prefers but can't really forget what is what, the profound physical meaning of each term! In this case, for example, how can one neglect the fact that $h(Itzykson)\xrightarrow{\beta\to 0} 0$? (and that is true even when the beta that goes to $0$ is the new $\beta(Itzykson)$) I'm sure there is a solid explanation for this but I can't figure it out. $\endgroup$ Jun 8, 2017 at 12:05
  • $\begingroup$ h(Itzykson) is not at all a field constant, it is a product of a field constant and $\beta$ $\endgroup$ Jun 8, 2017 at 12:17

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