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I am studying the renormalization group approach to the Ising model using as a reference Cardy's book "Scaling and renormalization in statistical mechanics". I cannot understand what happens in the zero temperature case (and possibly for $ T < T_c$) to the correlation length $\xi$. Here's my point:

Since the zero temperature is a fixed point, it should be either $\xi =0$ or $\infty$ in fact $\xi(\{K'\}) = 1/2 \ \xi(\{K\}) $, but at a fixed point $\{K'\} = \{K\}$ (I am using Cardy's notation where $\{K\}$ denotes the set of coupling for the theory).

Now if $\xi$ is finite below the critical temperature (as it is stated in some books), say at $T_0 < T_c$ or equivalently $K_0 > K^*$, it must be zero at zero temperature. This can be deduced in a similar fashion as it is done at the critical temperature (pag. 38). In short, if $n(K)$ is defined as the number of times you have to apply the renormalization group to get from $K_0$ to $K$, then $n(K)$ diverges as $K \to \infty$ (same to say $T \to 0$). Therefore $\xi(K)$ tends to zero (it is halved $n(K)$ times starting from $\xi(K_0)$ ).

On the other hand, it seems to me that it is possible to evaluate exactly the two-point spins correlation functions in the zero temperature limit as follows: $\langle\sigma(0) \sigma(r) \rangle = \sum_{\{\sigma\}} \sigma(0) \sigma(r) \exp(-\beta H[\sigma]) = 2$. In the last passage I have used that at zero temperature only the two spins configurations with lowest energy contribute to the sum, but those are with all spins up or all spins down and $\langle\sigma(0) \sigma(r) \rangle = 1$. Therefore the correlation length is infinite. (and for the argument above it should be infinite for every $T<T_c$).

So where is the error?

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  • $\begingroup$ I forgot about that thank you. But nevertheless, <\sigma(r)> = 0 for every r, since again at zero temperature only the above two configurations contribute and so on...Hence <sigma sigma> - <sigma><sigma> = <sigma sigma> = 2. (I had not the time to read your reference yet but I intend to do) $\endgroup$ – giulio bullsaver Jan 15 '15 at 8:30
  • $\begingroup$ Thank you very much, even tough I do not fully understand why one should say that the state must be pure (and therefore a boundary condition such as the one you said must be imposed), and not a mixture of the two pure, this must be the solution. I'd gladly upvote your answer if you wrote one. $\endgroup$ – giulio bullsaver Jan 15 '15 at 13:59
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The correlation length below the critical temperature can be defined using the rate of exponential decay of the truncated 2-point function, evaluated in a pure state (I'll choose the one induced by the $+$ boundary condition), namely $$ \xi_\beta(\vec n) = \lim_{k\to\infty} -\frac1k \log \langle \sigma_0 ; \sigma_{[k\vec n]}\rangle^+_\beta, $$ where $\vec n$ is a unit vector in $\mathbb{R}^d$ and $[k\vec n]$ is the point of $\mathbb{Z}^d$ closest to the point $k\vec n$ in $\mathbb{R}^d$. I used the standard notation $$ \langle \sigma_i ; \sigma_j\rangle^+_\beta = \langle \sigma_i \sigma_j\rangle^+_\beta - \langle \sigma_i \rangle^+_\beta\langle \sigma_j\rangle^+_\beta $$ for the truncated 2-point function (the covariance between the spins). Here, $\langle\cdot\rangle^+_\beta$ denotes expectation w.r.t. the (infinite-volume) Gibbs state obtained by taking the thermodynamic limit with $+$ boundary condition and inverse temperature $\beta$.

There are various ways of showing that the correlation length $\xi_\beta$ indeed goes to $0$ as $\beta\uparrow\infty$. Possibly one of the most straightforward (though a bit tedious) is through cluster expansion techniques (see, e.g., section 5.7.4 of this book).

The main point is that the average of the spins become nonzero below the critical temperature, but the fluctuations around this average value become completely uncorrelated in the limit. Morally, in order to correlate the fluctuations of two spins at $i$ and $j$, you must have a Peierls contour surrounding both $i$ and $j$, and the probability of this goes to $0$ exponentially fast in $\beta\|j−i\|$ (when $\beta>\beta_{\rm c}$).

There are several reasons why you must work with a pure state (or, more precisely, an extremal Gibbs measure) above. One is that these are the states corresponding to thermodynamic equilibrium: the only ones for which all macroscopic observables take deterministic values, for example. Note that if you were to consider the state obtained with free (or periodic) boundary condition, you would indeed see that the truncated 2-point function reduces to the usual 2-point function, $$ \langle \sigma_i; \sigma_j \rangle^{\rm free}_\beta = \langle \sigma_i \sigma_j \rangle^{\rm free}_\beta $$ and this quantities does not vanish as $\|j-i\|\to\infty$ when $\beta>\beta_{\rm c}$ (instead, it converges to the square of the spontaneous magnetization $m^*(\beta)$). Note also that, even though $\langle \sigma_i\rangle^{\rm free}_\beta = 0$, there is spontaneous magnetization in typical configurations: the expectation is zero only because this spontaneous magnetization is $m^*(\beta)$ or $-m^*(\beta)$ with probability $1/2$. Actually, a typical configuration under this measure will be typical of either the $+$ state or the $-$ state with probability $1/2$, so any natural way you decide to measure the correlation length from configurations under these pure states, should give you the same answer under the free state. The reason you cannot do it through the truncated 2-point function as above is that the expectation starts to mix up the contributions from the two possible macroscopic behaviors described by the pure states, but this does not correspond to anything physically relevant.

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