# Two dimensional Blume-Capel model with random crystal field

Hi, my name is Ofek and its my first post here. hope to be clear and if not I'll try to be more specific next time.

Writen by N. S. Branco

The model H = J*ƩSiSj + ƩΔi(Si)^2 - first sum over nearest neighbors and second sum (i) over all lattice sites.

The distribution of Δ (crystal field) is P(Δi)= pδ(Δi + Δ)+(1-p)δ(Δi - Δ)

In the article it says that the Flow in the parameter space on the critical surface is towords the fixed point of pure (p=0) ising spin 1/2 model, the fixed point value is (p*=0,Δ*=-∞, J*=finite irrelevant constant)

I have noticed that there is another fixed point of pure ising model spin 1/2 and its on the "other side" of the Δ scale at -
- (p=1, Δ=∞, J=finite const)

My question: how can it be that with an identical (pure ising model fixed point) to the one mentioned in the article (fixed point at p=0) we got a flow from all the points on the critical surface towords the p=0 fixed point and not to the (p=1) fixed point?

(I ignored another fixed point - random fixed point that only the critical line at zero temperature flows to it, and that is because i dont think its relevant to my question)

The solution of this model (i.e. finding the critical surface and the non-trivial fixed points) is rather tedious and involves 64 renormalization equations - (recursion relation of the renormalization group).
Despite these, my main goal of writing this question here is to get an idea or proposal about the nature of this other fixed point (p=1, Δ = +∞).

If it will make things easier ill mention that at Δ=-∞ and for Δ=+∞ in the Temperature - p plane we got a Dilute spin 1/2 model (perculation) with critical p=1/2

Thanks
Ofek

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Mute
Homework Helper
Hi, my name is Ofek and its my first post here. hope to be clear and if not I'll try to be more specific next time.

Writen by N. S. Branco

The model H = J*ƩSiSj + ƩΔi(Si)^2 - first sum over nearest neighbors and second sum (i) over all lattice sites.

The distribution of Δ (crystal field) is P(Δi)= pδ(Δi + Δ)+(1-p)δ(Δi - Δ)

In the article it says that the Flow in the parameter space on the critical surface is towords the fixed point of pure (p=0) ising spin 1/2 model, the fixed point value is (p*=0,Δ*=-∞, J*=finite irrelevant constant)

I have noticed that there is another fixed point of pure ising model spin 1/2 and its on the "other side" of the Δ scale at -
- (p=1, Δ=∞, J=finite const)

My question: how can it be that with an identical (pure ising model fixed point) to the one mentioned in the article (fixed point at p=0) we got a flow from all the points on the critical surface towords the p=0 fixed point and not to the (p=1) fixed point?

The Blume-Capel model is a spin-1 model. This is relevant for analyzing the fixed point behavior, as the two fixed points you quote are not identical. (I will assume for the moment that the fixed point you have found is indeed a fixed point; I have not double checked this myself).

The reason the two points are not identical is as follows:

For ##\Delta \rightarrow -\infty##, the energy of the system is minimized (as ##\sum \Delta_i (S_i)^2## is large and negative) so long as ##S_i \neq 0##, so it is favorable for the spins to be either up or down, but not zero. This is what gives the model spin-1/2 Ising-like characteristics at the fixed point.

However, for ##\Delta \rightarrow +\infty##, the energy of the system is maximized, as ##\sum \Delta_i (S_i)^2## is large and positive. Because the system wants to minimize its energy, this makes the ##S_i = 0## state the preferred spin orientation in order to counteract this large energy term. As a result, I would not expect this fixed point to be Ising-like at all.

You didn't take into account the distribution of the crystal field:
P(Δi)= pδ(Δi + Δ)+(1-p)δ(Δi - Δ)
which says that even for Δ=+∞ we get a limit for dilute ising spin 1/2 (or you can call it random site ising model) and you can see in the - (temperature - p) plane the same critical line as you see in Δ=-∞ but in mirror image because in this region the distribution is - P(-Δi)= pδ(Δi - Δ)+(1-p)δ(Δi + Δ)

so it says there is the same perculation point (critical p=0.5) but the pure ising model spin 1/2 at Δ=+∞ will be at p=1 (instead of p=0 like in the Δ=-∞ case)

Mute
Homework Helper
You didn't take into account the distribution of the crystal field:
P(Δi)= pδ(Δi + Δ)+(1-p)δ(Δi - Δ)
which says that even for Δ=+∞ we get a limit for dilute ising spin 1/2 (or you can call it random site ising model) and you can see in the - (temperature - p) plane the same critical line as you see in Δ=-∞ but in mirror image because in this region the distribution is - P(-Δi)= pδ(Δi - Δ)+(1-p)δ(Δi + Δ)

so it says there is the same perculation point (critical p=0.5) but the pure ising model spin 1/2 at Δ=+∞ will be at p=1 (instead of p=0 like in the Δ=-∞ case)

Ah, yes, you're right. There is a ##p \leftrightarrow 1-p## with ##\Delta \leftrightarrow -\Delta## symmetry. I flipped the sign on ##\Delta_i## by mistake earlier.

In that case, my best guess without digging into the details is that there is a region of parameter space that flows to your fixed point under the RG, but it mirrors the region which flows to the ##p^\ast = 0##, ##\Delta^\ast = -\infty## fixed point, so the authors only consider the one region. For example, Figs. 2-5 plot fixed p and positive ##\Delta/J##. The negative ##\Delta/J## region might show the ##1-p## behavior. (i.e., I would guess that the ##\Delta/J < 0## region of Fig. 2 would look like Fig. 5 and vice versa).

yes, you're right, I charted a 3D graph that contains what you have just depicted.

Back to my question-
there are two scenarios:

1. The author is right and the other pure fixed point at p=1 Δ=∞ isnt attracting, or acting in a way that make the flow on the critical surface be directed to the pure point at p=0 Δ=-∞.

2. The author was wrong, ignoring the other attracting fixed point (at p=1), which leads to a new problem - there are two attracting fixed points at both ends the critical surface, there is a need for something like a point or a line on the critical surface that RG will flow out of it towords the two points.

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