The Critical Scaling of the

Helicity Modulus of the O(3)

classical Heisenberg ferromagnet

Robert G. Brown

Acknowledgements: This work was done with my good friend and colleague, Dr. Mikael Ciftan. We gratefully acknowledge the support of the Army Research Office.

Model

Classical Heisenberg ferromagnet (CHF) (the ${\cal O}(3)$ model on a 3$d$ simple cubic lattice with periodic boundary conditions) in zero external field:

${\cal H} = - \sum_{a < b}^{nn} J_{ab} {\bf S}_a \cdot {\bf S}_b + {\cal H}_{\rm bath}(\{{\bf S}_i\},T)$

Goal

To compute and ``measure'' (with Monte Carlo) the critical exponents of the model, in particular the critical exponent $\nu_E$ of the Helicity Modulus $\Upsilon(t)$.

Methods

• Importance Sampling Monte Carlo (heat bath) ``with a twist'' to get $\bf E(L,T,\theta_{\rm twist})$ at high precision for $\bf L \in [8,64...]$.

• Finite size scaling used to get critical exponents at $T_c = 1.44295$ (accepted value, $\pm 0.00005$).

• Helicity studied by freezing and twisting the (previously periodic) boundary conditions in the (X,Y,1) plane.

Review of Theory

• Landau potential for a continuous ferromagnetic model is:
  

  $$V(S) = \frac{1}{2} r_0 S_\alpha S_\alpha + \frac{1}{4} u_0 S_\alpha S_\alpha S_\beta S_\beta,$$ (1)

  
  
  where $S_\alpha$ are the cartesian components of the coarse grain block spins.

• Define the block spin $\vec{S}(\vec{r})$ in terms of its mean value (the order parameter) plus a fluctuation:
  

  $$\vec{S}(\vec{r}) = \vec{m} + \Delta \vec{m}$$ (2)

  
  

• Further decompose the fluctuation into a longitudinal and transverse piece:
  

  $$\vec{S}(\vec{r}) = (m + \Delta m_{\vert\vert})\vec{n} + \Delta\vec{m}_\perp$$ (3)

  
  

• Derive the following general form for the free energy in terms of the transverse coarse grained spin fluctuation gradient $\nabla \vec{m}_\perp(\vec{r})$:
  

  $$F(\Delta \vec{m}) = {\rm longitudinal part} + \frac{1}{2} \int d^d\vec{r} b \cdot (\nabla \vec{m}_\perp(\vec{r}))^2.$$ (4)

  
  
  (with phenomenological parameter $b$, the ``spin wave stiffness''.

• One can relate a state of uniform twist angle to the gradient of the transverse spin fluctuation via $\nabla \phi = \vert\nabla \vec{m}_\perp\vert/m$. Substituting and differentiating to find the free energy density, one obtains the following two relations:
  

  $$\frac{dF}{dV} = \frac{1}{2} \Upsilon(T) (\nabla \phi)^2$$ (5)

  
  
  with
  

  $$\Upsilon(T) = b m^2(T).$$ (6)

  
  

• In Landau theory, $b$ approximately constant so $\Upsilon \sim m^2(T)$ as $T \to T_c$ from below. In detailed treatment one gets corrections:
  

  $$\Upsilon(t) \sim (-t)^{2\beta - \eta\nu}$$ (7)

  
  

• Finally, to use finite-size scaling theory (FSST) to extract the critical exponent, we must substitute $-t \to L^{-1/\nu}$, or
  

  $$\Upsilon(L) \sim (L)^{-2\beta/\nu + \eta}$$ (8)

  
  

In the last expression, $\Upsilon(L) \approx L^{-2\beta/\nu}$. $2\beta/\nu$ term from $m^2$ clearly dominant ($\eta$ is very small, $\sim 0.01$, for this model, while $\beta \approx 1/3$ and $\nu \approx 2/3$). The helicity modulus should vanish sharply near $T_c$ according to Landau theory.

But...

We cannot directly measure the free energy density $dF/dV$. We can directly measure the enthalpy density $E$. Following an identical argument:

$$\Delta E(\Theta) \approx \frac{1}{2} \Upsilon_E(T)(\nabla \phi)^2$$ (9)

where $\Delta E(\Theta)$ is the change in internal energy caused by twisting the boundary conditions through the angle $\Theta \le \pi/2$ with either helicity. From this obvious substitutions yield:

$$\Upsilon_E(T) = \frac{2L^2}{\Theta^2} \Delta E(\Theta)$$ (10)

With a page or two of algebra we can show that:

$$\Upsilon_E(t) \sim t^{v_E} \sim t^{-\phi}$$ (11)

with the critical exponent

$$v_E = -\phi = 1 - 2\nu - \alpha$$ (12)

This is what we wish to measure, in part to invert this equation and deduce the values of $\nu$ and $\alpha$.

Note that as before, if we make the finite size scaling hypothesis we will actually measure:

$$\Upsilon(L) \sim (L)^{-v_E/\nu} \sim (L)^{2 - \frac{1 - \alpha}{\nu}}$$ (13)

or

$$-v_E/\nu = 2 - \frac{1-\alpha}{\nu}$$ (14)

The enthalpy helicity should thus diverge at $T_c$.

It is easy to show that:

$$d-2 -v_E/\nu = d - \frac{1-\alpha}{\nu}$$ (15)

$$1 - v_E/\nu = \frac{1}{\nu}$$ (16)

$$\nu = \frac{1}{1 - \frac{v_E}{\nu}}$$ (17)

where the second step uses ``hyperscaling'' (widely believed but by no means proven for this model) to eliminate $\alpha$ for $d = 3$. With this we can compute $\alpha$ and $\nu$ given $-v_E/\nu$ and possibly check hyperscaling.

Measuring $\Upsilon(T,L)$ with Monte Carlo

• Calculations were performed on several generations of ``brahma'' (our beowulf compute cluster, also ganesh and rama).
• Heat bath only (cluster method a bit difficult if boundary layers are ``frozen'').
• Equilibrate $L \times L \times L$ lattice with periodic boundary conditions.
• ``Freeze'' (x,y,1) layer of spins.
• Rotate (x,y,1) spins through angle $\theta$ and store them in (x,y,L+1) layer (replacing PBC's in z-direction with frozen twisted PBC's).
• Re-equilibrate only the (x,y,2) to (x,y,L) spins with the heat bath (with PBC's in the x and y directions).
• Sample
• Repeat (easiest to restore PBC's, re-equilibrate, repeat).
• Sweep angles $\theta \in [0,\pi/2]$, $L \in [8,48]$ at $T_c$.
• Fit $\frac{E}{L^3} = \frac{1}{2} \Upsilon(L) (\nabla \phi)^2$ where $\nabla \phi = \theta/L$.
• Fit $\Upsilon(L) = L^{x/\nu}$
• Obtain $\nu$, $\alpha$ from hyperscaling.

Results

Figure 1: E per spin as function of interlayer twist angle $\Delta \phi$ for $L = 64$ (in progress). This is fit to obtain the helicity.

Figure 2: $\Upsilon (T)$ for $L = 8$ and $L = 16$, evaluated at low precision to get trend only.

Figure 3: The helicities for various $L$ at $T_c$. The nonlinear least squares fit of this yields $x/\nu = -v_e/\nu$.

Best result to date: $x/\nu = 0.353 \pm 0.02$

Conclusions

• The only direct measurement of this quantity to date.

• $\nu = \frac{1}{1 - \frac{v_E}{\nu}} = \frac{1}{1.353} = 0.739 \pm 0.01$. This is quite large compared to most other Monte Carlo results (which tend to yield $\nu \approx 0.705 \pm 0.01$) but is not inconsistent with the most recent renormalization predictions.

• The hyperscaling relation itself then yields $\alpha = -0.222$. This is a weakly singular quantity and is very difficult to measure. This is a major motivation of this work.

• For this particular talk, we emphasize that there are easily more than 30 ``GFLOP-years'' of effort in this result (whatever you consider a GFLOP to be). (32x400x3 = 38400) + (16x1300x2 = 41600) + (32x1600x1 = 51200) = 131.200 GHz-years, supercomputing indeed. Impossible without the beowulf/cluster model.