Mean Field Solution of Ising Model

Now that we understand the variational principle and the non-interacting Ising Model, we're ready to accomplish our next task. We want to understand the general d-dimensional Ising Model with spin-spin interactions by applying the non-interacting Ising Model as a variational ansatz. In other words, we want to use what we just derived for independent, uncorrelated spins as a trial Hamiltonian in a variational treatment of the full Ising model.

As we'll see, we'll be able to interpret our results as a mean-field model, where each of the spins on the Ising lattice experiences the ‘‘average’’ field produced by all its neighbors. We'll discuss whether or not this brutal averaging is justified, and talk about under what circumstances it's more or less reasonable.

As an added bonus, we'll also encounter our first example of a continuous phase transition. In this dramatic process, the spins go from a disordered random phase with no overall magnetization to an distinct low-temperature ordered ferromagnetic phase where a macroscopic fraction of the spins are all aligned in the same direction. It's really quite surprising that Nature exhibits these sorts of marvelous ‘‘collective conspiracies’’, and it's even more surprising that they occur at a single well-defined temperature called the critical temperature T_c

Finally, we'll talk about how to figure out the value of the critical temperature T_c

by drawing a picture and seeing where two graphs intersect.

Food for thought

We solved the 1D Ising model by using transfer matrices…but why can't you solve the general n-dimensional Ising model with transfer matrices?

Variational Solution to Ising Model

Here our full Ising model has an energetic coupling between neighboring spins sigma_i

and

and the external field

acting on all the spins. In our variational ansatz $H_{tr}$ , we pretend that the spins are actually decoupled from each other (that is, there's no sigma_i sigma_j

terms), and that each of the spins instead experiences an ‘‘effective external field’’ b_j

. (We'll discuss how to interpret b_j

later on.)

Our objective here is to find the values of b_j

that make our trial Hamiltonian the ‘‘best’’ possible approximation to the actual Hamiltonian. As we learned, the variational principle says us that the ‘‘best’’ value of b_j

is the one that minimizes the variational free energy

(Remember that the subscript

means that we're finding thermal averages using the weights from the trial ensemble. This statement is much fancier than it sounds; it just means that whenever we see something such as $langle sigma_j rangle_{tr}$ , we just replace it whatever the solution to the trial Hamiltonian was. For instance, in this example, we're using the non-interacting Ising model as a trial Hamiltonian, so we plug in the formulas at the bottom of the last page.)

Finding the variational free energy

We have no recourse but to plug everything into the formula for $F_{var}$ . We find that

And since expectations values are linear, we can pull the expectation values inside to go around the sigmas. (If you want to convince yourself of this, you can try writing out the difinition of langle ldots rangle

Before moving on, we can make one more simplification: since we picked a non-interacting trial Hamiltonian, the spins are uncorrelated, so $langle sigma_i sigma_j rangle_{tr}$ just factors into $langle sigma_i rangle_{tr} langle sigma_j rangle_{tr}$ . (We can't always do this; we just happened to pick a non-interacting trial Hamiltonian today.)

(I changed the labels of the indices in preparation for the next step where we differentiate w.r.t. b_j

.) At this point, we can expand out our expressions for $F_{tr}$ and langle sigma_j rangle

, but a lot of the terms will cancel out later on, so for now, we'll just leave this expression as it is. (Prof. Kivelson calls this ‘‘being clever,’’ but I call it ‘‘hindsight is 20-20.’’)

Finding the derivative

Since we want to minimize $F_{var}$ with respect to the variational parameters b_j

, we just take the derivative and set it equal to zero. Nothing particularly fancy here yet. For now, we'll leave everything in terms of langle sigma_j rangle

's because we're ‘‘clever.’’

where in the second sum, I'm summing over sites

that are nearest neighbors (n.n.) of site

Now time for some magic: the first and the third term cancel out, and then we can factor out a ${partial langlesigma_jrangle_{tr} over partial b_j}$ from what's left! And then when we set the derivative equal to zero, we're left with

Finally, at this point we finally need to plug in our expression for $langlesigma_irangle_{tr}$ from the previous page; we end up with a relation between the external field of a spin b_j

and the external field of its neighboring spins b_i

Translational invariance

Hooray, we finally have an expression for the b_j

's that satisfy our variational condition ${partial F_{var} over partial b_j} = 0$ . Unfortunately, all the b_j

's are related to each other in a nasty sort of way involving hyperbolic tanhs, so it all looks kind of messy.

So to proceed further, we make the assumption of translational invariance that all spins really experience the same ‘‘external’’ field; i.e., that b_j

equals the same

for all the sites

Remark

What's the motivation exactly for assuming b_j = b ? We tried to justify it in section by some argument about the zero-temperature limit, but I wasn't fully convinced, because the variational parameters depend on temperatures anyways.

A few weeks later in class, when we learned about Landau-Ginzburg theory, we were able to give a more convincing argument for why a spatially uniform solution for b_j is indeed the ‘‘best’’ solution.

Once we assume that b_j

is the same for all sites, we can simplify our result quite a bit, because we only need to find one unknown

now. The

on the LHS is the same as b_i

on the RHS. Our sum over the nearest neighbors of site

ends up being the same for each of its neighbors, so rather than summing, we just need to multiply by the number of nearest neighbors

Remark

The number of nearest neighbors in an Ising model on a -dimensional hypercube lattice is z = 2d . For instance, on a chain ( d=1 ), you have 2 neighbors to your left and right; on a square grid ( d=2 ), you have 4 neighbors (left, right, up, down); on a cubic lattice ( d=3 ), you have 6 neighbors ( pm hat x, pm hat y, pm hat z ), and so on and so forth.

Another Remark

It's confusing to write because the little looks like a subscript on , which is reminiscent of the J_z spin operator in quantum mechanics (and also reminiscent of a certain rapper…)

On the other hand, means a zeptojoule or $10^{-21}$ Joules! And wouldn't you know, the thermal energy k_b T is typically on the order of zeptojoules; at room temperature T sim 300textrm{K} , it's roughly 4 zJ….I swear there's a conspiracy going on here.

Solving for b

At this point, we have an expression involving

on the LHS and the RHS. If we think back to the bigger picture, we have just learned that the value of

which satisfies this equation is the ‘‘best’’ possible value of

in our trial Hamiltonian. Unfortunately, we can't explicitly solve for

in a closed-form way…so how do we proceed further?

We'll appeal to the maxim: when in doubt, simplify things and draw a picture. So let's set h=0

for now, and draw a picture, and see if we can gain any qualitative insight.

Our strategy here is to plot both sides of the equation and see where the graphs intersect. The intersection points will tell us the solutions for

. In this case, we want to where the graphs of y = b

and

intersect:

Surprisingly, the number of graph crossings depends on the temperature! This is by far the most interesting thing we've discovered in class so far: the behavior of the mean-field Ising model is completely different at low temperatures than at high temperatures. We've discovered a phase transition.

If you start out at high temperature, you'll only find a trivial solution at b=0

, but when you lower the temperature past a special critical temperature T_c

, all of a sudden, there will be nonzero solutions b=pm b_0 neq 0

as well. This is rather remarkable; it tells us the system behaves dramatically differently depending on whether T > T_c

. The high-temperature regime is called the paramagnetic phase and the low-temperature regime is called the (anti)ferromagnetic phase.

Physical interpreation

Let's think back to our main physical observable, the magnetization density m = langle sigma_j rangle

which tells us whether the spins tend to align up or down in thermal equilibrium. (To calibrate our intuition, it's helpftul to review how non-interacting spins behave at different temperatures.)

In our mean-field model, the magnetization density is given by m = tanh(beta b)

, which means at low temperatures when

is nonzero, there is a net nonzero magnetic moment in the system.

Wait a second. Let's think about how unusual this is. Since we dialed down the external field

to zero, there is no energetic incentive whatsoever for the spins to want to align one way versus another. There's nothing pulling on the spins to make them want to point in a particular. Yet we've just discovered that there's a nonzero net magnetic moment at low temperatures; in other words, that all the spins are preferring to point in some direction over the other! How is this possible?

Well, believe it or not, this is how nature behaves. It's what you might call a collective conspiracy – none of the individual spins feel an external field, but if enough of them happen to fluctuate and point in a particular way, they'll set up a strong enough ‘‘local field’’ to convince their neighbors to the same way – so strong, in fact, that a nontrivial fraction of the

spins also ‘‘get in on the conspiracy’’, and the material develops long-range order and gains a macroscopic magnetic moment. Truly remarkable, especially if you consider just how many spins there are ( $N sim 10^{23}$ ), and just how weak each individual magnetic moment is. Magical.

Behavior of Mean-field Ising Ferromagnet

At high temperatures , the spins exhibit a disordered paramagnetic phase, pointing in every which way. The average magnetization of the system is zero.
At low temperatures , the spins exibit an ordered ferromagnetic phase, where they all conspired to point in one direction (even when there's no external field telling them which way to point!) The average magnetization of the system is (shockingly) nonzero.

How come long-range order only develops at low T?

Here is a qualitative picture for why high temperature favors disorded phases. At sufficiently high temperatures, all the spins are jiggling around with so much thermal energy that any local conspiracies of all-up-zones or all-down-zones are quickly jiggled away out of existence before they have time to spread into a macroscopic fraction of the material.

There's also a nice argument that comes from the interplay between the two terms of the free energy F = E - TS

. Each phase has its own energy

and entropy

, and depending on the value of the temperature

, the free energy

of one phase might be higher than the other.

Which phase has the lower free energy? At high temperatures, when the entropic contribution is more heavily weighted, the paramagnetic phase flies off to an entropic heaven of possibilities, whereas at low temperatures, when all the system wants to do is to minimize energy, the orderly ferromagnetic phase wins out.

Solving for the critical temperature

Alright, enough qualitative remarks, I'm starting to sound like a chemist. Let's move on to solve for the critical temperature T = T_c

Our goal here is to find the special temperature T_c

that separates the nonzero-solution-regime from the no-nonzero-solution-regime (sorry, these words are confusing). There's a simple condition that tells you whether or not the green curve will hit the blue curve again after b=0

. Can you figure it out by staring at the picture?

Here's the secret: The green curve has to be steep enough at b=0

for the blue curve to ‘‘catch up to it’’ later on and intersect the green line again. If the green line's steeper than the blue line at b=0

, then there'll be a nonzero solution for

, and if it's shallower, there won't be. And if the green line is exactly as steep as the blue line at b=0

, then we're at the critical temperature that separates the two possibilities. Clever, eh?

This little observation gives us a mathematical procedure to find the critical temperature: we take the derivative of the LHS and RHS, set them equal to each other at b=0

, and solve for beta

. The slope of the blue curve y=b

is 1, and the slope of the green curve at b=0

Remark on thermodynamics

Why does taking a derivative of the free energy w.r.t. the external field give you the magnetization ${partial F_{tr} over partial b_j} = - langle sigma_j rangle_{tr}$ ?

Well, you can see this directly by plugging in F = -T log Z and then expanding out the partition function. It's a pretty fun exercise to work this out yourself…remember that (log f(x))' = f'(x) / f(x) …

If you're feeling more thermodyanmically inclined, you can also convince yourself that {partial F over partial H} = - M via the following thermodynamic argument. (Now I'm calling the external field and the magnetization in the true thermodynamic spirit.)

When we include the external magnetic field into a thermodynamic treatment, our internal energy function E = E(S,H) gains an extra parameter . The external field gives us another handle to perform work on the system; when we twiddle it by an amount , the energy of the system changes by an amount -M dH , where we recognize the magnetization as the conjugate variable to the external field . (Think back to E'n'M why magnetic dipole times magnetic field gives you an energy…)

Once we include a magnetic field, we can either change the internal energy by adding some heat T dS or by performing some magnetic work -MdH , which means that the total differential of the internal energy is

However, since we're considering the Ising model in a canonical ensemble at a fixed temperature , it doesn't make much sense to talk about derivatives of the (internal) energy , because is a natural function of the entropy rather than the temperature . So even though we can find the magnetization in principle as a derivative of the internal energy as $M = {partial E over partial H} |_S$ , it's pretty physically useless to do so, because that derivative holds entropy constant rather than the temperature.

To account for this issue, we perform a Legendre transformation to the free energy

which then becomes a natural function of the temperature as F = F(T,H) . The differential of free energy can then be expressed as

from which we can read off the magnetization

$M = - {partial F over partial H}bigg|_T$

where the derivative is now taken at a fixed temperature rather than at fixed entropy, meaning that it aligns with our treatment of the Ising Model in the constant-temperature canonical ensemble of statistical mechanics.

Okay, to be fair, I didn't really prove anything in the above, I just kind of pulled the -H dM term out of thin air and then waved my hands around for a sort of ‘‘plausibility’’ argument. But I do think it's a reasonable sort of thing to say. Hooray, thermodynamics.