What Happens When a Magnet Attracts Metal? (Part 4): How 10²³ Spins Agree, and a Bridge to the Higgs

The finale. Part 1 showed classical physics forbids magnets; Part 2 pulled the electron’s intrinsic moment ( $$g=2$$ ) out of Dirac’s equation; Part 3 traced spin back to the symmetry of spacetime itself. So we have explained one electron’s magnetism. But a fridge magnet has on the order of $10^{23}$ electrons, and the question that remains is the strangest of all: why do they agree? What lines them up — and why, as we are about to see, is the force that does the lining-up not a magnetic force at all?

Let me state the trap we are walking into before we spring it. After three parts of work we have earned a single electron that behaves like a tiny magnetic needle. That feels like victory. It is not. Take $10^{23}$ of these needles, drop them in a bar of iron, and the default outcome is that thermal jostling points them every which way and the bar is dead — no net field, nothing sticks to your refrigerator. To get a magnet, something has to reach in among all those spins and force them to agree. Part 4 is the story of who does the forcing. The punchline, which I will spoil now because it is too good to bury, is that the aligning agent is not magnetism. It is the Pauli exclusion principle wearing electrostatics as a costume.

8. Heisenberg Model

We now know that every electron is a tiny magnetic needle ( $$g=2$$ ). However, if you put a bunch of these tiny magnets together, thermal agitation at room temperature is sufficient to completely randomize their orientations, resulting in zero macroscopic magnetic moment (paramagnetism). To form ferromagnetism, there must be an extremely strong “coupling force” between spins that forces them to align. The classical magnetic dipole-dipole interaction is far too weak — only about one ten-thousandth of the thermal energy. So we can rule the obvious suspect out immediately: the little needles do not line up by tugging on each other magnetically; that force loses to room temperature by four orders of magnitude. The real power comes from the combination of the identical particle statistics (Spin-Statistics) we mentioned earlier and the Coulomb interaction. This is known as the Exchange Interaction.

To demonstrate the essence of this, let us consider the simplest model: a two-electron system (such as a Helium atom or electrons on two adjacent Iron atoms). Assume there are two electrons, 1 and 2, and two spatial orbitals $\psi_a(\vec{r})$ and $\psi_b(\vec{r})$ , where $\psi_a(\vec{r})$ is localized near atom A and $\psi_b(\vec{r})$ is localized near atom B. We assume these two orbitals are orthonormal: $\langle \psi_a | \psi_b \rangle = 0$ . Note that this orthogonality is a prerequisite assumption for the Heisenberg model we are about to derive. Even if they are not orthonormal, we can create two new orthonormal orbitals through a basis transformation. Assuming their overlap integral is small greatly simplifies the calculation of the two-electron system, but this does not mean there is no interaction between the two electrons, as the interaction involves exchange integrals which generally are not zero.

The total Hamiltonian of this system is $H = H_0 + H_{int}$ , where $$H_0$$ is the single-electron part (kinetic energy + nuclear potential energy), and $H_{int}$ is the Coulomb interaction between the two electrons: $H_{int} = \frac{e^2}{|\vec{r}_1 - \vec{r}_2|}$ . Note: there is absolutely no magnetic interaction term here, only pure electrostatic repulsion. Keep your eye on that absence — it is the whole point. Whatever ends up aligning the spins is going to have to come out of a Hamiltonian in which spin does not appear at all.

According to the Fermi statistics hypothesis for identical particles in quantum mechanics, the total wavefunction of the electrons $\Psi(1,2)$ must change sign (be antisymmetric) under the action of the particle exchange operator $P_{12}$ :

P_{12} \Psi(1,2) = -\Psi(1,2)

Since the total wavefunction consists of a spatial part $\phi(\vec{r}_1, \vec{r}_2)$ and a spin part $\chi(s_1, s_2)$ : $\Psi = \phi \otimes \chi$ . We first derive the spin part, and then obtain the spatial part based on the overall antisymmetry. We have two spin-1/2 particles (e.g., two electrons), so there are a total of $2 \times 2 = 4$ possible product states (uncoupled basis). We want to add their spins together to see what the total spin $\vec{S}_{tot} = \vec{S}_1 + \vec{S}_2$ looks like, finding the common eigenstates $|S, M\rangle$ of the total spin operator $\hat{S}^2$ and total magnetic quantum number $\hat{S}_z$ . According to angular momentum addition rules, the total spin $$S$$ formed by two spins of $$1/2$$ can be: $$S = 1/2 + 1/2 = 1$$ (Triplet, with 3 $$M$$ values: $$+1, 0, -1$$ ) and $$S = 1/2 - 1/2 = 0$$ (Singlet, with 1 $$M$$ value: $$0$$ ).

The Triplet corresponds to three components. The total magnetic quantum number $$M$$ is the sum of the magnetic quantum numbers of the two particles: $$M = m_1 + m_2$$ . To get $$M=1$$ , the only possibility is both electrons are spin-up: $$1/2 + 1/2 = 1$$ . So, the first member of the Triplet is determined: $|1, 1\rangle = |\uparrow\uparrow\rangle$ . Then, using the lowering operator ( $$S_-$$ ), we get the intermediate state ( $$M=0$$ ) from $|1, 1\rangle$ to derive $|1, 0\rangle$ . Using the total lowering operator $\hat{S}_- = \hat{S}_{1-} + \hat{S}_{2-}$ acting on the state $|j, m\rangle$ : $J_- |j, m\rangle = \hbar \sqrt{j(j+1) - m(m-1)} |j, m-1\rangle$ , acting on the coupled state on the left gives:

\hat{S}_- |1, 1\rangle = \sqrt{1(1+1) - 1(1-1)} |1, 0\rangle = \sqrt{2} |1, 0\rangle

Acting on the product state on the right gives:

\begin{aligned} (\hat{S}_{1-} + \hat{S}_{2-}) |\uparrow\uparrow\rangle &= (\hat{S}_{1-} |\uparrow\rangle_1)|\uparrow\rangle_2 + |\uparrow\rangle_1 (\hat{S}_{2-} |\uparrow\rangle_2) \ = |\downarrow\rangle_1|\uparrow\rangle_2 + |\uparrow\rangle_1|\downarrow\rangle_2 \ = |\downarrow\uparrow\rangle + |\uparrow\downarrow\rangle \end{aligned}

Thus we obtain:

|1, 0\rangle = \frac{1}{\sqrt{2}} (|\uparrow\downarrow\rangle + |\downarrow\uparrow\rangle)

The lowest weight state ( $$M=-1$$ ) is also simple; only both down can give $$-1$$ : $|1, -1\rangle = |\downarrow\downarrow\rangle$ . So the spin part of the Triplet is symmetric (does not change sign upon exchange), which means the spatial part must be antisymmetric.

The quantum numbers for the Singlet are $$S=0, M=0$$ . It must be some linear combination of the product states $|\uparrow\downarrow\rangle$ and $|\downarrow\uparrow\rangle$ with $$M=0$$ : $|0, 0\rangle = a |\uparrow\downarrow\rangle + b |\downarrow\uparrow\rangle$ . Since eigenstates with different quantum numbers must be orthogonal, the Singlet $|0, 0\rangle$ must be orthogonal to $|1, 0\rangle$ in the Triplet. Solving this gives:

|0, 0\rangle = \frac{1}{\sqrt{2}} (|\uparrow\downarrow\rangle - |\downarrow\uparrow\rangle)

The spin part of the Singlet is antisymmetric, so the corresponding spatial part must be symmetric. Summarizing:

Spatial Symmetric $\otimes$ Spin Antisymmetric (Singlet) Energy: $E_S$
- Spin Part (Antisymmetric): $$S=0$$ Singlet $\chi_S = \frac{1}{\sqrt{2}}(\uparrow\downarrow - \downarrow\uparrow)$ .
- Spatial Part (Symmetric): $\phi_S(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}} [\psi_a(\vec{r}_1)\psi_b(\vec{r}_2) + \psi_a(\vec{r}_2)\psi_b(\vec{r}_1)]$ .
Spatial Antisymmetric $\otimes$ Spin Symmetric (Triplet) Energy: $E_T$
- Spin Part (Symmetric): $$S=1$$ Triplet $\chi_T = \{\uparrow\uparrow, \frac{1}{\sqrt{2}}(\uparrow\downarrow + \downarrow\uparrow), \downarrow\downarrow\}$ .
- Spatial Part (Antisymmetric): $\phi_A(\vec{r}_1, \vec{r}_2) = \frac{1}{\sqrt{2}} [\psi_a(\vec{r}_1)\psi_b(\vec{r}_2) - \psi_a(\vec{r}_2)\psi_b(\vec{r}_1)]$ .

Pause on what this bookkeeping has quietly accomplished, because it is the secret engine of all magnetism. The two electrons want to align their spins or anti-align them, and the only thing forcing a choice is antisymmetry of the total wavefunction. If the spins are parallel (Triplet), the spatial part must be antisymmetric, which means $\phi_A$ vanishes whenever $\vec{r}_1 = \vec{r}_2$ — the electrons are forbidden from sitting on top of each other, so they keep their distance and pay less Coulomb repulsion. If the spins are anti-parallel (Singlet), the spatial part is symmetric and the electrons are perfectly happy to overlap — and so they pile up more Coulomb energy. The spin orientation, which the Hamiltonian never mentions, has been smuggled into the energy through the geometry the spins are forced to adopt. That is the whole trick. Now let us turn it into a formula.

Our goal is to find a mathematical expression $\hat{H}_{eff}$ containing only spin operators $\vec{S}_i$ and $\vec{S}_j$ such that when it acts on the Singlet and Triplet states, it automatically yields the corresponding energies $$E_S$$ and $$E_T$$ . To construct this Hamiltonian, the most natural building block is the dot product of the two spins $\vec{S}_i \cdot \vec{S}_j$ . We need to calculate the eigenvalues of this operator for the Singlet and Triplet states. Define the total spin operator for the two-electron system: $\vec{S}_{tot} = \vec{S}_i + \vec{S}_j$ . Squaring the total spin operator allows us to solve for the dot product term: $\vec{S}_i \cdot \vec{S}_j = \frac{1}{2} \left( \vec{S}_{tot}^2 - \vec{S}_i^2 - \vec{S}_j^2 \right)$ . Using the eigenvalue formula for the square of the angular momentum operator in quantum mechanics $\hat{S}^2 |s\rangle = s(s+1) |s\rangle$ (omitting $\hbar^2$ for brevity, or treating spin as dimensionless):

For a single electron ( $$s=1/2$$ ): $\vec{S}_i^2 = \vec{S}_j^2 = \frac{1}{2}\left(\frac{1}{2} + 1\right) = \frac{3}{4}$
For the Singlet ( $S_{tot}=0$ ): $\vec{S}_{tot}^2 |S\rangle = 0(0+1) |S\rangle = 0$
For the Triplet ( $S_{tot}=1$ ): $\vec{S}_{tot}^2 |T\rangle = 1(1+1) |T\rangle = 2 |T\rangle$ . Now, substituting these values back into the dot product formula to calculate the eigenvalues:
Dot product value for Singlet: $(\vec{S}_i \cdot \vec{S}_j) |S\rangle = \frac{1}{2} \left( 0 - \frac{3}{4} - \frac{3}{4} \right) |S\rangle = -\frac{3}{4} |S\rangle$
Dot product value for Triplet: $(\vec{S}_i \cdot \vec{S}_j) |T\rangle = \frac{1}{2} \left( 2 - \frac{3}{4} - \frac{3}{4} \right) |T\rangle = \frac{1}{2} \left( \frac{1}{2} \right) |T\rangle = \frac{1}{4} |T\rangle$

We assume the effective Hamiltonian $\hat{H}_{ij}$ has the following linear form (this is the most general rotationally symmetric form): $\hat{H}_{ij} = C_0 + C_1 (\vec{S}_i \cdot \vec{S}_j)$ , where $$C_0$$ is just a constant energy shift independent of spin configuration, which can be discarded (or the energy zero point redefined) when studying phase transitions and spin dynamics. Substituting the data for Singlet and Triplet, we have $E_S=C_1(-\frac{3}{4}), E_T=C_1(\frac{1}{4})$ . Defining the constant $J \equiv E_S - E_T$ , we obtain the two-particle Hamiltonian:

\hat{H}_{ij} = -J (\vec{S}_i \cdot \vec{S}_j)

If $$J>0$$ ( $$E_S > E_T$$ ): The coefficient is negative. The larger the dot product (parallel, +1/4), the lower the energy. This is Ferromagnetism.
If $$J<0$$ ( $$E_S < E_T$$ ): The coefficient is positive. The smaller the dot product (anti-parallel, -3/4), the lower the energy. This is Antiferromagnetism.

In other words: an interaction that was purely electrostatic has reorganized itself into a term that talks only about spin. No magnetic force ever entered the calculation. The energy scale of $$J$$ is set by Coulomb repulsion — electron-volts, not the microelectron-volts of dipole-dipole coupling — which is exactly why it can beat room temperature and hold $10^{23}$ spins in formation. This is the answer to the question we opened with: spins align because the Pauli principle ties their orientation to how much electrostatic energy they have to pay, and electrostatics is strong.

Now we generalize this two-particle interaction to the entire lattice. Assuming each electron $$i$$ only interacts with its nearest neighbors. We need to sum over all atoms in the lattice. To correct for double counting:

H_{exchange} = -\frac{J}{2} \sum_{i,j \text{ neighbor}} \vec{S}_i \cdot \vec{S}_j

Finally, we must consider the interaction of each electron spin with an external uniform magnetic field $\vec{B}$ . This is a single-body interaction and does not involve neighbors. Recalling our conclusion from the Dirac equation, the electron has a spin magnetic moment:

\vec{\mu}_S = -g \frac{e}{2m} \vec{S}

To make the formula more concise and universal, physicists define a natural combination of constants called the Bohr Magneton. The Bohr Magneton is the natural unit of magnetic moment in atomic physics. It is defined as:

\mu_B \equiv \frac{e\hbar}{2m_e}

This physical quantity contains three fundamental constants: elementary charge $$e$$ , Planck constant $\hbar$ , and electron mass $$m_e$$ . It represents the magnitude of the orbital magnetic moment generated by a classical electron moving in the ground state orbit of a hydrogen atom. This is in SI units; in Gaussian units, it also includes the speed of light: $\mu_B \equiv \frac{e\hbar}{2m_ec}$ . If we treat the spin operator $\vec{S}$ as a dimensionless operator (i.e., eigenvalues are $$1/2$$ instead of $\hbar/2$ ), then the true physical angular momentum is $\hbar \vec{S}$ . Extracting this $\hbar$ and combining it with the constants above:

\begin{aligned} \vec{\mu}_S &= -g \frac{e}{2m_e} (\hbar \vec{S}_{\text{dimensionless}}) = -g \left( \frac{e\hbar}{2m_e} \right) \vec{S}= -g \mu_B \vec{S} \end{aligned}

When an electron is in an external magnetic field $\vec{B}$ , its potential energy (Zeeman Energy) is given by the classical electromagnetism formula $U = -\vec{\mu} \cdot \vec{B}$ . Substituting the magnetic moment expression:

\begin{aligned} U_{Zeeman} &= - (-\vec{\mu}_S) \cdot \vec{B} \quad (\text{Note the sign: Potential energy is usually defined as } -\vec{\mu}\cdot\vec{B}) \\ &= - (-g \mu_B \vec{S}) \cdot \vec{B} \\ &= g \mu_B \vec{S} \cdot \vec{B} \end{aligned}

Physical Convention on Signs: In condensed matter physics, we usually want the Hamiltonian to reflect energy minima. Electrons are negatively charged, so the magnetic moment $\vec{\mu}$ is anti-parallel to the spin $\vec{S}$ . The lowest energy state is when the magnetic moment $\vec{\mu}$ is parallel to the magnetic field $\vec{B}$ . This means the spin $\vec{S}$ is anti-parallel to the magnetic field $\vec{B}$ . To avoid dealing with cumbersome negative signs, or to make spin look like it aligns “with” the field (defining $\vec{S}$ to point in the direction of the magnetic moment rather than angular momentum), literature sometimes adjusts the definition. However, the standard derivation (keeping the electron’s negative charge) gives the Zeeman term as $+ g\mu_B \vec{S} \cdot \vec{B}$ or $- \vec{\mu} \cdot \vec{B}$ . Nevertheless, in the customary notation of the Heisenberg model, for mathematical symmetry and ease of discussion (e.g., assuming $$g$$ is negative or redefining the spin direction), the Zeeman term is typically written with a negative sign, indicating that spins tend to align along the field (this is a phenomenological treatment):

H_{Zeeman} = -g \mu_B \sum_i \vec{S}_i \cdot \vec{B}

(This implies that energy is minimized when $\vec{S}_i$ is in the same direction as $\vec{B}$ . This implies we have redefined the spin direction, or taken $$g$$ to be negative. In phenomenological models, we only care about: which direction does the magnetic field tend to pull the spins.)

Now, we combine the two pieces of the puzzle: Internal Interaction (Exchange Energy generated by the Pauli principle and Coulomb force) and External Interaction (Coupling of magnetic moment with external field generated by relativistic quantum effects). Adding them together, we finally obtain the core Hamiltonian describing solid-state magnetism — the Heisenberg Model:

\boxed{H = -\frac{J}{2} \sum_{\langle i,j \rangle} \vec{S}_i \cdot \vec{S}_j - g \mu_B \sum_i \vec{S}_i \cdot \vec{B}}

This formula is the cornerstone of modern magnetism. The first term ( $$J$$ ) explains why magnets have magnetism (spontaneous magnetization, ordered alignment of spins). The second term ( $$B$$ ) explains how magnets are controlled by the outside world (magnetization process, hysteresis loop). $\mu_B$ and $$g$$ link microscopic quantum constants ( $\hbar, e, m_e$ ) with macroscopic observable magnetic fields. Notice how the two terms quietly close the arc of this whole series: the first term is the legacy of Part 1’s identical-particle statistics, the second is the legacy of Part 2’s $$g=2$$ . The fridge magnet sits at the meeting point of both.

9. Ising Model

We have completed the construction of the microscopic mechanism (Dirac $\to$ Spin $\to$ Exchange Interaction $\to$ Heisenberg Hamiltonian). Now, we must move from the microscopic to the macroscopic. We have a Hamiltonian that wants spins to align — but wanting is not the same as a magnet. A bar of iron at a billion degrees still has all the exchange coupling it ever had, yet it is not magnetic. Somewhere between “the spins prefer to align” and “the spins actually do align” sits temperature, and the transition between those two regimes is sudden, sharp, and one of the most beautiful things in physics. To see it, we need to handle the Heisenberg Model. However, solving the Heisenberg Model exactly in two or three dimensions is extremely difficult (because it contains non-commuting operators). Therefore, we need to introduce the Ising Model as an approximation and use Mean-Field Theory to demonstrate how symmetry is broken.

We first non-dimensionalize the Heisenberg model and analyze its structure. Assume the external magnetic field is along the $$z$$ -direction: $\vec{B} = (0, 0, B)$ . We expand the dot product of the spin operators $\vec{S}$ into longitudinal ( $$z$$ ) and transverse ( $$x, y$$ ) components:

\vec{S}_i \cdot \vec{S}_j = S_i^z S_j^z + (S_i^x S_j^x + S_i^y S_j^y)

To see the physical meaning of the transverse part more clearly, we introduce Ladder Operators:

S_i^+ = S_i^x + i S_i^y, \quad S_i^- = S_i^x - i S_i^y

Thus, the Heisenberg model can be rewritten in two parts:

H = \underbrace{\left[ -\frac{J}{2} \sum_{\langle i,j \rangle} S_i^z S_j^z - h \sum_i S_i^z \right]}_{\text{Ising Part}} + \underbrace{\left[ -\frac{J}{4} \sum_{\langle i,j \rangle} (S_i^+ S_j^- + S_i^- S_j^+) \right]}_{\text{Flip Part}}

(where $h = g\mu_B B$ ). These two parts have distinct physical meanings:

Ising Part (Longitudinal Term): This term involves only $$S^z$$ . Since $$S_i^z$$ on different lattice sites commute ( $$[S_i^z, S_j^z]=0$$ ), they behave like classical scalar variables. This describes the static alignment of spins along the $$z$$ -axis.
Flip Part (Transverse/Flip Term): This term involves $$S_i^+ S_j^-$$ . Its action is to flip the spin at site $$j$$ down ( $$S^-$$ ) while simultaneously flipping the spin at site $$i$$ up ( $$S^+$$ ): $S_i^+ S_j^- |\downarrow_i \uparrow_j\rangle = |\uparrow_i \downarrow_j\rangle$ . Physically, this represents the movement of spin excitations. Just like a spin-flipped state hopping through the lattice, this corresponds to Spin Waves or Magnons. This is a form of quantum fluctuation that imparts “kinetic energy” to the system, tending to destroy ordered alignment.

In many real magnetic materials, due to the symmetry of the crystal structure, there exists Magnetic Anisotropy. This means the energy of spins along certain directions (e.g., the $$z$$ -axis, the easy axis) is lower than in the $$x,y$$ plane. If the anisotropy is strong enough, or if we are only concerned with phase transition behavior in the classical limit, we can ignore the Flip Part (quantum fluctuations) and retain only the longitudinal term. This is the famous Ising Model.

At this point, we replace the operator $$S_i^z$$ with a classical variable $\sigma_i = \pm 1$ (absorbing the coefficients into $$J$$ ):

H_{Ising} = - \frac{J}{2} \sum_{\langle i,j \rangle} \sigma_i \sigma_j - h \sum_i \sigma_i

This is a massive simplification: we turn a non-commuting quantum matrix problem into a classical statistical combinatorial problem. The 2D Heisenberg model cannot be solved exactly to this day, though we can solve it using computational methods. However, the Ising model within it is relatively simpler. The solution to the 1D Ising model is very simple; there is no phase transition in 1D (Ising, 1924). But the 2D case is also very difficult to solve. It wasn’t until the brilliant Lars Onsager published a very simple paper stating that he had solved it — without providing the solution details, but giving the critical temperature and pointing out that it is ferromagnetic at low temperatures and paramagnetic at high temperatures — that we had the first exactly solvable model exhibiting a ferromagnetic phase transition. He withheld the solution process until C.N. Yang saw the paper and provided a solution, which was still famously difficult. To understand the physical picture intuitively, we will adopt the Mean-Field Approximation here.

The problem we face is many-body coupling: the state of $\sigma_i$ depends on $\sigma_j$ , and $\sigma_j$ depends on $\sigma_k$ … This chain reaction makes the partition function difficult to calculate. The Mean-Field idea is delightfully lazy, and that is its strength: when we focus on atom $$i$$ , we don’t care whether neighbor $$j$$ is flipping between $$+1$$ or $$-1$$ ; we only care about the average influence of the neighbors. We write $\sigma_j$ as an average value plus a fluctuation: $\sigma_j = \langle \sigma \rangle + \delta \sigma_j$ . Ignoring the second-order fluctuation term $\delta \sigma_i \delta \sigma_j \approx 0$ , the Hamiltonian can be linearized into a single-body form:

H_{MFA} = - \sum_i \sigma_i \underbrace{\left( J \sum_{j \in \text{neigh}} \langle \sigma \rangle + h \right)}_{h_{eff}}

This defines the Effective Molecular Field $h_{eff}$ :

h_{eff} = J z \langle \sigma \rangle + h

where $$z$$ is the coordination number (number of neighbors for each atom). Now, the problem becomes the statistical distribution of a single spin in an “external field” $h_{eff}$ . According to the Boltzmann distribution, the probability of this spin being up is proportional to $e^{\beta h_{eff}}$ , and the probability of being down is proportional to $e^{-\beta h_{eff}}$ (where $\beta = 1/k_B T$ ). Thus, the thermodynamic average of this spin $\langle \sigma_i \rangle$ is:

\langle \sigma_i \rangle = \frac{(+1)e^{\beta h_{eff}} + (-1)e^{-\beta h_{eff}}}{e^{\beta h_{eff}} + e^{-\beta h_{eff}}} = \tanh(\beta h_{eff})

Here is the lovely self-referential twist. Each spin responds to the average field set up by its neighbors — but each spin is one of those neighbors. Since the lattice is uniform, $\langle \sigma_i \rangle$ must equal the average value of the field source itself, $m = \langle \sigma \rangle$ . The spin has to agree with the very average it helps create. Substituting $h_{eff}$ , we obtain the famous Self-consistent Equation:

m = \tanh\left( \frac{J z m + h}{k_B T} \right)

Let us consider the most critical case: no external magnetic field (h=0). The equation simplifies to:

m = \tanh\left( \frac{T_c}{T} m \right)

where we package the constants into the definition of the Curie Temperature $$T_c=Jz/k_B$$ . This is a transcendental equation, and we can analyze the behavior of the solution graphically (by finding the intersection of $$y=m$$ and $y=\tanh(\frac{T_c}{T}m)$ ). Everything turns on a single comparison — the slope of the $\tanh$ curve at the origin, $$T_c/T$$ , against the slope of the line $$y=m$$ , which is just $$1$$ :

High-Temperature Phase ( $$T>T_c$$ ): The slope of the tanh curve at the origin is $$T_c/T<1$$ . The line and the curve intersect at only one point, $$m=0$$ . Physical meaning: Thermal agitation is intense; without an external field, there is no magnetism. This is the Paramagnetic Phase.
Low-Temperature Phase ( $$T<T_c$$ ): The slope of the tanh curve at the origin is $$T_c/T>1$$ . The origin $$m=0$$ becomes an unstable solution, and two new stable non-zero solutions $m=\pm m_0$ appear. Physical meaning: Even if $$h=0$$ , the system will spontaneously generate a non-zero magnetization $$m_0$$ . This is the Ferromagnetic Phase.

This is Spontaneous Symmetry Breaking: The Hamiltonian has a flip symmetry $\sigma\to-\sigma$ when $$h=0$$ . However, when the temperature drops below $$T_c$$ , nature is forced to choose between “all up” and “all down.” This choice is not imposed by an external force but is a collective decision made spontaneously by the system to lower its energy (driven by the exchange interaction $$J$$ ). This is precisely the statistical mechanical essence behind the macroscopic phenomenon of a magnet attracting iron. Sit with that word forced — we met it in Part 1, where relativity forced Dirac into matrices, and we are about to meet it one last time on a cosmic scale. The same logic keeps recurring: a symmetry that holds at high energy is not abandoned by choice, it is broken because staying symmetric becomes the more expensive option.

10. Magnetic Domains

Based on the Ising model and Mean-Field Theory, we concluded that when the temperature is below the Curie temperature $$T_c$$ , electron spins spontaneously align, producing a massive macroscopic magnetization $$M$$ . And now the theory hands us an embarrassment. If you go to a hardware store and buy an iron nail (room temperature is obviously far below Iron’s Curie temperature of $$1043K$$ ), it is not magnetic. It does not pick up other objects. By everything we just proved, it should be a magnet — and it sits there inertly in the bin. This is because we ignored one final energy competition.

Our previous Hamiltonian only considered Exchange Energy and Zeeman Energy. However, at the macroscopic scale, there is also the classical Magnetostatic Energy. If all $10^{23}$ atoms in a piece of iron were aligned upwards, this magnet would establish a huge magnetic field in the surrounding space. The magnetic field contains energy density $B^2/2\mu_0$ . Spreading out all those magnetic field lines costs a tremendous amount of energy. To reduce this Magnetostatic Energy, the material spontaneously splits into many tiny regions called Magnetic Domains. Although inside a domain, the exchange interaction keeps spins aligned (satisfying microscopic ferromagnetism), overall, the vector sum of the magnetic moments of the various domains is zero ( $\sum \vec{M}_i = 0$ ). There are no external magnetic field lines, thereby drastically reducing the magnetostatic energy. So the nail is not un-magnetized; it is perfectly magnetized, in millions of little patches that cancel.

The boundary between magnetic domains is called a Domain Wall. Inside the domain wall, spins do not flip abruptly but rotate gradually. This is another game of energy trade-offs: Exchange Energy wants spins to be parallel and resists their turning (preferring the wall to be as wide as possible); Magnetic Anisotropy wants spins to align along the easy axis and resists them pointing in intermediate directions (preferring the wall to be as narrow as possible). The balance between the two determines the thickness of the domain wall (typically hundreds of atomic layers). The formation of magnetic domains is not derived from “first principles” like spin, but belongs to the realm of Micromagnetics, involving energy minimization in continuum field theory, which we will not expand upon here.

Now, we can finally fully describe the macroscopic process of “a magnet attracting iron” — the very click that opened this entire series, now spelled out from the spacetime symmetry of one electron all the way up:

Initial State: The interior of the iron nail is filled with chaotic magnetic domains; the macroscopic magnetic moment is zero.
External Field Intervention: When you bring a magnet close to the nail, you provide an external magnetic field $\vec{B}_{ext}$ .
Domain Wall Motion: The equilibrium is broken. Domains whose direction aligns with $\vec{B}_{ext}$ have lower Zeeman energy ( $E = -\vec{M} \cdot \vec{B}$ ). Consequently, these “compliant” domains begin to swallow the surrounding “non-compliant” domains. The domain walls move.
Macroscopic Magnetization: The nail rapidly gains a huge net magnetic moment.
Gradient Force Work: This induced macroscopic magnetic moment $\vec{m}_{total}$ is pulled by the gradient force $\vec{F} = \nabla(\vec{m}_{total} \cdot \vec{B})$ generated by the magnet’s non-uniform field.
Click: The nail flies towards the magnet.

Conclusion: The Deep Symmetry of the Universe

When you play with two magnets in your hands, feeling the repulsion and attraction between them, you are feeling more than just a force. You are touching the essence of quantum mechanics and the secrets of cosmic evolution with your own hands. Let us review this journey and see how we rebuilt our physical intuition:

The Classical Collapse: We found that the Lorentz force does no work, and classical statistical physics forbids magnetism (Bohr-van Leeuwen Theorem).
The Relativistic Correction: The Dirac equation revealed that the electron must be a 4-component spinor carrying an intrinsic magnetic moment with g=2. Magnetism is the residue of relativistic effects in the low-speed world.
The Power of Quantum Statistics: The Pauli Exclusion Principle combined with Coulomb repulsion creates an equivalent “Exchange Interaction,” forcing spins to align parallel.
Symmetry Breaking: The Ising model taught us that when the temperature drops, the system, in order to survive (lower its energy), is forced to break rotational symmetry and choose a direction.

Finally, it is worth mentioning that the Spontaneous Symmetry Breaking (SSB) we saw in the Ising model has significance far beyond solid-state physics. It is a core paradigm for understanding the universe in modern physics. At the beginning of the Big Bang (extremely high temperature), physical laws possessed extremely high symmetry. All fundamental particles were massless, just like the iron block at high temperatures has no magnetism (paramagnetic phase). As the universe cooled, when the temperature dropped below a certain critical value, the Higgs Field filling the universe underwent a phase transition. Just like electron spins suddenly choosing to point in one direction, the Higgs field acquired a non-zero Vacuum Expectation Value in empty space. The self-consistent equation we drew at the origin is, with different labels on the axes, the same equation the early universe solved as it cooled.

In a ferromagnet, symmetry breaking endows the material with magnetism.
In the Standard Model, symmetry breaking endows fundamental particles with mass.

So, the next time you see a magnet pick up a paperclip, realize this: you are witnessing a miniature moment of cosmic creation. The mechanism that gives the nail its magnetism is the very same mechanism that gives the quarks and electrons in your body their mass, allowing this universe to exist.

Magnetic force does no work; it is the geometry of spacetime doing the work. The attraction of a magnet is the dance of a quantum ghost in the macroscopic world.