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PU-203 / L19 · lecture

Worked Case: Two-Level Systems and Paramagnets

~7 min read · Tier toggle changes depth (T1 plain · T2 standard · T3 rigour)

BEFORE YOU START

You must be comfortable with the canonical partition function \( Z \) and the relation \( F = -k_B T \ln Z \) derived in the previous lecture.

Why this matters

The two-level system is the simplest non-trivial quantum system, serving as the foundational model for paramagnetism.

While harmonic oscillators and ideal gases are common introductory examples, the two-level system (TLS) captures the essential physics of magnetic moments in a field. It demonstrates how discrete energy spectra lead to non-linear thermodynamic responses, such as saturation magnetization. By mastering this simple case, you gain the intuition to handle more complex spin systems and understand the limits of classical thermodynamics when quantum discreteness becomes significant.

The Physical Picture

Imagine a collection of \( N \) identical, non-interacting atoms, each possessing a magnetic dipole moment \( \boldsymbol{\mu} \). When placed in a uniform external magnetic field \( \mathbf{B} \), the interaction energy of a single dipole is given by \( E = -\boldsymbol{\mu} \cdot \mathbf{B} \). If we align the field along the \( z \)-axis, the energy depends only on the orientation of the dipole relative to this axis. For a simple spin-1/2 particle, there are only two possible orientations: parallel to the field (lower energy) and anti-parallel (higher energy). This creates a system with exactly two energy levels per particle. Because the particles do not interact, the total energy of the system is simply the sum of the energies of the individual particles. This independence is crucial; it allows us to calculate the partition function for the entire system by simply raising the single-particle partition function to the power of \( N \). This "factorization" is the key mathematical tool that makes the problem solvable and physically transparent.

Single-Particle Partition Function

Let the energy of the lower state be \( E_0 = -\mu B \) and the upper state be \( E_1 = +\mu B \). We define the energy gap as \( \Delta E = E_1 - E_0 = 2\mu B \). The single-particle partition function \( Z_1 \) is the sum of the Boltzmann factors for these two states:

\[ Z_1 = e^{-\beta E_0} + e^{-\beta E_1} = e^{\beta \mu B} + e^{-\beta \mu B} \]

Using the definition of the hyperbolic cosine, \( \cosh(x) = \frac{e^x + e^{-x}}{2} \), we can rewrite this compactly as:

\[ Z_1 = 2 \cosh(\beta \mu B) \]

Here, \( \beta = 1/k_B T \). This expression is valid for any temperature \( T \) and field \( B \). Note that \( Z_1 \) is always greater than or equal to 2, reflecting the fact that there are two accessible states.

Thermodynamic Potentials

For \( N \) independent particles, the total partition function is \( Z_N = (Z_1)^N \). The Helmholtz free energy \( F \) is given by \( F = -k_B T \ln Z_N \):

\[ F = -N k_B T \ln Z_1 = -N k_B T \ln \left[ 2 \cosh(\beta \mu B) \right] \]

From \( F \), we can derive the internal energy \( U \) and entropy \( S \). The internal energy is the negative derivative of \( \ln Z_N \) with respect to \( \beta \):

\[ U = -\frac{\partial \ln Z_N}{\partial \beta} = -N \frac{\partial \ln Z_1}{\partial \beta} \]

Calculating the derivative of \( \ln(2 \cosh(\beta \mu B)) \) with respect to \( \beta \) yields \( \mu B \tanh(\beta \mu B) \). Thus:

\[ U = -N \mu B \tanh(\beta \mu B) \]

The entropy \( S \) is obtained from \( S = -\left( \frac{\partial F}{\partial T} \right)_V \). Alternatively, using \( S = \frac{U - F}{T} \) is often simpler. Substituting the expressions for \( U \) and \( F \):

\[ S = \frac{N \mu B \tanh(\beta \mu B) + N k_B T \ln \left[ 2 \cosh(\beta \mu B) \right]}{T} \]

Dividing by \( N k_B \) gives the entropy per spin in units of \( k_B \), a dimensionless form that is useful for plotting; it rises from 0 at \( T = 0 \) to \( \ln 2 \) at high temperature.

Magnetization and Susceptibility

The magnetization \( M \) is the total magnetic moment per unit volume. For \( N \) particles, the total moment is \( N \langle \mu_z \rangle \). The average moment of a single particle is:

\[ \langle \mu_z \rangle = -\frac{\partial F_1}{\partial B} = -\frac{\partial}{\partial B} \left( -k_B T \ln Z_1 \right) \]

Since \( F_1 = -k_B T \ln(2 \cosh(\beta \mu B)) \), differentiating with respect to \( B \) gives:

\[ \langle \mu_z \rangle = \mu \tanh(\beta \mu B) \]

This is the Brillouin function for spin-1/2. At low temperatures (\( T \to 0 \)), \( \tanh(\beta \mu B) \to 1 \), so \( \langle \mu_z \rangle \to \mu \). The system is fully magnetized. At high temperatures (\( T \to \infty \)), \( \beta \mu B \ll 1 \), and using the approximation \( \tanh(x) \approx x \), we get:

\[ \langle \mu_z \rangle \approx \mu (\beta \mu B) = \frac{\mu^2 B}{k_B T} \]

This linear dependence on \( B \) and inverse dependence on \( T \) is Curie's Law. The magnetic susceptibility \( \chi = \frac{\partial M}{\partial B} \) is therefore \( \chi = \frac{N \mu^2}{V k_B T} \), confirming the \( 1/T \) behaviour characteristic of paramagnets.

The derivations

◆ DERIVATION EMBED · D-013 · not re-derived here

The Boltzmann Distribution

A small system exchanging energy with a large reservoir occupies each microstate with probability proportional to exp(-E/kT).

pi = e−Ei/kBT / Z ,    Z = Σi e−Ei/kBT
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◆ DERIVATION EMBED · D-156 · not re-derived here

The Partition Function and F = -kT ln Z

For a system in thermal contact with a reservoir at temperature T, normalizing the Boltzmann distribution pi ∝ e−βEi over microstates i forces the introduction of the canonical partition function Z = ∑i e−βEi (with β = 1/kT). The Helmholtz free energy is then F = −kT ln Z, and the internal energy, entropy, and every thermodynamic quantity follow as derivatives of ln Z.

Z = ∑i e−βEi    F = −kT ln Z    U = −∂(ln Z)/∂β    S = −(∂F/∂T)V,N
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Tier 3 · watch out

A common error is assuming the heat capacity \( C_V \) is constant. For a two-level system, \( C_V \) is zero at both \( T \to 0 \) and \( T \to \infty \), with a peak (Schottky anomaly) at an intermediate temperature. This arises because the entropy change is finite and bounded, unlike the unbounded entropy of a harmonic oscillator. C

Worked example

Problem. A paramagnetic salt contains \(N = 1.0 \times 10^{20}\) independent spin-\(\tfrac12\) moments, each \(\mu = 9.27 \times 10^{-24}\ \mathrm{A\,m^2}\) (one Bohr magneton), in a volume \(V = 1.0 \times 10^{-6}\ \mathrm{m^3}\) and a field \(B = 1.0\ \mathrm{T}\). Find the magnetisation \(M = (N\mu/V)\tanh(\mu B/k_BT)\) at 300 K and at 10 K, and compare with saturation.

  1. Saturation magnetisation: \(M_s = N\mu/V = 1.0 \times 10^{20} \times 9.27 \times 10^{-24}/1.0 \times 10^{-6} = 927\ \mathrm{A\,m^{-1}}\).
  2. At 300 K: \(x = \mu B/k_BT = 9.27 \times 10^{-24}/(1.38 \times 10^{-23} \times 300) = 2.24 \times 10^{-3}\); \(\tanh x \approx x\), so \(M = 927 \times 2.24 \times 10^{-3} = 2.08\ \mathrm{A\,m^{-1}}\).
  3. At 10 K: \(x = 0.0672\), \(\tanh x = 0.0671\), so \(M = 927 \times 0.0671 = 62.2\ \mathrm{A\,m^{-1}}\).
  4. The ratio \(62.2/2.08 = 29.9\) is almost exactly \(300/10 = 30\): both temperatures are in the Curie regime \(x \ll 1\), where \(M \propto B/T\).
  5. Saturation needs \(x \gtrsim 2\), that is \(T \lesssim \mu B/2k_B \approx 0.3\ \mathrm{K}\) at 1 T, which is why adiabatic demagnetisation works at millikelvin temperatures.
\[ M_{300\,\mathrm{K}} \approx 2.1\ \mathrm{A\,m^{-1}}, \qquad M_{10\,\mathrm{K}} \approx 62\ \mathrm{A\,m^{-1}}, \qquad M_s = 927\ \mathrm{A\,m^{-1}} \]

Units check. \(\mathrm{A\,m^2} \times \mathrm{m^{-3}} = \mathrm{A\,m^{-1}}\), the SI unit of magnetisation; \(\mu B/k_BT\) is (J T-1)(T)/(J) = dimensionless.

CHECK YOURSELF

Three quick tests

  1. What is the value of the single-particle partition function \( Z_1 \) as \( T \to 0 \)?
    AnswerAs \( T \to 0 \), \( \beta \to \infty \). \( Z_1 = e^{\beta \mu B} + e^{-\beta \mu B} \approx e^{\beta \mu B} \). It diverges exponentially.
  2. Why does the heat capacity vanish at \( T \to 0 \)?
    AnswerAt \( T \to 0 \), all particles are in the ground state. There are no excited states available to absorb energy, so \( \frac{\partial U}{\partial T} = 0 \).
  3. How does the magnetization behave for \( T \gg \mu B / k_B \)?
    AnswerIt becomes linear in \( B \) and inversely proportional to \( T \) (Curie's Law), \( M \propto B/T \).
PROBLEMS

Problem set for PU-203 →

Each derivation in this unit ends with five graded problems with full solutions.

NOTES

Lecture notes checked line by line by the editor; every number in the worked example was recomputed.

Derivations cited: D-013 · D-156 · Threads: energy, matter, chance