Part II: Quantum Gases | Chapter 2

Bose-Einstein Condensation

Critical temperature, condensate fraction, lambda transition in specific heat, and the physics of superfluidity

Historical Context

In 1924, Satyendra Nath Bose sent Einstein a paper deriving Planck's law by treating photons as indistinguishable particles. Einstein extended this to massive bosons and predicted that below a critical temperature, a macroscopic fraction of particles would collapse into the ground state -- a phenomenon now called Bose-Einstein condensation (BEC). This prediction was met with skepticism until Fritz London connected it to the superfluidity of helium-4 in 1938.

It took until 1995 for Eric Cornell, Carl Wieman, and Wolfgang Ketterle to achieve BEC in dilute atomic gases (rubidium-87 and sodium-23), earning the 2001 Nobel Prize. The lambda transition in liquid helium, first observed by Keesom in 1932, remains the most dramatic thermodynamic signature of BEC in nature.

1. The Critical Temperature

Derivation 1: Finding T_c from the Number Equation

For an ideal Bose gas, the average number of particles in excited states is:

\[N_{\text{ex}} = \frac{V}{\lambda_{dB}^3} g_{3/2}(z), \qquad g_{\nu}(z) = \sum_{l=1}^{\infty}\frac{z^l}{l^{\nu}}\]

where $z = e^{\beta\mu}$ is the fugacity and $\lambda_{dB} = \sqrt{2\pi\hbar^2/(mk_BT)}$is the thermal de Broglie wavelength. For bosons, $\mu \le 0$ so $0 \le z \le 1$. The function $g_{3/2}(z)$ has a maximum at $z = 1$:

\[g_{3/2}(1) = \zeta(3/2) \approx 2.612\]

The maximum number of particles that can be accommodated in excited states at temperature $T$ is:

\[N_{\text{ex}}^{\max} = \frac{V}{\lambda_{dB}^3}\zeta(3/2)\]

When $N > N_{\text{ex}}^{\max}$, the excess particles must occupy the ground state. The critical temperature $T_c$ is defined by $N_{\text{ex}}^{\max}(T_c) = N$:

\[\boxed{T_c = \frac{2\pi\hbar^2}{mk_B}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}}\]

where $n = N/V$ is the number density. This can be rewritten as$n\lambda_{dB}^3(T_c) = \zeta(3/2)$, stating that condensation occurs when the interparticle spacing becomes comparable to the thermal de Broglie wavelength.

2. Condensate Fraction

Derivation 2: Ground State Occupation Below T_c

Below $T_c$, we set $z = 1$ (since $\mu \to 0^-$) and the number of excited particles is:

\[N_{\text{ex}}(T) = \frac{V}{\lambda_{dB}^3(T)}\zeta(3/2) = N\left(\frac{T}{T_c}\right)^{3/2}\]

The condensate fraction is the number of particles in the ground state:

\[\boxed{\frac{N_0}{N} = 1 - \left(\frac{T}{T_c}\right)^{3/2}}\]

At $T = 0$, all particles are in the ground state ($N_0 = N$). The condensate fraction decreases continuously as $T$ increases, vanishing at $T_c$. This is a macroscopic quantum phenomenon: a single quantum state becomes occupied by an$O(N)$ number of particles.

Physical Significance

BEC is fundamentally a consequence of quantum statistics, not interactions. Even for an ideal gas (no interparticle potential), indistinguishability and the bosonic symmetry requirement force a macroscopic ground-state occupation. The $T^{3/2}$ scaling is specific to 3D; in 2D, the ideal Bose gas does not condense at finite temperature (Mermin-Wagner theorem), though the interacting 2D system can exhibit a Berezinskii-Kosterlitz-Thouless transition.

3. Energy and Pressure

Derivation 3: Internal Energy Below and Above T_c

The internal energy of the ideal Bose gas is:

\[U = \frac{3}{2}k_BT \frac{V}{\lambda_{dB}^3} g_{5/2}(z)\]

Below $T_c$ (where $z = 1$):

\[U = \frac{3}{2}Nk_BT\frac{g_{5/2}(1)}{g_{3/2}(1)}\left(\frac{T}{T_c}\right)^{3/2} = \frac{3\zeta(5/2)}{2\zeta(3/2)}Nk_BT\left(\frac{T}{T_c}\right)^{3/2}\]

Since $\zeta(5/2) \approx 1.341$ and $\zeta(3/2) \approx 2.612$:

\[U \approx 0.7703\,Nk_BT\left(\frac{T}{T_c}\right)^{3/2} \quad (T < T_c)\]

The pressure follows from $PV = \frac{2}{3}U$ (valid for all ideal quantum gases in 3D):

\[P = \frac{k_BT}{\lambda_{dB}^3}g_{5/2}(z)\]

Below $T_c$, $P$ depends only on $T$ (not on $V$), since $z = 1$. This means $(\partial P/\partial V)_T = 0$: the compressibility diverges, analogous to a first-order transition in the $P$-$V$ plane.

4. Specific Heat: The Lambda Transition

Derivation 4: C_V Below and Above T_c

Below $T_c$, differentiating $U$:

\[C_V = \frac{\partial U}{\partial T}\bigg|_V = \frac{15}{4}\frac{\zeta(5/2)}{\zeta(3/2)}Nk_B\left(\frac{T}{T_c}\right)^{3/2} \approx 1.926\,Nk_B\left(\frac{T}{T_c}\right)^{3/2}\]

Above $T_c$, the calculation is more involved because $z$ depends on $T$through the number equation. The result involves derivatives of $g_{\nu}(z)$:

\[C_V = Nk_B\left[\frac{15}{4}\frac{g_{5/2}(z)}{g_{3/2}(z)} - \frac{9}{4}\frac{g_{3/2}(z)}{g_{1/2}(z)}\right] \quad (T > T_c)\]

At $T = T_c$, $C_V$ from below equals $C_V$ from above:$C_V(T_c) \approx 1.926\,Nk_B$. However, the derivative $dC_V/dT$ is discontinuous -- it has a cusp. This is the famous lambda transition, named after the shape of the $C_V(T)$ curve which resembles the Greek letter $\lambda$.

The Lambda Point in Helium-4

In liquid helium-4, the lambda transition occurs at $T_{\lambda} = 2.17\,\text{K}$. The ideal BEC prediction gives $T_c \approx 3.1\,\text{K}$, which is too high because interatomic interactions deplete the condensate. At $T = 0$, only about 8% of helium atoms are in the zero-momentum state (compared to 100% for the ideal gas), yet the superfluid fraction is 100%.

The real lambda transition in He-4 has a logarithmic divergence in $C_V$, not a cusp. This discrepancy highlights the role of interactions in determining the universality class of the transition.

5. Connection to Superfluidity

Derivation 5: Landau Critical Velocity

Landau showed that a flowing superfluid can only lose energy by creating excitations. For the fluid moving at velocity $v$, the energy of creating an excitation with momentum $p$ in the lab frame is:

\[\Delta E = \epsilon(p) + \mathbf{p}\cdot\mathbf{v} \ge \epsilon(p) - pv\]

Excitations can only be created if $\Delta E < 0$, which requires:

\[v > v_c = \min_p \frac{\epsilon(p)}{p}\]

For an ideal Bose gas, $\epsilon(p) = p^2/(2m)$, giving $v_c = 0$ -- no superfluidity. Interactions are essential: in helium-4, the phonon-roton spectrum gives:

\[\boxed{v_c = \min\left(c_s, \frac{\Delta}{p_0}\right)}\]

where $c_s$ is the speed of sound and $\Delta/p_0$ comes from the roton minimum. In practice, $v_c \sim 58\,\text{m/s}$ from the roton minimum, though vortex nucleation often limits superflow at much lower velocities.

Two-Fluid Model

Below $T_{\lambda}$, helium-4 behaves as if it consists of two interpenetrating fluids: a superfluid component (density $\rho_s$) with zero viscosity and zero entropy, and a normal component (density $\rho_n$) carrying all the entropy. At $T = 0$, $\rho_s = \rho$; at $T_{\lambda}$,$\rho_s = 0$. This model, developed by Tisza and Landau, explains phenomena like second sound (temperature waves) and the fountain effect.

6. Applications and Modern BEC

Key Applications

Ultracold atomic gases: Laser cooling and evaporative cooling achieve$T \sim 100\,\text{nK}$ in alkali vapors. The condensate is directly imaged, and the condensate fraction follows $1 - (T/T_c)^3$ for a harmonic trap (different exponent from the box due to modified density of states).

Atom lasers: Coherent extraction of atoms from a BEC produces a matter-wave analog of a laser beam, with applications in precision interferometry.

Cosmological analogs: BEC of axions has been proposed as a dark matter candidate. Phonons in a BEC can simulate curved spacetime metrics, enabling laboratory analogs of Hawking radiation.

Exciton-polariton condensates: In semiconductor microcavities, coupled light-matter quasiparticles can form non-equilibrium condensates at much higher temperatures (up to room temperature in some systems).

7. Computational Exploration

This simulation explores the condensate fraction, specific heat, and thermodynamic properties of the ideal Bose gas through the BEC transition.

BEC: Condensate Fraction, Specific Heat, Chemical Potential, and Pressure

Python

script.py220 lines

import numpy as np
import matplotlib
matplotlib.use('Agg')
import matplotlib.pyplot as plt
def sp_zeta(s, a=1):
    """Riemann zeta function via direct summation."""
    result = 0.0
    for n in range(1, 10000):
        term = 1.0 / (n + a - 1)**s
        result += term
        if abs(term) < 1e-15: break
    return result
def brentq(f, a, b, args=(), xtol=1e-12, maxiter=200):
    fa, fb = f(a, *args), f(b, *args)
    for _ in range(maxiter):
        c = (a + b) / 2.0; fc = f(c, *args)
        if abs(fc) < xtol or (b - a) / 2.0 < xtol: return c
        if fa * fc < 0: b, fb = c, fc
        else: a, fa = c, fc
    return (a + b) / 2.0

# ============================================================
# Bose-Einstein Condensation: Thermodynamics
# ============================================================

def g_n(nu, z, nmax=500):
    """Polylogarithm / Bose function g_nu(z) = sum z^l / l^nu"""
    if z <= 0:
        return 0.0
    if z >= 1.0:
        return sp_zeta(nu)
    result = 0.0
    for l in range(1, nmax + 1):
        term = z**l / l**nu
        result += term
        if abs(term) < 1e-12:
            break
    return result

zeta32 = sp_zeta(3.0/2.0)  # ~2.612
zeta52 = sp_zeta(5.0/2.0)  # ~1.341

fig, axes = plt.subplots(2, 2, figsize=(12, 10))
fig.suptitle('Bose-Einstein Condensation', fontsize=16, fontweight='bold', color='white')
fig.patch.set_facecolor('#0a0a0a')

for ax in axes.flat:
    ax.set_facecolor('#1a0a0a')
    ax.tick_params(colors='white')
    ax.xaxis.label.set_color('white')
    ax.yaxis.label.set_color('white')
    ax.title.set_color('#fca5a5')
    for spine in ax.spines.values():
        spine.set_color('#333')

# --- Panel 1: Condensate fraction ---
ax1 = axes[0, 0]
t_ratio = np.linspace(0, 2.0, 300)
n0_frac = np.where(t_ratio < 1.0, 1.0 - t_ratio**1.5, 0.0)
n_ex_frac = np.where(t_ratio < 1.0, t_ratio**1.5, 1.0)

ax1.fill_between(t_ratio, n0_frac, alpha=0.4, color='#60a5fa', label='Condensate N_0/N')
ax1.fill_between(t_ratio, n_ex_frac, 1.0, alpha=0.0)
ax1.plot(t_ratio, n0_frac, color='#60a5fa', linewidth=2)
ax1.plot(t_ratio, n_ex_frac, color='#f87171', linewidth=2, label='Excited N_ex/N')
ax1.axvline(x=1.0, color='#fbbf24', linestyle='--', alpha=0.7, label='T = T_c')
ax1.set_xlabel('T / T_c')
ax1.set_ylabel('Fraction')
ax1.set_title('Condensate Fraction')
ax1.set_xlim(0, 2)
ax1.set_ylim(0, 1.05)
ax1.legend(fontsize=8, facecolor='#1a0a0a', edgecolor='#333', labelcolor='white')

# --- Panel 2: Specific heat (lambda transition) ---
ax2 = axes[0, 1]
# Below Tc
t_below = np.linspace(0.01, 0.999, 200)
cv_below = (15.0/4.0) * zeta52 / zeta32 * t_below**1.5

# Above Tc: need to solve for z(T) numerically
t_above = np.linspace(1.001, 3.0, 200)
cv_above = []
for t in t_above:
    # n * lambda^3 = zeta32 * (Tc/T)^(3/2) => g_{3/2}(z) = zeta32 * t^{-3/2}
    target = zeta32 * t**(-1.5)
    if target >= zeta32:
        z_val = 0.9999
    else:
        try:
            z_val = brentq(lambda zz: g_n(1.5, zz) - target, 1e-10, 0.9999)
        except:
            z_val = 0.5
    g52 = g_n(2.5, z_val)
    g32 = g_n(1.5, z_val)
    g12 = g_n(0.5, z_val)
    if abs(g12) < 1e-10:
        cv_val = 1.5
    else:
        cv_val = 15.0/4.0 * g52/g32 - 9.0/4.0 * g32/g12
    cv_above.append(cv_val)

cv_above = np.array(cv_above)
t_all = np.concatenate([t_below, t_above])
cv_all = np.concatenate([cv_below, cv_above])

ax2.plot(t_all, cv_all, color='#f87171', linewidth=2)
ax2.axvline(x=1.0, color='#fbbf24', linestyle='--', alpha=0.7, label='T_c')
ax2.axhline(y=1.5, color='white', linestyle=':', alpha=0.3, label='Classical 3/2 Nk_B')
ax2.set_xlabel('T / T_c')
ax2.set_ylabel('C_V / (N k_B)')
ax2.set_title('Specific Heat: Lambda Transition')
ax2.set_xlim(0, 3)
ax2.set_ylim(0, 2.5)
ax2.legend(fontsize=8, facecolor='#1a0a0a', edgecolor='#333', labelcolor='white')

# --- Panel 3: Chemical potential vs T ---
ax3 = axes[1, 0]
t_range = np.linspace(1.01, 4.0, 200)
mu_vals = []
for t in t_range:
    target = zeta32 * t**(-1.5)
    if target >= zeta32:
        z_val = 0.9999
    else:
        try:
            z_val = brentq(lambda zz: g_n(1.5, zz) - target, 1e-10, 0.9999)
        except:
            z_val = 0.01
    mu_vals.append(np.log(z_val))  # mu / (kB Tc)

mu_below = np.zeros(50)
t_below2 = np.linspace(0, 1.0, 50)
ax3.plot(t_below2, mu_below, color='#60a5fa', linewidth=2, label='Below T_c (mu = 0)')
ax3.plot(t_range, np.array(mu_vals), color='#f87171', linewidth=2, label='Above T_c')
ax3.axvline(x=1.0, color='#fbbf24', linestyle='--', alpha=0.7)
ax3.set_xlabel('T / T_c')
ax3.set_ylabel('mu / (k_B T_c)')
ax3.set_title('Chemical Potential')
ax3.set_xlim(0, 4)
ax3.legend(fontsize=8, facecolor='#1a0a0a', edgecolor='#333', labelcolor='white')

# --- Panel 4: Pressure vs T ---
ax4 = axes[1, 1]
# P / (n kB Tc) = (T/Tc) * g_{5/2}(z) / zeta32
t_range2 = np.linspace(0.01, 3.0, 300)
p_vals = []
for t in t_range2:
    if t <= 1.0:
        z_val = 1.0
    else:
        target = zeta32 * t**(-1.5)
        try:
            z_val = brentq(lambda zz: g_n(1.5, zz) - target, 1e-10, 0.9999)
        except:
            z_val = 0.5
    p_vals.append(t * g_n(2.5, z_val) / zeta32)

p_classical = t_range2  # ideal gas P/(nkBTc) = T/Tc
ax4.plot(t_range2, p_vals, color='#a78bfa', linewidth=2, label='Bose gas')
ax4.plot(t_range2, p_classical, 'w--', linewidth=1.5, alpha=0.4, label='Classical ideal gas')
ax4.axvline(x=1.0, color='#fbbf24', linestyle='--', alpha=0.7)
ax4.set_xlabel('T / T_c')
ax4.set_ylabel('P / (n k_B T_c)')
ax4.set_title('Pressure')
ax4.set_xlim(0, 3)
ax4.legend(fontsize=8, facecolor='#1a0a0a', edgecolor='#333', labelcolor='white')

plt.tight_layout()
plt.savefig('output.png', dpi=150, bbox_inches='tight', facecolor='#0a0a0a')
plt.close()
print("Plot saved to output.png")

# Numerical results
print("\n" + "=" * 60)
print("BOSE-EINSTEIN CONDENSATION: NUMERICAL RESULTS")
print("=" * 60)

print("\n--- Key Constants ---")
print("  zeta(3/2) = " + str(round(zeta32, 6)))
print("  zeta(5/2) = " + str(round(zeta52, 6)))
print("  C_V(T_c)/Nk_B = " + str(round(15.0/4.0 * zeta52/zeta32, 4)))

print("\n--- Condensate Fraction N_0/N ---")
for t in [0.0, 0.2, 0.4, 0.6, 0.8, 1.0]:
    frac = max(0, 1.0 - t**1.5)
    print("  T/Tc = " + str(t) + ": N_0/N = " + str(round(frac, 4)))

print("\n--- Specific Heat C_V/(Nk_B) ---")
for t in [0.2, 0.5, 0.8, 1.0, 1.5, 2.0, 3.0]:
    if t <= 1.0:
        cv = 15.0/4.0 * zeta52/zeta32 * t**1.5
    else:
        target = zeta32 * t**(-1.5)
        try:
            z_val = brentq(lambda zz: g_n(1.5, zz) - target, 1e-10, 0.9999)
        except:
            z_val = 0.5
        g52 = g_n(2.5, z_val)
        g32 = g_n(1.5, z_val)
        g12 = g_n(0.5, z_val)
        cv = 15.0/4.0 * g52/g32 - 9.0/4.0 * g32/g12
    print("  T/Tc = " + str(t) + ": C_V/(Nk_B) = " + str(round(cv, 4)))

print("\n--- BEC Critical Temperatures ---")
# Rubidium-87 example
m_Rb = 87 * 1.66e-27  # kg
hbar = 1.055e-34
kB = 1.381e-23
n_typical = 2.5e18  # m^-3 typical for ultracold experiments
Tc_Rb = 2 * np.pi * hbar**2 / (m_Rb * kB) * (n_typical / zeta32)**(2.0/3.0)
print("  Rb-87 (n = 2.5e18 m^-3): T_c = " + str(round(Tc_Rb * 1e9, 1)) + " nK")

# Helium-4 liquid
m_He = 4 * 1.66e-27
n_He = 2.18e28  # liquid density
Tc_He = 2 * np.pi * hbar**2 / (m_He * kB) * (n_He / zeta32)**(2.0/3.0)
print("  He-4 liquid (ideal gas est.): T_c = " + str(round(Tc_He, 2)) + " K")
print("  He-4 experimental lambda point: T = 2.17 K")
print("\n" + "=" * 60)

Click Run to execute the Python code

Code will be executed with Python 3 on the server

8. Summary and Key Results

Core Formulas

Critical temperature: $T_c \propto n^{2/3}\hbar^2/(mk_B)$
Condensate fraction: $N_0/N = 1 - (T/T_c)^{3/2}$
Specific heat cusp at $T_c$: $C_V(T_c) \approx 1.926\,Nk_B$
Landau critical velocity: $v_c = \min_p[\epsilon(p)/p]$

Physical Insights

BEC is a phase transition driven by quantum statistics
Occurs when $n\lambda_{dB}^3 \sim 1$
Ideal gas gives $v_c = 0$; interactions needed for superfluidity
Lambda transition in He-4 is modified by strong interactions

Practice Problems

Problem 1: BEC Critical TemperatureA gas of $N = 10^6$ rubidium-87 atoms ($m = 1.44 \times 10^{-25}$ kg) is confined in a box of volume $V = 10^{-15}$ m$^3$. Calculate the BEC critical temperature.

Solution:

1. The BEC critical temperature for an ideal gas in a box is:

$$T_c = \frac{2\pi\hbar^2}{mk_B}\left(\frac{n}{\zeta(3/2)}\right)^{2/3}$$

2. Compute the number density:

$$n = \frac{N}{V} = \frac{10^6}{10^{-15}} = 10^{21}\;\text{m}^{-3}$$

3. Using $\zeta(3/2) = 2.612$, $\hbar = 1.055 \times 10^{-34}$ J$\cdot$s, $k_B = 1.381 \times 10^{-23}$ J/K:

$$\frac{n}{\zeta(3/2)} = \frac{10^{21}}{2.612} = 3.83 \times 10^{20}\;\text{m}^{-3}$$

4. The prefactor is:

$$\frac{2\pi\hbar^2}{mk_B} = \frac{2\pi(1.055 \times 10^{-34})^2}{(1.44 \times 10^{-25})(1.381 \times 10^{-23})} = \frac{6.99 \times 10^{-68}}{1.99 \times 10^{-48}} = 3.52 \times 10^{-20}\;\text{K·m}^2$$

5. Combining:

$$\boxed{T_c = 3.52 \times 10^{-20} \times (3.83 \times 10^{20})^{2/3} = 3.52 \times 10^{-20} \times 5.28 \times 10^{13} \approx 1.86 \times 10^{-6}\;\text{K} \approx 1.86\;\mu\text{K}}$$

This is typical for dilute alkali BEC experiments. The first observation by Cornell and Wieman (1995) in Rb-87 achieved $T_c \approx 170$ nK at lower densities in a harmonic trap.

Problem 2: Condensate FractionAt temperature $T = 0.6\,T_c$, what fraction of the atoms occupy the ground state in an ideal Bose gas? How many atoms are in the condensate if $N = 5 \times 10^5$?

Solution:

1. For $T < T_c$, the condensate fraction in an ideal Bose gas is:

$$\frac{N_0}{N} = 1 - \left(\frac{T}{T_c}\right)^{3/2}$$

2. At $T/T_c = 0.6$:

$$\frac{N_0}{N} = 1 - (0.6)^{3/2} = 1 - 0.6\sqrt{0.6} = 1 - 0.6 \times 0.7746$$

3. Evaluating:

$$\frac{N_0}{N} = 1 - 0.4648 = 0.5352$$

4. The number of condensed atoms:

$$\boxed{N_0 = 0.535 \times 5 \times 10^5 = 2.68 \times 10^5\;\text{atoms}}$$

5. The remaining $N - N_0 = 2.32 \times 10^5$ atoms occupy excited states with a thermal (non-condensed) distribution. As $T \to 0$, $N_0/N \to 1$ (complete condensation). At $T = T_c$, $N_0/N = 0$. The $(T/T_c)^{3/2}$ law is a consequence of the 3D density of states $g(\epsilon) \propto \epsilon^{1/2}$.

Problem 3: Specific Heat Jump at $T_c$Calculate the specific heat $C_V$ of an ideal Bose gas just below and just above $T_c$. Show that there is a cusp (discontinuity in slope) at the transition.

Solution:

1. Below $T_c$ ($\mu = 0$), the internal energy is carried entirely by the thermal (non-condensed) component:

$$\frac{U}{Nk_BT} = \frac{3}{2}\frac{\zeta(5/2)}{\zeta(3/2)}\left(\frac{T}{T_c}\right)^{3/2}$$

2. The specific heat below $T_c$ is $C_V = dU/dT$:

$$\frac{C_V}{Nk_B} = \frac{15}{4}\frac{\zeta(5/2)}{\zeta(3/2)}\left(\frac{T}{T_c}\right)^{3/2}$$

3. At $T = T_c^-$ (from below): using $\zeta(5/2) = 1.341$, $\zeta(3/2) = 2.612$:

$$\frac{C_V(T_c^-)}{Nk_B} = \frac{15}{4}\times\frac{1.341}{2.612} = 3.75 \times 0.5134 = 1.926$$

4. Above $T_c$, the gas behaves as a normal Bose gas. At $T = T_c^+$, $C_V/(Nk_B) = 1.926$ as well — the specific heat is continuous at $T_c$.

5. However, the derivatives differ. Below: $dC_V/dT > 0$ (rising). Above: $dC_V/dT$ approaches $3/2$ (the classical value) from above. The result:

$$\boxed{C_V(T_c) = 1.926\,Nk_B \approx 1.28 \times \frac{3}{2}Nk_B}$$

The specific heat has a cusp (lambda-shaped peak) at $T_c$: it is continuous but its derivative $dC_V/dT$ is discontinuous. This is characteristic of a third-order phase transition in Ehrenfest's classification (though modern theory classifies ideal BEC as a continuous transition with $\alpha = -1$ critical exponent in 3D).

Problem 4: Chemical Potential vs TemperatureFor an ideal Bose gas with $T_c = 100$ nK, sketch and compute the chemical potential $\mu$ at $T = 50$ nK, $T = 100$ nK, and $T = 200$ nK. Explain the behavior in each regime.

Solution:

1. Below $T_c$: The chemical potential is pinned at zero (the ground state energy, set to 0):

$$T = 50\;\text{nK} < T_c: \quad \boxed{\mu = 0}$$

2. At $T = T_c = 100$ nK: $\mu = 0$ exactly. The fugacity $z = e^{\mu/(k_BT)} = 1$.

3. Above $T_c$: $\mu < 0$. The condition fixing $\mu$ is $N = V g_{3/2}(z)/\lambda_{dB}^3$ where $g_{3/2}(z) = \sum_{k=1}^\infty z^k/k^{3/2}$ and $z = e^{\beta\mu}$.

4. At $T = 200$ nK = $2T_c$, we need $g_{3/2}(z) = \zeta(3/2)(T_c/T)^{3/2} = 2.612 \times (0.5)^{3/2} = 2.612 \times 0.354 = 0.924$. Numerically inverting: $z \approx 0.55$, so:

$$\mu = k_BT\ln z = k_B(200\;\text{nK})\ln(0.55)$$

5. Evaluating:

$$\boxed{\mu(200\;\text{nK}) = 200\,k_B \times (-0.598)\;\text{nK} = -119.6\,k_B\;\text{nK} \approx -1.65 \times 10^{-30}\;\text{J}}$$

Summary: $\mu = 0$ for $T \leq T_c$ (condensate present, ground state macroscopically occupied); $\mu < 0$ and decreasing for $T > T_c$ (no condensate); in the classical limit $T \gg T_c$, $\mu \to -k_BT\ln(n\lambda_{dB}^3) \to -\infty$.

Problem 5: BEC in a Harmonic TrapFor $N = 10^6$ Rb-87 atoms in an isotropic harmonic trap with frequency $\omega = 2\pi \times 200$ Hz, calculate the critical temperature and compare with the box result.

Solution:

1. For a 3D isotropic harmonic trap, the density of states changes from $\epsilon^{1/2}$ (box) to $\epsilon^2$. The critical temperature is:

$$k_BT_c = \hbar\omega\left(\frac{N}{\zeta(3)}\right)^{1/3}$$

2. Using $\zeta(3) = 1.202$ and $\hbar\omega = 1.055 \times 10^{-34} \times 2\pi \times 200 = 1.326 \times 10^{-31}$ J:

$$\left(\frac{N}{\zeta(3)}\right)^{1/3} = \left(\frac{10^6}{1.202}\right)^{1/3} = (8.32 \times 10^5)^{1/3} = 94.1$$

3. Therefore:

$$k_BT_c = 94.1 \times 1.326 \times 10^{-31} = 1.248 \times 10^{-29}\;\text{J}$$

4. Converting to temperature:

$$\boxed{T_c = \frac{1.248 \times 10^{-29}}{1.381 \times 10^{-23}} = 904\;\text{nK} \approx 0.9\;\mu\text{K}}$$

5. Key differences from the box: (a) the condensate fraction scales as $N_0/N = 1 - (T/T_c)^3$ (exponent 3, not 3/2), because the trap changes the effective density of states; (b) the specific heat has a different shape near $T_c$; (c) the condensate has a finite spatial extent given by the harmonic oscillator ground state $a_{\text{ho}} = \sqrt{\hbar/(m\omega)} \approx 0.76\;\mu$m, which is directly imaged in experiments.

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