Critical Temperature of BEC in a 3D Exciton Gas

1. The Physical Picture of Condensation

Before diving into the mathematics, let's build a physical intuition for what triggers this phase transition. At its heart, BEC is a consequence of the wave-particle duality of matter. Every exciton has an associated thermal de Broglie wavelength, ${\lambda }_{dB}$, which increases as the temperature decreases.

Wolfgang Ketterle, a Nobel laureate for his work on BEC, provides a very clear picture of this process.

As Ketterle explains, at high temperatures, the excitons behave like classical billiard balls. As the system is cooled, their de Broglie wavelengths grow. The transition to a BEC occurs when the temperature is so low that ${\lambda }_{dB}$ becomes comparable to the average distance between excitons. At this point, the individual exciton wave packets overlap, lose their individuality, and merge into a single, coherent macroscopic quantum state—a giant matter wave.

Our goal is to translate this picture, ${\lambda }_{dB}\sim n^{-1/3}$, into a precise formula for the critical temperature, $T_c$.

2. The Bose-Einstein Distribution and the Chemical Potential

The starting point for any analysis of an ideal Bose gas is the Bose-Einstein distribution, which gives the average number of non-interacting bosons, $⟨n(ϵ)⟩$, in a quantum state with energy $ϵ$:

$⟨n(ϵ)⟩=\frac{1}{e^{\beta (ϵ-\mu )}-1}$

where $\beta =1/(k_{B}T)$ and $\mu$ is the chemical potential.

The chemical potential $\mu$ plays a pivotal role. Since the occupation number $⟨n(ϵ)⟩$ must be positive or zero for any state, the denominator must always be positive. For the ground state where $ϵ=0$, this imposes a strict constraint:

$e^{-\beta \mu }-1>0⟹e^{-\beta \mu }>1$

Since $\beta >0$, this can only be true if $\mu$ is negative. Thus, for a gas of bosons, the chemical potential must always be less than or equal to the ground state energy, which we set to zero.

$\mu \le 0$

What happens as we lower the temperature? The system tries to accommodate more and more particles in the lower energy states. To increase $⟨n(ϵ)⟩$ for a given $T$, the value of $\mu$ must increase, approaching zero from the negative side. This behavior is the key to identifying the critical point for condensation.

The following video provides an excellent and detailed explanation of the role of the chemical potential and sets up the problem of counting the total number of particles.

3. Deriving the Critical Temperature

As the video explains, a subtlety arises when we calculate the total number of excitons, $N$. In the thermodynamic limit, we can replace the sum over discrete states with an integral over a continuous density of states, $D(ϵ)$. However, for a 3D gas of particles, $D(ϵ)\propto \sqrt{ϵ}$, which means $D(0)=0$. The integral would therefore completely miss the population of the ground state.

Since BEC is defined by the macroscopic occupation of the ground state, we must treat it separately. We write the total number of excitons $N$ as the sum of the number in the ground state, $N_0$, and the number in all excited states, $N_{ex}$:

$N=N_0+N_{ex}$

The number in the excited states is given by the integral:
$N_{ex}={\int }_0^{\infty }D(ϵ)⟨n(ϵ)⟩dϵ={\int }_0^{\infty }D(ϵ)\frac{1}{e^{\beta (ϵ-\mu )}-1}dϵ$

At a given temperature $T$, this integral for $N_{ex}$ is a monotonically increasing function of $\mu$. Since $\mu \le 0$, the integral has a maximum value which occurs when $\mu =0$. This means there is a maximum number of particles, $N_{ex,max}(T)$, that can be accommodated in the excited states at temperature $T$.

What happens if the total number of particles $N$ is greater than $N_{ex,max}(T)$? The "excess" particles, $N-N_{ex,max}$, have nowhere else to go and must accumulate in the ground state, $ϵ=0$. This is the onset of Bose-Einstein condensation.

The critical temperature, $T_c$, is defined as the temperature at which the maximum number of particles the excited states can hold is exactly equal to the total number of particles, $N$. Below this temperature, $N_0$ must be macroscopic. At $T=T_c$, we therefore have two conditions:

1. $\mu =0$
2. $N_0\approx 0$, so $N\approx N_{ex}$

We can now find $T_c$ by setting $\mu =0$ and $N_{ex}=N$ in our integral equation at $T=T_c$:

$N={\int }_0^{\infty }D(ϵ)\frac{1}{e^{ϵ/k_B{T}_c}-1}dϵ$

The derivation now proceeds with this integral.

Let's summarize the key steps of the derivation you just saw.

1. Density of States: For a 3D gas of excitons with mass $M_{ex}$ and spin degeneracy $g$ in a volume $V$, the density of states is:
   $D(ϵ)=g\frac{V}{4{\pi }^2}{\left(\frac{2M_{ex}}{{\hslash }^2}\right)}^{3/2}\sqrt{ϵ}$
2. Integral at $T_c$: We substitute this into our critical condition integral:
   $N=g\frac{V}{4{\pi }^2}{\left(\frac{2M_{ex}}{{\hslash }^2}\right)}^{3/2}{\int }_0^{\infty }\frac{\sqrt{ϵ}}{e^{ϵ/k_B{T}_c}-1}dϵ$
3. Change of Variables: Let $x=ϵ/k_B{T}_c$. Then $dϵ=k_B{T}_{c}dx$ and $\sqrt{ϵ}=(k_B{T}_c{)}^{1/2}\sqrt{x}$. The integral becomes:
   $N=g\frac{V}{4{\pi }^2}{\left(\frac{2M_{ex}{k}_B{T}_c}{{\hslash }^2}\right)}^{3/2}{\int }_0^{\infty }\frac{x^{1/2}}{e^x-1}dx$
4. Standard Integral: The integral is a standard form related to the Riemann zeta function $\zeta (s)$ and the Gamma function $\Gamma (s)$:
   ${\int }_0^{\infty }\frac{x^{s-1}}{e^x-1}dx=\Gamma (s)\zeta (s)$
   For our case, $s-1=1/2⟹s=3/2$. Using $\Gamma (3/2)=\sqrt{\pi }/2$, our integral evaluates to $\zeta (3/2)\sqrt{\pi }/2$.
5. Solving for $T_c$: Substituting this result and rearranging for $k_B{T}_c$, we arrive at the expression for the critical temperature. Letting $n=N/V$ be the exciton density:
   $k_B{T}_c=\frac{2\pi {\hslash }^2}{M_{ex}}{\left(\frac{n}{g\zeta (3/2)}\right)}^{2/3}$
   where $\zeta (3/2)\approx 2.612$. This is the central result of our lesson. You can also find this formula presented clearly in the thesis by Keeling (resource LINK, Eq. 1.5) and derived in the dissertation by Som (resource LINK, section 1.7.1).

4. Condensate Fraction for $T<T_c$

What happens below the critical temperature? For $T<T_c$, the chemical potential remains pinned very close to zero ($\mu \approx 0$), and the number of excitons in the excited states is:

$N_{ex}(T)={\int }_0^{\infty }D(ϵ)\frac{1}{e^{ϵ/k_{B}T}-1}dϵ$

Following the same integration procedure, we find a simple relationship:
$N_{ex}(T)=N{\left(\frac{T}{T_c}\right)}^{3/2}$

The remaining excitons must be in the ground state. The fraction of excitons in the condensate, $N_0/N$, is therefore:

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

This is a famous result that characterizes the growth of the condensate as the system is cooled below $T_c$. The UNSW video explains this final step well.

The function $N_0(T)/N$ serves as the order parameter for this phase transition. It is zero above the transition and continuously grows from zero as the temperature is lowered below $T_c$, indicating that this is a continuous, or second-order, phase transition.

Conclusion

In this lesson, we performed a complete derivation of the critical temperature for Bose-Einstein condensation in a 3D ideal gas, a model directly applicable to a low-density exciton gas.

Key Takeaways:

• BEC occurs when the thermal de Broglie wavelengths of bosons become comparable to their average separation.
• The transition is driven by the saturation of excited states. Below $T_c$, the excited states cannot hold all the particles, forcing the remainder into the ground state.
• The derivation requires separating the ground state population $N_0$ from the integrated density of excited states $N_{ex}$.
• The critical temperature is found by setting the chemical potential $\mu =0$ and is given by:
  $k_B{T}_c=\frac{2\pi {\hslash }^2}{M_{ex}}{\left(\frac{n}{g\zeta (3/2)}\right)}^{2/3}$
• For $T<T_c$, the fraction of excitons in the condensate is given by the order parameter:
  $\frac{N_0}{N}=1-{\left(\frac{T}{T_c}\right)}^{3/2}$

Preview of the Next Lesson

Our entire discussion today was based on a 3D gas. However, many modern semiconductor systems confine excitons to two dimensions (e.g., in quantum wells). Does BEC still occur in 2D? The answer is more complex. In the next lesson, we will investigate this and address the learning outcome: Describe the Berezinskii-Kosterlitz-Thouless (BKT) transition for interacting 2D exciton systems. We will find that while true long-range order is forbidden, a unique phase transition to a "quasi-condensate" with algebraic order is still possible.