Bose-Einstein Condensation

Consider a gas of weakly-interacting bosons. It is helpful to define the gas's chemical potential,

$$\mu = k T \alpha,$$ (8.221)

whose value is determined by the equation

$$N= \sum_r \bar{n}_r = \sum_r \frac{1}{{\rm e}^{ \beta (\epsilon_r-\mu)}-1}.$$ (8.222)

Here, $N$ is the total number of particles, and $\epsilon_r$ the energy of the single-particle quantum state $r$ . Because, in general, the energies of the quantum states are very closely spaced, the sum in the previous expression can approximated as an integral. Now, according to Section 8.12, the number of quantum states per unit volume with wavenumbers in the range $k$ to $k+dk$ is

$$\rho_k(k) dk = \frac{k^{ 2}}{2\pi^{ 2}} dk.$$ (8.223)

However, the energy of a state with wavenumber $k$ is

$$\epsilon = \frac{\hbar^{ 2} k^{ 2}}{2 m},$$ (8.224)

where $m$ is the boson mass. Let $\rho(\epsilon) d\epsilon$ be the number of bosons whose energies lies in the range $\epsilon$ to $\epsilon+d\epsilon$ . It follows that

$$\rho(\epsilon) d\epsilon = V \rho_k(k) \frac{dk}{d\epsilon} d\epsilon,$$ (8.225)

where $V$ is the volume of the gas. Here, we are assuming that the bosons are spinless, so that there is only one particle state per translational state. Hence,

$$\rho(\epsilon) = \frac{V}{4\pi^{ 2}} \frac{(2 m)^{3/2}}{\hbar^{ 3}} \epsilon^{ 1/2},$$ (8.226)

and Equation (8.222) becomes

$$N=\int_0^\infty \frac{\rho(\epsilon)}{{\rm e}^{ \beta (\epsilon... ...nfty \frac{\epsilon^{ {1/2}}}{{\rm e}^{ \beta (\epsilon-\mu)}-1} d\epsilon.$$ (8.227)

However, there is a significant flaw in this formulation. In using the integral approximation, rather than performing the sum, the ground-state, $\epsilon=0$ , has been left out [because $\rho(0)=0$ ]. Under ordinary circumstances, this omission does not matter. However, at very low temperatures, bosons tend to condense into the ground-state, and the occupation number of this state becomes very much larger than that of any other state. Under these circumstances, the ground-state must be included in the calculation.

We can overcome the previous difficulty in the following manner. Let there be $N_0$ bosons in the ground-state, and $N_{\rm ex}$ in the various excited states, so that

$$N = N_0 + N_{\rm ex}.$$ (8.228)

Because the ground-state is excluded from expression (8.227), the integral only gives the number of bosons in excited states. In other words,

$$N_{\rm ex}= \frac{V}{4\pi^{ 2}} \frac{(2 m)^{3/2}}{\hbar^{ 3}... ...nfty \frac{\epsilon^{ {1/2}}}{{\rm e}^{ \beta (\epsilon-\mu)}-1} d\epsilon.$$ (8.229)

Now, because the ground-state has zero energy, its mean occupancy number is

$$N_0 = \frac{1}{{\rm e}^{-\mu/k T}-1}.$$ (8.230)

Moreover, at temperatures very close to absolute zero, we expect $N_0\simeq N$ , which implies that

$$N \simeq \frac{1}{{\rm e}^{-\mu/k T}-1}.$$ (8.231)

We conclude that

$$-\frac{\mu}{k T} \simeq \ln\left(1+\frac{1}{N}\right)\simeq \frac{1}{N},$$ (8.232)

for large $N$ . Hence, at very low temperatures, we can safely set $\exp(-\mu/k T)$ equal to unity in Equation (8.229). Thus, we obtain

$$N_{\rm ex} =\frac{2}{\sqrt{\pi}}\left(\frac{2\pi m k T}{h^{ 2}}\right)^{3/2}V\int_0^\infty \frac{x^{ 1/2}}{{\rm e}^{ x}-1} dx,$$ (8.233)

where $x=\epsilon/(k T)$ . The value of the integral is ${\mit\Gamma}(3/2) \zeta(3/2)$ , where ${\mit\Gamma}(s)=\int_0^\infty x^{ s-1} {\rm e}^{-x} dx$ is a Gamma function, and $\zeta(s)=\sum_{n=1,\infty}n^{-s}$ a Riemann zeta function. (See Exercise 4.) Furthermore, $\Gamma(3/2)=\sqrt{\pi}/2$ , and $\zeta(3/2)=2.612$ . Hence,

$$N_{\rm ex} = \zeta(3/2) V\left(\frac{2\pi m k T}{h^{ 2}}\right)^{3/2}.$$ (8.234)

The so-called Bose temperature, $T_B$ , is defined as the temperature above which all the bosons are in excited states. Setting $N_{\rm ex}=N$ and $T=T_B$ in the previous expression, we obtain

$$T_B = \frac{h^{ 2}}{2\pi m k}\left[\frac{N}{\zeta(3/2) V}\right]^{ 2/3}.$$ (8.235)

Moreover,

$$\frac{N_{\rm ex}}{N} = \left(\frac{T}{T_B}\right)^{3/2}.$$ (8.236)

Thus, the fractional number of bosons in the ground-state is

$$\frac{N_0}{N} = 1- \left(\frac{T}{T_B}\right)^{3/2}.$$ (8.237)

Obviously, the preceding two equations are only valid for $0<T<T_B$ . For $T>T_B$ , we have $N_0/N=0$ and $N_{\rm ex}/N = 1$ . Moreover, for $0<T<T_B$ we have $\mu/k T_B=0$ , whereas for $T>T_B$ , combining Equations (8.229) (with $N_{\rm ex}=N$ ) and (8.235) yields

$${\mit\Gamma}(3/2) \zeta(3/2) = \left(\frac{T}{T_B}\right)^{3/2}\int_0^\infty \frac{x^{ 1/2}}{{\rm e}^{ x-\alpha}-1} dx,$$ (8.238)

where $x=\epsilon/(k T)$ and $\alpha=\mu/(k T)$ . Expanding in powers of ${\rm e}^{-(x-\alpha)}$ , we obtain

$$\Gamma(3/2) \zeta(3/2)= \left(\frac{T}{T_B}\right)^{3/2}{\mit\Gamma}(3/2)\sum_{n=1,\infty}\frac{{\rm e}^{ n \alpha}}{n^{ 3/2}},$$ (8.239)

where $y=n x$ (see Exercise 4), which reduces to

$$\zeta(3/2) = \left(\frac{T}{T_B}\right)^{3/2}\sum_{n=1,\infty} \frac{{\rm e}^{-n \alpha}}{n^{ 3/2}}.$$ (8.240)

The previous equation can be solved numerically to give $\alpha=\mu/(k T_B)$ as a function of $T/T_B$ .

Figure: Variation with temperature of $N_0/N$ (solid curve) and $N_{\rm ex}/N$ (dashed curve) for a boson gas.

Let us estimate a typical value of the Bose temperature. Consider a boson gas made up of $N_A/10=6.02\times 10^{ 22}$ ${\rm He}^4$ atoms confined to a volume of 1 litre. The mass of a ${\rm He}^4$ atom is $6.65\times 10^{-27} {\rm kg}$ . Making use of Equation (8.235), we obtain

$$T_B = \frac{(6.63\times 10^{-34})^{ 2}}{2\pi (6.65\times 10^{-2... ....02\times 10^{ 22}}{(2.612) (1\times 10^{-3})}\right]^{ 2/3}=0.062 {\rm K}.$$ (8.241)

At atmospheric pressure, helium liquifies at $4.12 {\rm K}$ , long before the Bose temperature is reached. In fact, all real gases liquify before the Bose temperature is reached.

Figure: Variation with temperature of $\mu/k T_B$ for a boson gas.

Figure 8.7 shows the variation of $N_0/N$ and $N_{\rm ex}/N$ with $T/T_B$ . A corresponding graph of $\mu/k T_B$ [numerically determined from Equation (8.240) when $T>T_B$ ] is shown in Figure 8.8. The sudden collapse of bosons into the ground-state at temperatures below the Bose temperature is known as Bose-Einstein condensation. In 1995, E.A. Cornell and C.E. Wieman led a team of physicists that created a nearly pure condensate by cooling a vapor of rubidium atoms to a temperature of $1.3\times 10^{-7} {\rm K}$ . (Cornell and Wieman were subsequently awarded the Nobel prize in 2001.)

The mean energy of a boson gas takes the form

$$\overline{E}= \sum_r \frac{\epsilon_r}{{\rm e}^{ \beta (\epsilon_r-\mu)}-1}.$$ (8.242)

As before, we can approximate the sum as an integral, and write

$$\overline{E} = \int_0^\infty \frac{\epsilon \rho(\epsilon)}{{\rm... ...nfty \frac{\epsilon^{ 3/2}}{{\rm e}^{ \beta (\epsilon_r-\mu)}-1} d\epsilon.$$ (8.243)

In this case, we do not need to worry about the omission of the ground-state in the integral, because this state makes no contribution to the mean energy (because $\epsilon=0$ in the ground-state). For temperatures above the Bose temperature, all the bosons are in excited states, and we expect the mean energy to approach the classical value, $(3/2) N k T$ . However, below the Bose temperature, a substantial fraction of bosons are in the ground-state, so we expect the mean energy to fall well below the classical value.

As we have seen, the chemical potential of a boson gas is very close to zero for temperatures below the Bose temperature. Hence, setting $\mu=0$ in the previous expression, and making the substitution $x=\epsilon/(k T)$ , we obtain

$$\overline{E} =\frac{2}{\sqrt{\pi}}\left(\frac{2\pi m k T}{h^{ 2}}\right)^{3/2} V \int_0^\infty \frac{x^{ 3/2}}{{\rm e}^{ x}-1} dx.$$ (8.244)

The integral is equal to ${\mit\Gamma}(5/2) \zeta(5/2)$ , where ${\mit\Gamma}(5/2)=3\sqrt{\pi}/4$ , and $\zeta(5/2)=1.34$ . (See Exercise 4.) Thus, we obtain

$$\overline{E} = \frac{3}{2} \frac{\zeta(5/2)}{\zeta(3/2)} N k T\left(\frac{T}{T_B}\right)^{3/2}=0.770 N k T\left(\frac{T}{T_B}\right)^{3/2}$$ (8.245)

where use has been made of Equation (8.235). Note that $\overline{E}\propto T^{ 5/2}$ for temperatures below the Bose temperature, but that $\overline{E}$ becomes similar in magnitude to the classical value, $(3/2) N k T$ , as $T\rightarrow T_B$ . The molar specific heat below the Bose temperature is

$$\frac{c_V}{R} = \frac{1}{N k} \frac{d\overline{E}}{d T}=\frac{1... ...a(3/2)}\left(\frac{T}{T_B}\right)^{3/2} = 1.92\left(\frac{T}{T_B}\right)^{3/2}.$$ (8.246)

Likewise, the mean pressure is

$$\bar{p} = \frac{2}{3} \frac{\overline{E}}{V} = \frac{\zeta(5/2)}... ...{T}{T_B}\right)^{3/2}=0.513 \frac{N k T}{V}\left(\frac{T}{T_B}\right)^{3/2}.$$ (8.247)

(See Exercise 7.) In both cases, the quantities become similar in magnitude to their classical values as $T\rightarrow T_B$ , but fall far below these values when $T\ll T_B$ .

Figure: Variation with temperature of $\overline {E}/\overline {E}_{\rm classical}$ for a boson gas, where $\overline{E}_{\rm classical} = (3/2) N k T$ .

For $T>T_B$ , Equation (8.243) becomes

$$\overline{E} = \frac{3}{2} N k T\left(\frac{T}{T_B}\right)^{3/2}\frac{1}{\zeta(3/2)}\sum_{n=1,\infty} \frac{{\rm e}^{ n \alpha}}{n^{ 5/2}}$$ (8.248)

(see Exercise 4), where $\alpha$ is determined from the numerical solution of Equation (8.240). Figure 8.9 shows how the mean energy of a boson gas varies with temperature, and clearly illustrates that the energy approaches its classical value asymptotically in the limit $T\gg T_B$ . Finally, Figure 8.10 shows how the molar heat capacity of a boson gas varies with temperature. It can be seen that the $c_V$ curve has a change in slope at $T=T_B$ , reaching a maximum value there of $1.92 R$ . At higher temperatures, $c_V$ approaches the classical value, $(3/2) R$ , asymptotically.

Figure: Variation with temperature of $c_V/c_{V {\rm classical}}$ for a boson gas, where $c_{V {\rm classical}}=(3/2) R$ .