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Bose–Einstein Condensation (BEC)

1. Introduction

Bose–Einstein Condensation is a quantum phenomenon predicted by Albert Einstein in 1924–25 based on the work of Satyendra Nath Bose. It occurs when a dilute gas of bosons is cooled to extremely low temperatures, close to absolute zero, such that a large fraction of particles occupy the lowest quantum state.

At very low temperature, quantum effects become macroscopic.

2. Bose–Einstein Distribution

The average number of particles in a state of energy \( \epsilon \) is:

\[ n(\epsilon) = \frac{1}{e^{\beta(\epsilon - \mu)} - 1} \]

where:

For bosons, the chemical potential satisfies:

\[ \mu \le 0 \]

3. Condition for Bose–Einstein Condensation

Total number of particles:

\[ N = \sum_{\epsilon} \frac{1}{e^{\beta(\epsilon - \mu)} - 1} \]

In the continuum limit:

\[ N = \int_0^{\infty} \frac{g(\epsilon)\, d\epsilon}{e^{\beta(\epsilon - \mu)} - 1} \]

For a 3D ideal gas, density of states:

\[ g(\epsilon) = \frac{V}{4\pi^2} \left(\frac{2m}{\hbar^2}\right)^{3/2} \sqrt{\epsilon} \] When temperature decreases, \( \mu \to 0 \). At the critical temperature \( T_c \), condensation begins.

4. Critical Temperature \( T_c \)

Setting \( \mu = 0 \): \[ N = \frac{V}{\lambda^3} \zeta(3/2) \] where \[ \lambda = \sqrt{\frac{2\pi \hbar^2}{m k_B T}} \] is the thermal wavelength. Solving for \( T_c \): \[ T_c = \frac{2\pi \hbar^2}{m k_B} \left( \frac{n}{\zeta(3/2)} \right)^{2/3} \] where \( n = \frac{N}{V} \).

5. Fraction of Condensed Particles

For \( T < T_c \): \[ \frac{N_0}{N} = 1 - \left( \frac{T}{T_c} \right)^{3/2} \] where: At \( T = 0 \), all particles occupy ground state.

6. Physical Interpretation

At very low temperatures:

7. Experimental Realization

BEC was first observed in 1995 in dilute gases of alkali atoms (Rubidium-87) by Eric Cornell and Carl Wieman.

8. Important Properties

9. Applications

10. Summary

Bose–Einstein Condensation occurs when bosons occupy the same lowest quantum state below a critical temperature, producing a new state of matter with macroscopic quantum behavior.

WHAT IS BOSE-EINSTEIN CONDENSATION ? DERIVE BOSE TEMPERATURE

Ans : It is the fifth state of matter where all bosons are compressed share a common minimum energy state.\\ When BOSONS are cooled to extemely low temperature they fall or condednse into a lowest accessible quantum quantum state. This state is called BE condensate.\\ This Phenomenon can explain Superfluidity of liquid Helim.

EXPLANATION :

\[ \begin{align} f(E) &= \frac{1}{e^{\left(\frac{\alpha + \beta E}{kT}\right)} - 1} \\ &= \frac{1}{e^{\alpha} e^{\frac{\beta E}{kT}} - 1} \\ &= \frac{1}{A \, e^{\frac{\beta E}{kT}} - 1} \end{align} \]

Thus, \[ f(E) = \frac{1}{A \, e^{\frac{\beta E}{kT}}} \] for a Maxwell-Boltzmann gas at \( A \gg 1 \), where \[ A = e^{\alpha} = \frac{V}{N} \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \]

For a Bose-Einstein Gas:

\[ \begin{align} N_{\text{Total}} &= N_{\text{Ground}} + N_{\text{Excited}} \\ N_{\text{Total}} &= N_{\text{Excited}} \quad \text{for} \quad T > T_B \\ N_{\text{Total}} &= N_{\text{Ground}} \quad \text{for} \quad T < T_B \\ N_{\text{Total}} &\approx N_{\text{Excited}} \quad \text{for} \quad T = T_B \end{align} \]
\[ \Rightarrow N_{\text{Ground}} = N_{\text{Total}} - N_{\text{Excited}} = N_{\text{Total}} \left( 1 - \left( \frac{T}{T_B} \right)^{3/2} \right) \]
\[ \begin{align} A &= 1 = \frac{V}{N} \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \\ \Rightarrow N &= V \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \\ \Rightarrow \frac{N_{\text{Excited}}}{N_{\text{Total}}} &= \left( \frac{T}{T_B} \right)^{3/2} \\ \Rightarrow N_{\text{Ground}} &= N_{\text{Total}} \left( 1 - \left( \frac{T}{T_B} \right)^{3/2} \right) \end{align} \]

This shows that when \( T < T_B \), a large number of particles accumulate in the ground state. At \( T = 0\,K \), all particles are in the ground state.

(This was first achieved in a laboratory in 1995 by Eric Cornell and Carl Wieman at the University of Colorado at Boulder (NIST–JILA lab), for which they were awarded the Nobel Prize in 2001.)

WHAT IS BOSE-EINSTEIN CONDENSATION?

Bose-Einstein Condensation and Bose Temperature

Definition:

Bose-Einstein Condensation (BEC) is a quantum phenomenon in which a gas of bosons at low density, when cooled to a sufficiently low temperature, locks together to share a common minimum energy state.\\ This state of matter is called the \textbf{Bose-Einstein Condensate} — often referred to as the \textbf{fifth state of matter}.\\ In BEC, the bosons condense in momentum space, with their wavefunctions overlapping and becoming indistinguishable from each other. The theoretical prediction was given by \textbf{Satyendra Nath Bose} in 1924 and later extended by \textbf{Albert Einstein}. The phenomenon was experimentally observed in laboratories in \textbf{1995} (Rubidium-87) and in \textbf{2001} (Sodium-23).\\ This phenomenon explains key quantum behaviors like:
  1. Superfluidity in liquid Helium
  2. Superconductivity in materials

Derivation of Bose Temperature \( T_B \)

The number of bosons in the excited states is given by:
\begin{align*} N_e &= \int_0^\infty \frac{g(E)}{e^{\beta(E - \mu)} - 1} \, dE \end{align*}
At the \textbf{critical temperature} (Bose temperature), the chemical potential \( \mu = 0 \). So:
\begin{align*} N &= N_e = \int_0^\infty \frac{g(E)}{e^{\beta E} - 1} \, dE \end{align*}
The density of states in 3D is:
\begin{align*} g(E) &= 2\pi V \left( \frac{2m}{h^2} \right)^{3/2} E^{1/2} \end{align*}
So:
\begin{align*} N &= \int_0^\infty \frac{2\pi V \left( \frac{2m}{h^2} \right)^{3/2} E^{1/2}}{e^{\beta E} - 1} \, dE \\ &= V \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \int_0^\infty \frac{u^{1/2}}{e^u - 1} \, du \quad (\text{Substitute } u = \beta E) \\ &= V \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \Gamma(3/2) \zeta(3/2) \end{align*}
Since \( \Gamma(3/2) = \frac{\sqrt{\pi}}{2} \), and \( \zeta(3/2) \approx 2.612 \), we have:
\begin{align*} N &= V \left( \frac{2\pi m k T}{h^2} \right)^{3/2} \zeta(3/2) \end{align*}
Solving for \( T \) gives the \textbf{Bose temperature} \( T_B \):
\begin{align*} T_B &= \frac{h^2}{2\pi m k} \left( \frac{N}{V \zeta(3/2)} \right)^{2/3} \end{align*}

Proof of BEC Below \( T_B \)

When \( T < T_B \), the number of particles in excited states becomes:
\begin{align*} N_e &= N \left( \frac{T}{T_B} \right)^{3/2} \end{align*}
Then, the number of particles in the ground state is:
\begin{align*} N_0 &= N - N_e = N \left[ 1 - \left( \frac{T}{T_B} \right)^{3/2} \right] \end{align*}

Conclusion:

As \( T \to 0 \), almost all bosons fall into the ground state: \[ \lim_{T \to 0} N_0 = N \] This marks the formation of a Bose-Einstein Condensate — a macroscopic occupation of the lowest energy state, a purely quantum phenomenon.\\

Derive Bose Temperature and Prove that below Bose temperature The number of bosons will increase in the ground state to show the Bose-Einstein Condensation

ANS :\\ It is a quantum phenomenon in which a gas of bosons of low density when cooled to a suffiencently low temperature locked together to share a common minimum energy state. This state of matter is called the Bose-Einstein Condensate. It is called as the 5th state of matter. The bosons condense in the momentum space with their wavefunctions overlapping with each other. The theory for the possibility of such a state of matter was given by SN bose in 1924 but was made practically possible in the laboratory after 71 years in 1995 and 2001. This Phenomenon can explain Superfluidity of liquid Helim \& the phenomenon of superconductivity

BOSE EINSTEIN GAS AT ANY TEMPERATURE T

$$ \begin{aligned} N&=\int_{0}^{\infty} f(E) g(E) d E \\ & =\int_{0}^{\infty} \frac{1}{e^{(\alpha+\beta E)}-1} g(E) d E \\ & =\int_{0}^{\infty} \frac{1}{e^{\alpha} e^{\beta E}-1}\left(2 \pi V\left(\frac{2 m}{h^{2}}\right)^{3 / 2} E^{1 / 2}\right) d E \\ & =\int_{0}^{\infty} \frac{1}{z^{-1} e^{\beta E}-1}\left(2 \pi V\left(\frac{2 m}{h^{2}}\right)^{3 / 2} E^{1 / 2}\right) d E \\ & =2 \pi V\left(\frac{2 m}{h^{2}}\right)^{3 / 2}(k T)^{1 / 2}(k T) \int_{0}^{\infty} \frac{u^{1 / 2}}{z^{-1} e^{u}-1} d u \\ & (\beta E=u \Rightarrow E=u k T) \end{aligned} $$
$$ \begin{aligned} & =2 \pi V\left(\frac{2 m k T}{h^{2}}\right)^{3 / 2} \frac{\sqrt{\pi}}{2} \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{u^{3 / 2-1}}{z^{-1} e^{u}-1} d u \\ & =V\left(\frac{2 m \pi k T}{h^{2}}\right)^{3 / 2} \frac{2}{\sqrt{\pi}} \int_{0}^{\infty} \frac{u^{3 / 2-1}}{z^{-1} e^{u}-1} d u \\ & =V\left(\frac{2 m \pi k T}{h^{2}}\right)^{3 / 2} \frac{1}{\Gamma(3 / 2)} \int_{0}^{\infty} \int_{z^{-1} e^{u}-1}^{u^{3 / 2-1}} d u \\ & =V\left(\frac{2 m \pi k T}{h^{2}}\right)^{3 / 2} F_{3 / 2}(z) \\ & =\frac{V}{\left(\frac{h^{2}}{2 m \pi k T}\right)^{3 / 2}} F_{3 / 2}(z) \\ & =\frac{V}{\lambda^{3}} F_{3 / 2}(z) \end{aligned} $$
Hence $$F_{n}(z)=\dfrac{1}{\Gamma(n)} \int_{0}^{\infty} \dfrac{u^{n-1}}{z^{-1} e^{u}-1} d u$$

SIMPLIFICATION OF THE FUNCTION IN TERMS OF REIMAN ZETA FUNCTION

$$ \begin{aligned} & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \frac{u^{n-1}}{z^{-1} e^{u}\left(1-\frac{1}{z^{-1} e^{u}}\right)} d u \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \frac{u^{n-1}}{z^{-1} e^{u}}\left(1-\frac{1}{z^{-1} e^{u}}\right)^{-1} d u \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} z \frac{u^{n-1}}{e^{u}}\left(1+\frac{z}{e^{u}}+\frac{z^{2}}{e^{2 u}}+\ldots\right) d u \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} z \frac{u^{n-1}}{e^{u}} \sum_{r=1}^{r=\infty}\left(\frac{z}{e^{u}}\right)^{r-1} d u \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \sum_{r=1}^{r=\infty}\left(\frac{z}{e^{u}}\right)^{r} u^{n-1} d u \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \sum_{r=1}^{r=\infty} z^{r} e^{-r u} u^{n-1} d u \quad \text { Taking } r u=x \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \sum_{r=1}^{r=\infty} z^{r} e^{-x}\left(\frac{x}{r}\right)^{n-1} \frac{d x}{r}\\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} \sum_{r=1}^{r=\infty} z^{r} e^{-x} x^{n-1} \frac{d x}{r^{n}} \\ & =\frac{1}{\Gamma(n)} \int_{0}^{\infty} e^{-x} x^{n-1} d x \sum_{r=1}^{r=\infty} \frac{z^{r}}{r^{n}} \\ & =\sum_{r=1}^{r=\infty} \frac{z^{r}}{r^{n}} \end{aligned} $$
Hence,
\begin{align*} F_{n}(z)&=\sum_{r=1}^{r=\infty} \frac{z^{r}}{r^{n}}\\ \Rightarrow N&=\frac{V}{\lambda^{3}} F_{3 / 2}(z) \& \end{align*}
Since
\begin{align*} U &= \int_{0}^{\infty} f(E) g(E) E \, dE \\ &= \int_{0}^{\infty} \frac{1}{z^{-1} e^{\beta E}-1} \cdot 2 \pi V \left( \frac{2m}{h^2} \right)^{3/2} E \cdot \frac{1}{2\sqrt{E}} \cdot E \, dE \\ &= \pi V \left( \frac{2m}{h^2} \right)^{3/2} \int_{0}^{\infty} \frac{E^{3/2}}{z^{-1} e^{\beta E} - 1} \, dE \\ &= \pi V \left( \frac{2m}{h^2} \right)^{3/2} \frac{3\sqrt{\pi}}{2} \left( \frac{1}{\frac{3\sqrt{\pi}}{2}} \int_{0}^{\infty} \frac{E^{3/2}}{z^{-1} e^{\beta E} - 1} \, dE \right) \\ &= V \left( \frac{2\pi m}{h^2} \right)^{3/2} \frac{3}{2} \left( \frac{1}{\frac{3\sqrt{\pi}}{2}} \int_{0}^{\infty} \frac{u^{3/2}}{z^{-1} e^{u} - 1} \, du \cdot (kT)^{3/2} \cdot kT \right) \\ &= V \left( \frac{2\pi m kT}{h^2} \right)^{3/2} kT \cdot \frac{3}{2} \left( \frac{1}{\frac{3\sqrt{\pi}}{2}} \int_{0}^{\infty} \frac{u^{3/2}}{z^{-1} e^{u} - 1} \, du \right) \\ &= \frac{V}{\left( \frac{h^2}{2\pi m kT} \right)^{3/2}} kT \cdot \frac{3}{2} \left( \frac{1}{\frac{3\sqrt{\pi}}{2}} \int_{0}^{\infty} \frac{u^{3/2}}{z^{-1} e^{u} - 1} \, du \right) \\ &= \frac{V}{\lambda^3} \cdot \frac{3kT}{2} \cdot \frac{1}{\Gamma(5/2)} \int_{0}^{\infty} \frac{u^{5/2 - 1}}{z^{-1} e^{u} - 1} \, du \\ &= \frac{V}{\lambda^3} \cdot \frac{3kT}{2} \cdot F_{5/2}(z) \\ U &= \frac{3kT V}{2 \lambda^3} F_{5/2}(z) \end{align*}

BOSE-EINSTEIN GAS AT $T \geq T_{B}$

At sufficeinttly high temperature
  1. All the particles are in the exicited state
  2. The number of paricles in the ground state with energy $E=0$ is neglibly small
  3. $A$ is quite larger than 1
So neglecting the higher terms and considering first two terms.
\begin{align*} \Rightarrow \frac{U}{N} &= \frac{3kT}{2} \cdot \frac{f_{5/2}(z)}{f_{3/2}(z)} \\ &= \frac{3kT}{2} \cdot \frac{z + \frac{z^2}{2^{5/2}} + \cdots}{z + \frac{z^2}{2^{3/2}} + \cdots} \\ &= \frac{3kT}{2} \cdot \left( \frac{1 + \frac{z}{4\sqrt{2}}}{1 + \frac{z}{2\sqrt{2}}} \right) \\ &= \frac{3kT}{2} \cdot \left(1 + \frac{z}{4\sqrt{2}} \right) \left(1 + \frac{z}{2\sqrt{2}} \right)^{-1} \\ &= \frac{3kT}{2} \cdot \left(1 - \frac{z}{4\sqrt{2}} \right) \\ &= \frac{3kT}{2} \left(1 - \frac{1}{4\sqrt{2}} \cdot \frac{N}{V} \left( \frac{h^2}{2\pi m kT} \right)^{3/2} \right) \\ \Rightarrow U &= \frac{3NkT}{2} \left(1 - \frac{1}{4\sqrt{2}} \cdot \frac{N}{V} \left( \frac{h^2}{2\pi m kT} \right)^{3/2} \right) \\ &= \frac{3Nk}{2} \left(T - \frac{1}{4\sqrt{2}} \cdot \frac{N}{V \sqrt{T}} \left( \frac{h^2}{2\pi m k} \right)^{3/2} \right) \end{align*}

A. Specific heat :

$$ \begin{aligned} C_{V} & =\left(\frac{\partial U}{\partial T}\right)_{V}=\frac{3 N k}{2}\left(1-\frac{1}{4 \sqrt{2}} \frac{N}{V}\left(\frac{h^{2}}{2 \pi m k}\right)^{3 / 2} \frac{\partial}{\partial T}\left(\frac{1}{\sqrt{T}}\right)\right) \\ & =\frac{3 N k}{2}\left(1+\frac{1}{8 \sqrt{2}} \frac{N}{V}\left(\frac{h^{2}}{2 \pi m k T}\right)^{3 / 2}\right) \end{aligned} $$
The first term is for the classical ideal gas . Second term is for the first order correction due to the deviation from the classical behaviour. When the temperature will be sufficiently large the correction terms will be very small.\\

B. Pressure

$$ \begin{aligned} & P=\frac{2}{3 V} U=\frac{2}{3 V} \frac{3 N k}{2}\left(T-\frac{1}{4 \sqrt{2}} \frac{N}{\sqrt{T V}}\left(\frac{h^{2}}{2 \pi m k}\right)^{3 / 2}\right) \\ & =\frac{N k}{V}\left(T-\frac{1}{4 \sqrt{2}} \frac{N}{\sqrt{T} V}\left(\frac{h^{2}}{2 \pi m k}\right)^{3 / 2}\right) \\ & =\frac{N k T}{V}\left(1-\frac{1}{4 \sqrt{2}} \frac{N}{V}\left(\frac{h^{2}}{2 \pi m k T}\right)^{3 / 2}\right) \\ & =\frac{R T}{V}\left(1-\frac{1}{4 \sqrt{2}} \frac{N}{V}\left(\frac{h^{2}}{2 \pi m k T}\right)^{3 / 2}\right) \\ & \Rightarrow P V=R T\left(1-\frac{1}{4 \sqrt{2}} \frac{N}{V}\left(\frac{h^{2}}{2 \pi m k T}\right)^{3 / 2}\right) \end{aligned} $$

3. BOSE-EINSTEIN GAS AT TEMPERATURE $ T

Behavious of Chemical Potential;

$$ \begin{aligned} & NHence,=\sum n_{i}=n_{0}+n_{1}+n_{2}+n_{3}+\ldots \\ & =\frac{g_{0}}{A e^{\beta E_{0}}-1}+\frac{g_{1}}{A e^{\beta E_{1}}-1}+\frac{g_{2}}{A e^{\beta E_{2}}-1}+\ldots \quad \text { Where } \mathbf{g}=\text { degeneracy of the level. } \\ & =\frac{g_{0}}{A-1}+\frac{g_{1}}{A e^{\beta E_{1}}-1}+\frac{g_{2}}{A e^{\beta E_{2}}-1}+\ldots \end{aligned} $$

The first term is for the number of particles in the ground state. The above expression is valid for all temperatures.Since number of particles in a state at any temperature can not be negative,

$$ A \geq 1 \Rightarrow e^{-\mu / k T} \geq 1 \Rightarrow-\mu / k T \geq 0 \Rightarrow \mu \leq 0 $$
since
$$ A = e^{\alpha} = e^{-\frac{\mu}{kT}} $$

at sufficiently low temperature $\mu \approx-10^{-38} \mathrm{erg}$ which is very small and can be taken to be zero.Thus the chemical potential can be negative or zero for the bosons.
it is zero for photons.
It can be positive and negative for Fermions and as well as for Maxwell gas.
The chemical potential at temperatute greater than the Bose temperature is negative. So

$$ \text { A>1, when } T>T_{B} $$

But when the temperature is lowered below $T_{B}$, the chemical potential becomes 0 . So A=1, when $T

\begin{align*} \Rightarrow N &= \frac{V}{\lambda^{3}} F_{3/2}(A=1) = \frac{V}{\lambda^{3}} \sum_{r=1}^{\infty} \frac{1}{r^{3/2}} \\ &= \frac{V}{\lambda^{3}} \zeta(3/2) \\ &= \frac{V}{\lambda^{3}} (2.6124) \\ &= \frac{V}{\left( \frac{h}{\sqrt{2\pi m k T}} \right)^3} (2.6124) \\ &= V \left( \frac{2\pi m k T}{h^2} \right)^{3/2} (2.6124) \end{align*}
Here, \( \zeta \) is the Riemann zeta function.
\begin{align*} U &= \frac{3kT V}{2\lambda^3} \zeta(5/2) = \frac{3kT V}{2\lambda^3} (1.341) \end{align*}
At this temperature \( T_B \), the number of particles in the ground state is negligible. The total number of particles are in the excited states.
\begin{align*} & N_{e}=V\left(\frac{2 \pi m k T}{h^{2}}\right)^{3 / 2}(2.6124) \tag{2.6124}\\ & T_{B}=\left(\frac{h^{2}}{2 \pi m k}\right)\left(\frac{N}{2.6124 V}\right)^{3 / 2} \end{align*}
BOSE TEMPERATURE is the minimum temperature at which all the particles are in the excited state.\\ As \( N_{\text{total}} = N_{\text{ground}} + N_{\text{excited}} = N_0 + N_e \), we have:
\begin{align*} \Rightarrow \frac{N_e}{N_{\text{total}}} &= \left( \frac{T}{T_B} \right)^{3/2} \\ \Rightarrow N_e &= N_{\text{total}} \left( \frac{T}{T_B} \right)^{3/2} \end{align*}
Then, the number of particles in the ground state is:
\begin{align*} N_0 &= N_{\text{total}} - N_e \\ &= N_{\text{total}} - N_{\text{total}} \left( \frac{T}{T_B} \right)^{3/2} \\ &= N_{\text{total}} \left( 1 - \left( \frac{T}{T_B} \right)^{3/2} \right) \end{align*}

Cases

Case-1:When \( T > T_B \)

The system is entirely in the gaseous phase. All the particles occupy excited states only.

Case-2:When \( T = T_B \)

\begin{align*} N_{\text{ground}} &= N_{\text{total}} \left( 1 - \left( \frac{T_B}{T_B} \right)^{3/2} \right) \\ &= N_{\text{total}} (1 - 1) = 0 \end{align*}

So, no particles occupy the ground state. All particles are in the excited states.

Case 3:When \( T < T_B \)

The system may be regarded as a mixture of two phases:
1. A Bose-Einstein condensate (ground state occupation).
2. A thermal gas (excited states).
a gaseous phase in the exicited states with number of particles as $N_{e}=N_{\text {Total }}\left(\frac{T}{T_{B}}\right)^{3 / 2}$
a condensed phase with number of particles in the ground state $N_{0}=N_{\text {total }}-N_{e} \ll N_{\text {total }}$
A large number of particles are in the ground state and some are in the excites state too.

Case- 4:When $\mathrm{T}=0\left(T \rightarrow 0, N_{0} \rightarrow N_{\text {total }} \rightarrow N\right)$

:

Since $N_{\text {ground }}=N_{\text {Total }}\left(1-\left(\frac{0}{T_{B}}\right)^{3 / 2}\right)=N_{\text {total }}$\\ All the particles occupy the ground state. The condesation of the particles takes place in the momentum space not like the condensation of gas in real space. All the wavefunctons of particles overlap with each other due to increasing wavlength. The state is called Bose-Einstein Condensate and the process as Bose-Einstein Condensation\\ ( 1924 theoritical predication and practical realisation in laboratory in 1995 and 2001)

Graphical Explanation

$f(E)=\frac{1}{e^{\left(\frac{\alpha+\beta E}{k T}\right)}-1}=\frac{1}{e^{\alpha} e^{\frac{\beta E}{k T}}-1}=\frac{1}{A e^{\frac{\beta E}{k T}}-1}$
Thus

$f(E)=\frac{1}{A e^{\frac{\beta E}{k T}}}$ is For MB gas at $A \ggg 1$ where $A=e^{\alpha}=\frac{V}{N}\left(\frac{2 \pi m k T}{h^{2}}\right)^{3 / 2}$

For a Bose -Einstein Gas $A \geq 1$. Therefore The chemical Potential is negative .

For a Bose-Einstein Gas $A>1$ when $\mathrm{T}>\mathrm{T}_{\mathrm{B}} \quad \& \quad \mathrm{~A}=1$ When $T \leq T_{B}$.

$N_{\text {Total }}=N_{\text {Ground }}+N_{\text {exicited }}$ $N_{\text {Total }}=N_{\text {excited }} \quad$ for $T=T>T_{B}$ $N_{\text {Total }}=N_{\text {Ground }} \quad$ for $T

Which shows that When $T (It was first achieved in Laboratory in 1995 by Eric Cornell and Carl Wieman at the University of Colorado at Boulder NIST-JILA lab fow which they have been awarded Noble Prize in 2001)

What are Phonons

  1. Phonon is the quantum of lattice vibrational energy (elastic wave).
  2. It has integral spin.
  3. It obeys Bose-Einstein statistics and is made of indistinguishable particles.
  4. Phonons travel with the speed of sound in a solid, unlike photons which travel at the speed of light.
  5. The number of phonons is not conserved; phonons can be created and destroyed.
  6. They are neutral.
  7. Phonons exhibit both wave and particle characteristics.
  8. Vibrational spectrum of a phonon varies in the range \( 10^4 \) to \( 10^{12} \) Hz.
  9. Energy of a phonon is given by: \( E = \hbar \omega \)
  10. Phonons play a major role in many physical properties of condensed matter, such as thermal and electrical conductivity. The study of phonons is a crucial part of condensed matter physics.

    (The concept of the phonon was introduced by Igor Tamm (1895–1971, Vladivostok, Russia) in 1932, for which he was awarded the Nobel Prize in 1958.)