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(a) Compute the Bose-Einstein condensation temperature for a gas of \(N=4 \times 10^{4}\) rubidium atoms in an asymmetric harmonic trap with oscillation frequencies \(f_{1}=120 \mathrm{~Hz}, f_{2}=120 / \sqrt{8} \mathrm{~Hz}\), and \(f_{3}=120 / \sqrt{8} \mathrm{~Hz}\), (b) If condensation occurred at the same temperature in a cubic box, what is the volume of the box?

Short Answer

Expert verified
(a) \( T_c \approx 7.65 \times 10^{-10} \text{ K} \), (b) \( V \approx 1.98 \times 10^{-17} \text{ m}^3 \)

Step by step solution

01

- Calculate the geometric mean frequency

Determine the geometric mean frequency for the asymmetric harmonic trap using the formula: \[ \overline{f} = (f_1 f_2 f_3)^{1/3} \]Substitute the given frequencies: \[ \overline{f} = \left(120 \times 120/\sqrt{8} \times 120/\sqrt{8}\right)^{1/3} \]Simplify the expression inside the parentheses and compute the value.
02

- Calculate the geometric mean frequency value

Perform the calculation: \[ \overline{f} = \left(120 \times 120/2.828 \times 120/2.828\right)^{1/3} \]Continue simplifying: \[ \overline{f} = \left(120 \times 42.426 \times 42.426\right)^{1/3} \]Finally: \[ \overline{f} \approx 68.04 \text{ Hz} \]
03

- Compute the Bose-Einstein condensation temperature

Use the formula for the condensation temperature: \[ T_c = \frac{\hbar \overline{\omega}}{k_B} \left( \frac{N}{\zeta(3)} \right)^{1/3} \]where: \( \hbar \) is the reduced Planck's constant, \( k_B \) is Boltzmann's constant, \( \overline{\omega} = 2\pi \overline{f} \), and \( \zeta(3) \approx 1.202 \). Substitute the values: \[ \overline{\omega} = 2 \pi \times 68.04 \text{ Hz} \]\[ \overline{\omega} \approx 427.58 \text{ s}^{-1} \]\[ T_c = \frac{\hbar \times 427.58}{k_B} \left( \frac{4 \times 10^4}{1.202} \right)^{1/3} \]Simplify the expression and compute the numeric value.
04

- Calculate the Bose-Einstein condensation temperature value

Perform the final computation using known constants: \( \hbar \approx 1.054 \times 10^{-34} \text{ J} \cdot \text{s} \) and \( k_B \approx 1.38 \times 10^{-23} \text{ J/K} \). \[ T_c \approx \frac{1.054 \times 10^{-34} \times 427.58}{1.38 \times 10^{-23}} \left( \frac{4 \times 10^4}{1.202} \right)^{1/3} \]\[ T_c \approx 2.39 \times 10^{-11} \left( 3.33 \times 10^4 \right)^{1/3} \]\[ T_c \approx 2.39 \times 10^{-11} \times 32.05 \]\[ T_c \approx 7.65 \times 10^{-10} \text{ K} \]
05

- Determine the volume of the cubic box

For a cubic box, the condensation temperature is given by: \[ T_c = \frac{2 \pi \hbar^2}{k_B m} \left( \frac{N}{V} \right)^{2/3} \]Assuming the same temperature \( T_c \), solve for the volume \( V \). Rearrange the formula: \[ V = \left( \frac{2 \pi \hbar^2}{k_B m T_c} \right)^{3/2} \times N \]Now substitute the known values, using the mass of a rubidium atom \( m \approx 1.44 \times 10^{-25} \) kg.
06

- Compute the volume value

Perform the calculation:\[ V = \left( \frac{2 \pi (1.054 \times 10^{-34})^2}{1.38 \times 10^{-23} \times 1.44 \times 10^{-25} \times 7.65 \times 10^{-10}} \right)^{3/2} \times 4 \times 10^4 \]First simplify inside the parentheses: \[ V \approx \left( \frac{6.28 \times 1.11 \times 10^{-68}}{1.52 \times 10^{-57}} \right)^{3/2} \times 4 \times 10^4 \]Then compute the final value: \[ V \approx \left( 7.30 \times 10^{-12} \right)^{3/2} \times 4 \times 10^4 \]\[ V \approx 1.98 \times 10^{-17} \text{ m}^3 \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Geometric Mean Frequency
The geometric mean frequency is a crucial concept when working with an asymmetric harmonic trap. It's a single frequency that represents the combined effect of the individual frequencies of the trap. This mean frequency is calculated using the formula: \ \(\=f = (f_1 f_2 f_3)^{1/3} \). Here, \( f_1, f_2,\) and \( f_3 \) are the oscillation frequencies of the trap.
The geometric mean frequency simplifies the analysis by condensing the three different frequencies into one representative frequency. In our exercise, this involved using the given frequencies to find the mean frequency which turns out to be approximately 68.04 Hz.
Condensation Temperature Formula
To determine the Bose-Einstein condensation temperature, we use a specific formula. This exact condensation temperature, denoted as \( T_c \), is calculated with the formula: \ \( T_c = \frac{\hbar \overline{\omega}}{k_B} \left( \frac{N}{\zeta(3)} \right)^{1/3} \)
Variables such as \( \hbar \) (reduced Planck's constant), \( k_B \) (Boltzmann's constant), and \( \zeta(3) \approx 1.202 \) are standard constants in physics, while \( \overline{\omega} \) is derived from the geometric mean frequency \( \overline{f} \).
This formula helps in determining the temperature at which Bose-Einstein condensation occurs for a given system, aiding in understanding the behavior of quantum gases.
Asymmetric Harmonic Trap
An asymmetric harmonic trap isn't symmetrical, meaning it has different frequencies in each direction. In our problem, the trap had frequencies \( f_1 = 120 \) Hz, \( f_2 = 120/\sqrt{8} \) Hz, and \( f_3 = 120/\sqrt{8} \) Hz. To work with such a trap, we had to find the geometric mean frequency to reflect the varied oscillations.
Such traps are often used in experiments with ultra-cold gases because they provide a controlled environment to study quantum phenomena, like Bose-Einstein condensation.
Bose-Einstein Statistics
Bose-Einstein statistics describe the distribution of indistinguishable particles over different energy states in thermal equilibrium. Unlike classical particles, bosons (particles that obey Bose-Einstein statistics) tend to occupy the same state, leading to phenomena like Bose-Einstein condensation.
This condensation occurs when a significant number of these particles occupy the lowest energy state, typically at very low temperatures. This concept was key to understanding our problem as it helped determine the condensation temperature where these particles exhibit collective quantum behavior.
Cubic Box Volume Calculation
For a cubic box, the volume calculation involves understanding how a given number of particles behave at the condensation temperature. The formula used is:\ \( V = \left( \frac{2 \pi \hbar^2}{k_B m T_c} \right)^{3/2} \times N \)
Here, \( m \) is the mass of a single particle (like a rubidium atom), and \( T_c \) is the condensation temperature. By rearranging variables and substituting known values, you can calculate the volume of the box that can maintain the Bose-Einstein condensation at the computed temperature. In our exercise, the final volume was found to be approximately \( 1.98 \times 10^{-17} \) cubic meters.

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Most popular questions from this chapter

One mole of a dilute gas of \(\mathrm{He}^{3}\) atoms (which are spin- \(1 / 2\) fermions) at temperature \(T\) \(=140 \mathrm{~K}\) is contained in a box of volume \(\mathrm{V}=1.0 \mathrm{~cm}^{3}\). (a) Compute the lowest order (in density) correction to the classical ideal gas pressure. (b) What fraction of the total pressure is due to the Fermi statistics of the atoms?

The density of states of an ideal Bose-Einstein gas in a cubic box of volume \(V\) is $$ g(E)= \begin{cases}\alpha E^{3} & \text { if } \quad E>0 \\ 0 & \text { if } \quad E<0\end{cases} $$ where \(\alpha\) is a constant. Compute the critical temperature for Bose- Einstein condensation.

Liquid helium \(\left({ }^{4} \mathrm{He}_{2}\right.\) ) undergoes a superfluid transition at a temperature of \(T=2.16 \mathrm{~K}\). At this temperature it has a mass density of \(\rho=0.145 \mathrm{~g} / \mathrm{cm}^{3}\). Make the (rather drastic) assumption that liquid helium behaves like an ideal gas and compute the critical temperature for Bose-Einstein condensation.

A two-dimensional electron gas can be formed at the interface of GaAs/AlGaAs semiconductors. The effective mass of the electrons is \(m=0.067 m_{e}\), where \(m_{\mathrm{e}}\) is the mass of the, electron in free space. Treat the electrons like an ideal Fermi gas of particles with spin-1/2 and mass \(m\) in a two-dimensional box with area \(A=L^{2}\). (a) What is the density of states of the electron gas? (b) If the electron density is \(\mathrm{N} / \mathrm{A}=2.0 \times 10^{13} / \mathrm{cm}^{2}\), what is the Fermi energy of the two- dimensional electron gas?

An ideal Bose-Einstein gas consists of noninteracting bosons of mass \(m\) which have an internal degree of freedom which can be described by assuming that the bosons are two-level atoms. Bosons in the ground state have energy \(E_{0}=p^{2} /(2 m)\), while bosons in the excited state have energy \(E_{1}\) \(=p^{2} /(2 m)+\Delta\), where \(p\) is the momentum and \(\Delta\) is the excitation energy. Assume that \(\Delta \gg k_{\mathrm{B}} T\). Compute the Bose-Einstein condensation temperature, \(T_{c}\), for this gas of two-level bosons. Does the existence of the internal degree of freedom raise or lower the condensation temperature?

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