In the grand canonical ensemble, the variance in particle number is \(\left(N^{2}\right\rangle-\langle N\rangle^{2}=1 / \beta(\partial(N\rangle / \partial \mu)_{T, .}\) (a) Compute \(\left\langle N^{2}\right\rangle-\langle N\rangle^{2}\) for a classical ideal gas. (b) Compute \(\left\langle N^{2}\right\rangle-\langle N\rangle^{2}\) for a Bose-Einstein ideal gas at fairly high temperature (keep corrections to the classical result to first order in the particle density). How and why is your result different from the classical ideal gas result? (c) Compute \(\left\langle N^{2}\right\rangle-(N)^{2}\) for a Fermi-Dirac ideal gas at a fairly high temperature (keep corrections to the classical result to first order in the particle density). How and why is your result different from the classical ideal gas result?

Short Answer

Expert verified
Classical: \(\left\langle N^{2} \right\rangle - \left\langle N \right\rangle^{2} = \left\langle N \right\rangle \); Bose-Einstein: \(\left\langle N \right\rangle (1 + n\lambda^3)\); Fermi-Dirac: \(\left\langle N \right\rangle (1 - n\lambda^3)\). Variance differs due to particle statistics.

Step by step solution

01

Understanding the Equation

The variance in particle number is given by the formula: \[ \left\langle N^{2} \right\rangle - \left\langle N \right\rangle^{2} = \frac{1}{\beta} \left( \frac{\partial \left\langle N \right\rangle }{\partial \mu} \right)_{T} \]
02

Classical Ideal Gas

For a classical ideal gas, the number of particles \( \left\langle N \right\rangle \) is given by the equation \( \left\langle N \right\rangle = \frac{V}{\lambda^3} e^{\beta \mu}\), where \( \lambda \) is the thermal wavelength. Differentiating \(\left\langle N \right\rangle \) with respect to the chemical potential \( \mu \): \[ \left( \frac{\partial \left\langle N \right\rangle}{\partial \mu} \right)_{T} = \left\langle N \right\rangle \beta \] Substituting, \[ \left\langle N^{2} \right\rangle - \left\langle N \right\rangle^{2} = \left\langle N \right\rangle \]
03

Bose-Einstein Ideal Gas at High Temperature

At high temperatures for a Bose-Einstein gas, we consider corrections to first order in particle density, \( n = \frac{N}{V} \). The corrected number of particles is given by \( \left\langle N \right\rangle = \frac{V}{\lambda^3} e^{\beta \mu} \). The first-order correction term is \( \left\langle N \right\rangle (1 + n\lambda^3) \). Differentiating and substituting: \[ \left( \frac{\partial \left\langle N \right\rangle}{\partial \mu} \right)_{T} \approx \left\langle N \right\rangle \beta (1 + n\lambda^3) \] \[ \left\langle N^{2} \right\rangle - \left\langle N \right\rangle^{2} = \left\langle N \right\rangle (1 + n\lambda^3) \]
04

Fermi-Dirac Ideal Gas at High Temperature

Similar to Bose-Einstein gas, we consider first-order corrections to particle number density for a Fermi-Dirac gas. The corrected number of particles is given by \( \left\langle N \right\rangle = \frac{V}{\lambda^3} e^{\beta \mu} \). The first-order correction term is \( \left\langle N \right\rangle (1 - n\lambda^3) \). Differentiating and substituting: \[ \left( \frac{\partial \left\langle N \right\rangle}{\partial \mu} \right)_{T} \approx \left\langle N \right\rangle \beta (1 - n\lambda^3) \] \[ \left\langle N^{2} \right\rangle - \left\langle N \right\rangle^{2} = \left\langle N \right\rangle (1 - n\lambda^3) \]
05

Comparing Results

For the classical ideal gas, we found \( \left\langle N^{2} \right\rangle - \left\langle N \right\rangle^{2} = \left\langle N \right\rangle \). For the Bose-Einstein gas, the variance increases due to the positive correction term. For the Fermi-Dirac gas, the variance decreases due to the negative correction term. This highlights the difference in particle statistics: Bose-Einstein statistics lead to bunching, while Fermi-Dirac statistics prevent multiple occupancy due to the Pauli exclusion principle.

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Key Concepts

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

Classical Ideal Gas
The classical ideal gas provides a simple model where we can understand basic thermodynamic properties without quantum effects. In this context, we assume that gas particles do not interact and obey the laws of classical mechanics. The average number of particles \( \langle N \rangle \) in a given volume \( V \) at temperature \( T \) and chemical potential \( \mu \) is given by:

\[ \langle N \rangle = \frac{V}{\lambda^3} e^{\beta \mu} \]

Here, \( \lambda \) represents the thermal wavelength, and \( \beta \) is the inverse temperature. When differentiating \( \langle N \rangle \) with respect to the chemical potential \( \mu \):

\[ \left( \frac{\partial \langle N \rangle}{\partial \mu} \right)_{T} = \langle N \rangle \beta \]

Substituting back into the variance formula:

\[ \langle N^2 \rangle - \langle N \rangle^2 = \langle N \rangle \]

This result reflects that, for a classical ideal gas, the variance in particle number corresponds directly to the average number of particles. This is because particles are independently distributed.

Bose-Einstein Statistics
In a Bose-Einstein ideal gas, particles obey Bose-Einstein statistics, leading to interesting quantum effects not seen in classical gases. At high temperatures, we can approximate the number of particles with first-order corrections to the particle density \( n = \frac{N}{V} \):

\[ \langle N \rangle = \frac{V}{\lambda^3} e^{\beta \mu}(1 + n\lambda^3) \]

Here, the correction term \( (1 + n\lambda^3) \) accounts for quantum statistics. Differentiating:

\[ \left( \frac{\partial \langle N \rangle}{\partial \mu} \right)_{T} \approx \langle N \rangle \beta \left(1 + n\lambda^3\right) \]

Substituting into the variance formula:

\[ \langle N^2 \rangle - \langle N \rangle^2 = \langle N \rangle \left(1 + n\lambda^3\right) \]

The variance is greater than in the classical case, due to the positive correction term. This increase in variance reflects the bunching effect of bosons, unlike classical particles. At low temperatures, this feature can lead to Bose-Einstein condensation, where a macroscopic number of particles occupy the same quantum state.

Fermi-Dirac Statistics
Conversely, in a Fermi-Dirac ideal gas, particles adhere to the Pauli exclusion principle, meaning no two fermions can occupy the same quantum state. For a Fermi-Dirac gas at high temperature, we use a similar correction approach:

\[ \langle N \rangle = \frac{V}{\lambda^3} e^{\beta \mu}(1 - n\lambda^3) \]

The negative correction term \( (1 - n\lambda^3) \) takes into account the exclusion principle. Differentiating:

\[ \left( \frac{\partial \langle N \rangle}{\partial \mu} \right)_{T} \approx \langle N \rangle \beta \left(1 - n\lambda^3\right) \]

Substituting into the variance formula:

\[ \langle N^2 \rangle - \langle N \rangle^2 = \langle N \rangle \left(1 - n\lambda^3\right) \]

The variance here is lower than in the classical case. This decrease happens because of fermion properties that restrict the number of particles occupying the same state. As a result, the variance reflects this anti-bunching behavior, which is fundamental to Fermi-Dirac statistics. This is crucial for understanding properties of systems like electron gases in metals, where electron behavior is governed by these statistics.

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Most popular questions from this chapter

(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?

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.

To lowest order in the density, find the difference in the pressure and isothermal compressibility between an ideal boson and an ideal fermion gas. Assume that the fermions and bosons have the same mass and both are spinless. (Note: You are now considering fairly high temperature.)

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?

Electrons in a piece of copper metal can be assumed to behave like an ideal Fermi gas. Copper metal in the solid state has a mass density of \(9 \mathrm{~g} / \mathrm{cm}^{3}\). Assume that each copper atom donates one electron to the Fermi gas. Assume the system is at \(T=0 \mathrm{~K}\). (a) Compute the Fermi energy, \(\varepsilon_{\mathrm{F}}\), of the electron gas. (b) Compute the Fermi "temperature", \(T_{\mathrm{F}}=\varepsilon_{\mathrm{P}} / \mathrm{k}_{\mathrm{B}}\).

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