/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 73 Consider a gas of \(N\) identica... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider a gas of \(N\) identical spin-0 bosons confined by an isotropic three- dimensional harmonic oscillator potential. (In the rubidium experiment discussed above, the confining potential was actually harmonic, though not isotropic.) The energy levels in this potential are \(\epsilon=n h f,\) where \(n\) is any nonnegative integer and \(f\) is the classical oscillation frequency. The degeneracy of level \(n\) is \((n+1)(n+2) / 2\). (a) Find a formula for the density of states, \(g(\epsilon),\) for an atom confined by this potential. (You may assume \(n \gg 1\).) (b) Find a formula for the condensation temperature of this system, in terms of the oscillation frequency \(f\). (c) This potential effectively confines particles inside a volume of roughly the cube of the oscillation amplitude. The oscillation amplitude, in turn, can be estimated by setting the particle's total energy (of order \(k T\) ) equal to the potential energy of the "spring." Making these associations, and neglecting all factors of 2 and \(\pi\) and \(s o\) on, show that your answer to part (b) is roughly equivalent to the formula derived in the text for the condensation temperature of bosons confined inside a box with rigid walls.

Short Answer

Expert verified
(a) \( g(\epsilon) = \frac{\epsilon^2}{(hf)^3} \) for \( n \gg 1 \). (b) \( T_c = \frac{hf}{k} \left( \frac{N}{2 \zeta(3)} \right)^{1/3} \). (c) It is roughly equivalent to the box formula with approximation.

Step by step solution

01

Understanding the Energy Levels and Degeneracy

The energy levels of the bosons in the harmonic oscillator potential are given as \( \epsilon = n h f \), where \( n \) is a non-negative integer and \( f \) is the oscillation frequency. The degeneracy for each level is \( \frac{(n+1)(n+2)}{2} \). We will use these facts to find the density of states.
02

Calculating the Density of States, \( g(\epsilon) \)

Assuming large \( n \), we use the density of states approximation. For a 1D harmonic oscillator, the number of states with energy less than \( \epsilon \) is \( N = \frac{\epsilon^3}{6(hf)^3} \). The density of states is the derivative of this with respect to \( \epsilon \):\[ g(\epsilon) = \frac{dN}{d\epsilon} = \frac{\epsilon^2}{(hf)^3} \text{ for } n \gg 1. \]
03

Finding the Condensation Temperature, \( T_c \)

To find the condensation temperature, we set the chemical potential to zero at \( T = T_c \). The number of particles, \( N \), is related to the density of states by the integral:\[ N = \int_0^{\infty} \frac{g(\epsilon)}{e^{\epsilon/kT} - 1} d\epsilon. \]At condensation temperature, this integral gives \( N \approx 2 \zeta(3) \left( \frac{kT_c}{hf} \right)^3 \).Solving for \( T_c \), we find:\[ T_c = \frac{hf}{k} \left( \frac{N}{2 \zeta(3)} \right)^{1/3}. \]
04

Comparing with Box Condensation Temperature

The condensation temperature for bosons in a box is typically given by:\[ T_c \propto \frac{N^{2/3}}{V^{2/3}} \]where \( V \) is the volume. Using the volume equivalence with the harmonic potential, \( V \approx (E_spring/hf)^3 \), where \( E_{spring} \approx kT \), we can estimate:\[ T_c \approx \frac{h f}{k} \left( \frac{N}{V} \right)^{1/3}. \]This shows rough equivalence to the box case when factors of \( 2 \) and \( \pi \) are ignored.

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Key Concepts

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

Harmonic Oscillator Potential
Imagine a system where particles, such as bosons, are trapped inside a three-dimensional space. This space is defined by a force similar to that seen in a spring or pendulum, namely the harmonic oscillator potential. In simpler terms, this potential attempts to bring particles back to a central point whenever they move away. The potential is effective in studying many quantum systems due to its symmetrical and predictable nature. In the context of Bose-Einstein Condensation, the energy levels for particles in this potential are expressed as \( \epsilon = nhf \), where \( n \) is a non-negative integer and \( f \) represents the frequency of oscillation. A vital point here is that energy is quantized - it only takes specific values.
Density of States
The density of states \( g(\epsilon) \) is a critical concept that helps us understand how energy levels are occupied by particles within a given potential. Imagine it as a guideline showing how many different ways particles can have a particular energy within the system. In mathematical terms, when dealing with our harmonic oscillator potential, the density of states is determined for large \( n \) as \( g(\epsilon) = \frac{\epsilon^2}{(hf)^3} \). This derivation arises from understanding how energy levels disperse in the given potential and becomes especially simplified under the assumption that \( n \) is large. For the isotropic potential in question, these states give a clearer perspective on particle behavior.
Condensation Temperature
The concept of condensation temperature \( T_c \) tells us the sweet spot at which bosons suddenly start to condense into the lowest available energy state. This occurs when cooling a collection of bosons to exceedingly low temperatures. In our scenario of a harmonic oscillator potential, the condensation temperature can be calculated as \( T_c = \frac{hf}{k} \left(\frac{N}{2 \zeta(3)}\right)^{1/3} \), where \( h \) is Planck’s constant, \( k \) is Boltzmann's constant, and \( N \) is the number of particles. Essentially, \( T_c \) depends directly on parameters like particle number and the specific environmental potential they are under. It provides profound insight into the temperature realm where phase transition to Bose-Einstein Condensate occurs.
Degeneracy of Energy Levels
Energy levels can sometimes be degenerate, meaning that multiple quantum states share the same energy. In a simple harmonic potential, degeneracy acts as a numerical balance that allows these states to coexist at identical energy levels. In the isotropic 3D harmonic oscillator potential considered, the degeneracy of a level \( n \) follows the formula \( \frac{(n+1)(n+2)}{2} \). This reflects how the states are distributed within the system, allowing multiple configurations of particles at the same energy level. The concept of degeneracy is especially crucial in quantum systems as it directly affects the system’s entropy and can influence phenomena such as Bose-Einstein Condensation.

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Most popular questions from this chapter

For a system of fermions at room temperature, compute the probability of a single-particle state being occupied if its energy is (a) \(1 \mathrm{eV}\) less than \(\mu\) (b) \(0.01 \mathrm{eV}\) less than \(\mu\) (c) equal to \(\mu\) (d) \(0.01 \mathrm{eV}\) greater than \(\mu\) (e) \(1 \mathrm{eV}\) greater than \(\mu\)

Consider the electromagnetic radiation inside a kiln, with a volume of \(1 \mathrm{m}^{3}\) and a temperature of \(1500 \mathrm{K}\) (a) What is the total energy of this radiation? (b) Sketch the spectrum of the radiation as a function of photon energy. (c) What fraction of all the energy is in the visible portion of the spectrum, with wavelengths between \(400 \mathrm{nm}\) and \(700 \mathrm{nm} ?\)

Consider two single-particle states, \(A\) and \(B,\) in a system of fermions, where \(\epsilon_{A}=\mu-x\) and \(\epsilon_{B}=\mu+x ;\) that is, level \(A\) lies below \(\mu\) by the same amount that level \(B\) lies above \(\mu\). Prove that the probability of level \(B\) being occupied is the same as the probability of level \(A\) being unoccupied. In other words, the Fermi-Dirac distribution is "symmetrical" about the point where \(\epsilon=\mu\).

In addition to the cosmic background radiation of photons, the universe is thought to be permeated with a background radiation of neutrinos ( \(\nu\) ) and antineutrinos \((\bar{\nu}),\) currently at an effective temperature of 1.95 K. There are three species of neutrinos, each of which has an antiparticle, with only one allowed polarization state for each particle or antiparticle. For parts (a) through (c) below, assume that all three species are exactly massless. (a) It is reasonable to assume that for each species, the concentration of neutrinos equals the concentration of antineutrinos, so that their chemical potentials are equal: \(\mu_{\nu}=\mu_{\bar{\nu}} .\) Furthermore, neutrinos and antineutrinos can be produced and annihilated in pairs by the reaction $$\nu+\bar{\nu} \leftrightarrow 2 \gamma$$ (where \(\gamma\) is a photon). Assuming that this reaction is at equilibrium (as it would have been in the very early universe), prove that \(\mu=0\) for both the neutrinos and the antineutrinos. (b) If neutrinos are massless, they must be highly relativistic. They are also fermions: They obey the exclusion principle. Use these facts to derive a formula for the total energy density (energy per unit volume) of the neutrino- antineutrino background radiation. (Hint: There are very few differences between this "neutrino gas" and a photon gas. Antiparticles still have positive energy, so to include the antineutrinos all you need is a factor of \(2 .\) To account for the three species, just multiply by \(3 .\) ) To evaluate the final integral, first change to a dimensionless variable and then use a computer or look it up in a table or consult Appendix B. (c) Derive a formula for the number of neutrinos per unit volume in the neutrino background radiation. Evaluate your result numerically for the present neutrino temperature of \(1.95 \mathrm{K}\) (d) It is possible that neutrinos have very small, but nonzero, masses. This wouldn't have affected the production of neutrinos in the early universe, when \(m c^{2}\) would have been negligible compared to typical thermal energies. But today, the total mass of all the background neutrinos could be significant. Suppose, then, that just one of the three species of neutrinos (and the corresponding antineutrino) has a nonzero mass \(m\). What would \(m c^{2}\) have to be (in eV), in order for the total mass of neutrinos in the universe to be comparable to the total mass of ordinary matter?

The heat capacity of liquid \(^{4} \mathrm{He}\) below \(0.6 \mathrm{K}\) is proportional to \(T^{3},\) with the measured value \(C_{V} / N k=(T / 4.67 \mathrm{K})^{3}\). This behavior suggests that the dominant excitations at low temperature are long-wavelength phonons. The only important difference between phonons in a liquid and phonons in a solid is that a liquid cannot transmit transversely polarized waves - sound waves must be longitudinal. The speed of sound in liquid \(^{4} \mathrm{He}\) is \(238 \mathrm{m} / \mathrm{s}\), and the density is \(0.145 \mathrm{g} / \mathrm{cm}^{3} .\) From these numbers, calculate the phonon contribution to the heat capacity of \(^{4} \mathrm{He}\) in the low-temperature limit, and compare to the measured value.

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