/*! 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 52 Equation \((9-27)\) gives the de... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 52

Equation \((9-27)\) gives the density of states for a system of oscillators but ignores spin. The result, simply one state per energy change of \(\hbar \omega_{0}\) between levels. is incorrect if particles are allowed diff erent spin states at each level. but modification to include spin is easy. From Chapter 8 , we know that a particle of spin s is allowed \(2 s+1\) spin orientations, so the number of states at each level is simply multiplied by this factor. Thus, $$ D(E)=(2 s+1) / \hbar \omega_{0} $$ (a) Using this density of states, the definition $$ \begin{array}{l} \text { Nheud }(2 s+1)=\delta, \text { and } \\ \qquad N=\int_{0}^{\infty} \mathcal{N}(E) D(E) d E \end{array} $$ calculate the parameter \(B\) in the Boltzmann distribution \((9-31)\) and show that the distribution can thus be tewritten as $$ \mathcal{N}(E)_{\text {Bolu }}=\frac{\varepsilon}{k_{\mathrm{B}} T} \frac{1}{e^{E / \mathrm{L}_{\mathrm{B}} T}} $$ (b) Algue that if \(k_{\mathrm{B}} T \gg \delta\), the occupation number is much less than I for all \(E\).

Short Answer

Expert verified
The Boltzmann distribution can be rewritten as \[\mathcal{N}(E)_{\text {Bolu }}=\frac{\varepsilon}{k_{\mathrm{B}} T} \frac{1}{e^{E / \mathrm{L}_{\mathrm{B}} T}}\] under the consideration of spin of the particles. In the limit of high temperatures where \(k_B T >> \delta\), the equation implies the occupation number is much less than 1 for all \(E\).

Step by step solution

01

Understanding the density of states \[ D(E) = \frac{(2s+1)}{\hbar \omega_0} \]

In a system of oscillators where the particles can have different spin states at each energy level, each particle with spin \(s\) is allowed to have \(2s +1\) spin orientations. The equation represents the density of states which is the number of quantum states available to be occupied.
02

Applying the definition of total number of oscillators

The total number of oscillators \(N\) is given by \[N=\int_{0}^{\infty} \mathcal{N}(E) D(E) d E\] where \(\mathcal{N}(E)\) is the quantity characterizing the number of particles at each energy level, and \(D(E)\) is the density of states. Plug the definition of \(D(E)\) into the equation and integrate.
03

Calculating the parameter \(B\)

Even though not given directly in the exercise, we get the form of distribution function from the equation \(\mathcal{N}(E) = B e^{-E / k_{B}T}\). Here \(B\) can be calculated by setting the condition that the total population is conserved. Replace \(\mathcal{N}(E)\) with \(B e^{-E / k_BT}\) in the integral for \(N\) and solve it for \(B\). Substituting \(B\) in the equation for \(\mathcal{N}(E)\), we get the final formula.
04

Interpreting the distribution at high temperatures

It is argued that if \(k_B T >> \delta\), the occupation number is much less than 1 for all \(E\). This can be proved by expanding the exponential in the derived Boltzmann distribution and ignoring high order terms for small \(E / k_BT\).

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Key Concepts

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

Quantum Oscillators
In the realm of quantum mechanics, oscillators are a foundational concept used to describe and understand various physical systems. A quantum oscillator typically refers to a quantum system that exhibits wave-like behavior, and it includes the quantization of energy levels. The simplest example of a quantum oscillator is the quantum harmonic oscillator, which models the energy of a particle trapped in a potential well with forces that restore it to an equilibrium position.

A crucial aspect of quantum oscillators is that the energy levels are discrete, meaning particles can only inhabit specific energy states, with the difference between these states being proportional to the fundamental frequency of the oscillator, represented by \( \hbar \omega_0 \). This discrete nature of quantum systems stands in contrast to classical systems where the oscillators' energy levels can vary continuously.

Understanding quantum oscillators is essential as they serve as prototypes for more complex quantum systems, and their energy levels can be affected by factors like spin orientations. This influence on energy levels leads to an important concept called the density of states, which quantifies the number of quantum states available at each energy level, factoring in particle spin.
Spin Orientations
Spin, a fundamental property of quantum particles, describes the intrinsic angular momentum that particles possess. For a particle with spin \( s \), which can be a fraction or whole number, there exist a finite number, precisely \( 2s+1 \) possible spin orientations that the particle can exhibit. From an atomic perspective, these orientations can be perceived as the 'directions' in which a particle's spin can be aligned.

When considering systems such as quantum oscillators, recognizing the impact of spin is crucial because it multiplies the number of available states for a particle to occupy. Therefore, when calculating the density of states, \( D(E) \) for a system, each energy level must be modified by this factor to reflect the correct number of quantum states, leading to the modified density of states expression \( D(E) = \frac{(2s+1)}{\hbar \omega_0} \). This accounts for all possible orientations the spin can take, thereby impacting statistical properties like distribution functions in quantum statistics.

As such, spin plays a non-trivial role in many quantum phenomena and is vital to consider when predicting and explaining the behavior of systems at the quantum level.
Boltzmann Distribution
The Boltzmann distribution is a cornerstone of statistical mechanics which describes the distribution of particles over different energy states in thermal equilibrium. It is essential for understanding how energy is distributed among particles in a system at a given temperature. The basic assumption of the Boltzmann distribution is that the probability of finding a particle in a particular state is proportional to \( e^{-E / k_{B}T} \), where \( E \) is the energy of the state, \( k_B \) is the Boltzmann constant, and \( T \) is the temperature.

To compute any statistical mechanical property of a system, you need to determine the parameter \( B \) in the Boltzmann distribution, which ensures that the total number of particles \( N \) is preserved across all energy states, as per the principle of conservation of particles. The exercise involves integrating the distribution over all energy states, weighted by the density of states, to find \( B \), thereby obtaining the explicit Boltzmann distribution function for the system in consideration.

The relationship between temperature and energy levels, as described by the Boltzmann distribution, is particularly insightful. At high temperatures, where \( k_B T \) is much greater than the energy difference between states (\( \delta \)), the occupation numbers are significantly less than one. This implies that each state is sparsely occupied, which aligns with the classical prediction that energy is more evenly spread across states when the system has more thermal energy available.

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Most popular questions from this chapter

Consider a gas of atoms that might serve as a laser medium but that is in equilibrium, with no population inversions. A photon gas coexists with the atoms. Would a photon whose energy is precisely the difference between two atomic energy states be more likely to be absorbed or to induce a stimulated emission or neither? We expect that in equilibrium the numbers of atoms at different levels and the number of photons of a given energy should be stable. Is your answer compatible?

The exact probabilities of equation \((9-9)\) rest on the claim that the number of ways of adding \(N\) distinct nonnegative integers to give a total of \(M\) is \((M+N-1) !\) \([M !(N-1) !]\). One way to prove it involves the following trick. It represents two ways that \(N\) distinct integers can add to \(M-9\) and 5 , respectively, in this special case. $$ \begin{array}{c|cccccccccccccc} \hline \text { 1 } & \text { X } & \text { X } & \text { X } & \text { I } & \text { I } & \text { X } & \text { I } & \text { I } & \text { I } & \text { I } & \text { X } & \text { I } & \text { I } \\ \hline 2 & \text { I } & \text { X } & \text { X } & \text { I } & \text { I } & \text { I } & \text { I } & \text { X } & \text { I } & \text { I } & \text { I } & \text { X } & \text { X } \\ \hline \end{array} $$ The X's represent the total of the integers, \(M\) - each row has \(5 .\) The I's represent "dividers" between the distinct integers, of which there will of course be \(N-\) I -each row has 8 . The first row says that \(n_{1}\) is 3 (three \(X\) 's before the divider between it and \(n_{2}\) ). \(n_{2}\) is 0 (no \(X\) 's between its left divider with \(n_{1}\) and its right divider with \(\left.n_{3}\right), n_{3}\) is \(1, n_{4}\) through \(n_{6}\) are \(0, n_{2}\) is \(1,\) and \(n_{8}\) and \(n_{9}\) are 0\. The second row says that \(n_{2}\) is \(2, n_{6}\) is \(1, n_{9}\) is \(2,\) and all other \(n\) are 0 . Further rows could account for all possible ways that the integers can add to \(M\). Argue that. properly applied, the binomial coefficient (discussed in Appendix J) can be invoked to give the correct total number of ways for any \(N\) and \(M\).

In a certain design of helium-neon laser, the chamber contaning these gases has a perfect mirror at one end. as usual, but only a window at the other. Beyond the window is a region of free air space and then the second mirror, which is partially reflecting. allowing the beam to exit. The resonant cavity between the mirrors thus has a region free of the helium-neon gas - the "lasing material" - in which you can insert something. If you insert a sheet of clear plastic at any orientation in this region between the mitrors, the laser beam disappears. If the same sheet is placed in the beam ourside the partially reflecting mirror, the beam passes through it, regardless of the orientation. Why?

Consider the two-sided room, (a) Which is more likely to have an imbalance of five particles (ie., \(N_{\mathrm{k}}=\frac{1}{2} N+5\) ): a room with \(N=20\) or a room with \(N=60\) ? (Note: The total number of ways of distributing particles, the sum of \(W_{N_{k}}^{N}\) from \(O\) to \(N,\) is \(2^{N}\).) (b) Which is more likely to have an imbalance of \(5 \%\) (i.e., \(\left.N_{R}=\frac{1}{2} N+0.05 N\right) ?\) (c) An average-size room is quite likely to have a trillion more air molecules on one side than on the other. Why may we say that precisely half will be on each side?

Pruve that fur any sine function \(\sin (k x+\phi)\) of wavelength shonter than \(2 a\), where \(a\) is the atomic spacing. there is a sine function with a wavelength longer than \(2 a\) that has the same values at the points \(x=a .20,3 a\). and so on. Wotr: It is probably easier to work with wave number than with wavelength. We seck to show that for every wave number greater than \(\pi / a\) there is an equivalent une less than \(\pi / a\).)
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