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Chapter 9: Problem 8

Suppose we have a system of identical particles moving in just one dimension and for which the energy quantization relationship is \(E=b n^{2 \Omega}\), where \(b\) is a constantand \(n\) an integer quantum number. Discuss whether the density of states should be independent of E. an increasing function of \(E,\) or a decreasing function of \(E\).

Short Answer

Expert verified
The density of states is a decreasing function of the energy, \(E\), in this system.

Step by step solution

01

- Understand Energy Quantization Relationship

Let's look at the given relationship \(E = b n^{2\Omega}\). Here, \(E\) is the energy, \(b\) is a constant, \(n\) is the quantum number which is an integer, and \(\Omega\) is also constant. This equation will help us understand how the energy changes with the quantum numbers.
02

- Calculate the Change in Energy

The first derivative of the energy with respect to the quantum number \(n\) would give us how the energy \(E\) changes with a change in \(n\). Differentiating \(E\) with respect to \(n\), we get: \(\frac{dE}{dn} = 2\Omega bn^{\Omega - 1}\). This shows that the energy changes with the quantum number \(n\). The rate of change depends on \(2\Omega b\) and \(n^{\Omega - 1}\). Since \(n\) is always an integer and greater than 0, and \(\Omega\) is also constant and positive, the term \(n^{\Omega - 1}\) is always positive.
03

- Determine Density of States Relationship With Energy

The density of states \(g(E)\) gives us the number of states per unit energy interval. By definition, it is the derivative of the number of states \(N\) with respect to energy \(E\), or \(g(E) = dN/dE\). However we know that \(N\) increments by 1 when \(n\) increments by 1, hence \(dN = dn\). And from step 2, we found \(\frac{dE}{dn} = 2\Omega bn^{\Omega - 1}\). Therefore, our expression for the density of states becomes \(g(E) = \frac{1}{\frac{dE}{dn}} = \frac{1}{2\Omega bn^{\Omega - 1}}\). As both \(b\) and \(\Omega\) are constants, \(g(E)\) is a decreasing function of \(n\), hence a decreasing function of \(E\) as the value of \(E\) increases with increasing \(n\) for positive \(n, b, \Omega\).
04

- Conclusion

Based on the energy function and the calculation of the density of states, we can clearly see that the density of states is not independent of energy \(E\) because it decreases as \(E\) increases. Therefore, the density of states is a decreasing function of \(E\) in this system of identical particles moving in just one dimension.

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Key Concepts

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

Energy Quantization
Energy quantization is a fundamental concept in quantum mechanics describing how systems have discrete energy levels. In classical mechanics, energy can vary continuously, but in quantum systems, energy levels are quantized, meaning that particles such as electrons can only possess specific energy values. This happens because of the wave-like nature of particles at a microscopic scale.
In the exercise, the energy quantization relationship is given as \(E = b n^{2\Omega}\). Here, \(E\) represents energy, \(b\) is a constant, \(n\) is an integer known as the quantum number, and \(\Omega\) is also constant. This equation showcases how energy levels change based on quantum numbers in quantum mechanical systems.
The term \(n^{2\Omega}\) implies that the energy levels grow rapidly with increasing quantum number \(n\), especially if \(\Omega\) is greater than one. The larger the value of \(\Omega\), the more spread out the energy levels become, indicating that transitions between different energy states might require significant energy input.
Density of States
The density of states (DOS) is a crucial concept in quantum mechanics that describes how many quantum states are available per energy interval at each energy level. It helps predict properties like electronic behavior in semiconductors, metals, and other materials.
By examining the derivative \(g(E) = \frac{dN}{dE}\), where \(N\) is the number of states and \(E\) is energy, we can determine how states are distributed over energy. Using the given exercise's energy relation, we found that energy \(E\) changes with the quantum number \(n\) through \(\frac{dE}{dn} = 2\Omega bn^{\Omega - 1}\). Taking its reciprocal gives the density of states: \(g(E) = \frac{1}{2\Omega b n^{\Omega - 1}}\).
This equation shows that as \(n\) increases, the density of states \(g(E)\) decreases. This implies that fewer quantum states are available at higher energy levels, a behavior vital for understanding phenomena such as electron distribution in energy bands.
Quantum Number
Quantum numbers are essential to fully describe the state of a quantum mechanical system. Each quantum number provides distinct information about the system, such as energy levels, angular momentum, or spin.
In the given scenario, the quantum number \(n\) directly influences energy quantization through the formula \(E = b n^{2\Omega}\). As an integer, \(n\) signifies different energy levels that a particle can occupy. The choice of \(n\) determines which quantized energy state the particle will be in.
Moreover, the integer nature of quantum numbers results in the discrete and non-continuous representation of physical quantities, differing fundamentally from classical physics. This discrete nature forms the cornerstone of many quantum phenomena, such as quantized energy levels in an atom, which underpin the structure of the periodic table and chemical behavior of elements.

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Most popular questions from this chapter

Determine the density of states \(D(E)\) for a \(2 D\) infinite well (ignoring spin) in which $$ E_{A_{x+} n_{2}}=\left(n_{x}^{2}+n_{y}^{2}\right) \frac{\pi^{2} \hbar^{2}}{2 m L^{2}} $$

Given an arbitrary thermodynanic systemn, which is larger. the number of possible macrostates, or the number of possible microstates, or is it impossible to say? Explain your answer. (For most systems, both are infinite, but it is still possible to answer the question)

We based the exact probabilities of equation (9.9) on the claim that the number of ways of adding \(N\) distinct nonnegative integers/quuntum numbers to give a total of \(M\) is \(\\{M+N-1) ! /\left[M^{\prime}(N-1) !\right]\). Verify this claim (a) for the case \(N=2, M=5\) and (b) for the case \(N=5, M=2\)

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 \(M\).

A "cold" object, \(T_{1}=300 \mathrm{~K}\), is briefly put in contact with a "hot" object, \(T_{2} \simeq 400 \mathrm{~K}\), and \(60 \mathrm{~J}\) of heat flows from the hot object to the cold one. The objects are then separated, their temperatures having changed negligibly due to their large sizes. (a) What are the changes in entropy of each object and the system as a whole? (b) Knowing only that these objects are in contact and at the given temperatures, what is the ratio of the probabilities of their being found in the second (final) state W that of their being found in the first (initial) state? What does chis result suggest?
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